Baffour Ba Series

CORE

MATHEMATICS
for

Schools and Colleges

By:

BAFFOUR ASAMOAH
CERT „A‟– OTC, Dip(Edu) – UCC, B.Ed (Maths) – UEW, BAT(Admin) – CSUC, BSc (Maths & Stats) - PUC

Baffour – Ba Series, Core Maths for Schools and Colleges Page i

Copyright© 2015
Allrights reserved under copyright law.
No part of this publication may be published, produced or stored on a print or any retrievable media
without approval or authorization.

For suggestions & views / copies;

Please call any of the following numbers:
0208211966 / 0244282977

ISBN : 978 – 9988 – 3 – 0744 – 8

Baffour – Ba Series, Core Maths for Schools and Colleges Page ii

PREFACE
Baffour Ba Series, is a series of Mathematics, covering the contents of Senior High Schools and
Colleges sylabuses.

In this book, each chapter is broken down into short, manageable sections intended to completely
alleviate or reduce to a large extent, the canker that is termed in Ghanaian context as “Math‟s Phobia”
by providing in its content, concise notes and commentaries with well – illustrated diagrams on each
topic and subtopic of the Senior High Schools and Colleges Syllabuses.

The book reflects the authors experience that students work better from work examples than abstract
discussion of principles, hence the provision of numerous and comprehensive range of worked
examples. A numbered, step-by-step approach to problem solving, trial test and challenge problems
intended to cater for gifted students are greatly featured in this book. Not left out in this book is
“exercises” on each subtopic which contains an extensive selection of Multiple- choice, Fill-ins, True
or False, Essay – type questions similar in standard to the WASSCE questions. Tackling these exercises
is no doubt, an excellent form of revision. Answers to all exercises are also provided to help students
assess their level of progress.

Taking into accounts, full analysis of the pattern and level of difficulty of examination questions, the
last part of the book is “some solved past questions”,and “Trial Objective Test” with answers within its
contents.

Baffour – Ba Series, Core Maths for Schools and Colleges Page iii
ACKNOWLEDGEMENTS
A myriad of services including spiritual, moral, inspirational, motivational and financial support,
typing, editing, printing, designing and what have you, were deeply rendered by friends, family
members, students, staff members, some co - tutors of J.H.S, S.H.S and Colleges of Education to make
this work a success. To mention a few: Mrs. Joyce Mensah, Mr. Kwabena Agyeman (More Questions),
Mr. Asamoah Koduah Thick, Mr. Stephen Amoateng, Mr. Frank Owusu, Mr. Kwaku Boateng, Mr.
Augustine Boadi, Mr. Isaac Owusu Mensah, Mr. Owusu Blessing, Mr. K.D. Bosiako, Mr. Adjei
Domfeh Romeo, Mr. Bernard Kusi Knambers, Mr. Attakorah Jacob, Sampson Brobbey, George Osei
Mensah (Kin – George), Cherish Asamoah Baffour – Ba, Godbless Agyei Asamoah Baffour Ba Snr,
Godbless Agyei Asamoah Baffour Ba Jnr, Grace Adwoa Pomaa, Benjamin Osei – Mensah (Benjii),
Appiah Collins, Maybel Owusu Ansah, Antwi Gloria, Mr. Opoku John, Department of Mathematics,
AGASS, Department of Mathematics, Okess .

I also wishto acknowledge the fact that the book is not absolutely free from errors of typing, grammar
and inaccuracy. These occurred as a result of oversight but not ignorance and incompetence on the part
of the author and editors. These should therefore, not undermine the credibility of the book, for they say
“to err is human”. However, your comments, corrections, suggestions and criticism are warmly
welcomed for consideration and rectification in the next edition.

Baffour Asamoah
(Baffour Ba)

Baffour – Ba Series, Core Maths for Schools and Colleges Page iv

 TABLE OF CONTENTS
Chapter Page Chapter Page
1. Sets and Operations on Sets - - - - - 1 17. Percentages II - - - - - - - - - - - 458
2. The Real Number System - - - - - - 27 18. Variations - - - - - - - - - - - - - -- 494
3. Algebraic Expressions - - - - - - - ---61 19. Statistics II - - - - - - - - - - - - ---516
4. Properties of Surds - - - - - - - - - ---79 20. Probability - -- - - - - - - - - - - -539
5. Number Bases - - - - - - - - - - - --- 89 21. Quadratic Equations- - - - - - - -- 569
6. Relations and Functions - - - - - ---107 22. Mensuration I - - - - - - - - - - - - 616
7. Plane Geometry I - - - - - - - - - -----165 23. Plane Geometry II - - - - - - - - -641
8. Equations and Inequalities - - - - ---201 24. Trigonometry I - - - - - - - - --- --661
9. Bearings and Vectors - - - - - - - - 228 25. Sequence and Series - - - - - - - -701
10. Statistics 1 - - - - - - - - - - - - - - - 259 26. Geometrical Construction - - -- -718
11. Rigid Motion - -- - -- - - - - - - - -299 27. Mensuration II - - - - - - - - - -- -732
12. Ratio and Rates - - - - - - - - - - - 327 28. Logical Reasoning - - - - - - - - - 760
13. Percentages I - - - - - - - -- -- -- -363 29. Trigonometry II -- - - - - - - ---- 772
14. Modulo Arithmetic - - - - - - - - 392 Answers to Exercises - - - - - - - - -775
15. Indices and Logarithms - - - - - - 415 Tables -- -- -- - - - - - - - - - - - - - -- -785
16. Simultaneous Equations - - - - - 438 References - - - - - - - - - - - - - - - - -789

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1 SETS AND OPERATIONS ON SETS Baffour – Ba Series

Definition of a Set 2. B = {Integers from -2 to 5}

A set is the collection of objects based on a well- 3. C = {-3, -2, -1, 0, 1, 2, 3}
defined property. Thus, we can have table, chair,
pencil and book as objects made of wood. Representation of Sets
Similarly, we can have Monday, Tuesday… A set is always represented by a capital letter and
Sunday, as set of days of the week. its members are represented by small letters.
e.g. A = {m, e, n}
Describing a Set
A set may be described by; Set A Members (elements)
a. Listing the members of the set. For example, The sign “∈” is read as “belongs to” and the sign
the set of even numbers from 2 to 8 = { } “∉” is read as “does not belong to”. Thus, we
b. Stating the property, for example, {even can write m∈ A and n∈ A because „m‟ and „n‟ are
numbers less than 10} members of set A. On the other hand, we can
c. Using the set builder notation. For example, {x write b∉A and d∉A because “b” and “d” are not
: 2 ≤ x ≤ 10}, where x is an even number. members of set A.

Exercises 1.1 Worked Examples

A. Given that A = {1, 2, 3, 4, 5} list the Put in ∉ or ∈ in the following, given that A = {x :
elements of the following Sets. 1 < x < 10} and B = {x : x ≥ -5}.
1. {x2: x ∈ A} 2. { : x ∈ A} (i) -2 A (ii) 2 A (iii) -10 A
3. {2x: x ∈ A} 4. {4x + 1: x ∈ A}
Solution
B. List the elements of the following set; i. -2 ∉ A ii. 2 ∈ A iii. -10 ∉ A
1. {x: 9 < x < 21}
Exercises 1.2
2. {x: 4 ≤ x < 36, where x is a multiple of 4}
A. Put in ∈ or ∉ in the following
3. {x: -5 ≤ x ≤ 5}
1. 6 ……… {Perfect squares}
4. {x: 9 < x < 21}
2. 27……… {3, 6, 9……….99}
3. 10 ……… {x : x > 8}
C. Describe the following sets in words;
4. 19 ……… {Prime numbers}
1. A = {2, 4, 6, 8…100}
2. C = {a, e, i, o, u}
B. Rewrite the following using set symbols:
3. D = {6, 12, 18…42}
1. 64 is a perfect square
4. E = {2, 3, 5, 7, 11, 13, 17, 19}
2. 7 is a prime number.
D. Describe with set builder notation: 3. is not an integer.
1. A = {Even numbers between 10 and 20} 4. 41 is not an odd number less than 25.

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Types of Set Subset is denoted by ⊂ or ⊃. For example, if N =
1. Null set or empty set {1, 2, 3, 4, 5, 6} and M = {1, 2, 3}, then M is a
It is a set with no element or member. For subset of N, because all the elements of M can be
example, there is no triangle with 5 sides. This set found in N. This is written as M⊃N or N⊂M.
is therefore an empty set. It is denoted by { } or .
Worked Examples
2. Unit set 1. If Y = {house, tree} and X = {cat, house, tree.
It is a set that has only one element or member. Find the relationship between Y and X.
For e.g. A = {x: 3 < x < 5} = {4}.
Solution
3. Finite set Since all the elements in Y can be found in X, we
It is a set that can take all its members. Precisely, say Y is a subset of X, i.e. Y⊂X
a finite set is a set that comes to an end. For e.g, A
= {even numbers between 2 and10} 2. If P = {2, 4, 8} and Q = {even counting
⇒A = {4, 6, 8} numbers less than 12}. What is the relationship
between P and Q?
4. Infinite set
It is s set that cannot take all its members. Solution
Precisely, an infinite is a sets does not come to an P = {2, 4, 8} and Q = {2, 4, 6, 8, 10}
end. For example: B = {even numbers} Since all the elements of P can be found in Q, P is
⇒B = {2, 4, 6, 8, 10 . . .} a subset of Q. That is P⊂Q

Exercises 1.3
3. P = {x : 20 < x < 30, where x is odd} and Q =
A. Identify the empty and unit sets.
{23, 29}. Establish a relationship between sets P
1. {Odd factors of 16}
and Q.
2. {Even numbers between 16 and 18}
3. {Prime numbers between 16 and 19}
Solution
4. {A number that is odd and even}
P = {21, 23, 25, 27, 29} and Q = {23, 29}
Since all the members of Q can be found in P, we
B. Identify the finite and infinite sets;
say Q is a subset of P. i.e. Q⊂P
1. {Multiples of 3} 2. {1, 2, 3…….100}
3. {Factors of 12} 4. {Even prime numbers}
Listing the Subsets of a Set
Relationship between Sets The empty set and the set itself are the first two
Two or more sets can be related as; subsets, equal subsets of every set. The other subsets include
sets or equivalent sets. each element of the set and a combination of the
elements in twos, threes etc until each and every
Subsets element combines with each other and all other
A set “M” is said to be the subset of another set “N” elements.
if all the elements of “M” can be found in “ N”.

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Note: The number of subsets in a set can be If P = {x : 40 < x < 50, where x is prime},
found by the formula: 2n where n is the number Q = {47, 43}, R = {41, 47}, A = {1, 2, 3},
of elements in the set B = {3, 2}, C = {3} and D = {2, 1, 3}, then:
1. B⊂A 2. A ⊂ B 3. C⊂A 4. C ⊂ B
Worked Examples
1. If M = {x, y}. List all the subsets of M. Equal Sets
Two or more sets are said to be equal if they have
Solution identical or the same elements. For e.g, if A = {1,
Subsets of M = { }, {x, y}, {x} and {y} 2, 3} and B = {3, 1 , 2}, then A = B because A
Number of subsets = 4 and B have the same elements.

2. List all the subsets of B = {a, b, c}. Worked Examples

1. If P = {1, 2, 3, 8, 10}, Q ={8, 1, x, 3, 2} and
Solution P = Q. What is the value of x?
Subsets of B = { }, {a, b, c},{a}, {b},
{c}, {a, b}, {a, c} and {b, c}. Solution
Number of subsets = 8 P = {1, 2, 3, 8, 10} and Q = {8, 1, x, 3, 2}
By matching or comparing the two sets, x = 10
3. Find the number of subsets in the set because 10 is the only element that is not
A = {1, 2, 3, 4} in set Q.

Solution Exercises 1.5

Number of subsets = 2n, where n = 4 Which of the pair of sets are equal?
Number of subsets = 24 = 2 × 2 × 2 × 2 = 16 1. {5, 7, 8, 10} and {e, f, g, h}.
2. {even primes} and {2}.
Exercises 1.4 3. {4, 8, 12…} and {multiples of 4}.
A. Find the number of subsets in a set that 4. {multiples of 2} and {even numbers}.
contains the following number of elements; 5. A = {x : x is an integer, and – 1≤ x ≤ 4} and
(1) 5 (2) 7 (3) 8 (4) 16 B = {-1,0, 1, 2, 3, 4}.

B. 1. If A = {1, 2, 3}, B = {3, 2}and C = {3} Equivalent Sets

determine which of the sets is a subset of the Two or more sets are said to be equivalent if they
other. have the same number of elements. For e.g, If A =
2. Write the pair of subsets of the set A = {1, 2, 3, {1, 2, 3} and B = {a, b, c}, then A and B are
4, 6, 12} equivalent sets because there are three elements
3. How many pair of subsets are there in the set in each set.
of factors of 18.
Exercises 1.6
Challenge Problem Which of the pair of sets are equivalent?
C.Use True or False for the following: 1. P ={ factors of 24}and Q ={factors of 30}

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2. P = {2, 3, 4, 5} and Q = {64, 2, 16, 27, 12} Solution
3. P = {prime factors of 24} P = {2, 3, 5, 7, 11, 13, 17, 19}
Q = {prime factors of 30} Q = {1, 3, 5, 7, 9}
4. A= {x : 1 < x < 10, where x is even} P∩Q = {3, 5, 7}
B ={x : 1< x < 10, where x is odd}
Exercices 1.7
Intersection of Sets A. 1. Given that M = {1, 2, 3, 4, 5….20}
The intersection of sets A and B is the set of Q = {3, 4, 5, 6, 7, 8} and R = {2, 3, 5, 7}, where
elements that can be found in both sets. It is Q and R are subsets of M, find:
denoted by . Thus, A intersection B is written as i. Q∩R ii. What type of set is Q∩R?
A B.
2. If A = {1, 2, 4, 7}, B = {2, 6, 8} and
Worked Examples C = {4, 5, 6, 7, 8}. Find the following:
1. If A = {5, 6, 7, 8, 9} and B ={9, 10, 11}. Find: i. A ∩ B ii. B ∩ C iii.A ∩ C
i. A B
ii. What type of set is A B? 3. If A = {1, 2, 3, 4,}, B = {1, 2, 5, 6} and C = {3,
5, 7}, list all the members of the following sets:
Solution i. A ∩ (B∩C) ii. A ∪ (B∩C)
A = {5, 6, 7, 8, 9} and B = {9, 10, 11}
i. A B = {9} ii. A B is a unit set. 4. Let F = {1, 2, 3….}, G = {4, 14, 24} and H =
{1, 3, 5, 7}. Find the following sets:
2. Find the intersection of the two sets:
i) F ∩ G ii) G ∩ H iii) F ∩ H
K = {prime numbers less than 20}
L = {odd numbers less than 18} B. Find the intersection of the pair of sets.
1. A = {Even numbers less than 12}
Solution B = {Prime numbers less than 10}
K = {2, 3, 5, 7, 11, 13, 17, 19} 2. M = {Prime factors of 30}
L = {1, 3, 5, 7, 9, 11, 13, 15, 17} N = {Prime factors of 20}
K∩L = {3, 5, 7, 11, 13, 17} 3. P = {Multiples of 12} and Q = {Factors of 24}
4. K = {Factors of 6} and L = {Factors of 12}
3. If X = {prime numbers less than 12} and 5. Given that P = {Factors of 20},
Y = {odd numbers less than 13}. Find X ∩ Y. Q = {Factors of 16}
R = {Multiples of 5 less than 20}
Solution
a. List all the elements of P, Q and R
X = {2, 3, 5, 7, 11} and Y = {1, 3, 5,7, 9, 11}
b. find:
X∩Y = {3, 5, 7, 11}
i. P Q iii. P R iii. Q R iv. Q P
c. What type of set is P Q?
4. P = {prime numbers less than 20}
d. What is the relationship between?
Q = {odd numbers less than 10}.
i. P Q and Q P ii. P Q and P
Find P∩Q.

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Union of Sets Solution
The union of two or more sets is a combination of A = {1, 2, 3, 4, 6, 8, 12, 24} and B = {1, 2, 4, 8}
all the elements of the involving sets in a single Therefore A∪B = {1, 2, 3, 4, 6, 8, 12, 24}
set. Union is denoted by ∪. For example, If A =
{1, 2, 3, 4} and B = {3, 4, 5, 6}, then A union B, 3. If X = {x : x < 13, where x is prime}
written as: A U B = {1, 2, 3, 4, 5, 6} Y = {x : x < 13, where x is odd}.
i. List the elements of X and Y.
Worked Examples: ii. List the members of X∪Y.
1. Given that:
A = {x : 1 < x < 10, where x is odd}, Solution
B = { x : x is a factor of 12} and i. X = {2, 3, 5, 7, 11} and Y = {1, 3, 5, 7, 9, 11}
C = {multiples of 4 less than 10} Therefore, X∪Y = {1, 2, 3, 4, 5, 7, 9, 11}
(a) List all the elements of A, B and C.
(b) Find: 4. List the elements of each set.
i. A ∩ B ii. B ∩ C iii. A B = {whole numbers from 20 to 25}
iv. A U C v. B U C vi. A U B D = {factors of 63} and find the members of:
c. What type of set is? i. B D ii. B∪D
i. A∩B? ii. A∩C? iii. A∪C?
d. What is the relationship between A∩B Solution
and B∩C ? B = {20, 21, 22, 23, 24, 25}
D = {1, 3, 7, 9, 21, 63}
Solution i. B D = {21}
a. A = {3, 5, 7, 9}, B = {1, 2, 3, 4, 6, 12} ii. BUD ={1,3,7,9,20,21,22,23, 24, 25,63}
and C = {4, 8}
5. If P = {7, 11, 13} and Q = {7, 9, 11, 13}. Find
b. i. A B = {3} ii. B C = {4} P∪Q.
iii. A { }
iv. A ∪ C = { 3, 4, 5, 7, 8, 9 } Solution
v. B ∪ C = {1, 2, 3, 4, 6, 8, 12} P = {7, 11, 13} and Q = {7, 9, 11, 13}.
vi. A ∪ B = { 1, 2, 3, 4, 6, 7, 9, 12 } Therefore, P∪Q = {7, 9, 11, 13}

c. i. A B = {3} is a unit set Solution

ii. A C = { } is an empty set. A = {1, 2, 3, 4, 5} and B = {3, 4, 6}
iii. A∪C = {3, 4, 5, 7, 8, 9} is a finite set. i. A∪B = {1, 2, 3, 4, 5, 6}
ii. Number of elements in A∪B = n (A∪B) = 6
d. A { } { } therefore, they
are equivalent sets. Exercises 1.8
Given that E = {7, 9, 1}, F = {1, 5, 7, 9} and G =
2. Find A∪B, if A = {factors of 24} and {1, 7, 13}. Find the following sets:
B = {factors of 8} 1. E∪F 2. E∪G 3. F∪G 4. E∪ (F∪G)

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Properties of Sets Operations 3. Distributive property
1. Commutative property a. A (B ∪ C) = (A B) ∪ (A C)
a. A B = B A b. A ∪ B = B ∪ A b. A ∪ (B C) = (A ∪ B) (A ∪ C)
Consider the sets, A = {1, 2, 5, 8, 9}, B = {2, 5,
Consider the sets A = {1, 3, 6, 7, 8, 9} and 8, 9, 10} and C = {5, 9, 12, 13}
B = {2, 3, 7, 8, 10} A ∪ (B C) = {1, 2, 5, 8, 9} U {5, 9}
A ∪ B ={1, 2, 3, 6, 7, 8, 9, 10} = {1, 2, 5, 8, 9}
B ∪ A = {1, 2, 3, 6, 7, 8, 9, 10}
(A ∪ B) (A ∪ C)
A ∪ B = B ∪ A. = {1, 2, 5, 8, 9, 10} {1, 2, 5, 8, 9, 12, 13}
Hence, we say the union of sets is commutative = {1, 2, 5, 8, 9}
Similarly, A B = {3, 7, 8}, B A = {3, 7, 8} A ∪ (B C) = (A ∪ B) (A ∪ C)
A B=B A
Hence, we say the intersection of sets is Hence for any three sets A, B and C,
commutative A ∪ (B C) = (A ∪ B) (A ∪ C).
The union is said to be distributive over
2. Associative property intersection.
a. (A B) C = A (B C) In a similar manner, A (B ∪ C)
b. (A ∪ B) ∪ C = A ∪ (B ∪ C) = {1, 2, 5, 8, 9} {2, 5, 8, 9, 12, 13}= { 2, 5, 8, 9}

Consider the sets A = {1, 5, 6, 7}, B = {2, 5, 7, 6} (A B) ∪ (A C)

and C = {1, 7, 8, 10} = {2, 5, 8, 9} ∪ {5, 9}
(AUB)UC = {1, 2, 5, 6, 7}U{1, 7, 8, 10} = {2, 5, 8, 9}
= {1, 2, 5, 6, 7, 8, 10} A (B ∪ C) = (A B) ∪ (A C)
AU(BUC) = {1, 5, 6,7}U{1, 2, 6, 7, 8,10}
= {1, 2, 5, 6, 7, 8, 10} Hence for any three sets A, B and C,
(AUB)UC = AU(BUC) A (B ∪ C) = (A B) ∪ (A C). The
intersection is said to be distributive over the
Hence for any three sets A, B and C, (AUB)UC = union.
AU(BUC). The operation of union of sets is
associative. Exercises 1.9
Similarly, (A B) C = {5, 6, 7} {1,7, 8, 10} 1. Given A ={1, 3, 4, 5} B = {1, 3, 5, 8} and C =
= {7} {3, 5, 9, 10}, verify that:
A (B C) = {1, 5, 6, 7} {7}= {7} (AUB)UC = AU(BUC)
(A B) C = A (B C)
2. In A = 2, 4, 5, 7, 10}, B = {2, 3, 5, 8} and C =
Hence for any three sets A, B and C, {2, 5, 10, 12}, show that:
(A B) C = A (B C). The operation of (A B) C = A (B C)
intersection of sets is associative.

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3. Verify that A (B ∪ C) = (A B) ∪ (A C) B = {x : x is multiples of 3} and
for the sets A = {10, 12, 15}, B = {9, 10, 11, 12, C = {x : x is factors of 16}
13} and C = {8, 12, 15} a. List all the elements of U, A, B and C
b. Find: (i) AI (ii) BI (iii) CI
4. For the sets A = {2, 3, 5, 8}, B = {2, 5, 10,
12} and C = {2, 4, 5, 7, 10},verify the relation Illustration of Diagrams
A ∪ (B C) = (A ∪ B) (A ∪ C) A. Joint Sets
B U
A
Venn Diagrams
Sets can be represented in diagrams called Venn
diagrams. In doing so, the universal set and the
complement is very necessary. The shaded regions represent AUB
The non-shaded region represents (AUB) 1
Universal Set
It is the set within which all other subsets can be 2. A B U
found. It is denoted by ∈ Students are
restricted to work within the universal set and not
go beyond it.
The shaded portion represents:
Complement of a Set A ∩B1 = A only
It is the set of elements that can be found in the
A B U
universal set but not in a particular set. It is 3.
denoted by the symbol I. For instance, if = {1,
2, 3, 4, 5} and A = {1, 2, 3}, then the complement
of A, written as: AI = {4, 5}. The shaded portion represents A∩B and the non-
shaded portions represent (A∩B)1
Worked Examples
Given that U = {whole numbers from 1 to 10}, 4. A U
B
A = {first five even numbers} and B = {first five
odd numbers}, if A and B are subsets of U, find:
i. AI ii. BI
The shaded portion represents:
Solution A1∩B = B only
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
i. A = {2, 4, 6, 8, 10} AI = {1, 3, 5, 7, 9} A B U
5.
ii. B = {1, 3, 5, 7, 9 }. BI = {2, 4, 6, 8}

Exercises 1.10 The shaded portion represents:

Given that: U = {x : 1 ≤ x ≤ 12}, A1∩B1 = (AUB)1. That is outside A or B or
A = {x : x is numbers less than 10} both.

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B. Disjoint sets The non-shaded portions represent (A∩B)1
1. U 4.
A B A U
BB

The shaded regions represent AUB The shaded portion represents:

The non-shaded region represents (AUB) 1 A1∩B1 = (AUB)1. That is outside A or B or both

A B U D. Single Sets
2.
1. A U

The shaded portion represents: A∩B1 = A only

3. A B U The shaded region is AI.

The non-shaded region is A.

The shaded portion represents: A1∩B = B only Worked Examples

1. P and Q are subsets of the universal,
4. A B U U = {x: 1 ≤ x ≤ 10}
P = {x: x is even numbers}
Q = {x: x is odd numbers}
The shaded portion represents: i. List all the elements of U, P and Q
A1∩B1 = (AUB)1. That is outside A or B or both. ii. Represent U, P and Q in a diagram
iii. Find the number of subsets in Q.
C. Subsets (B ⊂ A)
1. U Solution
A B i. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
P = {2, 4, 6, 8, 10} Q = {1, 3, 5, 7, 9}

The shaded regions represent A∪B ii. P ∩ Q = { } disjoint diagram

The non-shaded region represents (A∪B) 1
Q U
2. P
A B U 2 4 1 3
6 8
5 7 9
10
The shaded portion represents: A∩B1 = A only
A U iii. n(Q) = 5
3. B
Number of subsets = 2n, but n = 5
25 = 32 subsets
The shaded portion represents A∩B and B only.

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2. Given that: ∪ = {x: 1≤ x ≤ 18} 4. M is the set consisting of all positive integers
A = {x : x is prime numbers} and B ={x : x between 1 and 10. P and Q are subsets of M such
is odd numbers greater than 3} that P = {factors of 6} and Q = {multiples of 2}
a. If A and B are subsets of the universal set, U, i. List the element of M, P and Q.
list all the elements of A and B. ii. Represent M, P and Q on a Venn diagram.
b. Find the intersection of sets A and B iii. Find P∩Q.
c. i Illustrate U, A and B on a Venn diagram.
ii. Shade the region for prime factors of 18 on the Solution
Venn diagram. i. M = {2, 3, 4, 5, 6, 7, 8, 9}
P = {2, 3, 6} and Q = {2, 4, 6, 8}
Solution
a. ∪ = {1, 2, 3, 4……..18} ii. P ∩Q = { 2, 6}
A = {2, 3, 5, 7, 11, 13, 17} Q P U
B = {5, 7, 9, 11, 13, 17}
3 2 4, 8
6
b i. A∩B = {5, 7, 11, 13, 17} ,
ii. A∪B = {2, 3, 5, 7, 11, 13, 15, 17}
U iii. P∩Q = {2, 6}
A B
5, 7
2, 3 11, 13, 9, 15 Exercises 1.11
17 1. If U = {x : 5 < x < 15, x ∈ integers},
1, 4,6,8,10,12,14,16,18
A = {x : x is factors of 14} and B = {x : x is
multiples of 3}, where A and B are subsets of ∪,
show this information on a Venn diagram.
3. a. E and F are subsets of the universal set,
∪ = {x : 1 ≤ x < 15, where x is a positive integer},
2. Given that ∪ = {x : x is factors of 24}, P = {x :
E = {x : 1 < x < 15, where x ∈ even numbers} and
x is odd numbers} and Q = {x : x is prime
F = {x : 9 < x < 15, x ∈ multiples of 4}.
numbers}, where P and Q are subsets of U
i. List the elements of U, E and F.
a. List all the elements of ∪, P and Q.
ii. Draw a Venn diagram to show the sets, U, E
b. Represent ∪, P, and Q in a Venn diagram.
and F.
3. Given that ∪ = {x : 18 < x ≤ 25},
Solution
A = {x : x is multiples of 2} and
i. ∪ = {1, 2, 3,4,5,6,7,8,9,10,11,12,13,14}
B = {x : x is multiples of 3}, where A and B are
E = {2, 4, 6, 8, 10, 12, 14} and F = {12}
subsets of ∪
ii. U
E F i. List the elements of A and B.
2, 4, ii. Show the information in a Venn diagram.
6, 8, 12
10, 14 iii. List the members of (A∪B)1
iv. What type of set is B only?
1, 3, 5, 7, 9, 11, 13

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4. Given that U = {x: x is factors of 36}, M = {x: French and 4 did not choose any of the languages.
x is factors of 10} and N = {x: x is multiples of 3 It is seen that the sum of the choices is 9 + 5 + 4
to 12} where M and N are subsets of U. = 18, which is more than 15 (the number of
i. Show ∪, M and N in a Venn diagram. students), creating a problem and impression of
ii. From the diagram, list the elements of {M∪N} an overlap somewhere. Cases of this nature are
and N only. called two set problems and are solved by first
representing them in Venn diagrams.
5. a. Show the following sets in a Venn diagram;
µ = {x : 1 ≤ x ≤ 10}, A = {x : 1≤ x ≤ 5} and B = Regions of the Venn Diagram
{x : 3 < x < 9} U
A B
b. Show by shading the following regions;
i. A B1 ii. A1 B II I II
I
IV
6. A and B are subsets of a universal set, µ = {x :
1 ≤ x ≤ 18} such that A = { x : x is even numbers} II + I + III + IV = U
and B = {x : x is multiples of 3}
i. List the elements of the sets, µ, A, B, (A B), A B U
(A∪B) and (A∪B)1 A only A B only
ii. Illustrate the information on a Venn diagram.
I
(A )
7. Given that µ is the set of positive integers less
than 100 and the set A and B are subsets of µ. A is n(A) – n(A∩B) + n(A∩B) + n(B) – n(A∩B) +
the set of multiples of 5 and B is the set of n(A∪B)1 = U
multiples of 7; n(A) only + n(A∩B) + n(B) only + n(A∪B)I = U
i. List the elements of A, B and A B. Exactly one = n (A) only + n(B) only
ii. Describe in words the elements of the set
A B. The first circle of the Venn diagram must be
iii. Write down the values of n(A), n(B), n(A B) . represented by the capital letter of the first item
Show that; n(A B) = n(A) + n(B) – n(A B) and the second circle, by the capital letter of the
second item. For e.g. „A‟ for agriculture and „B‟
Two Set Problems for business as shown above.
Two set problem arises when a number of people
are made to choose between only two items. The Region I or (A∩B) is represented by a variable
choice could be for one item only, or both items usually, x, when its value is unknown. When the
or none of the items. The problem occurs value is given, put it in that region.
whenever the choice made is either more than or
Region II or A only is represented by the
less than the number of people making the
number of people who opted for it, n(A), minus
choice. For instance, 15 students were made to
region II, x, (whether region II‟s value is given or
choose between English and French, as their
not). That is: n(A) – x. If n(A) only is given, do
favorite language. 9 chose English, 5 chose

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not subtract x, because n(A)– x = n(A) only, vi. Agriculture or Business or both?
meaning x has been subtracted already.
Solution
Region III or B only must be represented by the i. Agriculture but not Business = 7
number of people who opted for it, n(B), minus ii. Business but not Agriculture = 3
region II, x, (whether region II‟s value is given or iii. Neither Agriculture nor Business = 4
not). That is: n(B) – x . If n(B) only is given, do iv. Exactly two subject = 6
not subtract x, because n(B) – x = n(B) only, v. Exactly one subject = 7 + 3 = 10
meaning x has been subtracted already vi. Agriculture or Business or both
= 7 + 6 + 3 = 16
Region IV or (A∪B)1 represents the number of
people who refuse to make a choice between the 2. In a class of 60 students, 40 play Football and
two items. If the question goes along with the 25 play Hockey. Each student play at least one
phrase “each person chose at least one item”, game
then (AUB)1 or region IV is zero and cannot be a. Represent this information in a Venn diagram.
represented on the Venn diagram. b. How many students play both games?
c. How many students play Football only?
The sum of the numbers of all the regions must d. How many students play Hockey only?
be equal to the total number of people.
U = n(A) only + n(A∩B) + n(B) only + n(A∪B)I Solution
a. Let ∪ = {Students in the class},
Note: Venn diagrams for two set problems must F = {students who play football},
always intersect (Joint set). Before the diagram is H = {students who play hockey}
drawn, prepare a data that defines the variables n(∪) = 60, n(F) = 40, n(H) = 25, n(F
and the values of the variables used. F(40) H(25) U(60)
Worked Examples 40 - x x 25 - x
1. The number of students who offer Agriculture
or Business or both or none at a certain college is
represented in the diagram below; b. 40 – x + x + 25 – x = 60
65 – x = 60
A=13 B =9 U(20)
x = 65 – 60 = 5 students
7 6 3
4 c. n(F) only = 40 – x . But x = 5
n(F) only = 40 – 5 = 35students.
How many students offer; U(60)
F(40) H(25)
i. Agriculture but not Business? iv. n(H) only
ii. Business but not Agriculture? = 25 – x , but x = 5 35 5 20
iii. Neither Agriculture nor Business? = 25 – 5 = 20 students
iv. Exactly two subject? Therefore, 20 students played Hockey only.
v. Exactly one subject?

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3. There are 50 pupils in a class. Out of this ii. Kola but not Rola
= Kola only U(100)
number, speaks French only and of the
= x – 8, R(61) K(37)
remainder speak both French and English. If the
= 37 – 8 53 8 29
rest speak English only; Find the number of
= 29 10
students who speak
i. Both French and English. ii. English only.
iii. Draw a Venn diagram to illustrate the above 5. In a class of 50 students, of them study
information. Biology only, of them study Chemistry only
and 10% of them study neither subject.
Solution i. Represent this information in a Venn diagram
Let ∪ = {pupils in the class}, ii. How many students study both subjects?
E = {pupils who speak English} iii. How many students study Chemistry?
F = {pupils who speak French}
n(∪) = 50, n (F) only = × 50 = 5 Solution
Remainder = 50 – 5 = 45. i. Let U represent the pupils in the class,
n(E∩F) = × 45 = 36 B represents pupils who study Biology,
C represents pupils, who study Chemistry,
E(45) F(41) U(50) n(U) = 50,
ii. The rest
= 50 – (5 + 36) 9 3 5 n(B) only = × 50 = 10
= 50 – 41= 9 6 n(C) only = × 50 = 15
n (E) only = 9 n(BUC) =1
× 50 = 5
U(50)
n(B∩C) = x B
4. In a survey, 100 students were to choose C
between two new drinks, Rola and Koka. 61 liked 10 x 15
Rola, 8 liked both drinks and 10 like neither. 110 x
5
How many liked; 101
a. Koka b. Koka but not Rola ii. 10 + x + 15 + 50= 50
U(50)
x + 30 = 50 B C
Solution x = 50 – 30
x = 20 10 20 15
Let ∪ = {students in the class},
x
R = {students who like Rola} 5
K = {students who like Koka} iii. C = x + 15 = 20 + 15 = 35
n(V) = 100 n(R) = 61 n(K) = x
n(R∩K) = 8 n(R∪K)1 = 10 6. A class was asked to choose between Benz and
61 – 8 + 8 + x – 8 + 10 = 100 U(100) Toyota or both or none. 39% of them chose Benz,
63 + x = 100 R(61) K(x) 52% of them chose Toyota and 23% of them
x = 100 – 63 chose neither Benz nor Toyota. Show this in a
61 - 8 8 x - 8
x = 37 Venn diagram and find the percentage of students
10 who chose both brands of cars.

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Solution chemistry and 15% passed in neither subject.
Let ∪ = {students in the class}, i. Show this in a Venn diagram.
B = {students who chose Benz} ii. Find the percentage of students who passed
T = {students who chose Toyota} both subjects?
n(B∪T) = 100%, n(B) = 39%, n(T) = 52% iii. What percentage of students passed in at least
n(B∪T)1 = 23% , n(B∩T) = x one subject?
39% – x + x + 52% – x + 23% = 100%
114% – x = 100% 2. There are 30 students in a class. 17 of them
B = 39% T = 52% U
x = 114% – 100% belong to the hockey team, 9 belong to the
x = 14% 39% - x x 52% - x football team and 6 belong to both teams. How
x many students do not belong to any of the teams?

7. A and B are two sets and the numbers of 3. A survey was conducted among some people
elements are shown in the diagram below; to find out their favorite network as far as Airtel
and Vodafone were concern. 44% opted for Airtel
U
A BB only, 16% chose both networks and 17% did not
14 + x x 3x opt for any of the networks. Ifeach student opted
for at least one network;
i. what percentage of the people opted for
Given that n (A) = n (B), calculate: Vodafone only?
i. x ii. n(A∪B) ii. what percentage of the people opted for exactly
one network?
Solution
i. 14 + x + x = x + 3x 4. In a class of 35 boys, A is the set of boys who
14 + 2x = 4x take athletics and C is the set who play cricket. n
14 = 2x (A) = 15, n(C) = 16, n(A∩C) = 5. Using the
x=7 whole class as the Universal set, draw a Venn
diagram and mark the numbers in their
ii. n (A∪ B) appropriate regions.
= n(A) only + n(B) only + n(A B ) ii. How many boys take neither athletics nor
But n(A) only = 14 + x = 14 + 7 = 21 cricket?

n(B) only = 3x = 3(7) = 21 5. P and Q are two sets and the numbers of
n(A B ) = 7 elements are shown in the diagram below;
⇒n(A∪ B) = 21 + 21 + 7 = 49
P Q

Exercises 1.12 10 + 2x x 5x –8

1. A number of students wrote an examination in

physics and chemistry and had the following Given that n (P) = n (Q), calculate:
records; 60% passed in physics, 40% passed in i. n(P Q) ii. n(Q) iii. n(P∪Q)

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6. In an examination, x pupils take the history = A1 C1 B = (A∪C)1 B
paper and 3x pupils take the mathematics paper. 4. U
Given that 6 pupils take both papers, illustrate the A B
data on a Venn diagram indicating the number of
pupils in each region. If the number of pupils C
taking the examination is 46, find x.
The shaded region represents C only
7. In an examination, x pupils scored less than 51 = A1∩B1∩C = (AUB)1 C
marks, 2x pupils scored more than 49 marks and
4 pupils scored exactly 50 marks. Illustrate the B. Shading Two Regions
information on a labeled Venn diagram. Given 1. U
that the total number of pupils taking the A B
examination was 32, find x
C
Three Set Problems The shaded region represents A∩B
Three set problem arises when a number of U
people are made to choose between three items. 2. B
A
The choice could be for one item only, two items
only, all the three items or none of the three C
items. The shaded region represents B C

Diagrams for Three Sets 3. U

A B
A. Shading One Region
1. U
A B C

C The shaded region represents A C

The shaded region is A B C C. Shading Three Regions

1.
2. U U
A B A B

C C

The shaded region represents A only The shaded region represents A (B∪C)
= A B1 C1 = A (B C)1 U
A B
2.
3. A C
U
B
C
The shaded region represents B (A∪C)
The shaded region represents B only

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3. U 6. U
A B A B

C C

The shaded region represents C∩(A∪B) The shaded region represents (A∪C) ∩B1

4. U 7. U
B A B
A

C C

The shaded region represents (A∪B) ∩C1 The shaded regions represent exactly two items

5. U 8.
A B
U
A B
C
C
The shaded region represents (B∪C)∩A1
The shaded regions represent exactly one item

Solving Problems Involving Three Overlapping Sets

U U
B B
A A
V II VI A only (A∩ B) - x B only

I x
III IV (A∩C) - x (C∩B) - x

VII VIII C only

C 1
C
(AUBUC)

Fig. I Fig. II

For three sets A, B, and C,

1. a. n(AUBUC) = ( I + II + III + IV + V + VI + VII )
= A only + B only + C only + (A∩B) only + (B∩C) only + (A∩C) only + (A∩B∩C), if the values of
exactly two items only and exactly one item are given

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b. n(AUBUC) = n(A) + n(B) + n(C) – n(A∩ B) – n(B∩C) – n(A∩C) + n(A∩B∩C), if each person like
all the three items (without complement)
c. n(AUBUC) = ( I + II + III + IV + V + VI + VII + VIII)
= n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) + n(AUBUC)1, if some people do
not like all the three items (with complement)
2. Number of people in exactly one set: = (V + VI + VII)
= n(A) + n(B) + n(C) – 2n(A∩B) – 2n(A∩C) – 2n(B∩C) + 3n(A∩B∩C)
3. Number of people in exactly two of the sets = (II + III + IV)
= n(A B) + n(A C) + n(B C) – 3n(A B C)
4. Number of people in exactly three of the sets = n(A B C) = I
5. n(A B) only = (A B) C1 = II
6. n(A C) only = (A C) B1 = III
7. n(B C) only = (B C) A1 = IV
8. Number of people in set A only = A (B1 C1) = V
9. Number of people in set B only = B (A1 C1) = VI
10. Number of people in set C only = C (A1 B1) = VII
11. Number of people who do not like any of the three = n(A∪B∪C)1 = VIII
12. Number of people in two or more sets = (II + III + IV + I) = (at least 2 sets) :
= n(A B) + n(A C) + n(B C) – 2n(A B C)
13. n(A) = V+ I + II + III
14. n (B) = I + II + IV + VI
15. n(C) = I + III + IV + VII
16. n(A B) = I + II
17. n(A C) = I + III
18. n(B C) = I + IV

Note: Questions must be carefully read in order to place values at their respective regions in the
diagram. Any region without a given value must be represented by a preferred variable.

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Type 1 Type 2
It involves the situation whereby the values of all It involves finding the value of the intersection of
the regions of the Venn diagram is given to the three sets. This can be done by either using
answer some related questions. the formula, the diagram or the cover – up
method as shown in the examples below. Take
Worked Examples note of the fact that for all three sets A, B and C,
1. The number of students at Asaaman S.H.S 1A, 1. n(AUBUC) = n(A) + n(B) + n(C) – n(A∩ B) –
offering the various combinations of Arts, n(B∩C) – n(A∩C) + n(A∩B∩C), if each person
Biology and Chemistry is shown on the diagram like all the three items (No complement)
below; U= 65
A = 35 C = 43 2. n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) –
5 1 n(B∩C) – n(A∩C) + n(A∩B∩C) + n(AUBUC)1,
7
5 20 6 2 if some people do not like all the three items
3
(Complement)
2
B = 31
10 Worked Examples
How many students study: 1. In a survey of 200 students of a school, it was
i. Art and Biology but not Chemistry found that 120 study Mathematics, 90 study
ii. Biology but not Art Physics and 70 study Chemistry, 40 study
iii. Chemistry but neither Art nor Biology Mathematics and Physics, 30 study Physics and
iv. Biology and Chemistry Chemistry, 50 study Chemistry and Mathematics
v. Art or Chemistry (or both)? and 20 study none of these subjects.
vi. Art or Chemistry but not both? a. Find the number of students who study all three
vii. Neither Arts nor Biology nor Chemistry subjects.
vii. How many students are in the class? b. How many students study;
i. Physics only? ii. Chemistry only?
Solution
i. n(Art and Biology but not Chemistry) = 3 Solution
ii. n(Biology but not Art) = 2 + 6 = 8 Let U = {Students in the school}
iii. n(Chemistry but neither Art nor Biology) = 12 M = {Mathematics students}
iv. n(Biology and Chemistry) = 20 + 6 = 26 P = {Physics students}
v. n(Art or Chemistry or both) C = {Chemistry students}
= 3 + 7 + 20 + 5 + 6 + 12 = 53 n(MUPUC) = 200, n(M) = 120,
vi. n(Art or Chemistry (but not both) n(P) = 90, n(C) = 70,
= 3 + 7 + 6 + 12 = 28 (MUPUC)1 = 20, n(M∩P∩C) = x
vii. Neither Arts nor Biology nor Chemistry = 10 n(M∩P) = 40, n(P∩C) = 30, n(M∩C) = 50,
viii. Number of students in the class;
= 7 + 3 + 5 + 20 + 2 + 6 + 12 + 10 = 65

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Method 1 the values of the other regions uncovered and
U= 200 equate them to the value of the universal set.
M = 120 P= 90

g Covering set C, we have

f 40 - x
x 30 - x 70 + f + 40 – x + g + 20 = 200
50 - x But f = 30 + x and g = 20 + x
h By substitution,
20 C = 70 70 + 30 + x + 40 – x + 20 + x + 20 = 200
180 + x = 200
For three sets M, P, and C, n(MUPUC) x = 200 – 180 = 20
= n(M) + n(P) + n(C) – n(M∩P) – n(P∩C)
– n(M∩C) + n(M∩P∩C) + n(MUPUC)1 b. i. n(P) only = g = 20 + x
By substitution, = 20 + 20 = 40
200 =120 + 90 + 70 - 40 – 30 – 50 + x + 20
200 = 180 + x ii. n(C) only = h = - 10 + x
x = 200 – 180 = 20 = - 10 + 20 = 10

Method II 2. A class of 43 students was asked to choose

For set M, between offering Arabic or English or French or
f + 40 – x + x + 50 – x = 120 all the three languages. 28 of them chose Arabic,
f = 30 + x …………… (1) 30 chose English, 25 chose French, 11 chose
Arabic and English only, 9 chose Arabic and
For set P, French only and 10 chose English and French
g + 40 – x + x + 30 – x = 90 only. If each student chose at least one language;
g = 20 + x …………….( 2)
i. illustrate this information on a Venn diagram
ii. how many students chose exactly three
For set C,
languages?
h + 50 – x + x + 40 – x = 70
iii. how many students chose exactly two
h = -10 + x …………… (3)
languages?
For three sets M, P and C, n(MUPUC) iv. find the number of students who chose Arabic
= (I + II + III + IV + V + VI + VII + VIII) only or English only or French only.
= x + 40 – x + 50 – x + 30 – x + f + g +h + 20
200 = x + 40 – x + 50 – x + 30 – x + 30 + x + 20 + Solution
x – 10 + x + 20 Let A represents students who chose Arabic, E
200 = 180 + x represents people who chose English and F
x = 200 – 180 = 20 represents people chose French
n(AUEUF) = 63, n(A) = 28, n(E) = 30,
Method III: Cover up method n(F) = 25, n(A∩E) only = 11,
On the diagram, cover up one of the circles and n(A∩F) only = 9, n(E∩F) only = 10,
add the value of that covered circle to the sum of n(A∩E∩F) = x

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i. iv. Arabic only or English only or French only
U= 43
=a+b+c
A = 28 E= 30
= 8 – x + 9 – x + 6 – x, (But x = 5)
a 11 b
=8–5+9–5+6–5=8
x 10
9
c Type 3
F= 25 This involves the situation where the values of all
the regions are given to determine the value of
ii. Method 1 the complement (None of the three)
For set A,
a + 9 + x +11 = 28 Worked Examples
a = 8 – x ………. .(1) In a survey of 120 college students, it was found
that 47 like Voltic mineral water, 52 like Silver
For set E‟ mineral water, 58 like Deep mineral water, 25
b + 11 10 + x = 30 like both Voltic mineral and Silver mineral water,
b = 9 – x ………..(2) 18 like both Silver mineral and Deep mineral
water and 21 like both Voltic mineral and Deep
For set F, mineral water. If 13 of the students like all the
c + 9 + 10 + x = 25 three brands of water;
c = 6 – x ………..(3) i. Illustrate this information on a Venn diagram
ii. How many students like none of the three
But a = A only, b = E only and c = F only brands of water?
For three sets A, E, and F without complement,
n(AUEUF) = A only + E only + F only + (A∩E) Solution
only + (A∩F) only + (E∩F) only + n(A∩E∩F) Note : Because the question did not state that
43 = 8 – x + 9 – x + 6 – x + 11 + 9 + 10 + x every student like at least one brand of water, it
43 = 53 – 2x implies that there is a complement of the 3 sets.
2x = 10
x=5 Let U = college students
V = students who like Voltic mineral water,
E = students who like Deep mineral water
Method II
S = students who like Silver mineral water
Covering up set F,
n(VUDUS) = 120, n(V) = 47, n(D) = 58,
25 + 8 – x + 11 + 9 – x = 43
n(S) = 52, n(V∩S) = 25, n(V∩D) = 21,
53 – 2x = 43
n(D∩S) = 18, n(V∩D∩S) = 13, n(VUDUS)1 = x
2x = 10 U= 180
x=5 V = 47 D = 58
a 8 b
iii. Exactly two languages, 12 13 5
= 9 + 11 + 10 = 30
c
x S = 52

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For three sets V, D, and S, n(VUDUS) Type 4
= n(V) + n(D) + n(S) – n(V∩D) – n(D∩S) It involves finding the values for two or more
– n(V∩S) + n(V∩D∩S) + n(VUDUS)1 regions simultaneously.
I. Draw the Venn diagram and fill the given
By substitution, values in their appropriate regions
120 = 47 + 58 + 52 – 21 – 18 – 25 +13 + x II. Represent regions without values with any
120 = 106 + x preferred variables
x = 120 – 106 = 14 III. For each circle, derive an equation for each of
the variable represented
Method 2 IV. Solve the equations for the values of the
For set V, variables
a + 12 + 13 + 8 = 47 V. Draw a new diagram and substitute the values
a = 47 – 33 of the variables
a = 14 …………(1)
Worked Examples
For set D,
1. A survey conducted at Maase Hospital
b + 8 + 13 + 5 = 58
revealed that out of the number of 80 patients
b = 58 – 26
admitted in June last year, 33 suffered from
b = 32………….(2)
diarrhea, 41 suffered from malaria, 29 suffered
For set S, from typhoid, 9 suffered from diarrhea only, 12
c + 12 + 13 + 5 = 52 suffered from malaria only, 7 suffered from
c = 52 – 30 typhoid only and 11 were sick
c = 22………..(3) of all the three diseases.
a. Represent this information in a Venn diagram
For three sets V, D and S, b. Use your diagram to find the number of
n(VUDUS) patients who suffered from;
= (I + II + III + IV + V + VI + VII + VIII)
i. diarrhea and malaria only?
= 12 + 8 + 13 + 5 + a + b + c + x
ii. diarrhea and typhoid only?
By substitution,
iii. malaria and typhoid only?
120 = 12 + 8 + 13 + 5 + 14+ 32 + 22 + x
iv. none of the three diseases?
120 = 106 + x
x = 120 – 106 = 14
Solution
U = {patients at the hospital}
Method 3
D = {diarrhoea patients}
Covering up set S,
M = {malaria patients}
52 + a + 8 + b + x =120
T = {typhoid patients}
But a = 14 and b = 32
n(D) = 33, n(M) = 41, n(T) = 29,
52 + 14 + 8 + 32 + x = 120
n( D) only = 9, n(M) only = 12, n(T) only = 7,
106 + x = 120
n(D∩M∩T) = 11, n(DUMUT)1 = x,
x = 120 – 106 = 14
n(D∩M) only = a

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n(D∩T) only = b n(M ∩T) only = c U= 80
U= 80 D = 33 M = 41
D = 33 M = 41 9 10 1
9 a 1 11 8 2
3
11 c 2 7
b
x T = 29
7
x T = 29
b. i. n(D∩M) only = 10
From set D ii. n(D∩T) only = 3
a + b + 9 + 11 = 33 iii. n(M∩T) only = 8
a + b + 20 = 33
a + b = 13……………. (1) iv. None of the three diseases,
80 = 9 + 10 + 11 + 3 + 8 + 12 + 7 + x
From set M, 80 = 60 + x
a + c + 11 + 12 = 41 x = 80 – 60
a + c = 41 – 23 x = 20
a + c = 18……………..(2)
2. In a class of 50 students, their choice of the
From set T, colors Red, Yellow and Green were as follows;
b + c + 11 + 7 = 29 32 like Red, 28 like Yellow, 24 like Green, 10
b + c = 29 – 18 like Red only, 11 like both Red and Yellow
b + c = 11………………(3) colors and 4 like exactly three of the colors. If
each student like at least one of the colors, find
eqn (2) – eqn (1) the number of students like;
c – b = 5 ……………… (4) i. exactly one color , ii. exactly two colors.

eqn (3) + eqn (4) Solution

2c = 16 U = {students in the class}
c=8 R = {students who like red color}
Y = {students who like yellow color}
Put c = 8 in eqn (2) G = {students who like green color}
a + 8 = 16 n(U) = 50, n(R) = 32 , n(Y) = 28, n(G) = 24,
a = 18 – 8 = 10 n( R) only = 10, n(Y) only = a,
n(G) only = b, n(R∩Y∩R) = 4,
Put a = 10 in eqn (1) n(R∩Y) only = 7 n(R∩G) only = c
10 + b = 13 n(Y∩G) only = d
b = 13 – 10 = 3
Therefore, a = 10, b = 3 and c = 8

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 U= 50 U= 50
R = 32 Y = 28 R = 32 Y = 28

10 7 a 1 7 9
4 d 0 4 8
c 11
b 1
G = 24
G = 24
i. Exactly one color;
For set R, = n(R) only + n(Y) only + n(G) only
10 + 7 + 4 + c = 32 = 10 + 9 + 1 = 20
c = 32 – 21
c = 11…………………(1) ii. Exactly two colors;
n(R∩Y) only + n(R∩G) only + n(Y∩G) only
For set Y, = 7 + 11 + 8 = 26
7 + 4 + a + d = 28
a + d = 28 – 11 Type 5
a + d = 17…………… (2) It involves the situation where two sets intersects
with each other but not all the three sets.
For set G,
4 + c + b + d = 24
Worked Examples
4 + 11 + b + d = 24 (Put c = 11)
In a school, 50 students were asked their
b + d = 24 – 15
preference for three brands of soft drinks namely;
b + d = 9………………….(3)
Fanta, coke and malt. None liked all the three
brands, 8 liked only Fanta, 17 liked only coke and
For set U,
9 liked only malt. 10 liked two brands, 13 liked
10 + 7 + 4 + a + b + c + d = 50
malt and none liked Fanta and malt
i. Illustrate the information on a Venn diagram
But c = 11 and b + d = 9
ii. How many students liked;
10 + 7 + 4 + a + 11 + 9 = 50 (By substitution)
a. fanta, b. coke, c. at least one soft drink,
a = 50 – 41 = 9
d. one soft drink, e. none of the drinks.
Put a = 9 in eqn (2)
Solution
9 + d = 17
U = {Students}
d = 17 – 9 = 8
F = {students who like Fanta}
Put d = 8 in eqn (3) C = {students who like coke}
b+8=9 M = {students who liked malt}
b=9–8 n(U) = 50, n(F) only = 8, n (C) only = 17
b=1 n(M) only = 9, n(M) = 13, n(C M) = x
Therefore, a = 9, b = 1, c = 11 and d = 8 n(F C) = 10 – x n(FUCUM)1= y
n(F C M) = { }

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 took plantain only and 2 took rice only. 5 took all
U = 50 the three items of food.
C M =13 i. Draw a Venn diagram to illustrate this
F
information
8 10 – x 17 x 9
ii. Use your diagram to find the number of
y children who took:
a. Plantain and beans only;
b. Rice and beans;
For set M,
c. None of the three items of food.
x + 9 = 13
x = 13 – 9 = 4
Solution
i. Let U = {Number of children}
n(F C) = 10 – x
B = {childrenwholikebeans}
But x = 4
P = {Childrenwholike plantain}
n(F C) = 10 – 4 = 6
R = {children who like rice}
n(F) only + n(F C) + n (C) only + n(C M) +
n(U) = 60, n(B) = 21, n(P) = 24, n(R) = 18,
n(M) only + n(FUCUM)1= U
n( B) only = 3, n(P) only = 9,
⇒8 + 6 + 17 + 4 + 9 + y = 50
n(R) only = 2, n(B P R) = 5,
44 + y = 50 1
n(BUPUR) = x, n(B P) only = a
y = 50 – 4 4 = 6
U = 50 n(B R) only = bn(P R) only = c
U= 60
C M =13
F B = 21 P = 24

8 6 17 4 9 3 a 9
5 c
b
6
2
x R = 18
ii.a. Number of students who liked Fanta
= 8 + 6 = 14 ii. From set B
b. Number of students who liked coke; a + b + 5 + 3 = 21
= 6 + 17 + 4 = 27 a + b = 21 – 5 – 3
c. Number of students who liked at least one soft a + b = 13……………. (1)
drink ; = 8 + 6 + 17 + 4 + 9 = 44
From set P,
d. Number of students who liked one soft drink; a + c + 5 + 9 = 24
= 8 + 17 + 9 = 34 a + c = 24 – 5 – 9
e. none of the drinks = 6 a + c = 10……………..(2)

Some Solved Past Questions From set R,

1. 50 children had a choice of beans, plantain and b + c + 5 + 2 = 18
rice, of which 21 of them took beans, 24 took b + c = 18 – 5 – 2
plantain and 18 took rice. 3 took beans only, 9 b + c = 11………………(3)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 23

eqn (1) – eqn (2); of them. The following table gives further details
b – c = 3 ……………… (4) of the subjects studied:

eqn (3) + eqn (4); General Science only 4

2b = 14 Commerce only 5
b=7 Accounts only 7
All three subjects 3
Gen. Sci& Accounts 7
Put b = 7 in eqn (1);
Commerce & Accounts 8
a + 7 = 13
a = 13 – 7 = 6 a. Illustrate the above information in a Venn
diagram.
Put a = 6 in eqn (2); b. Find the number of students who studied:
6 + c = 10 i. General science or accounts or both, but not
c = 10 – 6 = 4 commerce,
ii. Commerce.
Therefore, a = 6, b = 7 and c = 4
U= 60 Solution
a. General science and accounts = x
B = 21 P = 24
U = 55
3 6 9 C
5 4 G
7 x–3
4 5
2 3
x R = 18 7–3 8–3

7 A
20
For set U;
3 + 6 + 7 + 5 + 9 + 4 + 2 + x = 50
4 + (7 – 3) + 3 + (x – 3) + 7 + (8 – 3) + 5 + 20 = 55
x = 50 – 36 = 14
4 + 4 + x + 7 + 5 + 5 + 20 = 55
x + 45 = 55
a. Plantain and beans only: x = 55 – 45 = 10
n(P∩B) only = 6
⇒ n(G C) only = x – 3 = 10 – 3 = 7
b. Rice and beans, U = 55
n(R∩B) = 5 + 2 = 7 G
C
7
4 5
c. None of the three items of food: 4 3 5
n(BUPUR)1 = x = 14 7
20 A

2. In a class of 55 students, some study at least b. i. Number of students who studied General
one of the following subjects: General science, science or accounts or both, but not commerce
Commerce and Accounts. 20 students study none = 4 + 4 + 7 = 15 students

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ii. Number of students who studied Commerce; 30% liked football and hockey, 20% liked hockey
= 7 + 3 + 5 + 5 = 20 students and athletics, 30% liked football and athletics and
10% liked all the three sports;
3. A, B and C are three intersecting sets. Seven a. What percentage of the house liked football
regions of the Venn diagram are P, Q, R, S, T, U and hockey, but not athletics?
and V. b. What percentage liked exactly two out of the
U= 200
three sports?
A B
r c. What percentage did not like any of the three
t u
q p sports?
s
v
C 4. In a survey of 60 people, it was found that 25
people read newspaper H, 26 read newspaper T,
Find; a. i. A∩B1 ii. A∩ (B U C) 26 read newspaper I, 9 read both H and I, 11 read
b. Write an expression for each of the regions in both H and T, 8 read both T and I, 3 read all three
terms of A, B and C newspapers. Find the number of people who read
i. P ii. Q iii. T at least one of the newspapers

Solution 5. In a survey of 200 students of a school, it was

a. i. A∩B1= {t} ii. A∩ (B U C) = {r, p, q} found that 120 study mathematics, 90 study
physics and 70 study chemistry, 40 study
b. i. P = (A B C) ii. Q = (A C) only
mathematics and physics, 30 study physics and
iii. T = A only chemistry, 50 study chemistry and mathematics
and 20 study none of these subjects. Find the
Exercises 1.13 number of students who study all three subjects.
1. A, B and C are sets. n(A) = 17, n(B) = 29 and
n(C) = 14. Also, n(A∪B∪C) = 45, n(A B C1) = 6. A class of 26 boys were each required to have
6, n(B C A1) = 1and n(A C∩B1) = 4. With the a certain text book in English, French and
aid of a Venn diagram, find n(A B C). Mathematics. 19 boys had the English book, 23
the French and 15 the mathematics. 16 had both
2. A school has rugby 15, a cricket 11 and a English and French books, 14 had French and
swimming 8. In all the three teams are 3 boys, 9 mathematics and 13 had mathematics and
are in the rugby team only and 5 are in both the English. How many boys in the class possessed
rugby and cricket teams; 2 boys are in the cricket all the three books?
team only. Show these facts in a Venn diagram,
and deduce the number of boys who represent the 7. In a class of 48 students, it is known that 24 of
school in the cricket team only. them do Arts, 20 do Chemistry and 22 do
Biology. All the students do at least three subjects
3. A survey of the members of a university hall while 7 do Art and Biology, 6 do Art and
revealed the following statistics; 60% liked Chemistry but not Biology and 8 do Chemistry
football, 50% liked hockey, 50% liked athletics, and Biology. Three do all the three subjects. How

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many of them do Chemistry only or Biology only 11. A examination was held for the filling of
or Art only? vacancies in three branches A, B and C of a
certain service. There were 75 candidates, all of
8. The universal set is the set of positive whom were asked to name the branch or branches
integers less than 13. The subsets are defined in which they were willing to accept a vacancy, if
as follows; T = {multiples of 2}, S = {multiples it were offered. The result of this enquiry was as
of 6} P = {prime numbers}. Copy the diagram follows: 3 candidates would accept A only, 3 B
below and write the numbers 1, 2, 3,…,12 in the only and 3 C only, There were all together 15
appropriate regions students who would not accept A, 10 who would
U
not accept B and 34 who would not accept C.
S
T With the aid of a Venn diagram, or otherwise,
6
Find how many were prepared to accept a
vacancy in any of the three branches.
1
P
12. In a certain school, there are 118 boys in form
9. In a secondary school, there are 124 students in three. Of these, 56 play table tennis, 67 play
form two. Of these 86 play table tennis, 84 play football and 44 play hockey. 23 play table tennis
football and 94 play volley ball: 30 play table and football, 18 play football and hockey and 20
tennis and volley ball, 34 play volley ball and play hockey and table tennis. Everybody plays at
football and 42 play table tennis and football. least one game and n boys play all the three
Each student play at least one of the three games games
and x student plays all the three games. i. Express these facts in a Venn diagram
a. Display these facts in a Venn diagram. ii. Find the value of n from the diagram
b. Write down an equation in x and hence find x
13. In a class of 50 students, 27 study French, 24
10. Out of a group of 20 persons, 7 like classical study History and 30 study Geography. Each
music, 12 like soul music, and 10 like highlife student studies at aleast one of the three subjects.
music; furthermore, 3 like both classical and 5 study all three subjects whilst 11 study French
highlife music, 2 like both classical and soul and History, 8 study History and Geography but
music and 2 like all the three kinds music. Draw a not French and 12 study French and Geography.
Venn diagram and find how many of the twenty i. Display these information on a Venn diagram
persons like soul and highlife music but not ii. How many students study only one of the three
classical music. (Assume all the 20 persons like subjects?
at least one of the three kinds of music) iii. How many study exactly two of the three
subjects?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 26

2 REAL NUMBER SYSTEM Baffour – Ba Series

The Real Number System Relating the Number Systems

The following set of numbers constitute the real At this point, the set of real numbers can be
number system: defined to include negative and positive integers,
1. Whole numbers (W): The system of whole zero, positive and negative fractions as well as
numbers consists of numbers in the set {0, 1, 2, decimals and irrational numbers, which can be
3…} represented by points on the number line.
2. Natural numbers (N): The system of natural
numbers consists of numbers in the set {1, 2, The numbers 1, 2, 3… are called counting
3…}. They are also called Counting numbers numbers or natural numbers or positive
3. Integers (Z): The system of integers consists integers.
of numbers in the set; {… -3, -2, -1, 0, 1, 2,
3…}. This is made up of both negative and The numbers 0, 1, 2, 3… are called whole
positive whole numbers including zero. numbers.
4. Rational numbers (Q): The system of rational
numbers consists of the set of both positive and If the set of rational numbers are represented by
negative fractions. In other words, they are Q, the set of integers by Z, the set of whole
numbers by W and the set of natural numbers by
numbers that can be expressed in the form, ,
N, then;
where a and b are integers but b 0.
Q={ }
5. Irrational numbers (H): The system of
Z={ }
irrational numbers consists of numbers that
W={ }
cannot be expressed in the form . This is
N={ }.
because their decimals are non-terminating. For
example, π = 3.13857… is said to be an irrational From the above, the following observations can
number because the digits are neither repeating be made;
successively nor terminating. Other examples of I. The set of natural numbers (N) is both a subset
irrational numbers include; √ , √ , √ , √ . of whole numbers (W) and integers (Z)
II. The set of whole numbers (W) is also a subset
Note: The root sign is not a characteristic of all of integers (Z).
irrational numbers because numbers like √ , √ , III. The set of integers (Z) is a subset of rational
etc are not irrational but rational numbers. numbers (Q),

6. Real numbers (R):The system of real numbers These are summarized as follows;
consists of the set of both rational and irrational N⊂W, N⊂ Z and N⊂ Q
numbers. This is represented by the set; Also, W⊂Z, W⊂Q and Z⊂Q These relationships
R = {… -2, -1, , 0 , , 1, √ …} are shown in the diagram below;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 27

 It follows that if a unit is divided into 8 equal
Q parts and 3 out of it is taken, we have a fraction
Z
W of , where 3 is the numerator, and 8is the
N
denominator and the remaining fraction is

The rectangle represents the set of rational Similarly, when 15 stars are put into groups of 3,
numbers Q. Since N, W and Z are inside the we have 5 groups of 3 as shown below;
rectangle, they are all said to be the subset of Q.
Likewise, each set of numbers that is contained in
another is said to be a subset
When 2 groups of 3 are selected out of the 5
Worked Examples groups, we have a fraction of and the
Find the following sets
a. N ∩Z b. QUZ remaining fraction is .

Exercises 2.2
Solution
a. N = {1, 2, 3…} and W { }. 1. In a test marked out of 10, Romeo scored .
Therefore, N∩W = { }=N What fraction of the test did he get wrong?

b. Z = { } and 2. Some oranges were shared between Cain and

Q={ } Abel. If Abel received of the oranges, what
Since Z⊂Q, then QUZ = Q fraction of the oranges wasreceived by Cain?

Exercises 2.1 3. In a 12-round boxing bout, Bukom Banku

A. Find the following; knocked out his Mexican opponent on round 8
1. NUZ 2. Q∩N 3.W∩Z 4. W∩N i. Express this as a fraction in its lowest term.
5. Q∩Z 6. WUN 7. NUW 8. ZUW ii. What fraction of the bout was not fought?

B. Clasify each as sets N, W, Z and Q Types of Fractions

1. -2.75 2. 0 3. 4. 6 5. 17 The various types of fractions are as follows.
1. Proper fractionIt is a fraction whose numerator
Common Fractions is less than the denominator .That is in the form
A fraction is defined as part of a whole or group. , where a < b or b > a. Examples are; , , ...
It consists of two parts: numerator and
denominator. Thus, if a unit is divided into„b‟
2. Improper fraction
equal parts and „a‟ part out of it is taken, we have It is a fraction whose numerator is greater than the
a fraction of , where a is the numerator and b is denominator. That is, where a > b, or
the denominator.
b < a. Examples are; …

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3. Mixed fraction Worked Examples
It is a combination of a whole number and a 1. Express 0.625 as a fraction to its lowest term
proper fraction i.e. A , where c > b or b < c
Solution
e. g. 1 , 15 , 3 etc
0.625 (3 d. p) =

4. Decimal fraction
It is a fraction whose denominator is a power of 2. Write 0.55 as a fraction in its lowest term
ten. Decimal fractions can be written with a
whole number, making it simpler to do Solution
0.55 (2 d. p)
calculations. Eg are; = 0.43, = 0.051…

5. Like fractions
They are two or more fractions with the same 3. Change 0.42 as a fraction in its lowest
term.
(equal) denominator. That is: and . Examples
are the pairs; and , and etc. Solution
0.42 =
6. Unlike Fractions
They are two or more fractions with different Exercises 2.4
(unequal) denominators. That is; and . Write the following decimals as fractions
Examples are; and , and , 1. 0.75 2. 1.5 3. 0.025
4. 0.15 5. 2.22 6. 0.035

7. A unit fraction Proper Fractions as Decimal Fractions

It is a fraction that has a numerator of 1. When the denominator of a fraction cannot be
Examples are; , , etc expressed as ten or a power of ten, then long
division is required. For example, can be
Decimals as Common Fractions
represented on the long division table as ; √ to
To write a decimal fraction as improper fraction;
mean . Since 7 is greater than 3, it cannot
I. Identify the number of decimal places;
divide 3 to give a whole number but a value less
II. Remove the decimal point to make it a whole
than one (i.e. zero point)
number;
III. Divide the whole by ten exponents the 0.
number of decimal places . For example, in 0.25, 7 3
there are 2 decimal places. If the decimal point is - Suffix 0 to the 3 to get 30.
removed, we have a whole of 25. Then divide 25
0.
by ten exponent the number of decimal places
7 30
which is 2 and express the fraction in its lowest
- Continue with the division .i.e. 7 into 30 to
form. That is; 0.25 = get 4.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 29
 II. Add the numerator to the product obtained.
0. 4
i. e. (A × c) + b
7 30
III. Write a division of the results by the
- multiply the quotient (answer) by 7 and write
denominator and simplify the numerator of the
the product under 30 and subtract. ( (
new result. That is;
0. 4
7 30
−28 Worked Examples
2 Express as improper fractions;
For all remainders, suffix 0 and divide by the 1. 2.
divisor
0. 485… Solution
7 30 (
−28 1. = =
(
20 2. = =
14
60
56 Exercises 2.5
40 A. Write the following as decimals;
Worked Examples 1. 2. 3. 4. 5
Express the following as decimal fraction.
1. = 0.3125 Improper Fractions as Mixed Fractions
Fractions of the form, , where e > c are called
16 50
-48 improper fractions.
20
-16 Improper fractions can be written as mixed
40 fractions of the form A . That is: = A+ .
-3 2
80
-80 Worked Examples
= 0.3125 Write the following as mixed fractions;
1. 2.
Mixed Fractions as Improper Fractions
A mixed fraction is a fraction written in the form
Solution
, where c > b. In A , A is a whole number,
1. =3+ =3
b is the numerator and c is the denominator.
2. =4+ =4
To write a mixed fraction as an improper
fraction, that is: in the form ; Exercises 2.6
I. Multiply the denominator by the whole number. Write the following as mixed fractions
i.e. A × c 2. 3. 5. 6.

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Terminating and Recurring Decimals the denominator. That is: , where m is
A decimal is said to be terminating when it a natural number.
comes to an end. For example, 0.75, 3.4, 0.625,
0.25, etc Worked Examples
Write the next four immediate equivalents
A decimal is said to be recurring when it repeats fractionsfor the following fractions;
digits and does not come to an end. Eg.
i. ii.
0.3333…, 0.181818…, 3.142142… etc

Recurring decimals are terminated by placing a Solutions

dot on the digit that repeats. For example, = = =
0.333…can be written as 0. ̇ whereas
4.181818… can also be written as 4. ̇ ̇ . for short.
ii. = = = = = = = =

If more than two digits repeat, place a dot on the

first and last digits that repeat or sometimes a Application of Equivalent Fractions
line over the pattern. For example, 0.657657… is Two or more equivalent fractions are said to
terminated as 0. ̇ ̇ . have the same value and can be equated as
such. Thus, if one of them contains a variable,
Exercises 2.7 the value of the variable can be found as follows:
Express the following as a decimal I. Simply equate the equivalent fractions;
II. Find the cross products;
1. 2. 3. 4. 5.
III. Solve for the value of the variable.

Equivalent Fractions Worked Examples

Two or more fractions that are written in
1. If 1 : x is equivalent to : 25, find x.
different forms but have the same value are
called equivalent fractions. For instance, if
Solution
and = , then and are said to be (
( *
equivalent fractions, because they have the
same value. ⇒

Equivalent fractions are formed by multiplying ,

the same number by the numerator and x=4
denominator of a fraction. That is; ,
2. If is equivalent to , find the value of x .
where n is a natural number.

Similarly, equivalent fractions can be formed by Solution

dividing the same number by the numerator and

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21(1 + x ) = 7 × 9 (By cross multiplication) II. Select the fractions with common denominator
21 + 21x = 63 (Expansion) as substitutes to the original fractions. That is
21x = 63 – 21 and
21x = 42
x=2 III. Compare the substitute fractions and select the
one with the larger numerator as the greater
3. If , find x fraction. ⇒
IV. Conclude that is less than . ⇒
Solution
(
1 ( *
2. Compare and

3x = 2 (By cross multiplication)

Solution

Exercises 2.8
A. Write four equivalent fractions:
1. 2. 3. 4.

B. Find the missing numbers;

1. 2.
Comparing Unit Fractions
4. 5. To compare two or more fractions each with
numerator 1 (unit fractions), the fraction with
smaller denominator is the greatest. For example,
Comparing Fractions
in and is greater than , written as:
Two or more fractions are compared by putting in
the symbols; ( or ( .
The act is intended to identify the fraction that is Exercises 2.9
greater or lesser than the other. Put in or = ;
1. 2. 3. 4. 5.
Worked Examples
1. Compare and Ordering Fractions
Two or more fractions are ordered by arranging
them either in ascending (increasing) order or
Solution
descending (decreasing) order.
I. Write equivalent fractions for each.
… The two common methods that can be used to
order fractions are;
1. The L.C.M Method

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Rewrite the fractions to have a common
denominator and order the numerators
accordingly
0.62
2. Multiplication by 100% Descending order = 63, 62, 56 = , 0.62,
Multiply each fraction by 100% and round the
product or answer to the nearest whole number.
Exercises 2.10
Then order the whole numbers in accordance with
A. Arrange in ascending order:
the fractions.
1. , 2. 4 , 4 , 4 3.
Worked Examples 4. 5. 6.
1. Arrange the fractions; in order of
ascendency.
B. Arrange in descending order;

Solution 1. 2.
Method 1 3. 4. 0.32, , 27%,
Find the L.C.M. of all the denominators and
workout as follows, C.1. Mr. Johnny used of a full tank of petrol to
=
travel to Kumasi and of the full tank of petrol
to travel to Accra. Which journey required more
Comparing the numerators, the ascending order
petrol?
becomes =

2. Mrs. Aku uses of a bag of flour formeat

Method 2
Multiply each fraction by 100 and round off the pies and of the same size bag of flour for
answer to the nearest whole number. cakes. For which item does she uses less flour?
%
Addition and Subtraction of Fractions
% To add and subtract fractions, consider whether
% they have like or unlike denominators. Like
denominators refer to the same denominator, e.g.
Comparing the product, the ascending order is
and and unlike denominators refer to different
60, 67, 75 =
denominators, e.g. and .
2. Arrange the following fractions in
Addition of Fractions with Like Denominators
descending order 0.62
This is done by adding the numerators of the
respective fractions and maintaining the common
Solution
denominator. i. e
Multiply each fraction by 100%

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Worked Examples Solution
Perform the following addition; Method 1
1. + 2. + L.C.M of 2 and 3 = 6

Solution = =
1. ⇒ = =

2. Method 2
=
Subtraction of Fractions with Like
Denominators Multiplication of Fractions
This is done by subtracting the numerators of the To multiply two or more fractions;
respective fractions and maintaining the common I. Change all mixed fractions if any.
II. Multiply numerators and denominators
denominator. i. e
respectively.
III. Reduce the product to the lowest terms, if
Worked Examples
possible. i.e. = =
Perform the following subtractions.
1. =
Worked Examples
2. 1. Perform

Addition and Subtraction of Fractions Solution

with Unlike Denominators =
Method 1
To add or subtract two or more fractions with ⇒ = =
unlike denominators, convert the given fraction
to fractions with the same denominator and add 2. Perform
or subtract accordingly.

Method 2 Solution
I. Make use of the facts that:
1. = 2. =
3. Find of
II. Simplify the final answer where possible.
Solution
Worked Examples
1.

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Exercises 2.13
Perform the following; To simplify or evaluate complex fractions, apply
of 3. of BODMAS/BEDMAS principles which defines the
order of carrying out the simplification or
evaluation as:
Division of Fractions Bracket → of / Exponent → Division →
Division of fractions, is performed by going Multiplication → Addition → Subtraction
through the following steps: Before BODMAS is applied, change all mixed
I. Change the division to multiplication sign. fractions to improper fractions and write the final
II. Find the reciprocal of the second fraction. answer as a mixed fraction if possible.
III. Multiply the respective numerators and
denominators. This is summarized as: Worked Examples
1. With out using tables or calculators evaluate:
37 of ( ) – 80

Worked Examples Solution

Perform the following; 37 of ( ) – 80
1.
of ( )– (change mixed fractions )
Solution
Method 1 of ( )– (Bracket first)
( *
of –
= =

Method 2 × – = 1× – (of)
= –
( *

= = = – = – = – (

2. 1 3 = =7 (Subtraction)

Solution 2. of (
3 =

Complex Fractions Solution

Complex fractions are usually fractional of ( )
expressions that involve a combination of two or
= of
more fractions, two or more operators and a
bracket.

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= of Solution
Consider the numerator:
= =

= = = =7 Consider the denorminator:

( )
( )
3. Simplify , without using tables or = = = × =

calculators
⇒
Solution ( )
( ) ( ) ⁄
= = = ÷ = × = × = =1
⁄

Consider the numerator; Exercises 2.15

Simplify the following.
( )= ( )= = =
1. ( ) ( )
Consider the denorminator,
2. ( +
= =
3. ( ) (
( ) ⁄
⇒ = = ÷ = × = =6
⁄ 4. ( ) 1 ×( )

4. Evaluate , without using tables or 5. 6.

( )
calculators.

Solution Spending in Terms of Fraction

(Application of Addition and Subtraction of
=
Fractions)
When part(s) of a whole is used, the parts put
Consider the numerator, together are equal to the whole. In other
= = words, since fraction is part of a whole, fractions
of the same whole sum up to one. For example,
Consider the denorminator
8 = =
Thus, Fraction spent + Fraction left = 1….(1)
⁄
⇒ = = ÷ = × = × = From equation (1), it can be deduced that;
⁄
1. Fraction spent = 1 – Fraction left
5. Simplify , without calculators. 2. Fraction left = 1− Fraction spent
( )

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In general, if the fraction spent out of a given Solution
whole is , then the fraction left or remaining Fraction travelled by lorry and taxi
=
fraction = 1− = =

Worked Examples Fraction travelled by train,

1. A boy spends of his pocket money on = 1− =
transport and of the money on sweets.
i. What fraction of his pocket money did he 4. Amina spent of her pocket money on
spend on transport and sweets? transport and food. If she spend on transport
ii. Calculate the fraction of his money left only, what fraction did he spend on food?

Solution Solution
i. Fraction spent on transport and sweets Let the fraction spent on food = x and fraction
= spent on transport only =
Fraction spent on transport and Food =
ii. Fraction left = 1 – Fraction spent +x=
=1– = – = x=
Therefore, the fraction spent on food =
2. A student spent of his pocket money on
transport and fruits. He spent of the remainder 5. A tank can hold 240 liters of water. How much
on sweets. What fraction of his pocket money is water is in the tank when it is full?
left?
Solution
Solution
If or 1 = 240 liters,
Fraction Spent + Fraction Left = 1
= × 240 liters
Therefore, fraction left = 1− fraction spent
= 192 liters
=1 − Therefore, 192 liters of water will be in the tank

of the remainder = when it is full.

Fraction left = – = Exercises 2.16

A. 1. Keziah shared Gh¢2,700.00 between his two
children. The older child received of the
3. Kwaku travelled from Accra to Kumasi. He
amount. How much did the younger child
travelled of the journey by Lorry, of the receive?
journey by taxi and the rest by train. What
fraction of the journey did he travel by train?

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2. Tobia did of his homework in the first 30 Total quantity = fraction/sum of fraction(s)of the
total quantity + remaining quantity
minutes and of the homework in the next 30
For instance, if the fractions of the total quantity
minutes. What fraction of the homework was left
for him to do? spent are and the remaining quantity is m ,
then the total quantity,
3. Lily fetched a bucket of water and used of x= +m
the water for scrubbing the bath house, of the Solve for x to get the total quantity.
water for bathing and the rest for drinking. What
fraction of the water did Lily leave for drinking? Worked Examples
1. Mr. Brown withdrew some money from the
4. Mr. Green has a piece of farm plot. He decided
bank. He gave of it to his son and to his
to use of the land to cultivate cassava, for
daughter. If he had Gh¢500.00 left, how much did
pineapple production and the rest for fish pond. he take from the bank?
Find the fraction of the land he reserved for fish
production. Solution
Method 1
Finding the Total Quantity given the Fraction(s) Fraction Spent + Fraction Left = 1
Spent and the Remaining Quantity
But fraction spent = =
To find the total quantity given the fraction spent
and the remaining quantity, make use of any of ⇒Fraction left = 1− Fraction spent
the following methods; = 1−

Method I But remaining fraction = Remaining amount.

I. Calculate the remaining fraction.
⇒ if = Gh¢500.00
II. Make use of the fact that the remaining fraction
is equal to the remaining quantity. ⇒1 = 6 × Gh¢500 = Gh¢3,000.00
i.e. Fraction left = Quantity left or The amount withdrawn was Gh¢3,000.00
Method 2
Remaining fraction = Remaining quantity Let the amount withdrawn be x
III. By simple proportion, find the value of the
whole (total quantity) which is 1. That is:
Multiply through by L.C.M = 6
If fraction left = quantity left,
⇒1 = =
⇒3x + 2x + 3000 = 6x
3000 = 6x − 3x − 2x
Method 2 3000 = x or x = 3000
I. Represent the total quantity by a variable, say x The amount withdrawn was Gh¢3,000.00
II. Write an equation of the sum of fractions of the
total quantity and the remaining quantity and 2. Mr. Brown spent of his monthly salary on
equate it to the total quantity, x. That is,

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rent, on food and on books. If he still had Fraction of water drawn =
Gh¢55.00 left, what was his monthly salary? Amount of water drawn = 190 liters

Solution But fraction of water drawn is equal to amount of

Let the monthly salary of the man be x water drawn.
⇒ = 190 liters
Multiply through by L.C.M = 60
Capacity of barrel = + = 1

15x + 12(2) x + 10x + 3300 = 60x If = 190 liters,

3300 = 60 x − 15x − 24x − 10x ⁄
3300 =11x ⇒ = ⁄
x = 300 ⁄
= = = = 240 liters
The monthly salary was Gh¢300.00 ⁄

3. After reading of a story book on the If = 240 liters,

⁄
first and second days respectively, Tamar had 10 ⇒ =
⁄
pages of her story bookunread. Find the total
⁄
number of pages of her story book. = = = = 60 liters
⁄

Solution Capacity of barrel = +

Let the total pages of the story book be x. = 240l + 60l = 300liters
=x
= 15x Exercises 2.17
1. Mr. White purchased some bottles of mineral
3x + 5(2) x + 150 = 15x
150 = 15x − 3x −10x water for his son. After a week, he used of the
150 = 2x bottles leaving 6 bottles.
x = 75 pages i. Find the fraction of the bottles of mineral water
that remain after a week.
4. A barrel is full of water and 190 liters ofwater ii. How many bottles of mineral water did he
purchased that week?
is drawn from it leaving it full.Find the capacity
of the barrel.
2. Jerry read of a story book which contained
Solution 24 pages.
Let x be the fraction of water drawn i. What fraction of the story book was not read?
–x= ii. How many pages of the story book were not
read?
x= – = =

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3. Adwoa came with three different vegetables Fractions as Decimal Fractions
from the market. Out of those items, were To write fractions with power of ten as decimals,
go through the following steps;
tomatoes, were eggs and the remaining, pepper.
I. Identify the power of ten within the fraction
i. What fraction of the items was pepper? II. Recognize the numerator as a whole number.
ii. If she went to the market with Gh¢4,800.00, III. Move the decimal point at the right of the
calculate how much money she spent on each numerator to the left according to the power of 10
item. to obtain the decimal fraction.

4. Lazarus spent of his pocket money on food, Worked Examples

on transportation and the rest on books,. 1. Write as a decimal.
i. What fraction of the money did he spent on
books? Ans = Solution
ii. If he had Gh¢66.00 left after spending on food .
and books, how much money did Lazarus have in
The power of 10 2
his pocket? Ans: 231
Therefore = 0.08
5. Araba used of his money to purchase a school
bag and of the money to purchase a school 2. What is the decimal for ?
uniform. Find her total money if she has
Solution
Gh¢25.00 left after her purchases.
= 0.25
Decimal Fractions
A decimal is a fraction with power of ten as Exercises 2.18
denominator. For example, ... Without using Calculculators, express each as
Study the pattern below carefully, a decimal.
0.7 (1d.p), 1. 2. 3. 4.

0.27 (2d.p),
Decimal Places
0.023 (3d.p) The number of digits at the right of a decimal
point in a numeral is called the decimal place of
It is observed that the number of decimal places that numeral. For example, 0.25 has two decimal
depend on the power of ten of the denominator. places because, there are two digits after the
The decimal point separates the whole number decimal point.
part on its right from the fraction (or less than a
whole part) on its left. For instance, in 8.75, the Worked Examples
digit 8 on the left of the decimal point is a whole Identify the number of decimal places
number whilst 75 on the right of the decimal 1) 5.3154 2) 5 3)
point is less than one.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 40
Solution Worked Examples
1. 5. 3154 (4d.p) 2. 5 (0 d.p) 3. = 0.25(2d.p) 1. Express 0.625 as a fraction in its lowest term.

Solution
Exercises 2.19
0.625 (3 d .p) = = =
Identify the number of decimal place(s)
1. 2. 3. 4. 5.
2. Write 0.011 as a fraction

Common Fractions as Decimal Fractions Solution

To express a fraction as a decimal, multiply the
0.011 (3 d.p) = =
numerator and the denominator by a common
natural number such that the denominator
becomes a power of 10. The power of 10 3. Write 0.55 as a fraction in its lowest term.
indicates the number of decimal places the
numerator should be expressed. Solution
0.55 (2 d.p) =
Worked Examples
Convert the following to a decimal fraction; Exercises 2.21
1. 2. Write the following as common fractions
1) 0. 14 2) 0.025 3) 0.01 4) 0.035
Solutions 5) 2.4 6) 0.11 7) 0.002 8) 1.15
1. = = 0.08
Ordering Decimal Fractions
It is the act of arranging decimals in either
2. = 0.25 ascending or descending order. The steps are as
follows:
Exercises 2.20 I. Express the given decimals as a fraction with a
Express as decimals common denominator.
1. 2. 3. 4. II. By comparison, identify the fraction with the
highest numerator, followed by the next, in that
Decimal Fractions as Common Fractions order.
To convert a decimal fraction to a common III. Arrange the given decimals in either
fraction; ascending or descending order.
I. identify the number of decimal place(s) in the
decimal fraction. Worked Examples
II. Remove the decimal point to make it a whole 1. Arrange 0.63, 0.59, 0.6, 0.5 in ascending order.
number.
III. Divide the whole number by ten exponents Solution
the number of decimal places. Highest decimal place = 2

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Express each decimal as a fraction with a 1. Add 3.50 + 2.83
denorminator of “10 exponent 2”
0.63 (2 d.p) , 0.59 (2 d.p) = Solution
3.50
0.6 (1d.p) , 0.5 (1d.p) = + 2.83
(Comparing numerators) 6.33
In ascending order : 0.5, 0.59, 0.6, 0.63
2. Find 2.83 15.69
2. Write the decimals fractions in descending
order: 0.167, 0.25, 0.5, and 0.33 Solution
2.83
Solution + 15.69
Highest d.p. = 3 18.52
⇒Express each decimal as a fraction with a
denorminator of “10 exponent 3”. Multiplication of Decimals
To multiply two or more decimals, follow the
0.167 = , 0.25 = steps below;
I. Find the sum of decimal places
0.5 = , 0.33 =
II. Remove the decimal point and multiply the
whole numbers.
III. After getting the product, move the decimal
In descending order; 0.5, 0.33, 0.25, 0.167 point to the left with respect to the sum of the
decimal places to get the product of the decimals.
Exercises 2.22
A. Write in ascending order; Worked Examples
1) 0.51, 0.1, 0.02, 0.05 2) 2.5, 0.5, 0. 25 1. Multiply 3.25 by 0.5
3) 0.17, 1.7, 0.0017, 10 4) 0.33, 0.81, 0.4, 4.4
Solution
B. Write in descending order;
1. 3.5, 0.63, 3.3, 0.33 Method 1
2. 0.69, 0.52, 2.8, 6.3 3.25 × 0.5
3. 2.19, 3.47, 0.02, 2.83 Sum of d.p 2 + 1 = 3
325 × 5 = 1625
Addition and Subtraction of Decimals Express 1625 to 3 d.p = 1. 625
To add and subtract decimal fractions, align or 3.25 × 0 .5 = 1.625 (3 d. p)
arrange the digits in a vertical column or on the
place value chart, such that the decimal points are Method 2
in line with each other. 3.25 0.5
= = = 1.625
Worked Examples

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2. Find 6. 801 5.13 A number written in standard form must have the
following features:
Solution 1. a decimal fraction,
6. 801 × 5. 13
2. a multiplication sign,
Sum of d. p = 3 + 2 = 5
3. a positive or negative power of ten.
6801 × 513 = 3488913
6. 801 × 5. 13 = 34 .88913 Rules
I. Identify the position or location of the decimal
Exercises 2.24 point
Perform the following: II. Relocate, or move or place the decimal point
1. 0.7 4.5 1.01 2. 0.4 0.12 0.6 between the first two natural numbers(standard
position).
Division of Decimals
III. Each movement of the decimal point to left is
To divide two decimals;
10 and each movement of the decimal point to the
I. Change all the decimals to fractions.
right is
II. Then change the division sign to
IV. After completion, ignore zeros before the first
multiplication.
counting number and after the last counting
III. Find the product of the first fraction and the
number.
reciprocal of the second fraction.
Worked Examples
Worked Examples 1. Write 38200 in standard form.
1. Divide 0.85 by 0.25
Solution
Solution
The standard form of 38200 = 3.82 ×104
0. 85 ÷ 0.25 = = = 3. 4
2. Write 0. 00053 in standard form.
2. Perform 0. 64 ÷ 0.2
Solution
Solution Place the decimal point between 5 and 3 by
moving 3 places to the right;
= = 3.2
0.00053 = 5.3
= 5.3 ×
Standard Forms
It is the process of writing a number as aproduct 3. Write 84.36 in standard form.
of a decimal and a power of ten. That is
n
where 1 and n is given by the Solution
number of places the point is displaced from the 84. 36 = 8. 436 10
standard position. For example, 124 can be
written as; 1.24 × and this is called the 4. Write 61.38 102 in standard form.
standard form of 124.

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Solution iii. Fill empty spaces with zeros.
2 3
61. 38 = 6.138 10 = 6.138 10 ⇒1.25 × = 0.000125

Exercises 2.26 Exercises 2.27

A. Write the following in standard form Write the numerals for the following:
1) 0.0014 2) 0.00003975 3) 349.3 1.342.5 × 102 2. 117.3
4) 98000 5) 342.5 × 102 6) 4.3 × 3. 5.3 4. 71.803 × 105
5. 402 × × 108 6. 9 × × 104
B. Express the square root of each of the
following in the form: a × Addition and Subtraction of Standard Forms
1. 0.0009 2. 9 × To add or subtract standard form numbers:
3. 0.36 × 4. 1.6 × I. Rewrite each standard form number as an
ordinary number
Numerals for Standard Forms II. Add or subtract these ordinary numbers
When writing numerals for standard forms, III. Convert the results to standard form
consider the following:
I. If the power of ten is positive, move the Work Examples
decimal point to the right according to the 1. Calculate 3.2 × 104 + 9.7 × 103
value of the power of ten.
II. If the power of ten is negative, move the
Solution
decimal point to the left according to the value of
3.2 × 104 = 3.2 × 10,000 = 32,000
the power of ten.
9.7 × 103 = 9.7 × 1,000 = 9,700
III. After the movement(s), write all the numerals
32,000 + 9,700 = 41,700
but occupy all empty space (s) with zero (s) to get
41,700 = 4.17 × 104
the numeral for the standard form.
 3.2 × 104 + 9.7 × 103 = 4.17 × 104

Worked Examples 2. Add 4.25 × and 8.3 ×

Write the numerals for the standard forms:
1) 9.06 × 103 2) 1.25 × 3) 14.4 × × 105 Solution
4.25 × = 0.00425
Solutions 8.3 × = 0.00083
1. In 9.06 × 103,
0.00425 + 0.00083 = 0.00508
i. Power of ten = 3,
⇒0.00508 = 5.08 ×
ii. Move decimal point, 3 times to the right
 4.25 × + 8.3 × = 5.08 ×
iii. Fill empty space with zero.
⇒9.06 × 103 = 9060
3. Calculate 5.5 × – 6.95 ×
2. In 1.25 × ,
Solution
i. Power of ten = −4,
5.5 × = 5.5 × 100,000 = 550,000
ii. Move decimal point 4 times to the left

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6.95 × = 6.95 × 10,000 = 69,500 Exercises 2.29
550,000 – 69,500 = 480,500 Perform the following:
480,500 = 4.805 × 1. 2 × ×6×
 5.5 × – 6.95 × 4.805 × 2. 6 × ×3×
3. 6.3 × ×4×
Exercises 2.28 4. 1.2 × × 5.2 ×
A. Perform the following:
1.5 × +3× Application of Standard Forms
2. 4.5 × + 3.5 × Sometimes, students are required to perform an
3. 7.5 × + 2.5 × operation and express, write or leave the answer
4. 7.5 × + 2.5 × in standard form. Students must therefore, make
use of the knowledge of writing numbers in
B. Perform the following; standard form, as well as writing the numerals for
1. 8 × – 6.95 × standard forms.
2. 5.4 × – 9.95 ×
3. 1 × –1× Worked Examples
4. 7.5 × – 2.5 × 1. Simplify and leave your answer
in standard form
Multiplying Standard Form Numbers
Numbers in standard forms are multiplied by Solution
finding the product of the whole numbers
(mantissas) and adding the exponents. That is :
× = =

Worked Examples =
1. Calculate 2 × ×3×
=
Solution
2× ×3× = × = 0.5 × =5
= (2 × 3) × ( × )
=6× 2. Evaluate (
( )(
, without using
(
=6×
calculation.
2. Multiply 3 × by 2.5 ×
Solution
( )(
Solution =( (
3× × 2.5 ×
( )(
(3 × 2.5) × ( × ) = ( (
= 7.5 ×
= 7.5 × =

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= (
=( )
= =6× =6×
=( )

3. Without using calculators, evaluate =(

√ =( ×( = 13 × 10 = 1.3 × 102
, leaving your answer in
standard form. Exercises 2.30
A. Without using calculators simplify and
Solution
leave the answer in standard form:
√
1.

=( ) 2.
( (
=( ) 3.

= ( ) C. Simplify without using calculators and

leave your answer in standard form:
=( ) (
1. √
=(
(
=( 2. √
=( ×(
=( ×( =6× 3. √

4. Given that a = 5.0 × 102, a = 12.0 × 102 and 4. √

c = 100, evaluate without using tables or
calculators √ and leave the answer in
D. 1. Given that m = 13.0 × 102, n = 5.0 × 102
standard form. and r = 10, evaluate without using tables or

Solution calculators √ , leaving your answer in

√ standard form.
( (
=√ 2. Given that a = 4.0 × , b = 4.0 × and
c = 1.0 × , evaluate without using tables or
=√
calculators , leaving your answer in
=√ standard form.

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Approximation viii. hundredth ix. tenth
Measures of length, mass, time, area, population,
moneyetc, should always be given to a reasonable Solution
degree of approximation, especially when they i. To the nearest whole number = 1,908,654
cannot be stated with exactness or precision. ii. Nearest ten = 1,908,600
iii. Nearest hundred = 1,908,700
There are three main ways by which iv. Nearest thousand = 1,909,000
approximations may be done. By rounding off to: v. Ten thousand = 1,910,000
1. a nearest appropriate unit; vi. Hundred thousand = 1,900,000
2. a given number of decimal places; vii. One million = 2,000,000
3. a given number of significant figures. viii. Hundredth = 1,908,653.730
ix. Tenth = 1,908,653.700
Rounding to the Nearest Multiples of Ten
In a given numeral, the place values are identified Exercises 2.31
from the left of a decimal point as; Ones 1. Round 51,624,362 to the nearest:
(Nearest whole number), Tens, Hundreds, i. tens ii. Thousands iii. hundred thousands
Thousands, Ten thousand, Hundred thousand, iv. Hundreds v. ten thousands vi. one million
One million etc. To the right of the decimal
point, the place values are identified as; Tenth, 2. Round 342.0532 to the nearest;
Hundredth, Thousandth, Ten thousandth etc a. Tenth b. Whole number
c. Hundredth d. Ten thousandth
Rules for Rounding – off a Number
Rounding a number to a given place, that is: Writing to a Given Number of Decimal Places
nearest ten or multiples of ten depends on the I. Identify the number of decimal places in the
following figure on the right (the next immediate given decimal fraction.
digit to the right). II. Identify the retain digit(s) that represents the
1. If the next figure is greater than 5, increase the decimal places to which you are approximating
round - off figure by 1. III. Consider the next immediate digit on the right
2. If the next figure is 5, round off to the nearest of the retain digit(s);
even number . a. If that digit is less than 5(0 to 4), maintain the
3. If the next figure is less than 5, leave the round retain digit(s) as the answer.
– off figure as it is. b. If that digit is more than 4 (5 to 9), increase the
last retain digit(s) by 1 and nothing more.
In each case, the digits after the round off figure
are replaced by zeros. Worked Examples
Round off 5.20735 to:
Worked Examples i. 4 decimal places ii. 3 decimal places
Round – off 1,908,653.727 to the nearest; iii. 2 decimal places iv. 1 decimal place
i. whole number ii. Tens iii. hundreds
iv. Thousands v. ten thousands Solution
vi. hundred thousands vii. one million i. 5.20735 = 5.2074 to 4 d. p.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 47
ii. 5.20735 = 5.20735 to 3 d. p. b. 6 3 0 5 1
iii. 5.20735 = 5.21 to 2 d. p.
1 2 3 4 5 ( 5 S.F)
2. Find 4.13 × 1.15 and correct your answer to 2
decimal places. 2. All non-zero digits are significant. For
example, 261542 has 6 significant figures.
Solution
4.13 × 1.15 3. All zeros between non-zero digits are
Sum of d .p = 2 + 2 = 4 significant. For e.g. 0.4002 has 4 significant
×115 = 47495 figures.
⇒4.13 1.15 = 4.7495 (4 d. p) = 4.75 (2 d. p)
4. Leading zeros to the left of the first non-zero
Exercises 2.32 digits are not significant; such zeros merely
Correct to the decimal places indicated: indicate the position of the decimal point: e.g.
1. 0.05431 (2d.p) 2. 20.556 (2d.p) 0.013 has 2 significant figures and 0.00407 has 3
3. 432.97 (1d.p) 4. 40.4563 (2d.p) significant figures
5. 5.9730 (3d.p) 6. 7.9994 (2d.p)
5. Trailing zeroes that are also to the right of a
Significant Figures decimal point in a number are significant:
Another convenient way to indicate the degree of 0.0230 has 3 significant figures,
approximation is by means of the number of 0.4200 has 4 significant figures.
figures used. Thus, we say 21.3 has 3 significant
figures and that 54 has 2 significant figures. 6. When a number ends in zeroes that are not to
All digits are significant but in the case of zero, it the right of a decimal point, the zeros are not
is insignificant when it used to indicate the necessarily significant: e.g. 190 may be 2 or 3
position of a decimal point. In other words, significant figures, 50600 may be 3, 4, or 5
zero(s) immediately after the decimal point is not significant figures.
significant.
Writing a Number to a Given Number of
Significant Figures
Finding the Number of Significant Figures
Observe of the following rules when correcting a
To identify the number of significant figures in a
number to a given number of significant figures:
given numeral observe the following guidelines:
I. Identify the retain digits representing the given
Steps:
number of significant figures to be converted to.
1. Count from left to right, the first counting
For e.g. to convert 6305 (4 s.f ) to 3 s.f, the retain
number, ignoring the zeros to locate the
digits are 630
significant figures. For e.g. 0.0045672 and 63051
have 5 significant figures as shown below;
II. Consider the digit to the immediate right of the
a. 0. 0 0 4 5 6 7 2 last retain digit(s):
a. If it is 5 or more, (i.e. 5, 6, 7, 8, 9), increase
1 2 3 4 5 ( 5 S.F) the last retain digit by 1.

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b. If it is less than 5 (that is 0, 1, 2, 3, 4), maintain Last retain digit + zero = 7 + 0 = 7
the last retain digit. In other words, the last retain Replaced the rest of the digits with zeros.
digit remains the same. 2719 = 2700(2 s.f)
4. Correct 143.6943 to 4 significant figures.
III. All numbers after the retain digits are
replaced by zero(s) to maintain the value of the Solution
given number, where necessary. This rule is Retain digits = 143.6
applicable to only decimal whole numbers like The next immediate digit = 9, (greater than 4)
45271. For e.g: n45271 (5 s.f) = 45300(3.s.f) Last retain digit + one = 6 + 1 = 7
Replace the remaining digits by zeros
Worked Examples
143.6943 = 143.7(4 s.f)
1. Correct 0.00479 to 2 significant figures.
Exercises 2.33
Solution A. Correct to the significant figures indicated;
Zeros before non-zero digit(s) are not significant. 1. 466901 to 4.s.f 4. 65004 to 3.s.f
Significant figures = 47 2. 107422 to 2.s.f 5. 17612761 to 5.s.f
Retain digits = 0.0047.
The next immediate digit = 9 (greater than 4) Last B. Write to the indicated significant figures:
retain digit + 1 = (7 + 1) = 8. 1. 0.005799 to 3.s.f 2. 0.099506 to 2.s.f
Therefore, 0.00479 (3 s.f) = 0.0048(2 s.f) 3. 55.61489 to 5.s.f 4. 176.904 to 3.s.f

2. Correct 62049 to 3 significant figures. Rational Numbers

Any number that can be expressed in the form, ,
Solution where b ≠ 0 and a and b are integers is called a
Retain digits = 620
rational number. For e.g, , - 4, 3 , 5.2, 0.3
The next immediate digit = 4 ( less than 5)
Last retain + zero = (0 + 0) = 0 Rational numbers can also be described as
Replaced the rest of the digits after the last retain numbers with terminating or repeating decimal
digit by zeros. representation. For e.g, = 0.25 and = 0.3333…
Therefore, 62049 (5 s.f ) = 62000 (3 s.f)
A mixed number can also be described as a
Note : The 3 significant figures in the above rational number. That is: A = . For
example (62000, 3 s.f. ) are 620. This zero is
significant because it occupies the position as last
example, 2 = = 2.33
significant figure (3rd digit of 3.s.f)

3. Correct 2719 to two significant figures. Exercises 2.36

A. Write in the form , b ≠ 0
Solution 1)12.6 2) − 0.04 3) − 3.3 4) 3 5) 5 6)12
Retain digits = 27
The next immediate digit = 1, (less than 5)

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Rational Numbers on a Number Line Worked Examples
All rational numbers can be represented on a 1. Represent -1 and 2 on a the same number
number line. Locate positive integers on the right
line.
of 0 and negative integers on the left of 0 on the
number line.
Solution
a. Divide each unit of the number line into 5
I. Representing Rational Numbers of the Form;
equal parts
or , b ≠ 0, on a Number Line
b. Count 2 from -1 to the left to locate -1
Draw a number line and divide each unit into the
number of times of the denominator (b) . Count c. Count 4 from 2 to the right to locate 2 as
from zero, to the value of the numerator (a) to shown below;
locate the position of the fraction. -1 2

Worked Example
1. Represent on the same number line; -2 -1 0 1 3
2

Solution 2. Represent -3 on a number line.

To locate on the number line;

Solution
a. Divide each unit of the number line into 7
Locating -3 and dividing each unit into 3 equal
equal parts
parts. From - 3, count 1 to the left to locate -3
b. Count 3 from 0 to locate
c. Count 11 from zero to locate

A
-4 -3 -2 -1
Decimal in the form A.B on a Number Line
0 1 2 3 I. Draw a number line and divide each unit into
10 equal segments.
What rational number represents A?
II. Identify the whole number (A) on the number
III. Representing Rational Number of the Form line and from A, count B times to the right (if A is
positive)
A , c ≠ 0, on Number Line
III. From A, count B times to the left (if A is
Draw a number line and divide each unit into c negative).
equal parts. Locate A on the number line from 0,
and if it positive, count b times from A to the Worked Examples
right and if it is negative, count b times from A to Represent the following on the same number
the left to locate the mixed fraction on the line 0.4, 1.6, and - 0.7
number line.

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Solution Recurring decimals are written in the short form
-0.7 0.4 1.6 by placing a dot on the digit(s) that repeats. From
example, 3, 15÷11=1. ̇ ̇ (Read as “one point
three, six, dot, dot”) and from example 4, 1÷ 3 =
-1 0 1 2 0. ̇ (Read as “zero point three, dot”)

Exercises 2.37 Exercises 2.38

A. On separate number lines, represent the Express as recurring decimals;
following group of rational numbers; 1. 2. 3. 4.
1. , , 3. ,

2. , 4. 1 , 3 , 4 Irrational Numbers
Consider π = = 3. 142857143…

B. Locate the following on a number line; It is crystal clear that it is non- terminating but
non-repeating decimal. That is to say that
2. - 2 5. 6.3 8. -3
although the digits are not terminating, they are
3. - 4.8 6. -6 9. -3.5 not repeating successively as well. Such numbers
are called irrational numbers. Other examples of
C. On the following number lines, identify the
irrational numbers are √ , √ and √
rational numbers represented by letters
1. A B C D
Note: It is not the root sign that determines
whether a number is irrational or not. This is
-3 -2 -1 0 1 2 3 because√ 5,√ 2, so √ and √ are
P R rational numbers.
2. Q M N

Exercises 2.39
A. 1. Give 3 examples of irrational numbers.
-2 -1 0 1 2
2. Decimals that do not come to an end are
Terminating and Non –Terminating Decimals called…
Consider the following examples, 3. Decimals that repeat digits are called…
1. 19 ÷ 8 = 2.373 2. 6 ÷ 5 = 1.2 4. Numbers that do not repeat digits but do not
3. 15 ÷ 11 = 1.3636… 4. 1† 3 = 0.3333… come to an end are called…..

In examples 1 and 2, the digits after the decimal

B. Identify the type of decimal, whether
point come to an end. Such decimals are called
terminating or non-terminating
Terminating decimals. On the other hand, in
examples 3 and 4, the digits after the decimal 1. 2. 3. 4. 5.
point keep repeating themselves and are therefore
called repeating or recurring or non-terminating Repeating Decimals as Fractions
decimals. To write repeating decimals as fractions;

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I. Represent the decimal by a variable say x, and Let x = 0.13333 ………..(1)
name it equation (1). eqn (1)
II. Identify the number of digits that repeat. 10x = 1.3333 ….………..(2)
If one digit repeats, multiply eqn (1) by 10. eqn (1) × 100
If two digits repeat, multiply eqn (1) by 100. 100x = 13.333 ………….(3)
If three digits repeat, multiply eqn (1) by 1000 in
that order to obtain eqn (2) eqn (3) – eqn (2);
III. Find eqn (2) − eqn (1) and simplify to get the 90x = 12
variable as representing the fraction. x=

Worked Examples Exercises 2.40

1. Write 0. ̇ as a fraction Express the decimals as fractions;
1) 0. ̇ 2) 4. ̇ ̇ 3) 0. ̇ ̇
Solution
4) 0. ̇ ̇ 5) 1. ̇ ̇ 6) 3.1 ̇ ̇
Let x = 0.333….............. eqn (1)
Because only one digit is repeating, multiply
Properties of Operations on Rational Numbers
1. Commutative Property of addition
eqn 1 by 10.
Addition is said to be commutative on the set of
eqn (1) × 10
rational numbers, if for any two rational numbers,
10x = 3.33….................. eqn (2)
and , b ≠ 0 and d ≠ 0, + = +
eqn (2) – eqn (1)
9x = 3 2. Commutative Property of Multiplication
x= Multiplication is said to be commutative on the
set of rational numbers if for any two rational
numbers, and , b ≠ 0 and d ≠ 0, × = ×
2. Express 0.424242… as a fraction

Solution 3. Associative Property of Addition

Let x = 0.424242….......... eqn (1) Addition is said to be associative on rational
numbers if for any three rational numbers, ,
eqn (1) × 100 and , b ≠ 0, d ≠ 0 and f ≠ 0;
100 x = 0.42.4242.…....... eqn (2)
eqn (2) – eqn (1); ( )+ = +( )

99x = 42 4. Associative Property of Multiplication

x= Multiplication is said to be associative on rational
numbers if for any three rational numbers, ,
3. Express 0.13333… as a fraction and , b ≠ 0, d ≠ 0 and f ≠ 0;
( )× = ×( )
Solution

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5. Distributive Property of Multiplication over 5*2=5–6=-1
addition
For all rational numbers, , and , b ≠ 0, d ≠ ii. 2 * 5 = 2 – 3(5)
2 * 5 = 2 – 15 = - 13
0 and f ≠ 0;
( )=( ) +( ) 2. A binary operation * is defined over R, the set
of real numbers by a*b = 3a + b – 1. Find
6. Distributive Property of Multiplication over a*b if;
Subtraction i. a = 2 and b = -3 ii.a = -1 and b = 4
For all rational numbers, , and , b ≠ 0, d
Solution
≠ 0 and f ≠ 0;
i. a * b = 3a + b – 1
( )= ( )–( ) 2 *-3 = 3(2) + (-3) – 1
2 *-3 = 6 – 3 – 1 = 2
Exercises 2.41 ii. a = -1 and b = 4
Given that × ( )=( ) × 3.2, find: -1 * 4 = 3(-1) + 4 – 1
-1 * 4 = -3 + 4 – l = 0
i. the values of a and b
ii. What property of multiplication
3. The operation ◙ is defined on the set of real
numbers R, by m ◙ n = √ , where the positive
Binary Operations
square root is taken. Find: 10 ◙ (4 ◙ 25)
A binary operation is any symbol or sign that is
used to combine two elements of a set according
Solution
to some clearly defined rule.
m◙n=√ ,
Apart from the usual operations, +, – , , ×, ∪, , 10 ◙ (4 ◙ 25)
there are other signs such as ∆, ◘, ◙, ‡, ○, *, etc 4 ◙ 25 = √ =√ = 10 (Bracket first)
that are used to combine two elements of a set 10 ◙ 10 = √ =√ = 10
according to some rule. For example, in
a * b = 2a – b, a and b are combined by the 3. A binary operation ○ is defined on the set Q of
operation * under the rule 2a – b. Therefore, the rational numbers by: x○y = x + y + 2xy.
symbol, * is called a binary operation a. Evaluate ○ and ○

Worked Examples b. Find the value of x given that x ○ 3 = 24

1. A binary operation is defined over the set of
real numbers, R, as a * b = a – 3b. Find: Solution
i) 5 * 2 ii) 2 * 5 a. ○ = + + 2 ( × )= + + =2

Solution
a *b = a – 3b ○ = + +2( ) = + + =2
i. 5 * 2 = 5 – 3(2)

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b. x○3 = x + 3 + 2 (3x) a. p * q = p2 + q2 – 2pq
x○3 = 3 + x + 6x √ *√ (√ ) + ( ) – 2(√ ) ( )
√ √
x○3 = 7x + 3 (But x ○ 3 = 24)
7x + 3 = 24 =3+ – 2(√ ) ( )
√ √
7x = 21 =3+ ( )
x=3 =3+ =
4. The operation ∇ is defined on the set of real
number s by a∇b = a + b + 2ab, where a, b∈R *√ ( ) + (√ ) – 2( ) (√ )
√ √ √
i. a. Calculate 2 ∇ 3 and find (2∇3)∇ 5; = + 3 – 2( ) (√ )
b. Find the truth set of a∇ 7 = (a∇ 5) + (a∇ 2) √ √
= +3–1=
Solution
i. a∇b = a + b + 2ab ii.(√ ) * (√ )
√ √
2 ∇ 3 = 2 + 3 + 2(2)(3) = 17
* = ( ) + ( ) – 2( ) ( ) = 0
(2∇3)∇ 5 = 17∇ 5
= 17 + 5 + 2(17)(5) = 192 6. The operation * is defined on the set of real
numbers by m*n = ,n
ii. a∇ 7 = (a∇ 5) + (a∇ 2)
a. Evaluate 3 * (5 * 2)
a∇ 7 = a + 7 + 2(a)(7)
b. Find the truth set of :
= 15a + 7
i. 8 * k = 12 * 3 ii. k * 8 = 12 * 3

a∇ 5 = a + 5 + 2(a)(5)
Solution
= 11a + 5 a. 3 * (5 * 2)
a∇ 2 = a + 2 + 2 (a) (2) m*n=
= 5a + 2
(5 * 2) = =
a∇ 7 = (a∇ 5) + (a∇ 2) ⁄ ⁄
⇒3 * (5 * 2) = 3 * = ⁄
= ⁄
=1
⇒15a + 7 = 11a + 5 + 5a + 2
15a – 11a – 5a = 7 – 7
b. i. 8 * k = 12 * 3
- a = 0, a = 0
=

5. The operation * is defined on the set of real =

numbers by p * q = p + q – 2pq, where p, q∈ R
2 2
=3
a. i. Evaluate √ * and *√
√ √ 8 – k = 3k
ii. Use your results in (i) to evaluate; 8 = 3k + k
(√ ) * (√ ) 8 = 4k
√ √
k= =2
Solution
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ii. k * 8 = 12 * 3 p2 – q2 = p2q2…………..(1)
= But p q= –

= p2 q2 = – =

=3 From eqn (1),

p2 – q2 = p2q2
k–8=3×8
p2 q2 = =1
k – 8 = 24
k = 24 + 8
Exercises 2.42
k = 23
1. A binary operation * is defined on the set real
7. Two binary operations are defined as numbers by x*y = x +y + xy. Evaluate:
follows: p * q = + and p q = – a. 3 * 5 b. -2 * -2

a. If p = and q = , evaluate ; 2. The operation ◙ is defined over the set of real

i. p * q ii. p q iii. numbers, R, by a ◙ b = a + b2. Find:
b. If p * q = , evaluate p2 q2 a. (3 ◙ 2) ◙ 5 b. k if 3 ◙ k = 7

3.A binary operation * is defined on R, the set

Solution real numbers by p*q = p2 + 2pq. Find;
a. i. p * q
i. 3 * 5 ii. 5 * 3
p*q= + iii. (3 * 5) * (5 *3) iv. p if p*4 = 9
* = + = + = =
⁄ ⁄
4. Given that u ∆ v = , u+v 0, where u,
ii. p q v∈ R, evaluate. (4 ∆ 7) ∆ (7 ∆ 5)
p q= –
5. If * means „square the first number and divide
= – = – = =– by the second‟, find the values of;
⁄ ⁄
a. 9 * 6 b. * c. 1.2 * 0.018
⁄
iii. = =– 6. The operation ∼ is defined by m∼n = (m – n)
⁄

i. a. Form a table of the operation ∼ on the set

b. From p * q = , {2, 4, 6}.
b. Is the operation ∼ commutative? Give a
+ = (By substitution)
⁄ ⁄ reason.
= = × = ii. If m = 2, n = 4 and p = 6, evaluate;
⁄
a. m∼ (n + p) b. (m∼n) + (m∼p)
=
(p + q) (q – p) = (pq)(pq) (cross multiplication) 7. Given that a * b denotes , evaluate:

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i. 2 * 1 ii. 1 * 2 iii. (1 * 2) * (2 * 1) b. Use the table to;
i. evaluate 2 ‡ 4 and 4 ‡ 2
8. Given that a * b denotes – , evaluate: ii. what can you say about 2 ‡ 4 and 4 ‡ 2?
i. 2 * 1 ii. Find a when a * 7 = 1 * a iii. find (1 ‡ 2) ‡ 4

Solution
Table of Binary Operation a. M = {1, 2, 3, 4}
Given a well-defined binary operation, a table of
x ‡ y = x2 + y2 – 4
valuescan be constructed for a given set of
1 ‡ 4 = 12 + 42 – 4 = 13
values.
2 ‡ 1 = 22 + 12 – 4 = 1
I. Identify the definition of the binary and the
2 ‡ 3 = 22 + 32 – 4 = 9
given set of values
II. If there is n number of elements in the set, 3 ‡ 3 = 32 + 32 – 4 = 14
prepare n × n square table 4‡ 2 = 42 + 22 – 4 = 16
III. Place the operator at the top left corner of the 4 ‡ 3 = 42 + 32 – 4 = 21
table
IV. Occupy the first row and column of the table ◊ 1 2 3 4
with the elements of the given set as shown 1 2 1 6 13
below; 2 1 4 9 16
* n1 n2 n3 3 6 9 14 21
n1 4 13 16 21 28
n2
n3 b. From the table;
i. 2 ‡ 4 = 16 and 4 ‡ 2 = 16
IV. Operate each element of the first row against
each element of the second row under the binary ii. 2 ‡ 4 = 4 ‡ 2 = 16.
The operation ‡ is commutative.
definition and record your answers in the cells or
boxes until it is completed. iii. From the table (1‡ 2) = 1
(1 ‡ 2) ‡ 4 = 1 ‡ 4 = 13
Worked Examples
1. The operation ‡ is defined on the set R of real 2. An operation is defined on the set of integers
numbers by x ‡ y = x2 + y2 – 4 Z, by, x * y = x + y + 3xy, where x and y ∈ Z.
a. Copy and complete the table below for the i. Construct a table for this operation on the set,
operation ‡ on the set M = {1, 2, 3, 4} S = {-1, 0, 1, 2}
ii. Find from your table, a number b∈S, such that
◊ 1 2 3 4 c * b = c for all c ∈ S.
1 2 1 6
2 4 9 16 Solution
3 6 9 21 1.x * y = x + y + 3xy
4 13 28 S = {-1, 0, 1, 2}

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x * y = x + y + 3xy b. Find from the table:
-1 * -1 = -1 – 1 + 3(-1)(-1) = 1 i. 4 * 5 ii. x * x = 20
-1 * 0 = -1 + 0 + 3(-1)(0) = -1
-1 * 1 = -1 + 1 + 3(-1)(1) = -3 2. i. Copy and complete the table below forthe
-1 * 2 = -1 + 2 + 3(-1)(2) = -5 operation* defined by a * b = b2 – ab on the set
N = {1, 2, 2.3, 3.5, 4.7}
0 * -1= 0 – 1 + 3(0)(-1) = -1
* 1 2 2.3 3.5 4.7
0 * 0 = 0 + 0 + 3(0)(0) = 0 2
0 * 1 = 0 + 1 + 3(0)(1) = 1 2.3
0 * 2 = 0 + 2 + 3(0)(2) = 2 3.5
4.7
1 * -1= 1 – 1 + 3(1)(-1) = -3
1 * 0 = 1 + 0 + 3(1)(0) = 1 ii. Use the table to evaluate 2.3 * 3.5
1 * 1 = 1 + 1 + 3(1)(1) = 5
1 * 2 = 1 + 2 + 3(1)(2) = 9 Properties of Binary Operation
2 * -1= 2 – 1 + 3(2)(-1) = -5 The closure Property
2 * 0 = 2 + 0 + 3(2)(0) = 2 Given that a and b are members of set S and the
2 * 1 = 2 + 1 + 3(2)(1) = 9 operation * is defined over S. If a * b always
2 * 2 = 2 + 2 + 3(2)(2) = 16 gives an answer which is also in S, then the set S
is said to be closed with respect to *. That is:
* -1 0 1 2 a * b∈S, a, b∈S
-1 1 -2 -3 -5
0 -1 0 1 2 Worked Examples
1 -3 1 5 9 The binary operation * is defined on the set A =
{1, 2, 3, 4} by x * y = 3x – xy. Show whether * is
2 -5 2 9 16
closed under A in each of the following;
i. 1 * 2 ii. 3 * 4
ii. c * b = c
1*0=1
Solution
2*0=2
i. x * y = 3x – xy.
Therefore b = 0
But x = 1 and y = 2
1 * 2 = 3(1) – (1)(2)
Exercises 2.43
=3–2=1
1. The binary operation * is defined by m * n =
1 * 2 = 1 ∈ A, therefore set A is closed under the
mn – m. Copy and complete the table below for *
operation*
on the set {3, 4, 5, 6}
ii. x * y = 3x – xy
* 3 4 5 6
3 6 3 * 4 = 3(3) – (3)(4)
4 = 9 – 12 = - 3
5 3 * 4 = - 3 ∉ A, therefore set A is not closed under
6 24 the operation*
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Exercises 2.44 Solution
The binary operation* is defined on the set M = The operation * is commutative if;
{5, 6, 7, 8, 9}by x * y = xy – 4y. Show whether * a*b=b*a
is closed under M in each of the following; a * b = a + b + ab
i. 5 * 6 ii. 5 * 8 iii. 7 * 9 iv. 6 * 9 b * a = b + a + ba
a + b + ab = b + a + ba
The Commutative Property Therefore a * b = b * a
Let * be any binary operation defined over the set The operation * is commutative
S. We say the operation * is commutative if for
any two numbers a and b belonging to S, Exercises 2.45
a*b=b*a 1. The binary operation ∆ is defined over the set
of real numbers as m ∆ n = . Find:
Worked Examples
1. Let a * b = a + 2b be defined over the set of a. 2 ∆ 3 b. 4 ∆ 8
real numbers R. Verify whether or not, * is
commutative 2. The binary operation ○ is defined on the set of
real numbers by x ○ y = x2 – y. Find:
Solution i. 3 ○ 5 ii. 5 ○ 3
Method 1 iii. What can you say about i and ii?
a * b = a + 2b
Let a = 3 and b = 4 3. If a * b = ab + a + b, solve the a * 3 = 19
3 * 4 = 3 + 2(4) = 3 + 8 = 11
4. The operation * on R is defined by a * b =
Let a = 4 and b = 3 (a + b)2 – a2 – b2
4 * 3 = 4 + 2(3) = 4 + 6 = 10 a. Simplify the right hand side and calculate:
3*4≠4*3 – 5 * (– 1) and * 0
Therefore the operation is not commutative b. Prove that the operation * is commutative.

Method 2
5. The operation * on the integers a and b is
For the operation to be commutative
defined by a * b = ab + b. Find x given that
a* b = b * a
4*x=x*4
a * b = a + 2b
b * a = b + 2a
The Commutative Property on a Table
a + 2b ≠ b + 2a
A binary operation, * defined over a set S, is said
a * b ≠ b * a.
to be commutative if;
Therefore, the operation is not commutative
1. A table of values constructed is symmetrical
about the leading diagonal.
2. If a * b = a + b + ab, show whether the
operation * is commutative or not.
Consider the table below;

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 * e f g h x○ y = x + y + 2xy.
e e f g h a. Find the values of t, k, r and s in the table for
f f f h h the operation ○ on the set P = {1, 2, 3, 4}
g g h g h
h h h h h ○ 1 2 3 4
1 4 7 10 13
It can be seen that the table is symmetrical about 2 t 12 17 s
the leading diagonal. That is to say that each set 3 10 k r 31
of entries is a “reflection” of the other in the 4 13 22 31 40
leading diagonal. Therefore the operation* is said
to be commutative. b. Show whether set P is closed or not with
respect to the operation ○
2. Test two values on the table by interchanging c. Is the operation ○ is commutative or not?
their positions to see if the operation gives the
same answer. For e.g. from the above table: The Associative Property
e*f=f If a, b and care members of a set of real numbers
f*e=f R, then the operation * defined over
⇒ e * f = f * e = f. R is associative if: (a * b) * c = a *(b * c)
The operation* is said to be commutative
Worked Examples
Worked Examples The operation ◙ is defined over the set of real
The operation * is defined over the set M ={1, 2, numbers, R, by a ◙ b = ab + a + b. Show whether
3, 4} as a * b = a + b – 4. Construct a table of ◙ is associative or not.
values and show whether the operation * is
commutative or not. Solution
Let a, b and c ∈ R, then
Solution a ◙ (b ◙ c) = (a ◙ b) ◙ c
a * b = a + b – 4, But a ◙ b = ab + a + b
Set M = { 1, 2, 3, 4}
L. H. S: (a ◙ b) ◙ c = a ◙ (bc + b + c)
* 1 2 3 4
Let b ◙ c = bc + b + c = m
1 -2 -1 0 1
a ◙ ( b ◙ c) = a ◙ m = am + a + m
2 -1 0 1 3
By substitution,
3 0 1 2 3
4 1 2 3 4
a ◙ (b ◙ c) = a(bc +b + c) + a + (bc +b+ c)
= abc + ab + ac + a + bc + b + c
The operation * is commutative because the = a + b + c + ab + bc + ac + abc
table is symmetrical about the leading diagonal . R. H.S: a ◙ (b ◙ c) = (ab + a + b) ◙ c
Let ab + a + b = n
Exercises 2.46 ⇒ (a ◙ b) ◙ c = n ◙ c
The binary operation ○ is defined as:

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By substitution, (5 ∆ 6) = 5 + 6 + 2 = 13
(a ◙ b) ◙ c =(ab + a + b) c + (ab +a + b) + c 4 * 13 = 2(4)(13) = 104
= abc + ac + bc + ab + a + b + c
= a + b + c + ab + bc + ac + abc ii. (4 * 5) ∆ (4 * 6),
Comparing L.H.S results to R.H.S results, it is (4 * 5) = 2(4)(5) = 40 (First bracket)
seen that a ◙ (b ◙ c) = (a ◙ b) ◙ c
Therefore, the operation ◙ is associative 4 * 6 = 2(4)(6) = 48 (Second bracket)

Exercises 2.47 ⇒(4 * 5) ∆ (4 * 6) = 40 ∆ 48

If a * b = ab + a + b, solve the equations 40 ∆ 48 = 40 + 48 + 2 = 90
a * 3 = 19 and ( a * 3) + ( 2 * a) = 4m
iii. The operation * is not distributive over ∆
The Distributive Property because 4 * (5 ∆ 6) ≠ (4 * 5) ∆ (4 * 6)
Ifa, b and c are three members of the set of real
numbers, R and the binary operations * and ◘ are Exercises 2.48
defined over S, then * is distributive over ◘ if a * 1. Two binary operations * and ○ are defined
(b ◘ c) = (a * b) ◘ (a * c) over the set, R, of real numbers by x * y = xy
However, if a * (b ◘ c) ≠ (a *b) ◘ ( a * c), then * and x ○ y = x + y. Find:
is not distributive over ◘ i. x * (y ○ z) ii. (x * y) ○ (x * z)
iii. Show whether the operation * is distributive
Worked Examples over ○ or not
Two binary operations * and ∆ are defined as
a * b = 2xy and a ∆ b = a + b + 2 for all a, b∈ R. 2. Let U = {1, 2, 3… 4}, A = {2, 3, 5, 7, 8}
B = {1, 3, 5, 9} and C = {4, 9, 10} where A,
Evaluate: i. 4 * (5 ∆ 6) ii. (4 * 5) ∆ (4 * 6) B and C are subsets of U. Find:
iii. What can you say about (i) and (ii) i. A ∩ (BUC), ii. (A∩B) U (A∩C),
iii. What can you say about the operations ∩ and
Solution U over the set?
i. a * b = 2xy and a ∆ b = a + b + 2
In 4 * (5 ∆ 6), taking the bracket first,

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3 ALGEBRAIC EXPRESSIONS Baffour – Ba Series

Definition Exercises 3.1

Any mathematical statement written without an Group the following into like terms and unlike
equal sign is called an expression. An algebraic terms;2y, 6m, 12a, 10m, 13x, 5x, 5y, 2k, 15n, 4p,
expression is therefore a mathematical statement 7q, 8r, 9a, 10v, 3r, 25x and 54u
that contains numbers and variables and / or an
operator, but not an equal sign. For e.g, 2m − n, Simplifying Algebraic Expressions
2p, 6y + 3x etc. The process of adding, subtracting, multiplying
and dividing two or more algebraic expressions is
Like Terms and Unlike Terms called simplification.
Likes terms are algebraic terms that have the
same variable factors. For example, 5x and 10x, Addition and Subtraction of Algebra
2m and 15m, 3y and 8yetc Like terms can be added and subtracted. For
instance, 3pens can be added to 7pens to get
Unlike terms are algebraic terms that have 10pens. That is, 3p + 7p = 10p. Also 3pens can be
different variable factors. For example, 5x and taken away from 7pens to get 4pens. That is: 7p −
10y, 3a and 8betc 3p = 4p

Worked Examples Unlike terms on the other hand, cannot be added

Group the like terms in the following: nor subtracted. This is explained by the fact that
1. 4a2 + 3ab – 2a2 + ab 3pens cannot be added to nor subtracted from
2. 2a + 4b2 – 3a + 3b2 – b2 7erasers. That is:
3. 3p + 5q + 2p – q 1. 3pens + 7erasers = 3pens + 7erasers
2. 7 erasers – 3pens = 7erasers – 3pens.
Solutions
1. 4a2 + 3ab – 2a2 + ab Thus under addition and subtraction of unlike
= 4a2 – 2a2 + 3ab + ab terms, the variable terms remain the same.

2. 2a + 4b2 – 3a + 3b2 – b2 Worked Examples

= 2a – 3a + 4b2 + 3b2 – b2 Simplify the following:
(1) 16y + 4y + 2y (1) 12a – 8a
3. 3p + 5q + 2p – q = 3p + 2p + 5q – q (3) 25t – 14t – t (4) 17x + 6y

4. Re - group 4x2 + 3x2y – xy2 + 2x2y Solutions

(1) 16y + 4y + 2y = 22y (2) 12a – 8a = 4a
Solution (3) 25t – 14t – t = 10t (4) 17x + 6y = 17x + 6y
4x2 + 3x2y – xy2 + 2x2
= 4x2 – xy2 + 3x2y + 2x2y Exercises 3.2
A. Simplify the following:

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1. 1.32a + 7.68a 2) 10k – k 3. 13m + 51n 3. Simplify 4p + 6p2 – 2p + 2p2
4. 9s – 4s 5. 8m – 3m – m 6. a – 3
Solution
B. Simplify the following; 4p + 6p2 – 2p + 2p2
1. 16m + 6n – 18m + 4n = 6p2 + 2p2 + 4p – 2p
2. 20y – 32x – 36n + 8y = 8p2 + 2p
3. 3x + (x + 5x)
4. (9y – 11y) + 10y Exercises 3.3
A. Simplify the following:
C. A = 3x + 4y and B = 2x – 5y. Express the 1. 5p – 6a – 7p + 10a
following in terms of x and y, in their simplest 2. 8x2 – 15a – 20x2 + 3a
forms: 3. 5w + 7p2 – 4w + 3p2
1. A + B 2. 2A + 3B 3. A – B 4. 3A – 2B 4. 3x2 + 6xy – 3y2 + 4x2 – 8xy + 2y2

Grouping Algebraic Expressions B. Re-group the following and simplify:

Like and unlike terms of algebraic expression are 1.4a + 3b + 3a + 2b 4. 3x2 – 4x2 – 2x3 + 4x3 – x2
regrouped to make addition and subtraction 2. p + 3q – p + 29
possible and simple. 3. 3h + 8 – 2h + 2 5. 7m + 6n + 3m + 2n

In regrouping, the operation sign is part of the Multiplication of Algebraic Expressions

algebraic term and therefore must be lifted This is done by expanding and regrouping the
alongside. For example, in regrouping 2x + 6y + terms of the expression.
3x − 2y we get; Reminder: × =
2x + 3x + 6y – 2y = 5x + 4y
Worked Examples
Worked Examples 1. Perform 3ab × 2a
Simplify the following;
1. 8m – 4n + 3m + 10n Solution
3ab × 2a
Solution = 3× a × b × 2 × a
8m – 4n + 3m + 10n =3 =6× 2
× = 6a2b
= 8m + 3m + 10n – 4n
2
= 11m + 6n 2. Find 3 ×

2. 5x – 8y – x – 2y Solution
3 2
Solution 3 2
= (3 × 2 × 4) c = 24
5x – 8y – x – 2y 2 3 3
3. Simplify (3a b ) (4a b)
= 5x – x – 8y – 2y
= 4x – 10y Solution

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3a2b3 × 4a3b = (3 × 4) = 12 I. Involving Monomial Denominators
For all operations involving algebraic fractions,
Exercises 3.4 follow the same rule as in arithmetic.
A. Simplify the following;
1. 5 2 × 3 2 2. 2
n× 4 3
To add or subtract algebraic fractions involving
3 2 2
3. × b 4. q ×4 monomial denominators, observe the following;
1. Find the L.C.M of the denominators.
B. Simplify the following: 2. Express each fraction in terms of the L.C.M.
1. 2. a2b × ab 3. t+ t and simplify.

4. a2b × (-ab3) 5. × -8 6. (8x2y3) ( xy4) Worked Examples

1. Simplify –
Division of Algebraic Expressions
This is done by expanding the dividend and the Solution
divisor and dividing the like terms.
−
Reminder: = (
= =
Worked Examples
(
Simplify the following; 2. Simplify −
1. 6 3b2 3
Solution
(
Solution −
6 3 2 3 =
( (
= =
2
= = = 2a b
II. Involving Binomial Denominators
2. 35x5 y2 5x2 y To add or subtract algebraic fractions involving
binomial denominators:
1. Factorize denominators (if necessary).
Solution
2. Find the L.C.M of the denominators.
35 5 2
3. Express each fraction in terms of the L.C.M.
= (35 = 7x3 y and simplify.

Exercises 3.5 Work Examples

Simplify; 1. Simplify –
1) 18k3j7 3j2k 2) 99a3b2m4 3a3bm3
3) 48 6 4) 3 27 Solution
–
Algebraic Fractions ( (
A. Addition and Subtraction = (
= (
=(
( ( (

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B. Multiplication and Division Solution
I. Involving Binomial Denominators × = = =
Before attempting to simplify when multiplying
or dividing algebraic fractions, factorize where
3. Perform
possible and divide top and down by common
factors. The contents of a bracket should be
considered as a single term. Solution
= × =
Work Examples
(
1. Simplify ( ( Exercises 3.6
A. Simplify the following;
Solution
( 1. + 2. + 3. –
(
=(
(
4. + – 5. – 6. ×

2. Simplify
B. Simplify;
Solution 1. × 2. ×
( (
= = ( ( ( (
(
3. (
4.
( ( (

3. Simplify
C. Simplify:
Solution 1. 2. 3. –
( (
= ( (
= 4. – 5. – 6. –
(
7. – 8. + 9.
II. Involving Monomial Denominators
I. Factorize denominators and numerators if
necessary D. Simplify the expressions:
II. Cross out common factors and leave remaining 1. 2. 3.
factors as answer
4. 5. 6.

Worked Examples
E. Simplify the expressions:
1. Find the product of and
1. – 2. –

Solution 3. – 4. –

× = = 5. – 6.
7. 4 + – 8. 2 + +
2. Multiply by

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Expansion of Algebra 1. 4(2r – 3a) – 5(6r + a) 2. 7(2x – 5) + 15x
It is the method of removing brackets from an 3. 6(2p – 1) + 4(3 + 5p) 4. 2c (4c – 2) + 5r + 7)
expression. In this process, the number or
variable outside the bracket multiplies each factor C. Expand and simplify the following;
or term inside the bracket. For example, the 1. 3 (x + 2y) – 5(2x + 4y)
expansion of 2(a + b) is carried out as; 2. 4(3a + 6b) – 3(2a + 4b)
2 (a + b) = 2 = 2a + 2b 3. 5(2c + 4d) + 2(3c + 2d)
4. - 6(2 + 3s) + 4(3r – 5s)
Worked Examples
1. Remove the brackets in a – 2(b – 3c) Binomial Expansion
A binomial is an algebraic expression with two
Solution terms.
a – 2(b – 3c) = a – 2b + 6c
Binomial expansion involves the act of removing
2. Expand and simplify: the brackets on binomials. The process is also
3(6b – 9a) + 7(6a – 5b) called multiplication of binomials. For example,
the expansion of ( ( is the same as
Solution the product or multiplication of (a + 4) and (a + 5)
3(6b – 9a) + 7(6a – 5b) ⇒ (a + 4) × (a + 5)
= 18b – 27a + 42a – 35b
= 18b – 35b – 27a + 42a The expansion is done by disintegrating the first
= -17b + 15a bracket and multiplying each of its terms by the
whole of the terms in the second bracket or vice-
3. Simplify 6(7a + 4) – 3(8a + 9) versa. That is:
( (
Solution = ( (
6(7a + 4) – 3(8a + 9) 2
= + 5a + 4a + 20
= 42a + 24 – 24a – 27 = 2+ 9a + 20
= 42a – 24a + 24 – 27
= 18a – 3 Worked Examples
Expand the following:
4. Simplify (5m + 3n) – (2m – n) 1. ( (
Solution
Solution
(5m + 3n) – (2m – n)
( (
= 5m + 3n – 2m + n
( (
= 5m – 2m + 3n + n 2
= 3m + 4n 2

Exercises 3.9
2. Remove the brackets in (p – 7) (p + 6)
A. Expand and simplify the following:

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Solution 2. (3m + n) (4m – n) + 2m – 3n) (m + 2n)
( ( 3. (a − 2b) (a + 3b) − (2a + b) (2a – b)
( ( 4. (6p – q)(p – 2q) – (3p – 2q)(2p – q)
2
5. (3p + 4q) (3p – 4q) – (p + q) (9p – 16q)
2

Factorization of Algebraic Expressions

3. Multiply (2a + b) by (a + 2b) Factorization is the process of writing expressions
in brackets. This is done by bringing out the
Solution highest common factor and/ or the common
(2a + b) (a + 2b) variable outside the bracket, whilst the rest of the
= 2a (a + 2b) + b(a + 2b) factors are put into bracket. For example, 4
= 2a2 + 4ab + ab + 2b2 is factorized as: ( ( = 2(2x + 3)
= 2a2 + 5ab + 2b2

4. Simplify the expression: Similarly, 12x + 4xy is factorized as :

(3x – y)(3x + y) – (3x + 2y)(3x – 2y) 12x + 4xy = (4 × 3x) + (4 × xy)
Common factor = 4
Solution Common variable = x
(3x – y) (3x + y) – (3x + 2y) (3x – 2y) 12x + 4xy = 4x(3 + y)
= 3x(3x + y) – y (3x + y ) – [3x(3x – 2y) + 2y(3x – 2y) ]
= [9x2 + 3xy – 3xy – y2] – [9x2 – 6xy + 6xy – 4y2 ] Worked Examples
= 9x2 – y2 – [9x2 – 4y2] 1. Factorize the following:
= 9x2 – y2 – 9x2 + 4y2 i. 2a – 2a2 ii. 5ax − 15 3 2
= 9x2 – 9x2 – y2 + 4y2 = 3y2 10x3y2 – 25x4y iv. 2m2 – 6mk
. 2+ vi. 7d + 14d4 – 7d3
5. Expand (x – √ ) (x + √ )
Solution
Solution i. 2 – 2 2 = 2 (1 – )
x(x + √ ) – √ (x + √ ) ii. – 15 3x2 = 5ax (1 – 2x)
= x2 + √ x – √ x – (√ )2 iii. 10 3 2 – 25 4 = 5 3y(2y – 5x)
2
iv. – 6mk = 2m(m –3k)
= x2 – 2 2
v. + pq = p (p + q)
vi. 7 + 14 4 – 7 3
= 7d (1 +2d3– d2)
Exercises 3.10
A. Expand the following;
2. Factorize 5xy + 10ny
1. ( ( 4. ( (
2.( ( 5.( (
Solution
3. (3a + c) (4b + d) 6. (2x – y) (x + t)
5xy + 10ny = 5y (x + 2n)
B. Expand and simplify;
3. Factorize 3r2s – 9rs2
1. (3y – 4) (4y – 5) + (2y – 3) (y – 5)

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Solution = (mp + np) – (mt – nt)
3r2s – 9rs2 = 3rs (r – 3s) = p(m + n) – t(m + n)
= (p – t ) (m + n)
4. Factorize 26x – 13y
2. Find the factors of ax + by + bx + ay
Solution
26x – 13y = 13 (2x – y) Solution
ax + by + bx + ay
5. Factorize a2b – 2ab2 – 3b3 = ax + ay + bx + by
= (ax +ay) + (bx + by)
Solution = a(x + y)+ b(x + y)
a2b – 2ab2 – 3b3 = b(a2 – 2ab – 3b2 ) = (a + b) (x + y)

Exercises 3.11 3. Factorize 3mx + 2nx – 3my – 2yn

A. Factorize the following:
1. 12yt2 – 16y2t 2. 21pq2 – 49p3q 3.6pr + 42p2r Solution
4. 16rs + 40rs3 5. 27k2m – 9m 6. 15cd2 + 25c2d 3mx + 2nx – 3my – 2yn
= (3mx + 2nx) – (3my – 2yn)
B. Factorize the following: = x (3m + 2n) – y (3m + 2n)
1. 4x2y – 6xy2 + 8xyt 2. 3abc + 99c2 – 15a2c2
= (3m + 2n) (x – y)
3. 12k2m + 18kmn + km2 4. 16p2q2 – 20pqr – 8pq
4. Factorize 3a2 + 2ab – 12ac – 8bc completely.
Factorization by Grouping
Algebraic expressions having four terms are
Solution
factorized by grouping. Observe the following
3a2 + 2ab – 12ac – 8bc
steps when factorizing a four term algebraic
= (3a2+ 2ab) – (12ac – 8bc)
expression;
= a (3a + 2b) – 4 (3a + 2b)
I. Group the terms such that each pair can have a
= (a – 4c)(3a + 2b)
common factor or common variable.
II. Factorize each group separately to have the
Factorisation by Re- grouping
same factor in the bracket.
Worked Examples
III. Bracket the factors outside the bracket and
1. Factorise 4wy + 3xz – 6wz – 2xy
multiply it by one of the common factors in the
brackets to complete factorization.
Solution
4wy + 3xz – 6wz – 2xy
Worked Examples = (4wy – 6wz ) – (2xy + 3xz) (re-qrouping)
1. Factorize mp + np – mt – nt
= 2w(2y – 3z) – x (2y – 3z)
= (2w – x) (2y – 3z)
Solution
mp + np – mt – nt
2. Factorize 6 + xy – 3x – 2y

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Solution bx = mx + nx …….(2)
6 + xy – 3x – 2y
= (xy – 3x ) – ( 2y + 6) (re-qrouping) Put eqn (2) in eqn (1)
= x (y – 3) – 2 (y – 3) x2 + mx + nx + c
= ( x – 2) (y – 3) Once, four terms are obtained, divide the
expression into two with a bracket leaving an
Exercises 3.12 operator in-between.
A. Factorize the following: ⇒(x2 + mx) + (nx + mn)
1. 2ap + aq – bq – 2bp 4. 2pr – 4ps + qr – 2qs
2. x2 – ax + bx – ab 5. pr + 3ps – 2qr – 6qs Factorize completely to get;
3. xy – 3xc + 2qy – 6qc 6. 4xy – 8y2 – 8xd – 16yd x(x+m)+n(x+m)
(x + n) (x + m)
B. Factorize the following:
1. 2yz + 5z – 8y – 20 2. 2cd – 2ce + d2 – e2 Worked Examples
3. 3a + 3b + a2 – b2 4. 3fd – 3fe – d2 + e2 1. Factorize x2 + 5x + 6

Quadratic Expressions of the form: Solution

ax2 + bx + c In x2 + 5x + 6, a = 1, b = 5 and c = 6
Any mathematical expression of the form: Find all the factors of the constant term, 6
ax2 + bx + c, where a, b and c are constants and 6 = (1, 6), (2, 3), (-1, -6) and (-2, -3)
a 0 is called a quadratic expression. Thus, in Select the pair of factors that sum up to the co-
ax2 + bx + c, efficient of x, which is 5
a is called the co-efficient of x2,
b is called the co-efficient of x , x2+ 5 x + 6
c is called the constant term.
2+3 2×3
Factors of Quadratic Expressions
1. Expressions of the Form: x2 + bx + c But 5x = 2x + 3x
To find the factors of expressions of the form: x2 Re-write the expression to obtain four terms as, x2
+ bx + c, find all the factors of the constant term, + 2x + 3x + 6
c, such that the product of a pair of the factors Divide the expression into two with a bracket
equal to the constant term, c, and the sum of that leaving an operation sign in- between
same pair of factors equal the co-efficient of x . (x2 + 2x) + (3x + 6)
Simply put, x (x + 2) + 3 (x + 2)
x2 – (sum of roots)x + (product of roots) (x+ 3) (x + 2)
x2 + b x + c……. (1)
2. Factorize completely x2 + 8x +15

m+n m×n Solution

Where m and n are pair of factors of c, and x2 + 8x + 15

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Factors of 15= (1, 15), (3, 5), (-1, -15) (-3, -5) (x – 2) (x – 5)
8x = 3x + 5x, 15 = 3 × 5 Factors of x2 – 7x + 10 = (x – 2) (x – 5)
x2 + 3x + 5x + 15
(x2 + 3x) + (5x + 15) 3. Expressions of the form: x2 + bx – c
x(x + 3) + 5(x + 3) When the co-efficient of x is positive and the
(x + 3) (x + 5) constant term is negative,i.e.x2 + bx – c, find the
pair of factors of the constant term (- c ), such
2. Expressions of the form: x2 – bx + c that one of the pair is negative and the other one,
When the coefficient of x is negative and the positive
constant term is positive, i.e. x2 – bx + c, find ⇒ – c = (− m n), b = (− m + n)
only the negative factors of the constant term, The form: x2 + bx − c is factorized as:
such that; x2 – mx + nx – mn
c = (- m - n) and b= (- m + - n) and follow the (x2 – mx) + ( nx – mn)
usual process. This is illustrated as follows: x (x – m) + n (x – m)
x2 – mx – nx + (-n × -m) (x + n) (x – m)
(x2 – mx) – ( nx + -n × -m)
x(x – m) − n(x – m) Worked Examples
(x – n)(x − m) 1. Factorize x2 + 5x – 14 completely

Worked Examples Solution

1. Find the factors of x2 – 5x + 6 x2 + 5x – 14
Factors of -14 = (-1, 14) (-2, 7),
Solution 5x = -2x + 7x, -14 = -2 7
x2 – 5x + 6 x2 – 2x + 7x – 14
Negative factors of 6 = (-1, -6) (-2, -3) (x2 – 2x) + (7x – 14)
-5x = -2x + -3x, 6 = -2 -3 x(x – 2) + 7(x – 2)
x2 – 2x – 3x + 6 (x – 2) (x + 7)
(x2 – 2x) – (3x + 6) Factors of x2 + 7x – 12 = (x – 2) (x + 7)
x(x – 2) – 3(x – 2)
(x – 2) (x – 3) 2. Factorize x2 + 7x – 18 completely
Factors of x2 – 5x + 6 = (x – 2) (x – 3)
Solution
2
2. Express x – 7x + 10 as factors x2 + 7x – 18
Factors of -18 = (-1, 18), (-2, 9) and (-3, 6)
Solution
x2 – 7x + 10 7x = -2x + 9x, -18 = -2 9
Negative factors of 10 = (-1, -10) (-2, -5) x2 – 2x + 9x – 18
-7x = -2x + -5x, 10 = -2 -5 = (x2 – 2x) + (9x – 18)
x2 – 2x – 5x + 10 = x(x – 2) + 9(x – 2)
(x2 – 2x) – (5x + 10) = (x – 2) (x + 9)
x(x – 2) – 5(x – 2)

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4. Expressions of the form: x2 – bx – c C. Completely factorize the following:
When the co-efficient of x is negative and the 1. x2 – 12x + 35 3. x2 – 15x + 44
constant term is negative, i. e.x2 – bx – c , find the 2. x2 – 19x + 48 4. x2 – 18x + 56
pair of factors of the constant term (-c), such that
one of the pair is positive and the other one, D. Find the factors of the following:
negative. 1. x2 – 7x – 44 3. x2 – 5x – 50
2
⇒ – c = (m – n), b = (m – n). Of the pair, the 2. x – 7x – 30 4. x2 – x – 20
first one is always bigger than the other.
5. Factors of Expressions of the Form:
Worked examples ax2 + bx + c, where x 1 or x 1
Factorize the following quadratic expression To factorize expressions of the form;
1. x2 – 8x – 9 2. x2 – 2x – 24 ax2 + bx + c, where x 1 or x 1,

Solutions Type I
1. x2 – 8x – 9 I. Find the product of the co-efficient of x2 and
Factors of -9 = (1, -9) (-3, 3) the constant term (c) i. e. (a×c)
-8x = x – 9x, -9 = 1 -9 II. Find all the pair of factors of ac
x2 + x – 9x – 9 III. Find the pair of factors of acthat sum up to
(x2 + x) – (9x – 9) the co-efficient of x.
x(x +1) – 9(x + 1) i.e. ax2 + bx + c
(x + 1) (x – 9)
Factors of x2 – 8x – 9 = (x + 1) (x – 9) a c=m n, bx = m + n

2. x2 – 2x – 24 IV. Substitute bx = mx + nxin ax2 + bx + cto

Factors of -24 = (4, -6) obtain four terms of the expression.
-2x = 4x – 6x, -24 = 4 -6 ⇒ax2+ mx + nx + c
2
x + 4x – 6x – 24
(x2 + 4) – (6x – 24) V. Factorize carefully by method of grouping to
x(x + 4) – 6(x + 4) obtain the factors of the expression.
(x + 4) (x – 6)
Factors of x2 – 2x – 24 = (x + 4) (x – 6) Note:
1. If b and c are positive, the pair must be
Exercises 3.13 negatives.
A. Factorize the following: 2. If b is positive and c is negative, negate the
1. x2 + 10 x + 21 3. x2 + 12x + 27 smallest factor of the pair.
2. x2 + 11x + 30 4. x2 + 18x + 72 3. If b and c are negative, negate the bigger
factor of the pair.
B. Factorize completely: 4. If only b is negative, negate both pair of
1. t2 + 6t – 16 3. m2 + 10m – 24 factors.
2. m2 + 10m – 39 4. m2 + 5m – 14

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Worked Examples (6x2 + 4x) – (9x – 6)
Factorize the following completely 2x(3x + 2) – 3(3x + 2)
1. 3x2 + 4x + 1 2. 5x2 + 7x + 2 (2x – 3) (3x + 2)
3. 4x2 + 4x – 3 4. 6x2 – 5x – 6 Factors of 6 x2 – 5x – 6 = (3x + 2) (2x – 3)

Solutions Exercises 3.14

1. 3 2 + 4x + 1 A. Factorize the following;
ax2 + bx + 1. 3x2 + 11x + 6 3. 5x2 – 12x + 4
a =3 and b = 1 2. 6x2 + 7x – 5 4. 7x2 + 9x + 2
a c=3×1=3
Factors of 3 = (1, 3) and 4x = x + 3x. Substitute in B. Find the factors of the following;
the expression to obtain four terms; 1. 4x2 – 21x + 20 4. 5x2 – 17x + 6
⇒3x2 + x + 3x + 1 2. 12x2 – 7x + 1 5. 2x2 – 7x + 3
(3x2 + x) + (3x + 1) 3. 3x2 – 20x + 12 6. 6x2 + 11x – 2
x(3x + 1) + 1(3x + 1)
(x + 1) (3x + 1) Type 2
Follow the same process as that of type 1
Factors of 3x2 + 4x + 1= (x + 1) (3x + 1)

Worked Examples
2. 5x2 + 7x + 2
Factorize the following:
5 × 2 = 10
1. 10x2 – 9xy + 2y2 2. 5x2+ 18xy + 9y2
Factors of 10 = (2, 5) and 7x = 2x + 5x.
5x2 + 2x + 5x + 2
Solution
(5x2 + 2x) + (5x + 2)
10x2 – 9xy + 2y2
x (5x + 2) + 1(5x + 2)
10 × 2 = 20
(x + 1) (5x + 2)
Pair of factors of 20 that sum up to – 9
Factors of 5x2 + 7x + 2= (x + 1) (5x + 2)
= (-5, - 4). ⇒- 9xy = - 5xy – 4xy
3. 4x2 + 4x – 3
Put -9xy = -5xy – 4xy in 10x2 – 9xy + 2y2 ⇒10x2 –
4 × -3 = -12
5xy – 4xy + 2y2
Factors of -12 = (-2, 6) and 4x = -2x + 6x.
(10x2 – 5xy) – (4xy + 2y2)
4x2 – 2x + 6x – 3
5x(2x – y ) – 2y(2x – y)
(4x2 – 2x) + (6x – 3)
(5x – 2y) (2x – y)
2x (2x – 1) + 3 (2x – 1)
(2x + 3) (2x – 1)
2. Factorize 5x2 + 18xy + 9y2
Factors of 3x2 + 4x – 3= (2x + 3) (2x – 1)
Solution
4. 6x2 – 5x – 6
5x2 + 18xy + 9y2
– 6 × 6 = – 36
5 × 9 = 45
Factors of –36 = (4, -9) and –5x = 4x – 9x.
Pair of factors of 45 that sum up to 18 = (3, 15)
6x2 + 4x – 9x – 6

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⇒18xy = 3xy + 15xy But 6x = 3x +3x and 9 = 3×3
5x2 + 3xy + 15xy + 9y2 x2 + 3x + 3x + 9
(5x2 + 3xy) + (15xy + 9y2) (x2 + 3x) + (3x + 9)
x(5x + 3y) + 3y(5x + 3y) x(x + 3) + 3(x + 3)
(x + 3y) (5x + 3y) (x + 3) (x + 3) = (x + 3)2
the factors of x2 + 6x + 9 = (x + 3)2
Exercises 3.15
Factorize the following: 2. What are the factors of x2 – 10x + 25?
1. 2x2 + 5xy + 3y2 2. 6x2 – 8xy + 2y2
3. 14x2 + 19xy – 3y2 4. 3x2 – 7xy + 4y2 Solution
5. 5x2 – 25xy – 30y2 6. a2 + ab – 6b2 x2 – 10x + 25
7. a2 + 12ab – 45b2 8. 3r2 + 7rs – 20s2 Factors of 9 = (1, 25), (-5, -5), (-1,-25) (5, 5)
9. a2 + ab – 6b2 10. 2p2 – 3pq – 2q2 But -10x = –5x –5x and 25 = -5 × -5
x2 – 5x – 5x + 25
Factors of Perfect Squares of the form: (x2 – 5x) – (5x + 25)
x2 + bx + c x(x – 5) – 5(x – 5)
In x2 + bx + c, if the pair of factors of the constant (x – 5) (x – 5) = (x – 5)2
term, c, is of the same kind and that same kind of Therefore factors of x2 – 10x + 25 = (x – 5)2
factors sum up to the co-efficient of x, we say the
expression is a perfect square. B. Factors of Perfect Squares of the form: ax2 +
i. e. x2 + b x + c bx + c
For perfect expressions of the form;
ax2 + bx + c, where x ≠ 1 or x > 1,
m+m m×m
Method I
Such that c = m m and bx = mx + mx. I. Find the product of the co-efficient of x2 and
the constant term, c. i. e. (a × c)
For all perfect squares, II. Write all the pair of factors of (ac)
1. x2 + bx + c III. Find the pair of factors of the same kind of
= ( x + m )( x + m) = (x + m)2 ac,that sum up to the co-efficient of x.
i.e. ax2 + bx + c
2. x2 − bx + c a c = m m, bx = mx + mx
= ( x – m ) ( x – m) = (x – m)2 IV. Substitute the pairs in place of bx to obtain
(Where x is the root of x2 and m is the root of c) four terms i.e. bx = mx + mx, so ax2 + mx + mx + c
V. Factorize the four term expression by method
Worked Examples of grouping to obtain the answer in the form
1. Factorize x2 + 6x + 9 [(√ ) √ ]

Solution Method II
x2 + 6x + 9 I. Try to find the square root of ax2and the
Factors of 9 = (1, 9) and (3, 3) constant term c

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II. If possible: (25x2 – 10x) – (10x + 4)
a. express the answer in the form [(√ ) 5x(5x – 2) – 2(5x – 2)
(5x – 2) (5x – 2) = (5x – 2)2
√ ] on condition that the coefficient of x is
positive
Method II
b. express the answer in the form [(√ ) 25x2 – 20x + 4
√ ] on condition that the coefficient of x is (√ ) = 5x and (√ ) = 2
negative Coefficient of x is negative
(5x – 2) (5x – 2) = (
Worked Examples
1. Factorize 4x2 + 12x + 9 Exercises 3.16
A. Factorize the following completely;
Solution 1. x2 + 10x + 25 3. x2 – 32x + 256
Method 1 2. x2 + 40x + 400 4. x2 – 26x + 169
4x2 + 12x + 9
4 × 9 = 36
B. Find the factors of the following;
Factors of 36 = (1, 36) (2, 18), (3, 12), (6, 6) 1. 9x2 – 12x + 4 2. 4x2 – 20x + 25
But 12x = 6x + 6x and 36 = 6 × 6 3. 4x2 – 36x + 81 4. 16x2 –16x + 4
4x2 + 6x+ 6x+ 9
(4x2 + 6x) + (6x+ 9) Unknowns of a Perfect Quadratic Expression
2x(2x + 3) + 3(2x + 3) A. Given the perfect square ax2 + bx + c to find
(2x + 3) (2x + 3) = (2x + 3)2 the value of b,
I. Write the pair of common factors for ax2 + bx + c
Method II II. Simplify the common factors and compare
4x2 + 12x + 9
(√ ) = 2x and (√ ) = 3 B. Given the perfect square ax2 + bx + c to find
Coefficient of x is positive the value of the constant term c,
(2x + 3) (2x + 3) = ( I.Use the fact that :
x2 – (sum of roots)x + (product of roots)
2. 25x2– 20x + 4 II. Identify the sum of roots and multiply to get
the product of roots.
Solution III. Then make comparison.
Method 1
25x2 – 20x + 4 Worked Examples
25 × 4 = 100 1. Find the integer k if 4a2 + ka + 81 is a perfect
square
Factors of 100 = (–10, –10)
But – 20x = –10x –10x and 100 = –10 × –10 Solution
25x2– 20x + 4 If 4a2 + ka + 81 is a perfect square,
25x2– 10x– 10x + 4 ⇒ 4a2 + ka + 81 = (2a + 9) (2a + 9)

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But (2a + 9) (2a + 9) x2 – 32y2 = x2 – (3y)2
= 2a(2a + 9) + 9(2a + 9) x2 – (3y)2 = (x + 3y) (x – 3y)
= 4a2 + 18a + 18a + 81 ⇒Factors of x2 – 9y2 = (x + 3y) (x – 3y)

4a2 + 36a + 81= 4a2 + ka + 81 3. Completely factorize 9p2 – 16

⇒ 36a = ka
k = 36 Solution
9p2 – 16,
2. Find the number k given that a2 – 12a + k is a But 9 = 32 and 16 = 42
perfect square. 32p2 – 42 = (3p)2 – 42
(3p)2 – 42 = (3p +4) (3p – 4)
Solution ⇒ Factors of 9p2 – 16 = (3p + 4) (3p – 4)
For all quadratic expressions:
x2 – (sum of roots)x + (product of roots) 4. Factorize 9 – (x + 1)2
Ifa2 – 12a + k is a perfect square
⇒ a2 – 12a + k = a2 – 6a – 6a+ (6 × 6) Solution
a2 – 12a + k = a2 – 12a+ 36 9 – (x + 1)2
By comparison, k = 36 = 32 – (x + 1)2
= {(3) + (x + 1)}{(3) – (x + 1)}
Factors of Difference of Two Squares = (3 + x + 1) (3 – x – 1) = (4 + x) ( 2 – x )
Type I: (a2 – b2)
5. (a + 3b)2 – (2a – b)2
Factors of expressions of the form: a2 – b2 = (a +
= {(a + 3b) + (2a – b)}{a + 3b) – (2a – b)}
b) (a – b). This process is called difference of two
= (a + 3b + 2a – b) (a + 3b – 2a + b)
squares. The process is not applicable when the
= (3a + 2b) (4b – a)
operator is “addition” but “subtraction”
1. a2 + b2 (a – b) (a + b)
Exercises 3.17
2. a2 + b2 (a + b) (a + b)
A. Find the factors of the following;
1. 81 – (5y)2 2. (8p)2 – 9 3. 121 – 100m2
Worked Examples
4. 16 – (3a)2 5. 9y2 – 4 6. 9x2t2 – 1
1. Factorize 4m2 – n2
B. Fcatorize the following:
Solution
1. (a + b)2 – 4r2 6. q2 – (p – 2n)2
4m2 – n2 = (2m)2 – n2
2. (x + y)2 – 9m2 7. (x + y)2 – 4
(2m)2 – n2 = (2m – n) (2m – n)
3. (n – 2p)2 – 64q2 8. 4r 2 – (s – r)2
Factors of 4m2 – n2 = (2m + n) (2m – n)
4. (3t + 5u)2 – (2t – 3u)2

2. What are the factors of x2 – 9y2?

Type 2
Expressions containing one quadratic
Solution
x2 – 9y2, but 9 = 32
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Here, there is always a quadratic expression = m (m – n) – n (m – n)
(usually perfect square) within the given = (m – n)2
expression ⇒m2 – 2mn + n2 – 9r2
I. Identify the quadratic expression within the = (m – n)2 – 9r2
given expression and the remaining term. = (m – n)2 – (3r)2
II. Find the factors of the quadratic expression = (m – n – 3r) (m – n + 3r)
III. Form a difference of two squares with the
common factors obtained from the quadratic 3. Factorise a2 – b2 – 10b – 25
expression and the remaining term of the given
expression. Solution
IV. Simplify the difference of two squares a2 – b2 – 10b – 25
⇒: a2 – b2 = (a + b) (a – b) = a2 – [b2 + 10b + 25]
= a2 – [b2 + 5b + 5b + 25]
Note: The expression can also contain two = a2 – [(b2 + 5b)+ (5b + 25)]
quadratic expressions = a2 – [b(b + 5)+ 5(b + 5)]
= a2 – [(b + 5)(b + 5)]
Worked Examples = a2 – (b + 5)2
1. Factorize x2 – 6x + 9 – 4y2 = [ a + (b + 5)] [a – (b + 5)]
= (a + b + 5) (a – b – 5)
Solution
Quadratic expression = x2 – 6x + 9 Exercises 3.18 A
Remaining term = – 4y2 Factorize the following;
1. x2 + 2xy + y2 – 4
Factors of quadratic expression 2. a2 – 12a + 36 – 16b2
x2 – 6x + 9 = (x – 3) (x – 3) = (x – 3)2 3. 4a2 + 36a + 81 – 25b2
⇒x2 – 6x + 9 – 4y2 4. 4 – 20b + 25b2 – a2
= (x – 3)2 – 4y2 5. a2 – 4ab + 4b2 – x2 + 6x + 9
= (x – 3)2 – (2y)2
= (x – 3 – 2y) (x – 3 + 2y) Challenge Problems
= (x – 2y – 3) (x + 2y – 3) Factorize the following:
1. x2 – 6x + 9 – 4y2 2. a2 – b2 + 12b – 36
2. Factorize m2 – 2mn + n2 – 9r2, completely:
Other Applications
Solution Finding the values of the variable(s) in a
m2 – 2mn + n2 – 9r2 quadratic expression given the factors of the
Quadratic expression = m2 – 2mn + n2 expression
Remaining term = – 9r2 I. Identify the given expression
II. Identify the factors of the given expression
Factors of m2 – 2mn + n2 III. Expand and simplify the factors to obtain an
(m2 – mn) – (mn+ n2) equivalent expression

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IV. Compare the two equivalent expressions to Worked Examples
obtain the values of the required variable(s) 1. Find the value of x for which is undefined.

Worked Examples
Solution
If (x – 3) and (2x + 3) are factors of
For to be undefined,
2x2 + mx – n, find the value of (m + n)
1 – 3x = 0 (Equate the denominator to zero)
Solution 1 = 3x, ( Solve for x)
(x – 3) (2x + 3) = 2x2 + mx – n x= .
x(2x + 3) – 3(2x + 3)
 The expression is undefined when x =
⇒2x2 + 3x – 6x – 9 = 2x2 + mx – n
2x2 – 3x – 9 = 2x2 + mx – n
2. For which values of x is the expression
Comparing L.H.S and R.H.S (
undefined.
(
m = – 3 and n = – 9
(m + n) = – 3 + (-9) = -12 Solution
For ( to be undefined,
(
Exercises 3.19
1. If (x – 5) and (x + 2) are factors of x2 + kx – 10, (x – 4)(x + 2) = 0
find the value of k. x – 4 = 0 or x + 2 = 0
x = 4 or x = – 2

2. If x2 + mx + = ( ) , find the value of m 3. Determine the values of x for which the

expression is undefined.
An Undefined Algebraic Fraction
An algebraic expression of the form, , also called
Solution
algebraic fraction, is said to be undefined (does
For to be undefined,
not exist), if the denominator is equal to zero. 2
x + 5x + 6 = 0
⇒; , b = 0
(x + 3) (x + 2) = 0
To find the value(s) for which an algebraic x + 3 = 0 or x + 2 = 0
fraction is undefined, x = -3 or x = - 2.
I. Equate the denominator of the expression to  The undefined values are x = -3 or x = -2
zero
II. Solve for the value of the involving variable.
4. For what values of y is + undefined?
III. The value(s) of the variable makes the
expression undefined. This means that the
variable can take all values except the value that Solution
makes the expression undefined For + to be undefined
y(y – 1) = 0

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y = 0 or y – 1 = 0 Solution
y = 0 or y = 1 For to be equal to zero;
(
(
( (factotize and set the numerator to zero)
5. Find the values of x for which is (

undefined. 1 – 2x = 0 (solve for x)

1 = 2x
Solution x=
( (
=( =
(
2. Find the value of x for which the expression
Now,
2x + 4 = 0 is zero.
2x = - 4
x = -2 Solution
For to be zero,
Exercises 3.7
Find the value or values of the variable for =0
which the expression is undefined; =0
( ( ( =0
1. 2. 3. ( ( x ( x + 11) – 4(x + 11) = 0
4. 5. 6. (x – 4) (x + 11) = 0
√
x – 4 = 0 or x + 11 = 0
x = 4 or x = -11
A Zero Algebraic Fraction
 The expression is zero when x = 4 or x = -11
An algebraic expression of the form, also called
algebraic fraction, is zero, if the numerator is Exercises 3.8
equal to zero. That is; , a = 0 Find the zeros of the following:
(
1. 2. 3. ( (
To find the value (s) for which an algebraic (
4. 5. ( 6.
fraction is zero or the zero(s) of an algebraic ( (

fraction:
I. Equate the numerator of the expression to zero. Algebraic Substitution
II. Solve for the value of the involving variable. It is the act of replacing variables with their
III. The value(s) of the variable makes the specific values in a given expression. The process
expression zero. This value(s) is called the zeros is also called evaluation of expressions.
of the expression, meaning, if the variable takes To evaluate algebraic expressions, substitute each
that value, the entire expression is zero. variable with its value/number and perform the
operations included.
Worked Examples
Worked Examples
1. Find the value of x for which is zero.
( 1. Evaluate 6z + 4x, if x = 3 and z = 2

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Solution –
Replace x with 3 and z with 2 to evaluate the ( ( ( (
expression as shown below; = – = – = =
6z + 4x = 6 (2) + 4(3) = 12 + 12 = 24
Exercises 3.20
2. Evaluate the expression 4x + (7 – z) – 6y for
x = 2, z = 4, y = 5; A. 1. Evaluate 2xy – 8x + 3y –12, when x = 5
and y = 7.
Solution
Substitute x = 2, z = 4, y = 5 in 2. If m = 3 and n = -3, evaluate (3m – n).
4x + (7 – z) – 6y
= 4(2) + (7 – 4) – 6(5) = 8 + 3 – 30 = - 19 3. Given that a = 2 and b = 3, evaluate;
(2a + b) (a – 2b)
3.If p = -2 and q = 3, evaluate – 4. Factorize 6a2 – ab – b2, and hence find the
value of the expression when a = 4 and b = 6
Solution
– 5. If x2 + y2 = 73 and xy = 12, find the value of
( ( (x + y)2
= – = – = = = -7
( (

4. Without using calculator, evaluate – B. If x = –1, y = –2, z = 3, t = 0 evaluate:

, where x = 2, y = -3 and z = 4 1. + + 3. – 2xy + 5xyz
(
2. x 4.
Solution

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4 SURDS Baffour – Ba Series

Meaning of Surds e.g. 2√ × 4√ = 2 × 4√ × √ = 8√

On a calculator √ 1.414213562…, a decimal
that does not terminate nor recur, hence cannot be Simplifying Surds
expressed in the form (as a fraction of two When a surd is expressed as product of a rational
integers). Such a root is called a surd. A surd is number and a surd, that is in the form a √ or √
therefore, the roots of rational numbers that , where b is a prime number, the surd is said to be
cannot be expressed as rational numbers. in its simplest form. E.g. 2√ √ . This means
Examples of surds are: √ , √ , √ ,√ ,√ that √ and √ are not in their simplest form.

In general, a number which cannot be expressed Rules

as fraction of two integers is called an irrational I. Break down the number in the root sign into
number. products of two factors, such that one of the
Most irrational numbers are not surds as it is in π. factors will be a perfect square (4, 9, 16, 25,
49…) and the other, a non – perfect square. e.g.
Note that the presence of a root sign does not √ =√
necessarily means that we are dealing with surds. II. Distribute the root sign for each factor.
Thus, √ and √ are not surds because e.g. 18 = √ =√ √
= 49 and = 1.728, and so √ = 7 and III. Simplify the perfect square and leave the
√ = 1.2. That is √ , and √ are surd.
rational numbers . e. g.18 = √ =√ √ = 3 × √ = 3√

Exercises 4.1 Worked Examples

Which of the following are surds? 1. Simplify √
1. √ 2. √ 3.√ 4. √
Solution
5. √ 6. √ 7. √ 8. √
1. √ = √ =√ √ = 2√
Properties of Surds 2. Simplify √
1. (√ =√ ×√ =a
e.g. √ × √ = (√ =7 Solution
2. (√ )=√ ×√ √ =√ × √ = 8√
e.g. (√ )=√ ×√ =√
√ √ 3. Simplify √
3. =√ e.g. =√ =√
√ √

4. a√ = a × √ = √ × a Solution
e.g. 3 × √ = √ × 3 = 3√ √ =√ ×√ ×√
5. a√ × c√ = a× c× √ × √ = ac√ = 2 × 7 × √ = 14√

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2. Simplify 5√ On the other hand, two or more surds without a
common factor or without the same number in the
Solution root sign are called unlike surds. E.g. 5√ and
5√ = × √ = 5 × 2√ = 10√ 4√ . Unlike surds cannot be added nor
subtracted.
Exercises 4.2
A. Express each in its simplest form; Worked Examples
1. √ 2. √ 3. √ 1. Simplify 5√ – 7√ + 4√
4. √ 5. √ 6. √
Solution
Surds of the Form a√ as a Single Square 5√ – 7√ + 4√ = (5 – 7 + 4) √ = 2√
Roots, √
Given a surd of the form a√ 2. Simplify √ + √ – 2√ +√
I. Find the square of the coefficient of the square
root. i.e. a2 Solution
II. Put the square of the coefficient of the square √ + √ – 2√ + √
root in a square root and multiply by the number
= √ × √ + √ – 2 ×√ ×√ + √ ×√
in the square root. i.e. √ × √
III. Express the product as a single square root. = 5 × √ + √ – 2 × 3× √ + 2 × √
√ ×√ =√ = √ –6√
= 2√
Worked Examples
3. Express 3√ –√ in the form a√
1. Express 6 √ as a simple square root

Solution Solution
6√ = √ ×√ =√ =√ 3√ – √
=3√ – √
Exercises 4.3 =3×√ ×√ –√ ×√
A. Express the following as square roots = 3 × 5√ – 2√
√ √ √
1. 2. 3. 4√ 4. 5. = 15√ – 2√ = (15 – 2) √ = 13√
√ √

B. Square the following; 4. Given that √ +√ = m√ , find the value

√ of m
1. √ 2. (√ × √ ) 3. 4.
√( √

Addition and Subtraction of Surds Solution

Two or more surds with a common factor or with √ +√ = m√
the same number in the root signs are called like =√ +√
surds. E.g. 5√ and 4√ . Like surds can be =√ ×√ +√ ×√
added and subtracted. = 3√ + 2√
Baffour – Ba Series, Core Maths for Schools and Colleges Page 80
 √ = m√ √ = ×√ ×√ = ×2×3 =
Therefore m = 5
Exercises 4.5
Exercises 4.4 Simplify the following;
A. Simplify each of the following; 1. √ 2. √ 3. √
1. 9√ – 5√ 3. 12√ – 7√ √ √ √
2. √ + 2√ – 3√ 4. √ – √ 4. 5. 6.

B. Simplify: B. Multiplying a surd by a Surd

√ +√ – 2√ 2. √ +2√ –√ To multiply two given surds:
3. √ +√ –√ 6. 4. √ +6√ Method
5. √ –√ – √ 6. √ +√ +√ I. Break each surd into a product of two factors
7. √ – 3√ –√ + √ such that one will be a perfect square and the
other, a non perfect square, if possible
Multiplication of Surds II. Simplify by grouping like terms.
A. Multiplying a Fraction by a Surd
Simplify the surd and cross out common factors if Method 2
possible. I. Find the product of the given surds to obtain a
single surd.
Worked Examples II. Break the single surd obtained into a product
of two factors such that one will be a perfect
1. Simplify √
square and the other, a non-perfect square, if
possible.
Solution III. Simplify by grouping like terms.
√
= √ = ×√ ×√ Worked Examples
√ 1. Simplify √ ×√
= ×2×√ =
Solution
2. Simplify √ Method 1
√ ×√
Solution = √ =√ =√ ×√ = 4√
√
Method 2
= ×√
√ ×√ =√ ×√
= ×√ ×√ = ×2×√ =5√
√ ×√ ×√ ×√
=2×√ ×2×√
3. Simplify √ = 2 × 2 × √ × √ = 4√
Solution

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2. Simplify √ √ Solution
√ (√ )
√
Solution
√
√ √ =√ ×√ =√ – =√ –1 =4–1 =3
√
√ ×√ ×√ ×√
= 4 √ × 4√ 2. Simplify √ (√ )
√
=4×4×√ ×√
= 16 √ Solution
√ (√ )
√
3. Simplify √ √
=√ √ =5+ =5 =
√
Solution
√ √ =√ ×√ 3. Simplify √ (√ )
√
=√ ×√ ×√ ×√
= 3 √ × 5√ Solution
=3×5×√ ×√ √ (√ )
√
= 15 √
=√ √ √
Exercises 4.6 =√ √
Simplify the following: √ √

1. 3√ √ 2. √ √ 3. 3√ √ =6+ =6 =
4. 5√ √ 5. √ √ 6. 5√ √
4. Evaluate √ ( √ )
√
C. Expansion of Surds
Reminders Solution
1. – = (a + b) (a – b) √ ( √ )
√
2. ( – 2ab +
=3×7 √ = 21 + 6 = 27
√
3. ( = + 2ab +
4. (a + b) (c + d) = a(c + d) + b (c + d) 5. Express √ (√ ) in the form p√ ,
√
5. a( b + c ) = ab + ac where p and q are real numbers.

Type 1: Expansion of the form Solution

a( b + c ) = ab + ac √ (√ )
√

Worked Examples =√ (√ √ )
√

1. Simplify √ (√ ) =√ ( √ )
√ √
= 3√ √
√

Baffour – Ba Series, Core Maths for Schools and Colleges Page 82

=3×√ ×√ √ √ √
= +
=3×3×√ √ √
=- +
=9√ √
= 7√ P=- and Q = (Not necessary)

Exercises 4.7
3. Simplify ( √ )( )
A. Simplify the following; √

1. 4 √ ( √ – 5√ ) 2. 3√ (4√ – 6√ )
Solution
3. 5√ ( √ –√ ) 4. 3√ (4√ – 6√ ) ( √ )( )
√

B. Simplify: = 6( ) √ ( )
√ √
√
1. √ ( √
√
) 2. √ ( √
√
) = 18 – + 6√ –
√ √
3. √ ( √ ) 4. √ ( √ ) = 18 – 4 – + 6√
√ √ √

5. √ ( √ ) 6.√ ( √ ) = 14 – + 6√
√ √ √
√
= 14 – × + 6√
√ √
Type 2: Expansion of the form
√
(a + b) (c + d) = a(c + d) + b (c + d) = 14 – + 6√

Worked Examples 4. Simplify ( √ – 5√ ) ( √ + 9√ )

1. Simplify (√ – √ (√ √
Solution
Solution = ( √ – 5√ ) ( √ + 9√ )
(√ – √ ) (√ + 3√ )
= √ ( √ + 9√ ) – 5√ ( √ + 9√ )
= √ (√ √ – √ (√ + 3√ )
= 6 + 3√ –√ –6 = 36√ + 27√ – 60√ – 45√
= 3√ –√ =36√ + 27√ – 60√ - 45√
= 2√ =2×√ ×√ =2×2×√ = 4√ = 36×5√ + 27√ – 60√ – 45×7√

2. Simplify ( √ )( √ ), giving your = 180√ + 27√ – 60√ – 315√

answer in the form P + Q√

Type 3: Expansion of the form:
Solution
( – 2ab +
( √ )( √ ),
= 1( √ ) –√ ( √ ) Worked Examples
√ 1. Simplify; ( √ √
= √ – –3
Solution
√
= –3+√ –
Baffour – Ba Series, Core Maths for Schools and Colleges Page 83
( √ √ Type 5: Expansions of the form:
=( √ –2( √ ( √ ) ( √ – = (a + b) (a – b)
= 25√ – 20√ + 4√
Worked Examples
= 25(7) – 20√ + 4(5)
Find the product of (√ √ ) (√ √ )
= 175 + 20 – 20√ = 195 – 20 √
Solution
2. Express ( √ √ in the form a + b√ , (a + b) (a – b) = –
where a and b are integers.
(√ √ ) (√ √ )

Solution =√ –√ =5–2=3
( √ √
Exercises
=( √ –2( √ ( √ ) ( √
A. Simplify:
= 9√ – 12√ + 4√
1. (√ ) (√ )
= 9 × 2 – 12√ + 4 × 3
2. (√ √ ) (√ √ )
= 18 – 12√ + 12
3. (√ √ ) (√ √ )
= 30 – 12√
4. (√ √ ) (√ √ )

Exercises 4.8
B.1. Given that x = 1 + √ and y = 1 – √
Simplify the following:
simplify; i. 5x + 5y ii. 2xy iii.
1. ( √ √ 2. ( √ √
2. Given thatp = √ √ and q = √ √ ,
3. ( √ √ 4. √ √ simplify; i. 2p – 2q ii. 4pq iii.
2
3. Evaluate p – 2p, if
Type 4: Expansion of the form
i. p = 2√ ii. p = √ + 1
( = + 2ab +
4. Given q = 3 + √ and ̅ = 3 – √ , find
Worked Examples q + ̅ and q ̅
Simplify (√ √
Some Solved Past Questions
Solution 1. √ –√ + 5√ (√ ), leaving
(√ √ your answer in the from + b√ ,
=(√ ) + 2(√ )(√ ) + (√ )2
2

Solution
= 5 + 2(√ )+2
=√ – √ + 5 × 5 – 25 √
= 7 + 2√
= 100√ – 5√ + 25 – 25 √
Exercises 4.9 = 100√ – 5√ + 25 – 25 √
Simplify the following: = 100√ – 5√ – 25 √ + 25
1. (√ 2. (√ √ 3. (3√ √ = 70√ + 25 = 25 + 70√

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2. Simplify √ –√ , leaving your answer in =√ ×√ √ ×√
√

surd form.
=8×2×√ ×

Solution = 16 × ×√
√ –√ = √ = (1.4142) = 2.2627
= ×√ × √ –√ ×√
2. Without using tables or calculator, evaluate
= × 8√ – 5√
3√ ( √ ), if √ = 2.646
= 3 × 4√ – 5√
= 12√ – 5√ Solution
= 7√ 3√ ( √ )
= 21√ –
3. Without using tables or calculator, simplify
= 21√ – 42
√ – 2√ (2√ – 5√ = 21( – 42 (Put √ = 2.646)
= 13.566
Solution
√ – 2√ (2√ – 5√ 3. Without using tables or calculator,
= √ × √ – 4 × 2 + 10√ – 5√ evaluate √ ( √ if √ = 2.236
= 5√ – 8 + 10√ – 5√
= 10√ – 8 Solution
√ ( √
Substitution = 12√ –
It is the act of replacing the given value of a surd
= 12√ – 20
in an equation. For instance, given the value of
= 12 (2.236) – 20 = 6.832
√
I. Break the surd (mother surd) into a product of 4. Without using tables or calculator, evaluate
two surds such that one of them will be √ 3√ ( √ ), if √ = 2.646
II. Substitute the value of the given surd and
simplify to complete the work.
Solution
Worked Examples 3√ ( √ )
1. Simplify √ , to five significant figures, 21√ – 6 × 7
given that √ = 1.4142 21√ – 42
⇒21(2.646) – 42 = 13.566
Solution
√ =√ 5. Without using tables or calculators evaluate
= √ ×√ =√ ×√ ×√ √ ( √ ) if √ = 2.236

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Solution 3. Simplify
√
√
√ ( √ √
= 12√ – 4 × 5 Solution
= 12√ – 20 √ √ √ √ √ √ √
= × = =
⇒12(2.236) – 20 (Put √ = 2.236) √ √ √
= 6.832
3. Given that √ = 1.4142, find correct to 5
Exercises 4.11 significant figures, the value of 3 –
√
Given that √ = 1.414 and √ = 1.732,
evaluate to three significant figures; Solution
1.√ 2. √ √ √
3– =3– × =3–
√ √ √

Rationalizing the Denominators of Surds 3–

√
=3– (Put √ = 1.414214)
Rationalization is the process of removing
irrational numbers or surds from the denominator =3– = 0.87870 (5. s.f)
of a fractional surd.
4. Given that tan 22.50 = and that √ =
√
A. Expressions of the form:
√
To rationalize the denominators of expressions of 1.4142, calculate tan 22.50 to four significant
the form: , multiply the denominator of the figures
√
fraction by the numerator and denominator of the
fraction. Reminder: √ × √ = a Solution
tan 22.50 =
√
Worked Examples
√ √
1. Simplify given that √ = 1.4142 But × = =√
√ √ √

⇒ tan 22.50 = √ , (Put √ = 1.4142)

Solution
0
√ √ tan 22.5 = 1.4142 – 1 = 0.4142 (4 s.f)
= × = =3√
√ √ √
But √ = 1.4142, Some Solved Past Questions
3 √ = 3 × 1.4142 = 4.243 (4 s. f) 1. Given that √ = 2.236068, find the value of
√
to five significant figures.
2. Simplify √
Solution
√ √
Solution = × =
√ √ √
√ √ √ √ √
√ = = × = = = 0.44721(5 s. f.)
√ √ √

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2. Given that √ = 1.414214, correct to five
1. ( √ 2.√ – √
significant figures the value of ( )
√ √
3. 4. tan 750 =
√ √
Solution
√ B. Expressions of the form:
( )= × √
√ √ √
To rationalize the denominators of expressions of
√
=3– (Put √ = 1.414214) the form:
√
=3– = 2.2928 (5.s.f.) I. Multiply the conjugate of the denominator of
the given fraction by the numerator and
3. Express
√
in the form p + q√ where p, q∈ R denominator of the given fraction.
√
II. Simplify numerators and denominators, where
Solution possible
√ √ √
= ×
√ √ √ Note; a + √ and a – √ , where a is rational
√ ( √ √ √ √
= = = + = and √ is a surd, are conjugate compound
surds, also called conjugate surds. Thus,
√ √
4. Express in the form a√ + b√ , where 1. a + √ is the conjugate of a – √
√
a, b∈ R 2. a – √ is the conjugate of a + √

Solution √
3. Therefore, = ×
√ √ √ √ √ √ √ √
= ×
√ √ √
√ (√ √ Worked Examples
=
1. Rationalize the denominator of
√ √ √
=
√ √ √ √
= + Solution
√
=
√
+
√ = ×
√ √ √

√ √ (√ (√
= + = = = 2(√ + 1)
⇒ a = and b = (Not necessary)
√
2. Simplify
√
Exercises 4.12
A. Rationalize the denominators:
Solution
√ √
1. 3. 5. 6. 7. √ √ √
√ √ √ √ √ = ×
√ √ √
√ √
B. Given that √ = 1.414 and √ = 1.732, = = = 2√ – 3
evaluate to four significant figures;

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3. Rationalize the denominator of 2. Simplify 2√ –√ + , leaving your
√ √
answer in surd form
Solution
√ Solution
= ×
√ √ √
√
2√ –√ +
√
=(
√ )( √ )
= 2√ –√ ×√ +
√
√
= =√ – √ ×√ +
√

=
√ =√ ×√ –√ ×√ +
√
√
= 10√ – 4 × 6 + ×
√ √
Some Solved Past Questions √
= 10√ – 24 +
1. If (√ ) = p√ , find p
√ = 10√ + 5√ – 24
= 15√ – 24
Solution
√ Exercises 4.13
√
√ √ √ √ √ Rationalizing the denominator;
=√ × =√ – = =
√ √ √
√ 1. 2. 3.
⇒ = p√ , p = √ √ √ √ √

√ √
4. 5. 6.
√ √ √ √ √ √

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05 NUMBER BASES Baffour Ba Series

Base Ten Numerals (Decimals) Every 10 – groups of tens equals hundred (100)
The decimal system or the base ten numeration producing a sequence of Hundreds, Tens and
system uses the following digits; 0, 1, 2, 3, 4, 5, Units.
6, 7, 8, 9
10 – groups of hundreds equals thousands
Writing Base Ten Numerals (1000) which in turn generates a sequence of
Writing base ten numerals is a matter of making a Thousands, Hundreds, Tens and Units.
sequential combination of the digits; 0, 1, 2, 3, 4,
5, 6, 7, 8, 9. For instance, to begin the writings, 10 – groups of thousands equals ten thousands
the first digit precedes or combines with itself and (10,000) which generates the sequence :
the rest of the digits to obtain; 00, 01, 02, 03, 04, 05, Ten thousands ←Thousands ← Hundreds ←
06, 07, 08, 09… Tens ← Units (Ones)

To proceed, the next digits (which is 1) Worked Examples

combines (precedes) with all the base ten A. Group the following base ten numerals into
digits to obtain: 10, 11, 12, 13, 14, 15, 16, 17, … ten thousands, thousands, hundreds, tens and
units
To proceed, the next digit (which is 2) 1) 312 2) 8 3) 4052 4) 51309
combines (precedes) with all the digits to
obtain; 20, 21, 22, 23, 24, 25, 26, 27, 28, 29… Solution
The process is followed sequentially to write all 1. 312 = 3 - Hundreds, 1- Tens and 2- ones
base ten numerals. 2. 8 = 8 - Ones
3. 4052 = 4 -Thousands, 0- Hundreds, 5 -Tens, 2-
Basis of Base Ten Numerals ones.
The base ten numeration system is based on 4. 51309 = 5 - Ten thousands, 1 -Thousands, 3 -
groupings in tens. For e.g., the stars below can be Hundreds, 0 - Tens and 9 - Units
grouped in tens as follows
* * B. Write the numerals for the following;
* * * * * *
* * **
*
* * * * * 1. Seventy – two thousand, six hundred and
* * * * * * * thirty – three.
2. Forty – five thousand three hundred and
2- groups often 5 twenty - three
u
This 2 – groups of tens and 5 units are
ni written as Solution
25. Similarly, 8 – groups of ten andts 6 units are 1. Seventy – two thousand, six hundred and
written as 86, giving the sequence tens and units. thirty – three

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= 72(1000) + 6(100) + 3(10) + 3(1) Writing base five, numerals is a matter of
= 72000 + 600 + 30 + 3 = 72633 combining the digits in a carefully, orderly
manner.
2. Forty - five thousand, three hundred and
twenty- three To begin, the first digit, (0) combines with itself
= 45(1000) + 3(100) + 2(10) + 3(1) and the other digits to obtain:00, 01, 02, 03, 04
= 45000 + 300 + 20 + 3 = 45323
The next digit, 1, takes its turn to combine with or
Exercises 5.1 precede the other digits to obtain
A. Write the numerals for the following; the sequence: 10, 11, 12, 13, 14
1. Sixty – four thousand, five hundred and
twelve To proceed, the next digit 2, takes its turn to
2. Seven hundred and eighty –nine thousand five combine with or precede the other digits to
hundred and fifty – six obtain: 20, 21, 22, 23, 24
4. Nine hundred and sixty – nine million, sixty - When the process is duly and carefully
four thousand, five hundred and thirty followed, as many base five numerals can be
5. Two hundred thousand and fifty – seven written

B. Write the following numerals in words Exercises 5.2

1) 54381 2) 6,897,678 3) 12, 675, 220 1. Write the base five numerals up to 400.

Base Five Numeration Place Value of Base Five Numerals

Base five (5) is based on groupings in fives. For The place value of base five numerals begins
example, the following stars can be grouped in from the right as;
fives as; 50, 51, 52, 53 etc
* * * * * * 50 = units (ones)
* * 51 = fives (5)
* * * * *
52 = twenty – fives (25)
53 = one hundred and twenty fives (125)
2 3 Therefore, the place value of the digits in 341five
Groups of fives and units can be identified as;
( less than 5) 3 4 1
This can be written as 23five and read as ↓ ↓ ↓
2 1
“Two - Three base five” 5 5 50
(Twenty-fives) (Fives) (One or unit)
Note: The base depicts the manner of groupings.
Worked Examples
Writing Base Five Numerals Find the place value of the digit “2” in the
The base five numeration system uses the following base five (5) numerals.
digits: 0, 1, 2, 3, 4. 1) 3112five 2) 23031five 3) 1024five

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Solution position as the answer. i.e. 0, 1, 2, 3, 4, 10, 11,
1. 3 1 1 2 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34,
↓ ↓ ↓ ↓ 40, 41, 42, 43, 44…
53 52 51 50 Therefore, 14 + 3 + 2 = 34five
The place value of 2 in 31125 is 50 or ones
Similarly, to perform 14 + 3 + 2 in base eight,
2. 2 3 0 3 1 find the sum of the numbers as 19. Then write the
↓ ↓ ↓ ↓ ↓ base eight numerals and identify the 19th numeral
4 3 2 1
5 5 5 5 50 as the answer. i.e. 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12,
The place value of 2 in 230315 is 54 or six 13, 14, 15, 16, 17, 20, 21, 22, 23, 24, 25,26, 28…
hundred and twenty fives (625) Therefore, 14 + 3 + 2 = 23eight

3. 1 0 2 4 2. The remainder principle

↓ ↓ ↓ ↓ In this method, the sum of the numbers is divided
53 52 51 50 by the base in which the operation is performed
The place value of 2 in 1024five is 5 or fives to get a whole number and remainder as answer.
For e.g. 14 + 3 + 2 = 19, in base five, 19 5 =
Exercises 5.3 3R4 = 34five, .
A. Identify the place value of the digit “3” in
the following base five numerals; Similarly, in base six, 19 6 = 3R1 = 31six and in
1) 3041five 2) 4034five 3) 11003five base eight, 19 8 = 2R3 = 23eight

B. What is the place value of the digits written Worked Examples

against the numerals; 1. Add 344 and 24 in base five
1. 132 (3) 2. 341 (1) 3. 4321 (3)
4. 1140 (4) 5. 208541 (0) 6. 275041 (2) Solution
Method 1
Addition in Base Five and other Bases Set out the work as follows;
1. Positional method 344 five
One‟s ability to perform operations in base five + 24 five
and other bases depends on one‟s ability to write
the numerals of that base, atleastthe first 40 I. Add 4 and 4 to obtain 8. Locate the number that
numerals. The sum of the numbers (in base ten) is occupies the 8thposition on the base five
numeration and that is 13.
used as a position to identify the numeral that 0, 1, 2, 3, 4, 10, 11, 12, 13 so write 3, leaving a
occupies that particular position as far as the remainder of 1
numerals of that base is concern. For e.g. to 344five
perform 14 + 3 + 2 in base five, add the numbers + 24five
to get 19. Then write the base five numerals and 3
identify the number that occupies the 19th

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II. Add 4, 2 and 1 (the remainder) to Therefore, the missing numeral is1133five
obtain 7. Locate the 7th number of the base five
numerals as 12 so write 2, remainder 1 That is; Exercises 5.4
344five
A. Perform the following in base five
+ 24five
23five 1. 2344five + 304five
2. 3001five + 444five + 2333five
III. Add 3 and 1(the remainder) to get 4. The 4th
numeral is 4, so record it to complete the 3. 4441five + 1441five + 4214five
344five 4. 4230five + 2421five + 4404five
+ 24five
423five 5. 322five + 122five + 420five

Method 2 B. Find the missing numerals;

4 + 4 = 8, 8 5 = 1R3 = 13, so write 3 R 1 1. 411five 2. 1012five
4 + 2 + 1(R) = 7, 7 5 = 1R 2 = 12, so record 2 – ***five – ****five
leaving a remainder of 1 134five 414five
3 + 1(remainder) = 4, 4 5 = 0 R 4 = 4, so
Subtraction in Base Five
record 4 to complete the work;
To perform subtraction in base five and other
344five
bases successfully, always set out the work in the
+ 24five
vertical form or on the place value chart as we did
423five
for the addition in base five.
2. Perform 4103five+ 2422five Take note of the fact that when a bigger numeral
is subtracted from a smaller one, the need arises
Solution to borrow from the next immediate digit on the
Set out the work in vertical form or arrangement left with possessions.
as follows;
4103five Generally, in base n, a borrowed digit is n. Thus,
+ 2422five in base five, each borrowed digit is 5. When
12030five properly added, subtraction becomes possible.

3. Find the missing numeral in 2031 – **** = 343 Worked Examples

if the operation was performed in base five 1. Perform 321five 24five

Solution Solution
I. Set out the work as follows;
2031 – **** = 343
321five
**** = 2031 – 343 24five
2031five
–343five
1133five II.1 – 4, but 4 1, so borrow 1 (five) from 2 and
add to 1 to make 6. Subtract 4 from 6 to get 2.

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That is:
*
321 *
*
– 24
2
B
This gives rise to 1 group of twos and 1 unit (less
III. Proceed by taking note of the fact that the
than two) written in base two as 112, and read as
digit “2” is 1 (because, 1(five) has been borrowed
“one- one base two”
already), so, 1 – 2, but 2 > 1, so borrow 1(five)
from 3 and add to 1 to get 6. Now 6 – 2 = 4
Writing Base Two Numerals
321five
The digits for base two numeration system are 0,
24five 1 only.
42five In writing base two numerals, make use of the
previous knowledge of combining the digits as
iv. The next column number 3 is left with 2, we did when writing the base ten and five
(because 1 has been borrowed already) so record numerals.
2 to complete the work.
321five To start, 0 precede itself to give, 00 and precedes
24five 1 to give 01. So we have: 00, 01,
242five The digit 1, takes its turn to precede 0 to give 10
and precede itself to give 11. So we have; 00,
2. Find i. 304five– 43five ii) 1010five– 323five 01, 10, 11.

Solutions To continue, watch carefully, that all the other

i. 304five ii. 1010five numerals precedes 0 and 1 only to generate all the
43five 323five other base two numerals. Watch it as;
211five 132five 10 precede 0 to give 100.
10 precedes 1 to give 101
Exercises 5.5 Now we have: 0, 1, 10, 11, 100, and 101
A. Perform the following;
1) 4002five – 312five 2) 3121five – 2042five The numeral that takes turn is 11.
3) 104five – 42five 4) 4000five – 444five 11 precedes 0 to give 110, 11 precedes 1 to give
111 so we have; 0, 1, 10, 11, 100, 101, 110, 111,
B. Find the missing addend in base five;
The number that takes turns is 100.
1) 432 + …. = 1000 2)…… + 442 = 4343
100 precedes 0 to give 1000, 100 precedes 1 to
3) …. + 242 = 1020 4) 231 + ….. = 1104
give 1001 giving the sequence; 0, 1, 10, 11, 100,
101, 110, 111, 1000, 1001…
Base Two (Binary) Numerals
The base two numeration system involves
As long as each numeral takes turn to precede 0
groupings in twos. For e.g., the following stars
and 1, all the base two numerals can be
can be grouped in twos as:
generated.

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Addition in Base Two Subtraction in Base Two
Write the base two numerals as follows: Similar to subtraction in base five, when the need
0, 1, 10, 11, 100, 101, 110, 111 … to serve as arises to borrow, do so from the next immediate
guide in performing addition in base two. For number on the left with possession.
example, to add 11 and 10 in base two , set out
the work as shown below: Be mindful of the fact that any number borrowed
11 in base two is 2. When the borrowed digit is
+ 10 added to wherever it is required, subtraction
becomes possible. For e.g. to perform 10two –
Add 1 and 0 to obtain 1 1two, set out the work as follows,
11 10two
+ 10 −1two
1

Continue by adding 1 and 1 to get 2; then count 1 has to be subtracted from 0, but 1 > 0, so
from 0 to locate the number that occupies the 2nd borrow from the next number on the left.
position. That is; 10 and write it to get the final Remember that the value of the borrowed number
answer as; is 2, added to 0 to obtain 2 subtraction is now
11two possible, 2 – 1 = 1. That is;
+ 10 two 10
101two −1
12
Worked Examples
Perform the following in base two. Worked Examples
1. 11 + 101 2. 10 +10 1. 110two – 11two 2. 101two– 1two
3. 1001 + 11 4. 11 + 11 3. 101two– 11two 4. 1101two −110two

Solution Solution
Set out the work in vertical form as follows; Set out the work in the vertical form as follows:
1. 11 two 2. 10 two
1. 110two 2. 101two
+ 101 two + 10 two
1000 two 100 two – 11two −1two
11two 100two
3. 1001two 4. 11two
+ 11two + 11two 3. 101two 4. 1101two
1100two 110two − 11two −110two
10two 111two
Exercises 5.6
Perform the following in base two; Exercises 5.7
1. 101 + 110 3. 110 + 111 + 10 Perform the following in base two;
2. 11010 + 10110 4. 11101 + 11011 1. 110 – 101 2. 1010101 – 111111

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3. 1101 – 111 4. 11011 – 10110 I. Find the product of the numbers.
5. 1101 – 1001 6. 10100 – 1101 II. Divide each product by the base in which the
operation is being performed.
Other Bases Less than Ten III. Record the remainders as answers.
1. Decimal or denary system (base ten) counting
is done in ones, tens, hundreds, thousands,etc and Worked Examples
the digits used in writing are: 0, 1, 2, 3, 4, 5, 6, 7, 1. Perform 234five × 4five, and leave the answer in
8, 9 base five.

2. Base five system, counting is done in ones, Solution

fives, twenty – fives, one twenty – fives, etc and Method 1
the digits used in writing are: 0,1, 2,3, 4 Base five numerals: 0, 1, 2, 3, 4, 10, 11, 12, 13,
14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40…..
3. Base two system (binary), counting is done in Set out the work as follows;
ones, twos, fours, eights etc and the digits
234five
used are in writing are: 0, 1
× 4five
Base three system (ternary), counting is done in
ones, threes, nines, twenty – sevens, etc and the I. Multiply 4 and 4 to get 16. The 16th number of
digits used in writing are 0, 1, 2 the base five numerals is 31 so write 1 leaving a
remainder of 3
Base eight system (octal), counting is done in
ones, eights, sixty – fours etc and the digits used 234five
×4
in writing are 0, 1, 2, 3, 4, 5, 6, 7
1

Multiplication of Bases II. Multiply 4 and 3 to get 12 and add 3(the

To multiply two or more numerals of the same remainder) to get 15. Identify the 15th numeral of
base base five as 30. Record 0, leaving a remainder of 3

234five
Method 1
×4
I. Write the numerals of that base (as many as
01
you can)
II. Multiply the numerals (in base ten) and III. Multiply 4 and 2 to get 8 and add 3(the
identify the product as a position. remainder) to get 11. The 11th numeral of base
III. Locate the numeral that occupies that five is 21 so record it to complete the work
position.
234five
Method II ×4
Use the “remainder principle” by going 2101five
through the following processes:

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Method 2 C. Find the areas of rectangles with lengths
Set out the work as follows; and breadths
1. 101two and 101two 2. 1010two and 11two
234five
3. 101two and 11two 4. 1101two and 110two
× 4five
Changing to Base Ten
4 × 4 = 16, 16 5 = 3 R 1 = 31 a. Numerals of the form: A B Cx
Write (R) 1 as the answer, leaving the quotient 3 To change a numeral of the formA B Cx to base
234five ten, where x is positive and x 0 follow the steps
below;
× 4five
I. Identify the base of the numeral as x
1 II. Multiply the identified base by each individual
digit of the numeral as shown;
4 × 3 = 12 + 3 (quotient) = 15, (A× x) (B × x) (C× x)
III. Find the sum of the product of the base and
15 ÷ 5 = 3R0 = 30 the individual digit as shown below;
Write 0 (R) as the answer, leaving the quotient, 3 (A × x) + (B × x) + (C× x)
IV. Assign exponents on the identified base from
234five
the right to the left beginning from 0. That is; (A
× 4five
× x2) + (B × x1) + (C × x0)
01 Simplify the equation, bearing in mind that
x0 = 1 and x1 = x
4 × 2 = 8 + 3(quotient) = 11,
11 5= 2 R 1= 21 Worked Examples
1. Change 132five to a number in base ten
Write 21 to complete the work as shown
Solution
234five A B Cx = (A × x2) + (B × x1) + (C × x0)
× 4five 132five
= (1 × 52) + (3 × 51) + (2 × 50)
2101
= 25 + 15 + 2 = 42ten

Exercises 5.8 2. Convert 2342five to a base ten numeral

A. Calculate the following;
1. 3214five × 3five 2.212five × 14five Solution
3.104six × 1530six 4. 231four × 33four 2342five
5. 47eight × 21eight 6. 123eight × 46eight = (2 × 53) + (3 × 52) + (4 × 51) + (2 × 50)
= (2 × 125) + (3 × 25) + (4 × 5) + (2 × 1)
B. Find the perimeters of squares with sides of = 250 + 75 + 20 + 2 = 347ten
length:
a. 101twocm b. 11011twocm c. 1111twocm 3. Change 10111two to base ten

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Solution Exercises 5.9
= (1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20) A. Convert the following to base ten;
= (1 × 16) + (0 × 8) + (1× 4) + (1×2) + (1×1) 1. 4133 five 2. 3420 five 3. 1042 five 4. 13022 five
= 16 + 0 + 4 + 2 + 1 = 23 ten
B. Change the following to base ten;
4. Convert 233four as a base ten numeral. 1. 1011two 2. 11011 two 3. 1100 two 4. 100101 two

Solution C. Convert the following to base ten;

233four 1) 2122 three 2) 103.51eight 3) 12000 three
= (2 × 42) + (3 × 41) + (3 × 40) 4) 22.112 three 5) 126eight 6) 700 eight
= (2 × 16) + (3 × 4) + (3 × 1) 7. If 29ten = xeight = ysix = zfive = wthree, find x, y, z
= 32 + 12 + 3 = 47 ten and w
8. If 28nine = 35a = 101b = 122c, find the values of
b. Numerals of the form: AB.CDEx
a, b and c
To convert numerals of the form: AB . CDEx
to base ten, where x is positive and x 0, follow D. Perform the following in base three;
the steps below; 1. 102 + 212 2. 2102 + 21 3. 2102 – 1021
I. Multiply the base by each digit : 4. 1212 – 1121 5. 221 × 21 6. 1000 121
(A× x) (B × x) (C× x) (D× x) (E× x)
E. Perform in base eight:
II. Find the sum of the product of the 1) 123 + 25 2) 256 + 127 3) 235 – 172
(A× x) + (B × x) + (C× x) + (D × x)+ (E× x) 4) 1000 – 77 5) 32 × 6 6) 346 ×5

III. Assign exponents on the base (x) from the F. In which bases have the following
point to the left increasing from 0 and from the calculations been done?
point to right decreasing from 0. 1. 12 + 3 = 21 2. 12 × 3 = 41 3. 231 + 132 = 413
(A × ) + (B × x0) + (C× ) + (D × ) + (E × ) 4. 12 – 3 = 6 5. 12 6. 432 – 234 = 165
IV. Simplify the equation bearing in mind that
x0 = 1 and x1 = x G. Change the following to base ten:
1. 12.34five 3. 3001.232four
Worked Example 2. 414.105six 4. 10110.1001two
1. Convert 341.25six to a decimal numeral.
Changing from Base Ten to Other Bases
Solution The method of repeated division is used to
341.25six change a base ten numeral to other bases. The
= (3 × 62) + (4 × 61) + (1 × 60) + (2 × ) + (5 × steps are:
) I. Identify the base ten numerals (decimal) and
= (3 × 36) + (4 × 6) + (1 × 1) + (2 × 0.17) + (5 × the base being converted to
0.03) II. Prepare a three column table with the base
= 108 + 24 + 1 + 0.33 + 0.14 = 133.47 being converted to, in column 1, the decimal

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numeral in column two and Remainder (R) in Exercises 5.10
column 3. A. Write as a number in base five;
III. Use the base being converted to (divisor), to 1)162 2) 102 3) 167 4) 244 5) 30
divide decimal numeral (dividend) and the
answer (quotient) repeatedly, until you get a B. Convert to a number in base two;
number that is less than itself (divisor). 1) 59 ten 2) 77 ten 3)102 ten 4)66 ten
IV. Carefully record the quotient in column 2 and
the corresponding remainder (R) in column 3, C. Change the following:
after each division. 1. 119 ten to a base three numeral
V. Write the numbers at the column of the 2. 205 ten to a base three numeral.
remainder from the bottom upward as the answer. 3. 336 ten to a number in base four
4. 175 ten to a base six numeral.
Worked Examples
1. Write 259ten as a number in base eight. Conversion Between Non Decimal Bases
It involves changing from one base to another,
Solution other than base ten. For e.g. to change 123four to
8 259 R a number in base five, there is the need to change
8 32 3 123four to a number in base ten, and then change
8 4 0 further from the base ten numeral to base five.
4

⇒259ten = 403eight Worked Examples

1. Convert 432 five to a number in base three.
2. Change 19 to a binary numeral.
Solution
Solution Change 432five to base ten
2 19 R = (4 × 52) + (3 × 51) + (2 ×50)
2 9 1 = (4 × 25) + (3 × 5) + (2 × 1)
2 4 1 = 100 + 15 + 2 = 117 ten
2 2 0
2 1 0 Change 117 ten to base three
0 1
⇒19 =100112 3 117 R
3 39 0
3. Change 53 ten to a number in base four 3 13 0
3 4 1
Solution 3 1 1
4 53 R 0 1
4 13 1
4 3 1 ⇒432 five = 11100three
4 0 3
⇒53 ten = 311four
2. Express 231 four as a number in base six

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Solution Worked Examples
231four to base ten 1. Convert 3t4twelve to base ten
= (2 × 42) + (3× 41) + (1× 40)
= (2 × 16) + (3 × 4) + (1 × 1) Solution
= 32 + 12 + 1 = 45 ten 3t4twelve
Then Change 45 ten to base 6 = (3 ×122) + (10×121)+ (4 × 120)
= 432 + 120 + 4 = 556ten
6 45 R
6 7 3 2. Convert 659ten to base twelve
6 1 1
0 1 Solution

⇒231 four = 113six 12 659 R

12 54 11(e)
Exercises 5.11 12 4 6
A. Convert the following; 0 4
1. 232 five to a base seven numeral.
⇒ 659ten = 46e
2. 1011two to a number in base four.
3. 131five to binary numeral.
3. Convert 546eight to a base twelve numeral.
4. 133six to a number in base three.
Solution
C. Change the following to base five
Change 546eight to a number in base ten
1. 1110two 2. 101111two 3. 10101two
= (5 × 82) + (4 × 81)+ (6 × 80)
= (5 × 64) + (4 × 8) + (6 × 1)
D. Write as a number in the bases indicated.
= 320 + 32 + 6 = 358
1. 422 five to base six 3. 1102three as base four
2. 113four as base three 4. 314five as base three
Now change 358 to base 12

Base Twelve 12 358 R

The Latin word for twelve is duodecim giving 12 29 10
rise to the duodecimal numeration system. 12 2 5
The duodecimals system require twelve digits but 0 2
we have ten already, so we invent two more
namely t and e to write and count; 358ten = 25ttwelve
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, t, e ⇒546eight = 25ttwelve
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1t, 1e
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 2t, 2e.. Exercises 5.12
A. Convert from base twelve to base ten
The numeral 5t2etwelve is read as “five, tee, two, ee, 1) 53 2) 90 3) 8t 4) ett 5) 2t9e
base twelve” and it is converted to base ten as; B. Convert from base ten to base twelve
(5 ×123) + (10 ×122) + (2 ×121)+ (11 × 120) 1. 27 2) 100 3) 180 4) 1000 5) 3587

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C. Calculate the following in base twelve Solution
1. 42e + 9tt 2. t894 + e97e 3. 357 – 319 a.
4. 896 × 3 5. tet × 7 6. 5tt1 × e 1 2 3 4 5
1 2 3 4 5 10
Tables for Number Bases 2 3 4 5 10 11
Table of values are usually constructed under 3 4 5 10 11 12
addition and multiplication ⊗ for a given base. 4 5 10 11 12 13
5 10 11 12 13 14
Given base n, the table is constructed using the
numbers 0 to (n – 1). Thus, for base 5, the set of b.From the table;
values are {0, 1, 2, 3, 4} and for base 6, the range (2 3 5 = 14
of values are 0 , 1, 2, 3, 4, 5. For large bases, a set
of values may be given for operation. c. From the table;
i. 5 x = 12
On the table, the range of values for the base n or 5 3 = 12
the 0 to n – 1, values occupy the first column and Therefore { }
the first row, whilst the operator, whether + or ×
is circled to look like or ⊗ respectively, and ii. x x = 12
placed at the left top corner of the table as shown 4 4 = 12
below:
Therefore{ }
0
0 d. It is commutative because the principal
diagonal is symmetrical.

⊗ 0 2. a. Copy and complete the table below for

0 addition base eight.

Thereafter; 1 3 5 7
I. Perform the operation. 1 2 4 6
II. Divide the answer by the base. 3 6 10 12
III. Record only the remainder in the boxes or 5 6 10
cells to complete the table. 7 12 14 16

Worked Examples b. Use the table to find the truth set of;
1. a. Draw an addition table for base six on the set i. 7 x = 12 ii. m m = 12
P = 1, 2, 3, 4, 5
b. Use the table to find (2 3 5 Solution
c. Use the table to solve; a. 1 7 = 8, 8 in base 8 = 10
i. 5 x = 12 ii. x x = 12 5 5 = 10, 10 in base 8 = 12
d. Why would you say addition is commutative 5 7 = 12, 12 in base 8 = 14
on the table? 7 1 = 8, 8 in base 8 = 10

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 Worked Examples
1 3 5 7 Arrange the following in descending order of
1 2 4 6 10 magnitude; 103four, 1011two and 43five
3 4 6 10 12
5 6 10 12 14 Solution
7 10 12 14 16 Change all the numerals to base ten
103four = (1 × 42) + (0 × 41) + (3 × 40)
b. From the table;
i. 7 x = 12 = 16 + 0 + 3 = 19
7 3 = 12
⇒x = 3 1011two = (1 × 23) + (0 × 22) + (1 × 21) + (1 × 20)
= 8 + 0 + 2+ 1 = 11
ii. m m = 12
5 5 = 12 43five = (4 × 51) + (3 × 50) = 20 + 3 = 23
⇒m = 12
⇒ 23 > 19 > 11,
Exercises 5.13 43five, 103four and 1011two in descending order.
1. a. Draw addition table for base seven on the
set P = { } Exercises 5.14
b. Use the table to find; (2 3) (4 2) A. Arrange in ascending order;
c. Find the value of n if; 1. 3011four, 1110two, 221three
i. n n = 11 ii. 3 4 = 2 n 2. 13six, 40five, 10110two
3. 230five, 332six, 232four, 2202three
2. i. Copy and complete the table below for
multiplication in base five.
B. Arrange in descending order:
1. 244five, 55six, 2002three, 14ten
⊗ 1 2 3 4
1 1 2 4 2. 11011two, 11five, 101three
2 2 13 3. 403six, 11eight, 2201three
3 3 22
4 4 13 22 Division of Bases
To perform division of numerals of the same base
b. Use the table to find: or different bases other than base ten,
i. (2 ⊗2)⊗4 ii. 3 ⊗ 3 I. Change the involving numerals to base ten.
c. Use the table to find the truth set of; II. Perform the division on the base ten numerals.
i. (1 ⊗m) ⊗m = 14 ii. m⊗m = 31 III. Change the answer from base ten to the
required base.
Ordering Numerals in Different Bases
To order two or more numerals in different bases, Worked Examples
convert all the numerals to base ten and order 1. Perform 2134five ÷ 3five, leaving the answer in
them accordingly. base five

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Solution 4n + 2 = 14
2134five ÷ 3five, 4n = 14 – 2
Change the numerals to base ten 4n = 12
( ( ( ( n=3
=

= = = 98 2. If 102n = 51, find the value n.

Change 98 to a base five numeral
Solution
5 98 R (1 × n2) + (0 × n1) + (2 × n0) = 51
5 19 3 n2 + 0 + 2 = 51
5 3 4 n2 + 2 = 51
5 0 3 n2 = 51 – 2
n2 = 49
98 = 343five
n=√ =7
Therefore, 2134five ÷ 3five, = 343five
3. What is the value of A if 3A1five = 91
2. Perform 201four ÷ 23four
Solution
Solution
(3 × 52) + (A × 51) + (1 ×50) = 91
201four ÷ 23four
(3 × 25) + (A × 5) + (1 × 1) = 91
Change the numerals to base ten
75 + 5A + 1 = 91
( ( (
= 5A + 76 = 91
( (
5A = 91 – 76
= = =3 5A = 15
A= =3
Change 3 to base four to get 3
Therefore, 201four ÷ 23four = 3four
6. Find the value of x if 142x = 113four
Equations Involving Number Bases
Solution
Here, a numeral with unknown base (variable) is
142x = 113four,
equated to another numeral with a known base.
To solve such problems, express each side of the
Change each side of the equation to a decimal
equation as a base ten numeral, and solve for the
(1 × x2) + (4 × x1) + (2 × x0) = (1 × 42) + (1× 41) +
value of the variable. (3× 40)
x2 + 4x + 2 = 16 + 4 + 3
Worked Examples x2 + 4x + 2 = 23
1. If 42n = 12ten, find the value of n x2 + 4x + 2 – 23 = 0
x2 + 4x – 21 = 0
Solution
(x2 + 7x) – (3x – 21) = 0
(4 × n1) + (2 × n0) = 14
x (x + 7 ) – 3(x + 7) = 0
(4 × n) + (2 × 1) = 14
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(x – 3) (x + 7) = 0 Exercises 5.15
x – 3 = 0 or x + 7 = 0 A. Find the value of the variables
x = 3 or x = –7, but x ≠ -7 1. 63x = 45 2. 43n = 23 3. 34n = 19ten
Therefore x = 3 4. 12n = 5 5. 11n = 3 6. 33n = 21ten

7. Given that 4(13)n = 54n, find the value of the B. Find the value of the variables:
base n. 1. 62 = 2M2five 2. K30five = 90 3. 3Pfive = 19
4. M4five = 14 5. 10Ktwo = 5 6. 2K3four = 39
Solution
Change each side to base ten; C. Find the value of the variable;
4(13)n= 54n 1. 2(25)x = 54x 2. 3(12)x = 2(22)x
4[( ( ] = (5 × n1) + (4 × n0) 3. 1100three = 121x 4. 302x = 122six
4(n + 3) = 5n + 4
4n + 12 = 5n + 4 D. Determine the base;
12 – 4 = 5n – 4n 1. 52x + 153seven = 134 3. 414six + 33x = 251eight
8=n 2. 416seven – 72x = 151 4. 221y + 302four = 75

8. Find the base x such that 3657 + 43x = 217 Word Problems
Given a word problem involving number bases,
Solution I. Write a mathematical equation.
3657 + 43x = 217
II. Express each side of the equation as a number
(3 × 72) + (6 × 71) + (5 ×70) + (4 × x1) + (3 × x0) =
217 in base ten.
III. Solve for the value of the involving variable,
147 + 42 + 5 + 4x + 3 = 217
if any.
4x + 197 = 217
4x = 217 – 197
Worked Examples
4x = 20
1. A number is written as 37 in base x. Twice the
x=5
number is written as 75 in base x. Find the value
9. If 123five – 42x = 12, find the base x of x.

Solution Solution
123five – 42x = 12 2(37)x = 75x
(1 × 52) + (2 × 51) + (3 × 50) – (4 × x1) + (2 × x0) = 2[( ( ] = (7 × x1) + (5 × x0)
12 2( 3x + 7) = 7x + 5
25 + 10 + 3 – [4x + 2] = 12 6x + 14 = 7x + 5
25 + 10 + 3 – 2 – 4x =12 6x – 7x = 5 – 14
36 – 4x =12 -x=-9 x=9
- 4x = 12 – 36
- 4x = - 24 2. If 123y = 83, obtain an equation in y, hence
x=6 find the value of y.

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Solution ii. largest possible replacements for x and y.
123y = 83, iii. possible replacements for x and y such that
(1 × y2) + (2 ×y1) + (3 × y0) = 83 x=y
y2 + 2y + 3 = 83
y2 + 2y – 80 = 0 Solution
(y2 + 10y) – (8y – 80 ) = 0 27x = 32y,
y (y + 10) – 8(y + 10) = 0 Express each side of the equation in base ten
(y – 8 ) ( y + 10) = 0 (2 × x ) + (7 × x0) = (3 × y ) + (2 × y0 )
⇒y – 8 = 0 or y + 10 = 0 2x + 7 = 3y + 2
y = 8 or y = -10
⇒y = 8 , ignore the negative answer, y -10 Since each side is in base ten, substitute the base
ten values: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
3. Evaluate √ as a number in the decimal L.H.S R.H.S
system. 2x + 7 3y + 2
2(0) + 7 = 7 3(0) + 2 = 2
Solution 2(1) + 7 = 9 3(1) + 2 = 5
2(2) + 7 = 11 3(2) + 2 = 8
√ = √( (
2(3) + 7 = 13 3(3) + 2 = 11
=√ =√ =7 2(4) + 7 = 15 3(4) + 2 = 14
2(5) + 7 = 17 3(5) + 2 = 17
4. Find the base two number that is equivalent to
2(6) + 7 = 19 3(6) + 2 = 20
49seven.
2(7) + 7 = 21 3(7) + 2 = 23
2(8) + 7 = 23 3(8) + 2 = 27
Solution
2(9) + 7 = 25 3(9) + 2 = 30
Change 497 to base ten;
= (4 × 71 ) + (9 × 70 ) = 28 + 9 = 37 From the investigations,
2x + 7 = 2(8) + 7 = 23, ⇒x = 8
Express 37 as a number in base two;
3y + 2 = 3(7) + 2 = 23, ⇒y =7
2x + 7 = 2(2) + 7 = 11, ⇒x = 2
2 37 R
2 18 1 3y + 2 = 3(3) + 2 = 11, ⇒y =3
2 9 0 2x + 7 = 2(5) + 7 = 17, ⇒x = 5
2 4 1 3y + 2 = 3(5) + 2 = 17, ⇒y =5
2 2 0 Smallest possible replacements for x and y is
2 1 0
0 1
x = 2 and y = 3
Largest possible replacements for x and y is
49seven = 100101two x = 8 and y = 7

5. If 27x = 32y, find the Possible replacements for x and y such that
i. smallest possible replacements for x and y. x = y is x = 5 and y = 5

Exercises 5.16
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1. Make up tables for addition and multiplication Number Bases and Binary Combined
of ternary (base three) numerals. To solve problems involving a combination of
binary operation and number bases:
2. The length and breadth of a rectangle are I. Identify the binary operation and the given
1101two cm and 11two cm. Find its perimeter and base.
area as numbers in the base two. II. Perform the binary operation taking note
that answer obtained is in base ten.
3. Given that 12 is the equivalence of twice of 14y III. Convert the answer to the given base.
, what is the possible scale of y. IV. Complete the table of values if any, with the
answers obtained from the conversion.
4. If 120x = 21y, find the smallest and largest
possible replacements for x and y. Worked Examples
1. The operation ∆ is defined on the set of real
5. Show that 121m, where m is a natural number numbers by a ∆ b = in base four. Evaluate:
greater than 2, must be a square number.
i. 9 ∆ 3 ii. 84 ∆ (10 ∆ 5)
Challenge Problems
Solution
1. Show that no whole number replacements
exist for 43x = 26y i. a ∆ b =
(
9∆3= = = = 11
2. Copy and complete the following table. Each
row represents the same number expressed in
Change 11 to base four;
different ways.
11 ÷ 4 = 2 R 3
11 = 23four
Base 12 10 8 5 3 2
27
Alternatively,
2e
11101
4 11 R
3. Multiply the binary number 101 and 1011 and 4 2 3
express your answer in the scales of ten. 0 2
11 = 23four
4. A square has a side of length 120three mm. Find ii. a ∆ b =
its perimeter and area. 84 ∆ (10 ∆ 5)
(
(10 ∆ 5) = = = =7
5. Show that 43 – 34 is a multiple of 7, if the
84 ∆ (10 ∆ 5) = 8 ∆ 7
numbers are in the scale of eight, and is a (
multiple of 5, and is a multiple of 5 if the 84 ∆ 7 = = = = 47
numbers are in the scale of six. Check if it is also
true for 52 – 25, and make a general statement. Change 47 to base four;

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 4 47 R
4 11 3 Exercises 5.17
4 2 3
1. The operation * is defined by m * n = .
0 2
Evaluate the following:
47 = 233four i. 12 *18 in base 8 ii. (3 * 2) * 9 in base 6

2. If a * b = 2ab – 2, is the definition of the 2. a. Copy and complete the table below for *
operation * in base 8; defined by m * n = 7mn – m in base five for
a. Draw a table for * on the set { 2, 3, 4, 5, 6} the set {1, 2, 3, 4}
b. Use the table to evaluate the following;
i. 6 * 5 ii. 2 * 2 * 2 iii. (2 * 2) * 4 * 1 2 3 4
c. From the table, find the truth set of: 1 40 102
2 22 101
i. n* n = 36 ii. n * (2 * 2) = 106
3 33 124 220
4 44 310
Solution
a. b. Use the table to evaluate the following:
* 2 3 4 5 6 i. 2 * 4 ii. 4 * 4 iii. 1 * 2
2 6 12 16 22 26
c. Use the table to solve the following;
3 12 20 26 34 42
i. x * x = 413 ii. 2 * x = 130
4 16 26 36 46 56
5 22 34 46 60 72
6 26 42 56 72 106 3. a. Complete the table below for ◙ defined by x
◙ y = x + y – 2 in base nine.
b. From the table;
i. 6 * 5 = 60 ◙ 3 4 5 6 7
ii. 2 * 2 * 2 = 26 3 4 5 6 8
iii. (2 * 2) * 4 = 56 4 5 7 8
5 6 8 10 11
6 7 10 12
c. From the table;
7 8 10 12
i. n * n = 36
4 * 4 = 36 b. From the table, evaluate the following;
{n : n = 4} i. 4 ◙ (4 ◙ 4) ii. (3 ◙ 6) ◙ (5 ◙ 4)
c. From the table, find the truth set of the
ii. n* (2* 2) = 106 following;
6 * (2* 2) = 106 i. n ◙ n = 11 ii. (n ◙ 3) ◙ 6 = 8
{n : n = 6}

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6 RELATIONS AND FUNCTIONS Baffour – Ba Series

Definition of Relations x represents the Domain and y represents the Co-

A relation is a formula or statement (called rule) domain.
which enables one to match elements of two sets.
Among the two sets, the first set is called domain In the diagram above, set x contains “regions”
and the second set is called co-domain. and set y contains “regional capitals”. As such,
the two sets can be described by the relation “is
Relations can be described by means of; the regional capital of”.
1. Set of ordered pairs.
2. The solution set of open sentences. e.g x is the Worked Examples
square of y. Identify the relation that exist between the
3. Arrow diagrams and Cartesian graphs. following sets
1. x y
Study the diagram below;
A B ( , ( ,

( , ( ,
Solution
Domain Co-domain If the diagram is studied carefully, you will
Set A which contains the elements, Ghana, realize that each element of the domain
Nigeria, Togo, Mali occupies the first position multiplied by 2, equals the corresponding image
and is called the domain and set B with elements in the co-domain. Therefore, the relation “is twice
Accra, Abuja, Lome, Bamako, occupying the of” befits the description of both sets
second position is called the co-domain.
2. x y
The domain and the co-domain are always related
such as we have in the diagram above;Set A ( , ( ,
consist of countries and set B consist of capital
towns. Thus, the association between the two sets
is described by the relation “is the capital of”.
Similarly, x y Solution
Study the diagram carefully. Observe that;
( , ( , 1 + 3 = 4, 2 + 3 = 5, 3 + 3 = 6, 4 + 3 = 7
Therefore, the relation is “is 3 more than”
Domain Co-domain

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3. x y
( ) ( ,
( , ( ,

3. Many – to – one Relation

Solution It is the type of relation in which more than one
Observe that element of the domain match to one element of
7 – 5 = 2, 8 – 5 = 3, 9 – 5 = 4, 10 – 5 = 5 the co-domain. This is represented in the diagram
Therefore, the relation is “5 less than” below for the relation “is a product of”
x y
4. x y
( , ( )
( + ( +
Domain Co-domain

Solution
Observe that 4. Many – to – many Relations
× 24 = 12, × 18 = 9, × 10 = 5, Therefore, the It is the type of relation in which more than one
member of the domain match to many members
relation is “is half of”
of the co-domain as shown below for the relation,
“is less than”
Types of Relations
1. One – to – one Relation x y
It is the type of relation in which each element of ( + ( +
the domain matches to only one element of the
co-domain. This is represented as in the diagram Domain Co-domain
below for the relation “is increased by four”
x y Exercises 6.1
Identify the type of relation:
1.
( , ( , x y

Domain Co-domain ( , ( ,

2. One – to – many Relations

It is the type of relation in which one or more 2. x y
element(s) of the domain matches to many
elements in the co-domain. This is ( , ( +
diagrammatically represented as shown below for
the relation “is square root of”;

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3. ii. Plug in the equation, the given values of the
x y domain (x) to obtain the set of values of the
( ) ( , codomain (y)

4. x y When the domain is unknown:

i. Express the rule as mathematical equation
involving x and y.
( ) ( ,
ii. Change the subject to x and put in, the given
values of the codomain (y) to obtain the set of
values of the domain (x)
B. Identify the rule of relations;
1. x y
Note: It is not always possible to express the
( + ( + rules as a mathematical equation.

2. x y Worked Examples
1. The rule of a relation is “is one-third of”.
Given the domain, {12, 15, 18, 24}, find the co-
( , ( ,
domain. Draw a diagram of the relation.

3. Solution
x y
Rule =“is one-third of”
Domain ={12, 15, 18, 24},
( , ( ,
According to the rule, y = ( x),
when x = 12, y = × 12 = 4,
4. x y
when x = 15, y = × 15 = 5,
when x = 18, y = × 18 = 6,
( , ( ,
when x = 24, y = × 24 = 7,
The co-domain = {4, 5, 6, 7} respectively.
Unknown Domain or Co-domain x y
The three main components of a relation are
domain , codomain and the rule. Given any two ( , ( ,
of them, the third one can be found.

When the codomain is unknown: 2. The rule of a relation is “is four less than
i. Express the rule as mathematical equation twice”. Given the codomain {6, 8, 18, 18, 22},
involving x and y. find the domain. Draw a diagram of the relation.

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Solution The range is always in the co-domain (is a subset
Rule : “is four less than twice” of the co-domain).
Codomain = {6, 8, 10, 18, 22}
Domain = Unknown Consider the diagram below
From the rule, y = 2x – 4 X Y
x= (Change subject to x)
( , ( ,
When y = 6, x= =5
When y = 8, x= =6 The range = { }.
When y = 10, x = =7
Worked Examples
When y = 18, x = = 11
1. The rule of a relation is “is seven more than
When y = 22, x = = 13 twice”, the domain is { } and the co-
domain is {8, 9, 10, 11, 12, 13, 14, 15, 17, 20}
Domain = {5, 6, 7, 11, 13} what is the range of the relation?

Solution
The rule : “is 7 more than twice”
y = 2x + 7
( ) ( ) Domain = { }
Let x represent the elements of the domain
Exercises 6.2
From the rule, y = (2x + 7),
1. Given the domain {2, 4, 6, 8, 10}, find the co- When x = 1, y = 2(1) + 7 = 9
domain, if the rule of the relation is “five more When x = 2, y = 2(2) +7 = 11
than thrice”. Represent the relation in a diagram. When x = 3, y = 2(3) + 7 = 13
When x = 4, y = 2(4) + 7 = 15
2. Given the co domain {13, 14, 15, 17} for the When x = 5, y = 2(5) + 7 = 17
relation “is seven more than of”; Range of the relation = { }
i. find the respective elements of the domain,
ii. show the diagram of the relation
2. Given the domain = prime factors of 15} and
3. The co-domain of a relation is {25, 49, 81, the co domain = {factors of 30}, draw an arrow
100}. If the rule of the relation is “is a square diagram for the rule “is a multiple of” and find
of”, determine the domain of the relation, for all the range of the relation.
positive integers and draw a diagram of the
relation. Solution
Domain = {3, 5}
The Range of a Relation
Codomain = {1, 2, 3, 5, 6, 10, 15, 30}
The set of numbers in the co-domain that has
Rule : “is a multiple of”
corresponding members in the domain is called
the range.

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 OR x y

( , ( ,
( )

Since 1 → 3, 2 → 4, 3 → 5, and 4 → 6, the

ordered pairs formed are:(1, 3), (2, 4),(3, 5) and
( )
(4, 6). As ordered pairs, the first number is an
Range = {3, 5, 6, 10, 15, 30}
element of the domain and the second number,
Exercises 6.3 the corresponding element in the range.
1. For the domain = {prime factors of 30} and the
co domain = {factors of 12}. Draw an arrow Worked Examples
diagram for the rule “is increased by one” and 1. Which of the following ordered pairs do not
find the range of the relation. satisfy the rule of a relation “is twice as”:
(2, 4), (5, 10), (11, 22), (10, 5)
2. Domain = {prime factors of 24} and the co
domain = {factors of 36}. Draw an arrow Solution
diagram for the rule “is square of” and find the Rule is “is twice as”y = 2x
range of the relation. Domain { }
From the rule,
When x = 2, y = 2(2) = 4
3. Domain = {factors of 12 that are multiples of Therefore, (2, 4) satisfies the rule
3} and the co domain = {odd factors of 63}. When x = 5, y = 2(5) = 10
i. Draw an arrow diagram for the rule “is divisible Therefore, (5, 10) satisfies the rule
by” and find the range of the relation. When x = 11, y = 2(11) = 22
Therefore (11, 22) satisfies the rule
Relations as Ordered Pair When x = 10, y = 2(10) = 20
In relations, each elements of the domain has a Therefore (5, 10) does not satisfies the rule
corresponding element in the range. When each
element in the domain is paired with its 2. Which pair of ordered pairs is either true or
respective element of the range, we have ordered false for the rule “is the square of” (1, 1), (4, 16),
pairs: For instance, if the domain = { } (5, 20), (7, 49)
and range ={ } satisfy the rule “is 2 more
than”, then it can be represented Solution
diagrammatically as; Rule is “is the square of”
y = x2
Domain (x) : 1 2 3 4
When x = 1, y = (12) = 1
Range ( y ) : 3 4 5 6 Therefore, (1, 1) is true
When x = 4, y = (42) = 16

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Therefore (4, 16) is true be referred to as a mapping, all elements of the
When x = 5, y = (52) = 25 domain must be assigned.
Therefore (5, 20) is false
When x = 7, y = (72) = 49 On the other hand, in diagram C, c in the domain
Therefore (7, 49) is true has two images in the co- domain so it is not a
mapping but a relation. Similarly, in diagram D, 4
Exercises 6.4 in the domain do not have a corresponding
A. Show which of the following ordered pairs element in the co-domain. This relation is not a
do not satisfy the given rule of the relation in mapping. It follows that all mappings are
each case relations but not all relations are mappings.
1. Rule: is one-third of”, (9, 3), (12, 4),
(15, 6), (21, 7)
Types of Mapping
2. Rule “is a multiple of” (15, 5), (3, 9), (12, 36),
1. One – to – one mapping
(7, 49)
It is the type of mapping in which each element
3. Rule: “is nine plus of” (2, 10), (4, 11), (3,
of the domain matches to only one element of the
10), (6, 12), (10, 14). co-domain and vice – versa. It is also called one –
to - correspondence. This is represented as;
Mapping
x y
A mapping is a special type of relation in which
each member of the domain matches to a member
in the co-domain. In other words, mapping is an ( , ( ,
association between two sets, say A and B, such
that each element of Set A is associated with a
unique element of set B. 2. Many – to – one mapping
In a mapping, elements of the domain are It is the type of relation in which more than one
represented by x and the elements of the co- element of the domain match to one element of
domain are represented by y. the co-domain. This is represented by the diagram
Consider the diagrams below; below;
A. B.
( , ( )
( , ( , ( + ( ,

C. D. x y Exercises 6.5
x y
Identify whether it is a mapping or/and a
( ) ( + ( , ( + relation and the type of mapping / relation;
1. x y

In diagrams A, and B, each element of the

domain matches to an element in the co-domain. ( , ( ,
These are called mappings. Thus, for a relation to

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2. I. Determine the value of m (the gradient) by the
x y formula; m =
( , ( + Pick two values of the domain (x) and label them
as x1 and x2 and label their corresponding y values
as y1and y2and use it to determine the gradient, m
3. x y
II. Substitute the value of m in y = mx + c
III. Find the value of c by substituting any
( ) ( , ordered pairs of x and y in y = mx + c
IV. Finally, substitute the values of m and c in y
= mx + c to get the rule of the mapping.
4. x y
For linear reciprocal mappings, work without the
( ) ( , constant numerator, k, to determine y = mx + c
and write the rule as; y = or x→

Rule of a Mapping Method 2: (Simultaneous Approach)

The rule of a mapping is the statement or formula I. Write two linear equations of the form y = mx +
that shows how the domain and co-domain are c, using the values of x and y of the mapping.
related. Before finding the rule of a mapping, II. Solve the two equations simultaneously to
identify whether it is: determine the values of m and c.
1. Linear mapping or linear reciprocal mapping III. Substitute the values of m and c in y = mx + c
2. Exponential mapping or exponential reciprocal to determine the rule of the mapping
mapping
3. Quadratic mappings or quadratic reciprocal For linear reciprocal mappings, work without the
mappings constant numerator, k, to determine y = mx + c
and write the rule as;
Rule of Linear Mappings and Linear y= or x →
Reciprocal Mappings
A linear mapping and its reciprocal mapping are
Worked Examples
kinds of mapping in which the elements of the
1. What is the rule for this mapping?
domain have a common difference and the
elements of the co-domain also have a common x 1 2 3 4 55
difference.
y 1 3 5 7 9
Method 1: (Gradient Approach)
Solution
All linear mappings and its reciprocals follow the
Method 1
rule; y = mx + c or x → mx + c where m is the
Difference in x – values = 1
gradient, c is the y – intercept and m and c are
Difference in y – values = 2
constants.
This means that the mapping is linear. Thus,

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y = mx + c. When x = 3, y = 4
But m = 4 = 3m + c……………….(1)
If x1 = 3 and x2 = 4, y1 = 5 and y2 = 7
When x = 6, y = 2
m= = =2 2 = 6m + c ……………….(2)

Put m = 2 in y = mx + c to obtain y = 2x + c eqn (1) – eqn (2);

When x = 3, y = 5, substitute in y = 2x + c 2 = -3m
5 = 2(3) + c m=–
5=6+c
5–6=c
Put m = – in eqn (1);
–1 = c or c = – 1
Put c = – 1 in y = 2x + c to obtain y = 2x − 1 4 = 3(– ) + c
The rule is: y = 2x − 1 or x → 2x – 1 4 = -2 + c
4+2=c
Method 2 c=6
y = mx + c
Substitute m = – and c = 6 in y = mx + c, to
When x = 2, y = 3
3 = 2m + c……………….(1) obtain y = – x + 6 or x → – x + 6

When x = 3, y = 5 3. Determine the rule of the mapping:

5 = 3m + c ……………….(2) x 2 3 4 5

eqn (2) – eqn (1);

2=m y

Put m = 2 in eqn (1); Solution

3 = 2(2) + c For all linear reciprocal mappings, work without
3=4+c the constant numerator, k = 1
3–4=c y = mx + c
c = -1 When x = 2, y = 13
Substituting m = 2 and c = -1 in y = mx + c. 13 = 2m + c…………… (1)
The rule is: y = 2x – 1 or x → 2x – 1
When x = 1, y = 9
2. In the mapping below, find the rule 9 = m + c ……………….(2)
x 3 6 9 12 15
eqn (1) – eqn (2);
4=m
y 4 2 0 -2 -4

Solution Put m = 4 in equation (1)

y = mx + c 13 = 2(4) + c
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13 = 8 + c y= x–3
13 – 8 = c
For linear reciprocal mappings, y =
c=5
Substitute m = 4 and c = 5 in y = mx + c, to get But m = , c = –3 and k = 2.
y = 4x + 5. ⇒y= = =
⁄

The rule of the mapping is expressed in the form The rule is : y = or x →

y= or x →
Exercises 6.6
Substitute m = 4, c = 5 and k = 1. Determine the rule of the mappings;
1.
The rule is: y = or x → x 1 2 3 4 5

4. Determine the rule of the mapping below; y 12 9 6 3 0

x -5 0 5 10 15 2.
x -5 -3 -1 1 3 5
7 9
y 2 y -5 -0 5 10 15 20

3. x -5 -4 -3 -2 -1
Solution
From the mapping, k = 2
y = mx + c y 2 5 8 11 14

4. x 1 2 3 4 5
When x = -5, y = -5
-5 = -5m + c…………… (1)
y
When x = 5, y = -1
5.
- 1 = 5m + c ……………….(2) x 3 2 1 0 -1 -2

eqn (1) + eqn (2)

- 6 = 2c y -3
c= -3
6. x -6 -3 0 3 6
Put c = - 3 in equation (1)
- 5 = - 5m – 3
- 5 + 3 = –5m
y
- 2 = - 5m
7.
m= x 0 1 2 3 4

Substitute m = and c = –3 in y = mx + c, to get y 1 1 2 2

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8. x -1 0 1 2 3
Where k is the constant numerator, r is the
common ratio of the values of the co – domain
and (a, b) = (y1, x1).
y 5 3 1 -1 -3
9. x 1 2 3 4 5
Worked Examples
y -6 -8 -10 -12 -14
1. Find the rule of the mapping;
x 0 1 2 3 4
Rule of Exponential Mappings and Exponential
Reciprocal Mappings y 1 3 9 27 81
An exponential mapping is a kind of mapping in
which the elements of the co-domain differs from Solution
each other by a common ratio, usually denoted by “r” Method 1
For exponential mappings y =
If the elements of the co – domain are fractions But r = 3 1 = 3
with a constant numerator, whilst the Put r = 3 in y =
denominators differ by a common ratio, the ⇒y = 3x
mapping is said to be an exponential reciprocal The rule is y = or x →
mapping.
Method 2
All exponential mappings follow the rule; From the mapping, r = 3 1 = 3 and a = 1 and b
Rule 1 = 0 substitute in y = a
y = or x → ⇒y = (1)( =
where r is the common ratio. The rule is y = or x →

Rule 2 2. x 1 2 3 4
y=a or x →a
Where r is the common ratio and (a, b) = ( y1, x1) y 4 16 64 256
Solution
All exponential reciprocal mappings follow the Method 1
two rules; For exponential mappings, y =
But r = 4 1 = 4
Rule 1 Put r = 4 in y =
y= or x → ……… (2) ⇒y =
Where k is the common numerator and r is the The rule of the mapping is y = 4x or x → 4x
common ratio of the elements of the co – domain.
Method 2
Rule 2 From the mapping, r = = 4 and a = 4 and b = 1

y= or x → substitute in y = a
⇒y = (4)(

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The rule is y = (4) or y →(4)( 5. x -3 -1 1 3

3. Determine the rule of the mapping

y
x 1 2 3 4

The Rule of Quadratic Mappings

y
A mapping is said to be of quadratic form if the
Solution first difference of the values of the co- domain is
Method 1 not constant but the second difference. The rule
For exponential reciprocal mappings, of quadratic mappings is in the form: y = a +
y= bx + c
k =3 (common numerator) and r = =5
To determine the rule of quadratic mapping pick
Put k =1 and r = 5 in y = x any three points (x, y), ( , ) and ( , ) and
⇒y = x substitute each point into the relation y = a +
The rule for the mapping is y = bx + c to get three equations and solve for the
x
values of a, b and c simultaneously.
Method 2
From the mapping, k = 3, r = 25 5 = 5, Worked Examples
a = 5 and b = 0, substitute in y = 1. What is the rule for the mapping?
⇒y=( or x → ( x 0 1 2 3 4
( (

y 2 6 12 20 30
Exercises 6.7
Determine the rule of the mappings;
1. Solution
x 3 4 5 6 7 First difference = 6 – 2 = 4, 12 – 6 = 6,
20 – 12 = 8, 30 – 20 = 10
y 8 16 32 64 128 Second difference = 6 – 4 = 2, 8 – 6 = 2
2. x 0 1 2 3 4 10 – 8 = 2

y 1 Because the second difference gives a common

difference, d = 2, it is a quadratic
3. x 1 2 3 4
mapping following the rule;
y = a + bx + c or x → a + bx + c
y 2 4 8 16
For any 3 points (1, 6), (2, 12) and (3, 20)
4. x 1 3 5 7
When x = 1, y = 6, by substitution

y 6= ( + (1) +
6= + + ………………..(1)

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When = 2, = 12, by substitution
x 1 2 3 4 5
12 = ( + (2) +
12 = 4 + 2 + …………….(2)
y 0 11 28 51 80

When x = 3, y = 20 Solution
20 = a( + b(3) + c First difference = 11 – 0 = 11, 28 – 11= 17,
20 = 9a + 3b + c …………..(3) 51 – 28 = 23, 80 – 51 = 29

The three equations are: Second difference = 17 – 11 = 6,

a + b + c = 6……………….(1) 23 – 17 = 6, 29 – 23 = 6
4a + 2b + c = 12…………..(2) Because the second difference gives a common
9a + 3b + c = 20…………..(3) difference, d = 6, it is a quadratic mapping
following the rule;
eqn (2) – eqn (1) y = a + bx + c or x → a + bx + c
(4a – a) + (2b – b) + (c – c) = (12 – 6)
3a + b = 6………………….(4) For any 3 points (1, 0), (2, 11) and (3, 28)
When x = 1, y = 0, by substitution
eqn (3) – eqn (2) 0 = a( + (1) + c
(9a – 4a) + (3b – 2b) + (c – c) = (20 – 12) 0 = a + b + c………………..(1)
5a + b = 8…………………(5)
When x = 2, y = 11, by substitution
eqn (5) – (4) 11 = a( + b(2) + c
(5a –3a) + (b – b) = (8 – 6) 11 = 4a + 2b + c…………….(2)
2a = 2
a=1 When x = 3, y = 28
28 = a( + b(3) + c
Put a = 1 into eqn (5) 28 = 9a + 3b+ c ………….. (3)
5 (1) + b = 8
b=8–5=3 The three equations are:
a + b + c = 0………………. (1)
Put a = 1 and b = 3 into eqn (1) 4a + 2b + c = 11………….. (2)
1+3+c=6 9a + 3b + c = 28…………. .(3)
4+c=6
c=6–4=2 eqn (2) – eqn (1);
(4a – a) + (2b – b) + (c – c) = (11 – 0)
Substitute a = 1, b = 3 and c = 2 in 3a + b = 11………………… .(4)
y = a + bx + c to get y = + 3x + 2
The rule is: y = + 3x + 2 or x → + 3x + 2 eqn (3) – eqn (2);
(9a – 4a) + (3b – 2b) + (c – c) = (28 – 11)
2. Find the rule of the mapping below: 5a + b = 17………………… (5)

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eqn (5) – (4) ; The Image Under A Given Mapping
(5a –3a) + (b – b) = (17 – 11) Sometimes, mappings are given with some
2a = 6 missing images. For e.g., in the mapping below,
a=3 the image of 8 is given as m;
x 0 1 2 3 4…… 8
Put a = 3 into eqn (5)
5 (3) + b = 17 y 5 7 9 11 13 m
15 + b = 17
b = 17– 15 = 2 To find the image of8 or the value of m under the
mapping:
Put a = 3 and b = 2 in eqn (1) I. Find the rule of the mapping.
3+2+c=0 II. Use the rule of the mapping to find the value
5+c=0 of y, when x = 8.
c=-5 III. The value of y is the image of 8.

Substitute a = 3, b = 2 and c = –5 in For example, the rule of the mapping above

y = a + bx + c to get y = 3 + 2x – 5 can be deduced as y = 2x + 5;
The rule is: y = + 2x – 5 or x → + 2x – 5 When x = 8, y = 2(8) + 5
y = 16 + 5 = 21
Exercises 6.8 Since 8 → 21, m = 21
Find the rule of the mappings;
1. x 1 2 3 4 5 Worked Examples
↓ ↓ ↓ ↓ ↓ ↓ 1. In the mapping below, determine the image of
y 3 8 15 24 35 16
x 2 4 6 8 … 16
2. x -3 -1 1 3 5

y 0 -6 4 30 72 y 5 6 7 8… k

Solution
3. x 2 3 4 5 6 For linear mappings, y = mx + c
Let = 2, = 5, = 4 and = 6
y 4 14 30 52 80
m= = =

4. x -4 -3 -2 -1 0
Put m = in y = mx + c to obtain y = x + c
y -73 -41 -19 -7 -5
When x = 2, y = 5, put in y = x + c
5. x -2 0 2 4 6
5 = (2) + c
5=1+c
y -10 -6 -2 10 30

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5–1=c Word Problems
c=4 Given the rule of a mapping to find the image of
a number under the mapping, substitute the
Put c = 4 in y = x + c to get y = x + 4 number in place of x in the rule and workout to
get the image of the number. That is; if y = mx +
The rule is; y = x + 4 or x → x + 4
c, then the value of y when x = a, is the image of
a.
To find the image of 16 under that mapping,
Put x = 16 in y = x + 4 Worked Examples
y = (16) + 4 1. Find the image of -4 under a mapping defined
by the rule x → 5x + 3
y=8+4
y = 12, so k = 12
Solution
Therefore, the image of 16 is 12
In x → 5x + 3
When x = -4, - 4 → 5(- 4) + 3
Method II
- 4 = -20 + 3
The image of 16 under the mapping is;
- 4 → -17
16 → (16) + 4 The image of - 4 under the mapping is -17
16 → 8 + 4
16 → 12, k = 12 2. A mapping is defined by x→ 4x + 1, find the
Therefore, the image of 16 is 12 images of , and under the mapping.

Exercises 6.9 Solution

In the mappings below, find the image of the
values without images; i. → 4( ) + 1 = 2 + 1 = 3
1. x 9 12 15 18 … 33 ii. → 4( ) + 1 = -1 + 1 = 0
iii. → 4( ) + 1 = - 2 + 1 = -1
y 4 6 8 10 … k
2. x 1 2 3 4 32 3. Find the value of m in the mapping below;
x -2 -1 0 2 3
y 6 14 28 48 k
3. 10 y m -1 1 5 7
x 1 2 3 4… 9

y 5 15 45 135… k
Solution
Since 7 – 5 = 2
4.
-1 – m = 2
-70 1 2 3 4 6 x
-1 – 2 = m,
-3 = m or m = -3
m - -1 - -3 3 y

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4. Find the image of -4 under the mapping: is the image of 1, 4 is the image of 2, 9 is the
x → 2 + 5x – 3 image of 3 and 16 is the image of 4.

Solution In a function, two distinct members of the domain

In x → 2 + 5x – 3, when x = - 4 can have the same image, as seen in fig. II below;
– 4→ 2( + 5(-4) – 3 but not the other way round.
– 4 → 2(16) – 20 – 3 Domain Range
– 4 → 32 – 20 – 3
– 4→ 9
The image of – 4 is 9 ( , ( +
Fig. II
Exercises 6.10
1. A mapping is defined by the rule:x → 5 – 3x The Function Notation
– 2. Find the images of the following under the The usual letters used to denote functions are f, g,
mapping; h and their corresponding capital letters, but other
i. - 1 ii. - 4 iii. -7 letters may be used as well. For e.g., the
statement „f is function which maps x onto ‟ is
2. Find the image of the following under the mapping expressed in function notation as f(x) = or
defined by the rule: n →12n + 4 f:x→
i. ii. iii.
Types of Function
3. Find the images of the domain {1, 3, 8} under 1. One – to - one function
a mapping defined by the rule, x → . Show When each member of the range has exactly one
corresponding member of the domain the
this mapping in a diagram and state the type of
function is called a one – to – one function as in
mapping.
fig. I for f : x →
Functions Domain Range
A function is a special kind of relation in which
each element of the first set called domain is ( , ( ,
paired with one and only one element of the
Fig. I
second set called range
Domain Range 2. Many – to – one function
When a member of the range has two or more
( , ( , corresponding members in the domain the
function is called many – to – one function, as in
Fig. I fig. II for f : x →
Domain Range
A member of the range which corresponds to a
certain member of the domain is called the image ( + ( )
of that member. For e.g. in the diagram above, 1 Fig. II

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The Image under a given Function i. f(4) = 2(4) – 3 = 8 – 3 = 5
For any set of ordered pairs such as (x, y) and (a,
ii. g(-30) = + 5 = – 15 + 5 = – 10
b), f(x) = y and f(a) = b, means that the value of
the function at x is y and the value of the function iii. g(-30) – f(4) = – 10 – 5 = – 15
at a is b respectively. iv. f(4) + g(-30) = 5 + (- 10) = 5 – 10 = - 5
In f(x) = y or f : x → y, the image of x is y. In
(
other words, f : a → f(a), f(a) is called v. = =–2
(
theimageof a under f, or the value of a at f. The
set of values of f(a) is called the range of the
4. Find the element whose image under the
function, f
function, f : x → 2x + 3 is -9
Worked Examples
1. Given that f(x) = , find f(–10) Solution
(x, -9) under f : x → 2x + 3 means
Solution f(x) = -9
If f(x) = , 2x + 3 = –9
2
f(–10) = (-10) – (- 10) 2x = – 9 – 3
= 100 + 10 2x = – 12
= 110 x=–6
2. The domain of the function g(x) = 5x + 1 is {0,
5. A function f is defined as f (x) = 3x2 – 5x
1, 2, 3, 4, 5} find its range.
i. Evaluate f(-3)
Solution ii. Find the values of x for which f(x) = –
If g(x) = 5x + 1, then
g(0) = 5(0) + 1 = 1 Solution
g(1) = 5(1) + 1 = 5 + 1 = 6 i. f (x) = 3x2 – 5x
g(2) = 5(2) + 1 = 10 + 1 = 11 f (-3) = 3(-3)2 – 5(-3)
g(3) = 5(3) + 1 = 15 + 1 = 16 f (-3) = 3(9) + 15 = 42
g(4) = 5(4) + 1 = 20 + 1 = 21
g(5) = 5(5) + 1 = 25 + 1 = 26 ii. If f(x) = –
The range of the function is:
R={ } ⇒ 3x2 – 5x = –

(4)3x2 – (4)5x = – 3
3. If f(x) = 2x – 3 and g(t ) = + 5, find:
12x2 – 20x = -3
i. f(4) ii. g(-30) iii. g(-30) – f(4)
(
12x2 – 20x = -3
iv. f(4) + g(-30) v. 12x2 – 20x + 3 = 0
(
12x2 – 2x – 18x + 3 = 0
Solution (12x2 – 2x) – (18x + 3) = 0
If f(x) = 2x – 3 and g(t ) = +5 2x(6x – 1 ) – 3(6x – 1) = 0

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2x – 3 = 0 or 6x – 1 = 0 f(x) = x2 – 2,
x= or x= f(-2) = (-2)2 – 2 = 2
f(-1) = (-1)2 – 2 = -1
f(0) = (0)2 – 2 = -2
6. The functionf and g are defined as f : x → x – 2
f(1) = (1)2 – 2 = -1
and g : x → 2x2 – 1. Solve:
f(2) = (2)2 – 2 = 2
i. f(x) = g( ) ii. f(x) + g(x) = 0 Q = {-2, -1, 2}

Solution ii. P Q
i. f : x → x – 2 and g : x → 2x2 – 1
f(x) = g( )
( +
x – 2 = 2( )2 – 1
x – 2 = 2( ) – 1 ( )
x–2= –1 2. A function f is defined as f : x → 3x2 – 5x
2x – 4 = 1 – 2 i. Evaluate f(-3)
2x = -1 + 4
ii. Find the values of x for which f(x) = –
2x = 3
x = 1.5
Solution
ii. f(x) + g(x) = 0 i. f : x = 3x2 – 5x
( x – 2 ) + (2x2 – 1) = 0 f(-3) = 3(-3)2 – 5(-3)
x + 2x2 – 2 – 1 = 0 f(-3) = 18 + 15
2x2 + x – 3 = 0 f(-3) = 32
2x2 + 3x – 2x – 3 = 0
(2x2 + 3x) – (2x – 3) = 0 ii. f(x) = –
x ( 2x + 3 ) – 1( 2x + 3) = 0 ⇒3x2 – 5x = –
x – 1 = 0 or 2x + 3 = 0
12x2 – 20x = -3
x = 1 or 2x = -3
12x2 – 20x + 3 = 0
x = 1 or x = – 12x2 – 2x – 18x + 3 = 0
(12x2 – 2x) – (18x + 3) = 0
Some solved Past Questions 2x(6x – 1) – 3(6x – 1 ) = 0
1. The set P = {-2, -1, 0, 1, 2} maps onto Q by the ⇒(2x – 3)(6x – 1) = 0
function f(x) = x2 – 2, where x ∈ P. 2x – 3 = 0 or 6x – 1 = 0
i. Find the elements of Q. x = or x =
ii. Draw a diagram showing the mapping between
P and Q.
Exercises 6.11
Solution A. Given that g(x) = 1 + andf(x) = 2 , find
i. P = {-2, -1, 0, 1, 2} the values of;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 123
1. g(5) 2. g( ) 3. g(- 2) 9. A function f is defined by f : x→8 – 1. Find
the image of ⁄ and under f
4. f( ) 5. f (2) – g(- 2) 6. g(3) + f( )
10. Given that f(x) = 2x2 – 11x + k and that f(3) =
B. 1. A function f is defined by f(x) = , x∈R, 0, find the value of k and hence the value of f(-1)
but x ≠ 1
11. Given that the function f(x) is defined by
a. Calculate the values of f at 2, -3 and
f(x) = (x – 2) (x – 3), solve f(x) = 6
b. What element of the domain has the image 2?
12. Given that f(x) = (x + 2) (x – 3), find:
2. A function f is defined by f(x) = 3x – 1, where
x∈R. i. f(15)
a. What is the image of -3? ii. the value of x for which f(x) = 0
iii. the range of values of x for which f(x) < 0
b. What is the value of f ( ) ?
c. Find the element of the domain whose image is 20 13. A function f is defined by f : x → – 3,x ∈ R.

d. If f(x) = -19, find x i. Evaluate f (0), f (–1) and f( )

ii. Find a such that f (a) = 6
3. A function g is defined by g : x → + 1, x∈ R,
a. Calculate the value of; 14. The set P = {-2, -1, 0, 1, 2} maps onto Q by
i. g(-1) ii. g at 2 the function f(x) = x2 – 2, where x ∈ P
b. Find the image of zero under g i. Find the elements of Q
c. If g(r) = 101, find r ii. Draw a diagram showing the mapping between
P and Q
4. A function, f is defined by the f(x) = (x
4), find the values of the function when: C. Find the range of the function;
i. x is 6 ii. x is - 4 iii. x is 0 iv. x is 1 1. f: x → x – 13, Domain = { }
2. f : x → + 2x – 8, D = { }
5. Given that f(x) = + and that f(5) = 8, 3. f : x → 20 + , D={ }
calculate the value of: i. kii. f(4)
Finding the Constants of a Function Given the
6. A function t is defined by t(x) = a(x+ 2), Value of the Function at a Point
x ∈ R, a being a real number. If the image of 1 Given the values of a function at two or more
under t is -30, calculate the value of t(4) given points, the constants of the functions
represented by variables are found as follows:
7. If f(x) = , find a such that f(3) = 7 I. Substitute the value of the function at the given
points and equate to the value of the function at
8. Find the value of x which gives 6 for the value these points.
of the function defined by y = II. Obtain two or more equations of the same
function and name them accordingly.

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III. Solve the equations simultaneously to obtain eqn (2) × 3
the values of the variables (constants). 12a + 6b = - 12…………….(4)

eqn (3) + eqn(4)

Worked Examples
30a = 24
1. Given that f(x) = px + q, find the values of p
a=
and q, if f(2) = 4 and f(4) = 10

Solution Put a = in eqn (1)

f(x) = px + q,
9 ( ) – 3b = 18
f(2) = 4
⇒f(2) = 2p + q = 4 36 – 15b = 90
2p + q = 4 ……………(1) - 15b = 90 – 36
- 15b = 54
f(4) = 10 b=-
⇒f(4) = 4p + q = 10
4p + q = 10 …………..(2) f(x) = x2 + x

eqn (2) – eqn (1)

Exercises 6.12
2p = 6
1. A function h is defined by h(x) = ax + b, where
p=3
x ∈ R, and a and b are real numbers. If h(0) = – 8
and h(6) = 22, find the values a and b, and hence
Substitute p = 3 in eqn (1)
calculate h(–1)
2(3) + q = 4
2. f(x) is defined by f(x) = bx2 + cx, where b and c
6+q=4
are constants. Given that f(1) = 8 and f(2) = 22,
q = 4 – 6 = -2
calculate the values of b and c. Hence solve the
The function is f(x) = 3x – 2
equation f(x) = 2
3. f(x) = . Find a and b if f(-1) = 2.5 and f(2) = 1
2. Given that f(x) = ax2– bx, find the values of a
and b, if f(3) = 18 and f(-2) = - 4 4. f(x) = . Find b if (-2, 1) is an ordered pair
of the function
Solution 5. Given that the function f : x →ax2 – 2bx – 4
f(x) = ax2– bx maps 1 → 3, and 3 → 35, show that a = 3 and
f(3) = a(3)2– 3b, but f(3) = 18 find the value of b
⇒9a – 3b = 18 …………….(1)
An Undefined Function
f(-2) = a(-2)2– (-2)b, but f(-2) = - 4 A function is said to be undefined when the
⇒4a + 2b = - 4 …………….(2) denominator of the function is zero. i.e. , b = 0.

eqn (1) × 2 Therefore, for a function to be defined the

18a – 6b = 36 …………….(3) denominator must not be equal to zero. i.e. , b ≠ 0

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To determine the value for which a function is 3x = 2
undefined, x=
I. Equate the denominator of the function to zero.
The function is undefined if x =
II. Solve for the value of the variable.
III. The value of the variable makes the function
undefined. Exercises 6.13
IV. If the value of the variable is a, then we state State the value(s) of x for which the function is
x a. undefined;
1. f(x) = 2. f(x) =
Worked Examples
3. f(x) = 4. f(x) =
1. Determine the value of x for which the function ( (

f:x→ is undefined.
The Zeros of a Function
The zeros of a function is the value(s) of the
Solution variable that makes the numerator of the function
For f : x → to be undefined, zero, hence the whole function equals to zero. i.
⇒2x + 4 = 0, e. f(x) = , a = 0 ⇒f(x) = 0
2x = – 4 (Solve for x)
x=–2 To determine the zeros of a function:
The function is undefined if x = – 2 I. Equate the numerator of the function to zero
II. Solve for the value of the variable.
2. What values of x makes the function III. The value(s) of the variable is/are the zeros of
the function.
f:x→( undefined?
( IV. If the value of the variable is a, then we state
x a
Solution
Worked Examples
For f : x → ( to be defined,
( 1. Determine the zeros of the function:
⇒(x + 2) (x – 3) = 0,
f:x→
x + 2 = 0 or x – 3 = 0
The values of x that makes the function undefined
is x = – 2 or x = 3 Solution
f:x→
3. For what value of x is the function f (x) = =0
undefined? (x – 6) ( x + 6) = 0
x = 6 or x = - 6
Solution
For f : x → to be undefined, 2. For what value of x is the function, f(x) =
⇒3x – 2 = 0, equal to zero

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Solution Worked Examples
If f(x) = equal to zero 1. What is the inverse rule of the function,
f : x → 3x + 2?
⇒1 – 5x = 0
1 = 5x
Solution
x= Let f (x) = y
⇒y = 3x + 2
3. Find the zeros f(x) = In y = 3x + 2,
Switch the positions of x and y,
x = 3y + 2.
Solution
Make y the subject of x = 3y + 2,
f(x) = x – 2 = 3y
=0 =
( ( )=0
y=
x (x + 5) + 2(x + 5) = 0
(x + 2) (x + 5) = 0 The inverse function is:
x +2 = 0 or x + 5 = 0 y= or :x→
x = -2 or x = -5
2. What is the rule for the inverse of the
4. What are the zeros of f : x → 3x + 2?
function f : x → ,x≠5
Solution
3x + 2 = 0
3x = -2 Solution
x=– Let y = ,x≠5
switch the positions of x and y
Exercises 6.14 x= ,x≠5
State the zeros of the functions
1. f : x → 2. f : x →
( ( x (y – 5 ) = y + 2
xy – 5x = y + 2
3. f : x → 4. f : x → xy –y = 2 + 5x
5. f : x → x – 2 6. f : x → x + 10 y(x – 1) = 2 + 5x
y=
The Inverse of a Function The inverse rule of the function is:
The inverse rule of a given mapping is found by: y= , x ≠ 1or :x→ ,x≠1
I. Interchange the positions of x and y in the
function Exercises 6.15
II. Let y equal the inverse of the function
A. Find the inverse rule:
III. Make y the subject of the relation, to get the
1. f : x → 5x + 13 3. f : x → 5x – 8
inverse of the function.
2. f : x → 10 + 4x 4. f : x → 9 – 6x

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B. Determine the inverse rule:
1. f : x → 11x + 2. f : x → ,x≠
2ndQuadrant 1st Quadrant
3. f : x → x – 1 4. f : x → , x≠ 2

The Number Plane or Cartesian Plane

3rd Quadrant 4th Quadrant
A number plane, also known as Cartesian plane
is formed when a vertical line intersects a
horizontal line. The point of intersection is called
Naming the Axes
the origin, usually denoted by O.
The vertical co-ordinate axes is called the y – axis
and the horizontal co-ordinate axis is called the x
– axis.
origin

y-axis
x-axis

Co-ordinate Axes
The lines that intersect at right angles forming the
number plane are called the co- ordinate axes of Numbering the Axes
the plane. The x and y axes are divided into equal segment
C
by a given scale. The point at which the x - axis
intersects with the y - axis is called the origin and
Right angle
is labeled „0‟ (zero).
A B
The numbers above the x – axis are positive. The
Co-ordinate
numbers below the x – axis are negative.
axes D
The numbers to the right of the y – axis are
positive and the numbers to the left of y – axis are
AB and CD are called co-ordinate axes of the negative.
plane y
+ve nos

Quadrants
The co-ordinate axes divide the number plane
−ve nos +ve nos
−ve nos

into four segments in section called Quadrants. x

The quadrants are viewed in the anticlockwise
direction as shown in the diagram below;

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The diagram below illustrates the four main that axis and b units implies that starting from the
ideas: origin, number on each of the marked intervals of
that axis, leaving b space(s) or intervals between
2ndQuadrant 1st Quadrant the numbers. The first number on the positive
( +,+) axes is b for b units.
( −, +)

Plotting Points on the Number Plane

( −, − ) ( +,− ) Each point on the number plane can be described
3rd Quadrant 4th Quadrant by an ordered pair in which the first
element is the x – coordinate and the second
element, the y – coordinate. It is always a good
Scale of a Graph idea to start at the origin (0, 0) when plotting
The scale of a graph is the ratio of the points on the number plane..
space/distance/ interval between the lines to the
space/distance/ interval between the numbers. The ordered pair A (-2, 4), is found in the
coordinate system when you move 2 steps to the
On the graph sheet, each line is separated from left on the x – axis and 4 steps upwards on the y –
the other by a distance of 0.1cm. axis

Each axis of the graph is numbered according to a Similarly, the ordered pair A (3, –6), is found in
given scale. Thus, a scale of a cm : b units on a the coordinate system when you move 3 steps to
particular axis implies that starting from the the right on the x – axis and 6 steps downwards
origin, mark an equal space or interval of a cm on on the y – axis

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 y
10Scale : 2cm : 2units on both axes

6
A(-2,4)
4
2

x
-10 -8 -6 -4 -2 2 4 6 8 10
-2

-4

-6
B(3,-6)
-8

-10

Exercises 6.16 Identifying the Coordinate of a Point on a

Draw a number plane on a graph sheet for -10 Number Plane
and values and -12 and To find out the coordinate of a point in the
plot the following points on it: coordinate system, begin at that point and follow
1. C( ,3) 2. P(1,7) 3. Q(3 , 6 ) a vertical line either up or down to the x – axis.
There is your x – coordinate. And do the same but
4. R(-5, 2) 5. S(4, 5) 6. T(-5,-5) following a horizontal line to find the y –
7. V(0,-3) 8. U(4,-7) 9. W(-2,-1) coordinate. Write the point as ordered pair (x, y)

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 y

Scale : 2cm : 2units on both axes

10

M 8

6
V
4

x
-10 -8 -6 -4 -2 2 4 6 8
S -2
P
-4 N

-6
Q
-8

R -10

In the diagram above. the coordinate of the various points are P(4,-3), Q(1,-6), N(1,-3), M(-3,8),
S(-5,-2), R(-4,-10) and V(7, 5)

Graph of a Linear Function

The word “linear” refers to relations without I. Represent the function by y and the elements of
product of a variable and a variable. A linear the domain by x. That is f(x) = y
function is therefore a function in which the II. Substitute the values of x into the function to
degree of exponent of the variable is not more obtain the values of y.
than one. III. Prepare a two column table of values for x
and y as shown below:
A function is said to be linear if it written in the
x
form f(x) = mx + c or f : x → mx + c, where m
y
and c are constants. Graphs obtained from linear
functions are called graphs of linear function IV. Plot the points described by the table of
values on a standard graph sheet using a given or
Drawing a Linear Graph appropriate scale.
To draw the graph of a linear function,

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V. Join the points with a rule to form astraight y = 3x – 4, Domain { }
line.
When x = 1 1 → 3(1) – 4 = -1
Worked Examples When x = 2 2 → 3(2) – 4 = 2
When x = 3 3 → 3(3) – 4 = 5
1. i. Make a table of values for f : x → 3x – 4 for
When x = 4 4 → 3(4) – 4 = 8
the domain {1, 2, 3, 4, 5}
When x = 5 5 → 3(5) – 4 = 11
ii. Plot the points on a standard graph using a
scale of 2cm to 1 unit on x –axis and 2cm to 2
Table of values is:
units on y – axis and join the points with a ruler to
form a straight line
x 1 2 3 4 5
3x – 4 -1 2 5 8 11
Solution
i. Let f(x) = y

x = 2cm : 1 unit
12 y = 2cm : 2 units
10

6
4

-5 -4 -3 -2 -1 1 2 3 4 5
-2

-2 -4

-6
-8

2. i. Make a table of values for the function scale of 2cm to 1 unit on x – axis and 2cm to 2
y = 5 – 3x for the domain { -2, -1, 0, 1, 2, 3, 4} units on y – axis. Join the points with a rule to
and plot the set of points on a graph sheet using a form a straight line.

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Solution When x = 1, y = 5 – 3(1) = 2
f : x → 5 – 3x, When x = 2, y = 5 – 3(2) = -1
When x = 3, y = 5 – 3(3) = - 4
Let f (x) = y When x = 4, y = 5 – 3(4) = -7
⇒y = 5x – 3 Table of values is:
Domain{ }
When x = -2, y = 5 – 3(-2) = 11 x -2 -1 0 1 2 3 4
When x = -1, y = 5 – 3(- 1) = 8 y 11 8 5 2 -1 -4 -7
When x = 0, y = 5 – 3(0) = 5

x = 2cm to 1 unit 14
y = 2cm to 2 units
12
f(x) = 3x – 4
10

6
4

2 x

x
-4 -3 -2 -1 1 2 3 4 5

Exercises 6.18 -3 x -3, using a scale of 2cm to 2 units on

1. Draw a graph of y = x – 13, for the domain = both axes.
{ }, using a scale of 2cm to 5
3. Copy and complete the table below for the
units on the x – axis and 2cm to 2 units on the y –
relation x + y = 180;
axis

x 0 30 60 90 120 150 180

2. Draw the graph of the relation y = 2x + 1 and y
= x + 2 on the same graph sheet for the range y 180 90 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 133
ii. Using a scale of 2cm to 20 units on both axes, Graph of a Quadratic Function
mark both axes from 0 to 180. A function f, for which f(x) = ax2 + bx + c, where
iii. Plot all the points on the graph sheet and join a, b, c ∈ R and a 0, is called a quadratic
them with a ruler to form a straight line. function
iv. Use your graph to find:
a. y when x = 100 b. x when y = 70 The Cartesian graph of every quadratic function
is a Parabola of U or ∩ - shape
4. Copy and complete the table below for the
y
relation y = 2x + 3

x 0 1 2 3 4 5 6 Max Turning Point

y 5 7 15
Turning Value
i. Using a scale of 2cm to 1 unit on x – axis and
x
2cm to 2 units on y – axis on a graph sheet, mark o
2
y = – ax + bx + c
the x – axis from 0 to 5 and y – axis from 0 to 12.
ii. Plot on the graph sheet, the ordered pairs (x, y) y
from the table.
iii. With the help of a ruler, draw a straight line Range y = ax2+ bx + c
through all the points.
iv. From your graph, find y when x = 3.5 Domain
x
o
5. i. Copy and complete the table of the values for
Min. Turning Point
the relations; y1 = 2x + 5 and y2 = 3 – 2x for x
from - 4 to 3.
The Graph of a Quadratic Function
x -4 -3 -2 -1 0 1 2 3 To draw the graph of a quadratic function,
y1= -3 3 7 4
make sure the function is in the form:
2x + 5
y2 = 11 9 5
f(x) = ax2 + bx + c, a ≠ 0
3− 2x
Method I
ii. Plot the graph of each relation on the same I. Prepare a table of values for the given domain
graph sheet using a scale of 2cm to 1 unit on the (x) of the function to obtain the values of the co –
x – axis and 2cm to 2 units on the y – axis. domain (y) of the function as shown below:
iii. Find the coordinates of the point of x x1 x2 x3 x4
2
intersection of y1 and y2. ax
iv. Determine the shape of y1 = 2x + 5. + bx
+c
y y1 y2 y3 y4

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II. Plot the points (x, y) on a graph sheet, using a When x = 1, 1 → (1)2 + 1 = 2
given scale, or a convenience scale and join the When x = 2, 2 → (2)2 + 1 = 5
points to make a free hand sketch of the required When x = 3, 3 → (3)2 + 1 = 10
parabola
Table of values is
Method II
I. Substitute the values of the domain (x) into the x -1 0 1 2 3
function to obtain the values of the co – domain x2 + 1 2 1 2 5 10
(y) and prepare a two column table of values for
2. For the function f (x) = + 2x – 8 prepare a
the ordered pairs (x, y) as shown below:
table of values for -3 x 3
x
y Solution
Method 1
II. Plot the points (x, y) on a graph sheet, using a f (x) = + 2x – 8
given scale, or a convenience scale and join the Let f (x) = y
points to make a free hand sketch of the required ⇒ y = x2 + 2x – 8
parabola
x -3 -2 -1 0 1 2 3
Worked Examples x2 9 4 1 0 1 4 9
1. Prepare a table of values for f (x) = x2 + 1 and 2x -6 -4 -2 0 2 4 6
the domain = {-1, 0, 1, 2, 3} -8 -8 -8 -8 -8 -8 -8 -8
y -5 -8 -9 -8 -5 0 7
Solution
Method I Method II
f (x) = + 2x – 8
f (x) = x2 + 1
Let f (x) = y
Let f (x) = y
⇒y= + 2x – 8
⇒ y = x2 + 1
When x = -3, y = ( + 2(-3) – 8 = -5
x -1 0 1 2 3
When x = -2, y = ( + 2(-2) – 8 = -8
x2 1 0 1 4 9
+1 1 1 1 1 1 When x = -1, y = ( + 2(-1) – 8 = -9
y 2 1 2 5 10 When x = 0, y = ( + 2(0) – 8 = -8
When x = 1, y = ( + 2(1) – 8 = -5
Method II When x = 2, y = ( + 2(2) – 8 = 0
f (x) = x2 + 1 When x = 3, y = ( + 2(3) – 8 = 7
Let f (x) = y The table of values is
y = x2 + 1
Domain = {-1, 0, 1, 2, 3} x -3 -2 -1 0 1 2 3
When x = -1, -1 → (-1)2 + 1 = 2 y -5 -8 -9 -8 -5 0 7
When x = 0, 0 → (0)2 + 1 = 1

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Exercises 6.19 2. Copy and complete the table below for
Prepare a table of values for the domain f (x) = x2 – 5x + 2 for -2 x 4
1. x2 – 4x + 3 = 0, D = {-4. -2, 0, 2, 4, 6}
2. - x2 + 2x + 3 = 0 D = -5, -4, -3, -2, -1, 0} x -2 -1 0 1 2 3 4
3. x2 – 4x + 3 = 0 D = {-3, 0, 3, 6, 9} f (x) 16 2 -4 -2

Completing a Given Table of Values Solution

To complete a given table of values for a given f (x) = x2 – 5x + 2
quadratic function, When x = -1
I. Identify the function f (-1) = (-1)2 – 5(-1)+ 2 = 8
II. Substitute the values of the domain without When x = 1
images into the function f (1) = (1)2 – 5(1) + 2 = -2
III. Workout for the images of the domain as When x = 2,
described by the function f (2) = (2)2 – 5(2) + 2 = - 8
IV. Copy the table and fill in all gaps with their
respective images to complete the table x -2 -1 0 1 2 3 4
f (x) 16 8 2 -2 -8 -4 -2
Worked Examples
1. Copy and complete the table of values for
f (x) = 3x2 – 5x + 4 for -3 x 3 Exercises 6.20
1. The following is an incomplete table of values
x -3 -2 -1 0 1 2 3
3x2 12 3 3 12 27 for the relation y =
-5x+ 4 9 -1 -11
f(x) 46 26 4 x -3 -2 -1 0 0.2 0.5 1 2 3
y 0.2 5 1.4
Solution
In 3x2, when x = -3, 3(-3)2 = 27 Copy and complete the table
When x = 0, 3(0)2 = 0
In -5x + 4, when x = -3, -5(-3) + 4 = 19 2. Copy and complete the table below for the
When x = -2, -5(-2) + 4 = -6 relation x2 – 12x + 3 = 0
When x = 0, -5(0) + 4 = 4
When x = 2, -5(2) + 4 = -6 x -1.5 -1.0 -0.5 0 0.5 1.0
y 3.00
f (-1) = 3 + 9 = 12 f (1) = 3 – 1 = 2
f (2) = 12 – 6 = 6 f (3) = 27 – 11 = 16
3. Copy and complete the following table of
x -3 -2 -1 0 1 2 3 values for the relation y = x2 – 3x + 7
3x2 27 12 3 0 3 12 27
-5x+ 4 19 14 9 4 -1 -6 -11 x 0.5 1 2 3 4 5.0
f(x) 46 26 12 4 2 6 16 y 7.00 17.00

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4. Copy and complete the table of values for the y
relation y = x2 – 5x – 2 in the interval -1 x 6 y

x -1 0 1 2 3 4 5
y 4 -8 -6 x x

The Roots or Zeros of Quadratic Function

The roots of a quadratic function is the value(s) In both diagrams, the functions have no roots or
of x for which the function f(x) = 0 or y = 0. The no zeros
root of the quadratic function is also called the
zerosof the function. Equation of Axes or Line of Symmetry
Parabolas can be described as being symmetrical,
From the graph, the roots of a function is meaning that a line can be drawn through a
determine with cognizance to the nature of the parabola, to divide it into two equal parts,
parabola in relation to the x – axis creating mirror images of each other. The straight
I. Whether U or ∩ – shaped, the points at which line bisecting the parabola is called a line of
symmetry.
the parabola cuts the x – axis is the roots or zeros
of the function as shown below, For all quadratic functions of the form: f(x) = ax2
+ bx + c, a ≠ 0, the line of symmetry has the
y
equation, x = –

x The Minimum and Maximum Points

a
b The vertex of a parabola is the point of
intersection of the line of symmetry and the
parabola itself. The vertex is the turning point:
The zeros or truth set of the function is x = a or
either maximum (highest) or minimum (lowest)
x=b y
point of the parabola.

When the parabola is U – shaped, it is said to

have a minimum or least turning point and when
x
it is ∩ – shaped, it is said to have a maximum or
a b
greatest turning point

The zeros or truth set of the function is x = a or

Values of x and y at the Turning Point
x=b
For all functions of the form:
f(x) = ax2 + bx + c, a 0, f(x) = y, at the turning
II. Whether U or ∩ – shaped, if the parabola does
not cut the x – axis, the function is said to have no point, x = – .
roots or no zeros as shown below;

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To get the value of y at the turning point, The Domain of a Function from the Graph
substitute x = – 2
in y = ax + bx + c The domain is the interval from the least value of
x to the greatest value of x. When stating the
The value of y is called the maximum or
domain, the lower value (L.v) is written first,
minimum value, depending on the nature of the
followed by the greatest value (G.v):
parabola.
i. Using the brackets = [L.v, G.v]
ii. Using an inequality = L.v f (x) G.v
Worked Examples
iii. Listing from the lowest value to the greatest
1. What is the value of x and y at the turning point
value e.g. Domain = {a, b, c, d}
of f(x) = 2x2 – 8x + 3

Worked Examples
Solution
1. Draw the graph of f(x) = 3x2 – 5x + 4 for the
In f : x →2x2 – 8x + 3, a = 2 and b = – 8
( domain {x: -3 4} using a scale of 2 cm : 1
At the turning point, x = – =– =4 unit on x – axis and 2cm : 5 units on y – axis. Use
Let f(x) = y your graph to estimate:
⇒y = 2x2 – 8x + 3 a. the coordinates of the turning point of the
graph,
Put x = 4 in y = 2x2 – 8x + 3 b. the maximum or minimum value of the
y = 2(4)2 – 8(4) + 3 = 3 function,
At the turning point of 2x2 – 8x + 3 = 0, c. the zeros of f,
x = 4 and y = 3 d. the range of f for the given domain,
e. the equation of the axis of symmetry of the
The Range of a Function from the Graph parabola.
The “range” is the interval from the least value
of y to the greatest value of y. When stating the Solution
range, the lower value (L.v) is written first, Method I
followed by the greatest value (G.v): f(x) = 3x2 – 5x + 4
i. Using the brackets = [L.v, G.v] Let f (x) = y
ii. Using an inequality = L.v f (x) G.v ⇒y = 3x2 – 5x + 4
iii. Listing from the lowest value to the greatest x = {-3, -2, -1, 0, 1, 2, 3, 4}
value e.g. Range = {a, b, c, d}

x -3 -2 -1 0 1 2 3 4
3x2 27 12 3 0 3 12 27 48
-5x 15 10 5 0 -5 -10 -15 -20
+4 4 4 4 4 4 4 4 4
y 46 26 12 4 2 6 16 32

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 y
50

45 x – axis = 2cm : 1unit and

2
y = 3x – 5x + 4 y – axis = 2cm: 5units
40

35

30

25

20

15

10

0
-4 -3 -2 -1 0 1 2 3 x4

a. The coordinates of the turning point e. The equation of the axis of symmetry :
At the turning point, x = – . x=– .
But in y = 3x2 – 5x + 4, a = 3, b = -5 In y = 3x2 – 5x + 4, a = 3, b = -5
By substitution, x=– =– = = 0.8
(
x=– =– = = 0.8 The equation of the axis of symmetry is x = 0.8
(

Put x = 0.83 in y = 3x2 – 5x + 4, Exercises 6.21

y = 3(0.8)2 – 5(0.8) + 4 1. a. Draw the graph of y = x (x – 4) for x =
y = 1.92 – 4 + 4 -1, 0, 1, 2, 3, 4, 5 with scales of 2cm to 1 unit on
y = 1.92 the x - axis and 2cm to 2 units on the y – axis
Therefore, the turning point is (0.8, 2) b. Find from the graph:
i. the coordinates of the turning point of y,
b. The function has a minimum value because it ii. the maximum or minimum value of y,
is U – shaped. The minimum value of y = 3x2 – iii. the equation of the axis of symmetry of y,
5x + 4 is 2.
2.a. On the same graph sheet, draw the graphs f
c. The zeros of f is the value(s) of x that makes y (x) = x2+ 3 and g (x) = 3x + 1 on a scale of 2cm
= 0. The curve does not cut the x – axis. to 1 unit on x – axis and 2cm : 5 units on y – axis,
Therefore, f has no zeros. for -4 4
b. From the graph, find:
d. The range of f for the given domain is i. the coordinates of a andb themeeting points of
2 y 46. f(x) and g(x)
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ii. the coordinates of the turning point of f(x) Worked Example
iii. the maximum or minimum value of the Given the parabola, y = 2x2 – 5x – 3, find:
function f i. the coordinates of the point of intersection with
iv. the zeros of f the y –axis,
v. the range of f for the given domain ii. the coordinates of the point of intersection with
vi. the equation of the axis of symmetry of the the x –axis,
parabola iii. the equation of the axis of symmetry,
iv. the coordinates of the vertex,
3. a. Draw the graphs of f(x) = 5x – x2 and g(x) = v. Sketch the parabola.
8 – x on the same axes for the domain
-1 6, and the scales 2cm to 1 unit on x – Solution
axis and 2cm: 5 units on y – axis i. At the y – axis, x = 0
b. From the graph, find: y = 2x2 – 5x – 3
i. P, the coordinates of the point of intersection of y = 2(0)2 – 5(0) – 3 = -3
f (x) and g (x)
ii. the coordinates of the turning point of f ii. At the x – axis, y = 0
iii. the maximum or minimum value of the When y = 0
function f, 2x2 – 5x – 3 = 0
iv. the zeros of f, (2x2 – 6x) – (x – 3) = 0
2x(x – 3) + 1 (x – 3) = 0
v. the range of f for the given domain,
(2x + 1) (x – 3) = 0
vi. the equation of the axis of symmetry of the 2x = -1 or x = 3
parabola.
x = – or x = 3
The curve intersects the x – axis at x = -0.5 or x = 3
4. a. On the same graph sheet, draw the graphs f
(x) = 6 – x – x2and g (x) = x + 3 on the same
iii. Equation of the axis of symmetry; x = –
scales of 2cm to 1 unit on x – axis and 2cm : 5
units on y – axis, for the domain From y = 2x2 – 5x – 3, a = 2 and b = -5
(
-4 3 x=– = =1.25
(
b.If f(x) and g(x) meet at the points a and b,find
the coordinates of a andb
iv. Coordinates of the vertex
When x = 1.25, y = (1.25)2 – 5(1.25) – 3 = -7.7
Sketching the Graph of a Function
(x, y) = (1.3, -7.7)
To sketch the graph of a quadratic function, x
I. Find the intercept on the y – axis, noting that at
v.
the y – axis, x = 0
II. Find the intercept on the x – axis, noting that at y
3
the x – axis, y = 0 -0.5
III. Draw a curve on the x –y plane to pass
through the points of intersection of x and y axis -3

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Exercises 6.22 (
M=( )= ( )
Given the following parabolas,
1. y = 2x2 – x – 15 3. y = (2x + 3)(2x – 5)
Exercises 6.23
2. y = (x + 4)(x – 5) 4. y = (3 – x )(x + 2)
Find the coordinates of the mid – point of the
lines joining the pair of points:
find:
1. C (- 4, -2) and D(-10, -10)
i. the coordinates of the point of intersection with
2. M( 2, -3) and N (4, 3)
the y – axis,
3. S(- 6, 1) and T ( 6, 6 )
ii. the coordinates of the point of intersection with
4. M( 0, -2) and N(- 4, 0)
the x – axis,
iii. the equation of the axis of symmetry,
The Coordinate of the End point of a Line
iv. the coordinates of the vertex,
Given the Mid-Point and One End point
v. Sketch the parabola.
I. Identify the given end point, the given midpoint
and the unknown end point
The Mid – point of a Straight Line
II. Represent the unknown end point by the
If A (x1, y1) and B(x2, y2) are any two points, then
ordered pair (x, y).
the coordinate of the midpoint, M of the straight
II. Add the respective x and y components of the
line joining the points A and B, such that AM =
end points divide each sum by 2 and equate to the
MB is calculated as; M = ( ) respective x and y component of the midpoint.
III. Form two separate equations and solve for the
Worked Examples values of x and y to obtain the coordinates of the
1. Find the midpoint of the straight line joining unknown end point.
the points A( 2, 1) and B (6, 5)
Worked Examples
Solution 1. M(4, 7) is the midpoint of the line joining A(x,
A( 2, 1) and B ( 6, 5) y) and B (6, 11). Find the coordinates of the point
Let = 2, = 1, = 6 and =5 represented by x and y
Mid – point M = ( )
Solution
M=( ) = (4, 3)
M(4, 7), A(x, y ) and B (6, 11)
Midpoint of AB = M
2. Find the coordinates of the mid – point of the
( ) = (4, 7)
straight line joining E(-6,-1) and F(3, -4)
= 4……………(1)
Solution = 7…………..(2)
E( -6, -1) and F( 3, -4)
Let = -6, = -1, = 3 and =–4 From eqn(1)
Mid – point M = ( ) x+6=4×2
By substitution, x+6=8

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x=8–6 or distance or length of the straight line joining
x=2 the points A and B, is calculated as;
( – ( –
From eqn (2)
√( – ( –
y + 11 = 7 × 2
y + 11 = 14
y = 14 – 11 = 3 Worked Examples
x = 2 and y = 3 1. Find the length of the line joining A (1, 3) and
Therefore the coordinates of A is (2, 3) B (4, 7)

Exercises 6.24 Solution

1. The point M(-2, -1) is the mid – point of . If Let (1, 3) B (4, 7)
the coordinate of P is (4, -5), find the coordinates ↓ ↓ ↓ ↓
of Q. x1 y2 x2 y2

√( – ( –
2. The point (5, 15) is the mid – point of ̅̅̅̅. If A is
the point (3, 14), what is the coordinates of B? = √( (
= √ =√ = √ = units
3. The point ( ) is the mid – point of ̅̅̅̅. If
C is the point (11, 9), what is the coordinate of D? 2. Find the length or magnitude of the line joining
the points P(-5, -4) and Q(8, 7).
4. If the coordinates of P is (-4, 7) and the
coordinates of the midpoint of PQ is (3, 2), what
Solution
is the coordinate of Q?
P( (
Let x1 = -5 y1 = – 4 x2 = 8, y2 = 7,
The Magnitude of a Straight Line
The magnitude of a straight line is also called the By substitution,
length or distance or modulus of a line. /PQ/ = √( – ( –
Consider the diagram below; /PQ/ = √( ( ( (
y
B( ) /PQ/ = √
/PQ/ = √ =√ = 17.03 units
1

y2 – y1 3. Find the distance of the point (8, -6) from the

origin.
A( ) x2 – x1

x Solution
1
The coordinates of the origin is (0, 0),
If A (x1, y1) and B(x2, y2) are any two points, then The distance of the point (8, - 6) from the origin is:
by applying Pythagoras theorem, the magnitude

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 4. P(1, 6), Q(-3, -1 ) and R(2, k) are three points.
= √( – ( – If = , find the two values of k.
√( (
Determining the Type of Triangle Using the
√ (
Distance Formula
√ =√ = 10 units
Given the vertices of a triangle as A, B and C, the
type of triangle is verified by finding the
Exercises 6.25
magnitude of AB, the magnitude of BC and the
A. Find the lengths of the straight lines joining
magnitude of AC. Thereafter, observe the
the following pair of points.
following:
1. A (3, 2) and B (7, 5)
I. If the magnitudes of the three sides are equal,
2. M(0, -2) and N(- 4 0)
then ∆ ABC is an equilateral triangle.
3. U(-1, -3) and V(-3, - 6 )
II. If the magnitudes of two sides are equal, then
4. S(-5, 6) and T (8, 9)
∆ABC is an isosceles triangle.
III. If the magnitudes of the three sides are
B.1. Find the length of the line joining K (-1, 2)
unequal, then ∆ ABC is a scalene triangle.
and L (5, 9).
IV. If the magnitudes of the three sides form a
2. If P (2, 7) and Q (2, 3) are two points in the
Pythagorean triples, the three points form the
OXY plane. Find .
vertices of a right – angled triangle.
3. Find the distance of the point (-15, 8) from the
origin.
Worked Examples
1. A, B and C are the points (3, 5), (7, 2) and (3, -
Challenge Problems 1) respectively. What type of triangle is ∆ ABC?
1. Show that the lines L : x + y + 4 = 0. K : 9x –
5y – 20 = 0 and M : 5x – 9y + 20 , form an
Solution
isosceles triangle.
A(3, 5) B(7, 2) and C(3, -1)

2. A is the point (4, 0) and B is (0, 4) . A point C /AB/ √( – ( –

is such that CA = CB
i. Write down the coordinates of four possible /AB/ √( (
positions of C.
/AB/ = √ (
ii. What is the equation of the locus of C?
/AB/ =√ =√ = 5 units
iii. When < ACB = 400, state the sizes of the
other two angles of ∆ ACB iv. What are the
coordinates of C if A, C, B do not form a triangle. /AC/ √( – ( –

3. A(2t, 0) and B(0, –t ) are two points. If = /AC/ = √( (

√ , find the two values of t /AC/ = √ (
/AC/ = √ =√ = units

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 /UV/ = √( (
/BC/ = √( – ( –
/UV/ = √ ( = √ = 5 units
/BC/ = √( (
/BC/= √( (
/VW/ = √( – ) ( –
/BC/ = √ = √ = units
Since = 5, ∆ABC is isosceles /VW/ = √( (
/VW/ = √( (
2. The triangle PQR has vertices at P(-3, 4), Q(3, /VW/ = √ = √ = 7.6 units
4 ) and R(0, 9). Show that ∆ PQR is no other
triangle than an equilateral.
/UW/ =√( – ( –
Solution
/UW/ = √( (
P(-3, 4), Q(3, 4 ) and C(0, 9).
/UW/ √( (
/PQ/ = √( – ( – /UW/ = √ =√ = 3.6 units
/PQ/ = √( ( . Therefore, ∆UVW is a
/PQ/ = √ ( =√ = 6 units scalene triangle

Exercises 6.26
/QR/ = √( – ( – 1. P, Q and R are the points (5, -3), (-6, 1) and (1,
/QR/ = √( ( 8) respectively. Show that the triangle PQR is an
isosceles triangle and find the mid - point of the
/QR/ = √ (
base.
/QR/ = √ =√ = 5 .8 6 units
2. The three points of a triangle are at A(1,1), B(4,
/PR/ = √( – ) ( – 5) and C(5, -2). Find the lengths of the side of
the triangle and show that it is isosceles. What
/PR/ = √( (
else can you say about the triangle?
/PR/ = √ ( = √ = √
/PR/ = 5 .8 6 units 3. Given the points A(4, 4), B(- 4, 1) and C(1, -4),
= = 6 units show that triangle ABC is not any other triangle
Therefore, ∆ PQR is an equilateral triangle than an isosceles triangle.

3. What type of triangle is ∆ UVW with vertices 4. The coordinates of the vertices A, B, C of the
U(-2, 2), V(3, 2) and W(- 4, 5)? triangle ABC are (–3, 7), (2, 19) (10, 7)
respectively, prove that the triangle is isosceles.
Solution 5. The three points of a triangle are at A(1, 1),
U(-2, 2), V(3, 2) and W(- 4, 5) B(4, 5) and C (1, 6). What type of triangle is ∆
ABC?
/UV/ = √( – ) ( –

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Three Points Forming a Right Triangle Method 2
To determine whether three given points form the √ , =√ , =√
vertices of a right triangle; By Pythagoras theorem
I. Find the distance between each pair of points
(√ ) (√ ) = (√ )
using the magnitude formula.
100 + 225 = 325
II. If the distances between the points form
325 = 325
Pythagorean triples, the three points form the
The points form the vertices of a right triangle
vertices of a right triangle.

Exercises 6.27
Worked Examples Three points that form the vertices of a
Determine whether the following points ( – 8, 1), triangle are given below. Show whether any of
(-2, 9) and (10, 0) form the vertices of a them form a right triangle;
right triangle 1. (5, 2), (0, -3), (4, -4)
2. (7, 0), (-1, 0), (7, 4)
Solution 3. (-4, 3), (-7, -1), (3, -2)
Name the points asA(-8, 1), B (-2, 9) and C(10, 0) 4. (-3, 2) , (-1, 5), (-6, 4)
√( – ( –
= √( ( The Gradient of a Line
=√ =√ = 10 units The gradient of a straight line, also known as the
slope of a straight line is a measure of the
√( – ( – steepness of the line. Steepness means, how an
angle is falling or rising. The gradient is usually
represented by the variable m.
√( (
√ ( The gradient of a line can be determined by the
√ = 15units following means:
1. from two given points,
√( – ( – 2. from any linear relation,
√( ( 3. from a linear graph.

√ ( √ Gradient of a Line from Two Points

18.02 units If two points are given as A(x1, y1) and
B(x2, y2), then the gradient of the line AB,
Check to see if the lengths,10, 15 and 18.02 form is determine by the formula;
Pythagoras triples: – –
m= or m = , where m is the gradient
102 + 152 = 325 – –
(18.02)2 = 325
⇒ 102 + 152 = (18.02)2 Worked Examples
The points form the vertices of a right triangle. E is the point (4, 2) and F is the point (2, 1).
Calculate the gradient of the straight line EF.

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Solution
y = mx + c
E (4, 2) and F (2, 1)
Let x1 = 4, y1= 2, x2 = 2 and y2 = 1
y = (slope)x + (where the line cuts the y – axis)
Gradient of EF
–
m= = = =
–
Method 2
If the equation of the line is in the form:
2. The gradient of a line passing through the the
ax + by + c = 0, then the gradient m = –
points P(6, 7) and Q(x, 8) is . Find the value of x.
In words, Slope, m = –
Solution
P(6, 7), Q(x, 8), m = Note:
Let x1 = 6, y1= 7, x2 = x and y2 = 8 1. when using this method, make sure that every
term is on the left hand side of the equation
Gradient of PQ; 2. The equation of a line through the origin, with
–
m= = = gradient m, is y = mx
–

Worked Examples
=
1. What is the gradient of y = x − 10?
=
Solution
⇒x – 6 = 3 y = x − 10 compared to y = mx + c,
x=3+6 =9
m= or in the equation y = x – 10,

Exercises 6.28 The number attached to x is – . ⇒m = –

Find the gradient of the line which passes
through the pair of points; 2. Find the gradient of line 4x + 2y – 5 = 0.
1. (-7, - 10) and (-5,-2)
2. (-3, - 8) and (- 4,-1) Solution
3. (-12, - 6) and (- 2,-1) 4x + 2y – 5 = 0 compared to ax + by + c = 0
4. (9, 14) and (- 2, 10) a = 4 and b = 2.
m = – = – = –2
Gradient of a Line given its Equation
To find the gradient of a line given its equation: 3. Find the gradient of 5y = 8 + 4x.
Method 1
Get y on its own such that the equation will be of Solution
the form, y = mx + c, where the gradient is m or Re – arrange 5y = 8 + 4x to take the form:
the coefficient of x. The gradient can also be ax + by + c = 0 . ⇒– 4x + 5y – 8 = 0
identified simply as the number in front of x. By comparison, a = – 4 and b = 5

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m=– =– = y

4. Find the gradient of the line7x + 4y + 2 = 0, and B (x2,y2)

its intercepts on the x – axis and y – axis. y2
y2−y1
A(x1,y1 )
Solution y1 x2−x1
Express 7x + 4y + 2 = 0 in the form:
y = mx + c by making y the subject, x
x1 x2
4y = – 7x – 2
y=– x–
To find the gradient of the line AB;
y = – x – compared to y = mx + c I. Choose any two points on the line AB.
The gradient , m = – and the intercept on the y – II. Draw a triangle to join the two points as
shown in the diagram above.
axis, c = – III. Calculate the gradient by,
–
m= = =
To find the intercept on the x – axis, substitute –

y = 0 in 7x + 4y + 2 = 0 and solve for x.

7x + 4(0) + 2 = 0 Note:
7x + 2 = 0 If the line rises from left to right, the gradient is
7x = – 2 positive.
x = – . The intercept on the x – axis is –
If the line falls from left to right, the gradient is
negative.
Exercises 6.29
What is the gradient of the straight lines;
1. x – y – 1 = 0 5. 2y – 6x – 1 = 0
Positive gradient
2. 3y – 7x + 6 = 0 6. y = − 12x – 6 Negative gradient
3. y − 8x + 1 = 0 7. y + x – 2 = 0
Summary
4. x – y = - 3 8. x + 7y + 5 = 0
zero gradient
Gradient of a Line from a Linear Graph
Study the diagram below carefully; +ve gradient -ve gradient

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Worked Examples
1. Using a scale of 2cm to 1 unit on both axes, plot the points A (0, −2) and B (3, 5) on a graph sheet.
Join the points with a ruler to form a straight line and find the gradient of the line.

Solution

y
6
Scale: 2cm : 1 unit on both axes
5 B (3, 5)

4
3
2

5 4 3 2- 1 1 1 2 3 4 5x

A(0, -2) - 2
-3
-4
-5
-6

–
Gradient (AB) = = = . Therefore, m= 1
–

2. i. Using a scale of 2cm to1 unit on the x – axis and 2cm to 2units on y – axis, plot the points P(3,-7)
and Q(-1,2). Join P and Q with a ruler
ii. Find the gradient of the line PQ

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Solution
y
12
10 x – axis = 2cm : 1unit
y-axis = 2cm : 2units
8
6
4

Q(-1, 2) 2

-5 -4 -3 -2 -1 1 2 3 4 5 x
-2
-4
-6
-8
P(3, -7)
-10

–
From the graph, m = = =
–
ii. Draw a straight line through the points with the
Exercises 6.30 help of a ruler.
A. 1. i. Using a scale of 2cm to 1 unit on both iii. Find the slope of the line.
axes of a number plane, plot the points P (4, 5) iv. From the graph, determine the value of y when
and Q (−2, 3). x = − 3.
ii. Join the points with a ruler to form a straight 4. i. Using a scale of 2cm to 1 unit on both axes,
line. locate the following points; A (0, 2), B (1 , 1 )
iii. Determine the gradient of the line.
and C (3, −3) on a graph sheet.
2. i. On a graph sheet, plot the pointsC (2, −3) ii Connect the points with a straight line using a
and D (−4, 3), using a scale of 2cm to 1 unit on ruler.
both axes. iii Find the slope of the line.
ii. Use a ruler to join the points to form a straight
line. 5. i. Using a scale of 2cm to 1 unit on x – axis and
iii. Determine the slope of the line. 2cm to 2 units on y – axis, mark on a graph sheet,
the x – axis from – 5 to 5 and y – axis from −12 to
3. i. Using a scale of 2cm to 1 unit on both axes 12.
of a graph sheet, plot the points Q (1, −1), R (−1, ii. Plot the points E(3, − 6), F(0, 2), G(1 , 6) and
1), S (−2, 2) and T (− 4, 4) H(−6, 10) and draw a straight line through them.

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iii Determine the slope of the line. iv. Use your graph to find :
iv. Use your graph to find: a. x when y = − 6 b. y when x = − 7
a. y when x = 2.5 b. x when y = −2.
9. i. Copy and complete the table of the values for
6. i. On a graph sheet, with a scale of 2cm to 2 the relations; y1 = 2x + 5 andy2 = 3 – 2x for x from
units on both axes, mark the x – axis from −10 to -4 to 3
10 and the y – axis from −12 to 12.
ii. Plot the set of ordered pairs: (9, 11), (3, 3), (8, x -4 -3 -2 -1 0 1 2 3
0) and join the points with a rule to form a y1 = -3 3 7 4
2x +
straight line.
5
iii. Determine the gradient of the line drawn. y2 = 11 9 5
iv . From your graph, find: 3 −2x
a. x when y = 9, b. y when x = −3.
ii. Plot the graph of each relation on the same
7. i. Using a scale of 2cm to 2 units on both axes, graph sheet using a scale of 2cm to 1 unit on the x
draw a straight line through the points M (-5, 5) – axis and 2cm to 2 units on the y – axis.
and N (4, 10) iii. Find the coordinates of the point of
ii. Determine the gradient of line MN. intersection of y1and y2.
iii. Find the co – ordinate of the point at which iv. Determine the slope of y1 = 2x + 5.
the line cuts the x – axis.
Drawing a Line at a Given Point on the
8. i. Copy and complete the table below for the
x and y – axes
function f(x) = x – 5 Sometimes students are required to draw a line at
a given point on the x or y – axis.If the point is on
x -9 -3 0 3 6 the x – axis, say x = a, draw a
f(x) -8 -5 vertical/perpendicular line to pass through the
point, a, on the x – axis.
ii Plot the values of x against f(x) on a graph If the point is on the y – axis, say y = c, draw a
sheet, using a scale of 2cm to 2 units on both horizontal line to pass through the point, c on the
axes. y – axis.
iii. What is the slope of the line? These are illustrated in the diagrams below;

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On the x – axis On the y- axis
y
y
c y=c

b
a
x
-c -b -a a b c x
-a

-b y = -b
x = -b x=a -c

Worked Examples ii On the same graph sheet, draw the line x = −3

1. i. Using a scale of 2cm to 1 unit on both axes, to meet line AB at E.
plot the points A (4, −4), B (−4, 5) and join the iii. Write the co – ordinate of E
AB with a ruler.
Solution

x = -3 y
B (-4, 5)
6
Scale: 2cm: 2 units on both axes
5
E = (-3, 4) 4

3
2
1

x
-5 -4 -3 -2 -1 1 2 3 4 5
-1
-2
-3
A(4, -4)
-4
-5

2. i. Using a scale of 2cm to 2 units on the x – ii. Draw the line y = 1.5 to meet line PQ at R.
axis and 2cm to 1 unit on the y – axis, plot the Indicate the coordinate of R
points; P (-5, -5) and Q (9, 4). Draw a straight
line to join PQ.

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 y
6 x− axis=2cm : 2 units
y – axis = 1cm : 2units
5
Q(9,4)
4

2 R(4.5, 1.5
y = 1.5
1

x
-10 -8 -6 -4 -2 2 4 6 8 10
-1

-2

-3

-4
P (-5, -5)
-5

-6

Exercises 6.31 them with a long straight line.

1. Using a scale of 2cm to 2 units on both axes, ii On the same graph sheet, plot the point C (4, 2)
plot the points A (0, 10), B (−6, −2), C (4, 3) and and D (−2, 3) and join them with a long straight
D (−3, −11). Use a ruler to join the points A to B line extended through C to meet line AB. Name
and C to D. the meeting point M.
ii. Draw the line x = −2 to meet AB at P and CD Measure the angle between the lines through AB
at Q. and CD.
iii. Use a protractor to measure angles BPQ and iii. Find the co – ordinates of the point M.
PQC. iv. On the same graph, draw the line x = −2.5.
iv. What common name is given to the angles
BPQ and PQC? 3. i. Plot on a graph sheet, the points A (0, 5) and
v. State the relationship between lines AB and B (4, 5) using a scale of 2cm to 1 unit on both
CD. axes.
ii Measure the acute angle the line AB makes with
2. i. Using a scale of 2cm to 1 unit on both axes, the x – axis using a protractor
plot the points A (2, 3) and B (−3, 4) and join

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Collinear Points Exercises 6.32
Three or more points are said to be collinear A. Show that the three given points in each
when they lie on the same straight line. Two or case are collinear.
more points that lie on the same straight line have 1. (2, -3), (3, 1) (5, 9)
the same gradient. Therefore, to show whether or 2. (- 3, 4), (1, 2) (7, - 1)
not, two or more points are collinear: 3. (- 3, 1), (1, 2) (9, 4)
I. Find the gradient of the lines 4. (1, 2), (0, - 1) (- 2, - 1)
II. If the gradients are equal, conclude that they 5. (6, 1), (3, 3), (- 3, 7)
are collinear, if not, conclude that they are not
collinear. Perpendicular Lines
Two lines are said to be perpendicular when the
Worked Examples product of their gradient is – 1.That is if is
Show that the three points (0, 0), (3, 5) and (21, perpendicular to , then: × = – 1. In other
35) are collinear. words, for any two lines and , if the
gradient of one is m, the gradient of the other is
Solution .
Let the points be A(0, 0), B(3, 5) and C(21, 35)
–
Gradient of AB = = = To prove whether or not two lines and are
–
– perpendicular:
Gradient of BC = = = =
– 1. Find the slope of each line.
The gradient of AB = the gradient of BC 2. Multiply both slopes.
 A(0, 0), B(3, 5) and C(21, 35) are collinear. 3. a. If the product in step 2 is -1, the lines
are perpendicular.
2. The points (2, -3), (3, -1) and (4, k) lie on a b. If the product is not – 1, the lines are not
straight line. Find k perpendicular.

Solution Note:
Let the points be A(2, -3), B(3, -1) and C(4, k) Knowing the slope/gradient of a line, the
If A, B and C are collinear, then gradient of AB is slope/gradient of a line perpendicular to it is
equal to gradient of BC found by turning the known gradient upside
– ( ( down and changing its sign.
Gradient of AB = = =
–

– (
For e.g. if a line has a gradient of – , then the
Gradient of BC = = =
– gradient of a line perpendicular to it is
=
Worked examples
2=k+1 1. Write down the gradients of lines
2–1=k
perpendicular to the lines of gradients 3, ,
k=1
– 6 and respectively

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Solution  is not perpendicular to
Let = 3, = , = – 6 and =–
The gradient of the line perpendicular to: Parallel Lines
= = y
B
= = ⁄
=–1× =–4 D
= = =
( θ θ
x
= = ( ⁄ )
=–1× = A
C
2. Given A( – 1, – 1), B (0, 4) P(– 4, 3), Q(6, 1), From the diagram above, line AB is parallel to
find whether or not AB is perpendicular to PQ line BC. Since parallel lines make equal
corresponding angles θ with the x – axis, parallel
Solution lines are said to have equal gradients (and vice
A( - 1, - 1), B (0, 4) P( - 4 , 3), Q(6, 1), versa)
Let the gradient of AB be
– –( In general, if two or more lines have the same or
= = = =5
– –( equal gradients, they are said to be parallel. That
is , if is parallel to then =
Let the gradient of PQ be
– – To prove whether or not two lines are parallel, do
= = = =–
– –( the following:
=5×– =–1 1. Find the slope of each line
 AB is perpendicular to PQ 2. a. If the slopes are the same, the lines are
parallel
3. Given Q(1, 4), R(6, 6), S(2, -1) and T(12, 3), b. If the slopes are not the same, the lines
show whether or not, is perpendicular to are not parallel

Solution Worked Examples

Q(1, 4), R(6, 6), S(2, -1) and T(12, 3) If A(0,3), B(7,2), P(6, -1) and Q(-1, -2) show
whether or not is parallel to
Let the gradient of QR be

=
–
= = Solution
– A(0,3) and B(7,2), P(6, -1) and Q(-1, -2)
Let the gradient of ST be Let the gradient of AB be
–
=
–
=
(
= = = = =–
–
–

× = × = Let the gradient of PQ be

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 – ( B. In the following cases, show whether or
= = = =
–
not AB is parallel or perpendicular to PQ
1. A(4, 3) B(8, 4) P(7, 1) Q(6, 5)
Therefore, is not parallel to 2. A(-2, 0) B(1, 9) P(2, 5) Q(6, 17)
3. A(8, -5) B(11, -3) P(1, 1) Q(-3, 5)
2. Given that Q(1, 4), R(6, 6), S(2, -1) and T(12, 4. A(4, 3) B(-7, 3) P(5, 2) Q(5, -1)
3), show whether or not is parallel to .
C. Two points on L1 and two points on L2 are
Solution given. Determine whether L1 and L2 are
Q(1, 4), R(6, 6), S(2, -1) and T(12, 3) parallel, perpendicular or neither
Let the gradient of QR be 1. L1(-2, 0) and (0, 6), L2(-3, -4) and (0, 5)
– 2. L1(-2, 0) and (0, 1), L2(0, 0) and (-4, 4)
= = =
– 3. L1(6, 3) and (8, 7), L2(7, 2) and (6, 0)
4. L1(1, 10) and (-1, 7), L2(0, 3) and (1, 5)
Let the gradient of ST be
– (
= = = = D.1. Show that the two lines 3x – 2y + 8 = 0 and
–
2x + 3y – 1 = 0 are perpendicular.
= =
Therefore, is parallel to 2. If the line ax + 3y = 0 is perpendicular to the
line 3x – 5y – 4 = 0, find the value of a.
3. Show that the two lines 3x + 5y – 8 = 0 and 5x 3. P(2, -3), Q(3, 1) and R(-1, k) are three points. If
– 3y – 11 = 0 are perpendicular. PQ is perpendicular to QR, find the value of k.

Solution Application to Parallelograms

Let L1 = 3x + 5y – 8 = 0 and L2 = 5x – 3y – 11 = 0 Given the vertices of a quadrilateral ABCD, it
From L1 = 3x + 5y – 8 = 0, can be verified whether quadrilateral ABCD is a
m1= = rectangle, a square or a parallogram.

First make a sketch of the figure as shown below:

From L2 = 5x – 3y – 11 = 0,
D C
m2 = = =

If L1 is perpendicular to L2, then m1 × m2 = - 1

× =-1 A B
Take note of the fact that for all parallelograms
(rectangles and squares ) with vertices ABCD;
Exercises 6.33
I. AB is parallel to CD , meaning the gradient of
A. Write the gradient of the line perpendicular
AB is equal to that of CD.
to the lines with the following gradients
II. AD is parallel to BC, meaning the gradient of
1. 4 2. – 3. – 12 4. 5. – AD is equal to that of BC.

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Worked Examples 3. A quadrilateral has vertices at P(2, 1) , Q(6, 3),
Show that A(6, - 2), B(- 2, - 4), C(0, 2) and D(8, R(5, 5), S(1, 3). Show that PQRS is a
4) are vertices of a parallelogram. parallelogram. Find the length of the diagonals
PR and QS.
Solution
Method 1 4. A quadrilateral has vertices at A(0, 0) , B(7,
A(6, - 2), B(- 2, - 4), C(0, 2) and D(8, 4) 24), C(22, 44) D (15, 20). Find the length of the
If ABCD is a parallelogram, then; sides. What type of quadrilateral is ABCD?
gradient of AB = gradient of DC
Gradient of AB; The Equation of a Straight Line
– ( The general form of the equation of a straight line
mAB = = = =
–
is ax + by + c = 0. On the other hand, the
intercept form: y = mx + c is equally acceptable.
Gradient of DC;
–
The equation of a straight line can be determined
mDC = = = = from:
–
mAB = mDC 1. Any two given points on the line
Therefore, ABCD is a parallogram 2. The gradient of the line and a point on the line
3. The gradient of the line and the intercept on the
Method 2 y – axis
A(6, - 2), B(- 2, - 4), C(0, 2) and D(8, 4)
If ABCD is a parallelogram, then; Equation of a Line given any Two Points
gradient of AD = gradient of BC If A( ) and B( ) are any two points on a
straight line or any two points joined by a straight
Gradient of AD; line then, the equation of the line is determined
– ( by any of the following methods:
mAD= = = =3
–
Method 1
Gradient of BC; I. Find the gradient using the formula,
– ( –
mBC=
–
=
–(
= =3 m= ……….(1)
–
II. Substitute the value of m and (x1, y1) or m
mAD = mBC and (x2, y2) in the formulas below respectively:
Therefore, ABCD is a parallogram y– ( ) or y – ( )
III. After substitution, express the equation
Exercises 6.33B y– ( ) or y – ( ) in the
1. Show that A(-3, 1), B(1, 2), C(0, -1) and form: ax + by + c = 0 or y = mx + c
D(- 4, -2) are vertices of a parallelogram.
Method 2
2. Show that P(1, 7), Q(7, 5), R(6, 2) and S(0, 4) I. Find the gradient using the formula,
are vertices of a rectangle. –
m= ……….(1)
–

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II. Using a point on the line, either (x1, y1) or (x2, = 3 and = – 4
y2), substitute the values in y = mx + c to Express the equation in the form y = mx + c,
determine the value of c (the intercept on y – – –
m= = =–
axis) – –(

III. Substitute the values of m and c in y = mx + c

to obtain the equation of the line Substitute m = – in y = mx + c
y=– x+c
Worked Examples
1. Find the equation of the straight line joining
Using the point (- 5, 2) (Because it lies on the line).
the points (– 5, 2) and (3, – 4)
When x = – 5, y = 2, put in y = – x + c
Solution 2 = – (–5) + c
In (– 5, 2) and (3, – 4), = -5, = 2,
4 × 2 = – 3(-5) + 4c (Multiply through by 4)
= 3 and = – 4
8 = 15 + 4c
– –
m= = =– 8 – 15 = 4c
– –(
y– ( ) – 7 = 4c
y– ( ( ) (By substitution) c=–
4(y – 2) = – 3(x + 5)
4y – 8 = – 3x – 15 (Expansion) Substitute c = – in y = – x + c
4y + 3x – 8 + 15 = 0 y=– x– or y = –
3x + 4y + 7 = 0
The equation is y = – x – or y = –
The equation of the line is 3x + 4y + 7 = 0

Method 2 2. Find the equation of the line joining the points

In (– 5, 2) and (3, – 4), = -5, = 2, M(6, 3) and N(5, 8)
= 3 and = – 4
– –
Solution
m= = =– In M(6, 3) and N(5, 8), = 6, = 3,
– –(
y– ( ) = 5 and = 8
– –
y–( ( ) (By substitution) m= = = -5
– –
y+4= ( y– ( )
4(y + 4) = – 3 (x – 3) (Multiply through by 4) y– ( ) (By substitution)
4y + 16 = – 3x + 9 y–3=
4y + 3x + 16 – 9 = 0 y + 5x – 3 – 30 = 0
3x + 4y + 7 = 0 5x + y – 33 = 0
The equation of the lineis 5x + y – 33 = 0
Method 3
In (– 5, 2) and (3, – 4), = -5, = 2, 3. Find the equation of the line which passes
through the points (2, -3) and (1, 3)

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Solution The Equation of a Line given the Gradient and
In (2, -3) and (1, 3), = 2, = -3, a Point on the Line
= 1 and = 3 Given the gradient of a line (m) and a point (x, y)
– – through which the line passes, the equation of the
m= = =- 6
– –
line is found as follows:
y– ( )
y–( ( ) Method I
y + 3 = -6x + 12 I. Substitute the value of the gradient (m) and the
y + 6x + 3 – 12 = 0 values of the point (x, y) in the relation y = mx +
6x + y – 9 = 0 c, to determine the value of c (intercept on the y –
The equation of the lineis 6x + y – 9 = 0 axis)
II. Substitute the value of m and c in the equation
Exercises 6.34 y = mx + c to determine the equation of the
A. Find the equation of the straight line straight line.
joining the following pair of points:
1. (3, 2) and (7, - 3) Method II
2. (- 3, 4) and (8, 1) I. Substitute the value of the gradient (m) and the
3. (- 1, - 4) and (4, -3) values of the point (x1, y1) in the relation,
4. (- 2, 5) and (3, - 7) y– ( ).
II. Expand and re- arrange the equation in the
B. 1. Find the equation of the straight line joining
form ax + by + c = 0 or y = mx + c, to obtain the
A(10, 0) and B(0, –7)
equation of the line.
2. i. Write down the equation of a line L, joining
Worked Examples
the origin to the point A(4, 2).
ii. What is the gradient of this line? 1. Find the equation of a line with gradient
which passes through the point (5, - 2)
3. The points (– 4, 2) and (1, –3) lie on the line
with y = mx + c. Find m and c, and hence Solution
the equation of the line . Method I
m= and (x, y) = (5, – 2)
4. Given that the line y = mx + c has a gradient of
Substitute in y = mx + c
-1, and passes through the point (2, 3), find the
values of m and c – 2 = (5) + c
– 2 × 2 = 3(5) + 2c
5. The points A, B and C have co-ordinates (7, 0), – 4 = 15 + 2c
(3, -3) (-3, 3) respectively. Find the coordinates – 4 – 15 = 2c
of D, E, F, the mid-points of BC, CA, AB 2c = – 19
respectively. c=–

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Put c = – and m = in y = x+c 4y – 20 = – 3x + 21
4y + 3x – 20 – 21 = 0
= x–
3x + 4y – 41 = 0
The equation of the line is = x– The equation of the line is 3x + 4y – 41 = 0

Method II Exercises 6.35

A. Write down the equations of the lines
m= and (x1, y1) = (5, – 2)
through the following points, and having the
Substitute in y – ( )
following gradients: express your answers in
y–( ( ) the form ax + by + c = 0, where a, b and c are
y+ ( ) integers
2 (y + 2) = 3(x – 5) 1. (- 2, 5), 3 2. (4, 3), - 3. (0, -3), 2
2y + 4 = 3x – 15 4. (2, - 5), 5. - 5, (- 3, -7) 6. (0, 4), -
3x – 2y – 15 – 4 = 0
3x – 2y – 19 = 0 Challenge Problems
The equation is 3x – 2y – 19 = 0 1. PQRS is a parallelogram in which the opposite
vertices are P(2, 1) and R(4, 4). If the slope of PQ
2. The gradient of a line is 4. If the line passes is and the slope of PS = -2, find:
through the point (1, 3), find the equation of the i. the equation of PQ
line. ii. the equation of QR
iii. hence, or otherwise, find the coordinates of Q
Solution and S
m = 4, (x1, y1) = (1, 3)
Substitute in y – ( )
2. i. A line of gradient – is drawn through the
y– ( )
origin O. Write down the equation of this line.
y – 3 = 4x – 4
ii. A point A is chosen on the line with first
4x – y – 4 – 3 = 0
coordinate 3. If B is the point (5, 0), show that the
4x – y – 7= 0
triangle is an isosceles and calculate it area
The equation of the line is 4x – y – 7= 0
iii. If C is the point (3, 4), calculate the area of
OABC
3. Find the equation of a straight line that passes
through (7, 5) with gradient – The Equation of a Line given the Gradient and
the Intercept on the y – axis
Solution If a straight line cuts the y – axis at the point (0,
m=– and (x1, y1) = (7, 5) c), the distance of this point from the origin is
Substitute in y – ( ) called the intercept on the y – axis.

y–5=– (x – 7) The equation of a straight line of gradient m,

4(y – 5) = -3 (x – 7) making an intercept c on the y – axis is: y = mx + c

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Given the values of m and c, the equation of the The equation of the straight line parallel to
straight line is found by substituting them in another line of gradient m, passing through the
y = mx + c point ( , ) is found by making y the subject of
the formula: y – = m( )
Worked Examples
1. A straight line which has a gradient of 7 cuts Worked Examples
the y – axis at – 2. Find the equation of the line 1. What is the equation of the line parallel to the
line 2x + 7y – 8 = 0, which passes through the
Solution point (– 1, 3)?
Gradient, m = 7,
Intercept on y - axis, c = – 2 Solution
Substitute in y = mx + c to get y = 7x – 2 Method 1
The equation of the line is y = 7x – 2 In 2x + 7y – 8 = 0
m=– = –
2. Find the equation of a straight line with
The line parallel to 2x + 7y – 8 = 0 has a gradient
gradient – which cuts the axis at the point (0, 8) of – and passes through (- 1 , 3)
⇒m = – , ,
Solution
m=– . In (0, 8), the line cuts the x – axis at (0, Substitute in y – = m( )
8). Hence c = 8. y – 3 = – (x – (–1)
y = – x + 8. y – 3 = – (x + 1)
The equation of the line is y = – x + 8 7(y – 3) = – 2(x + 1)
7y – 21 = – 2x – 2
Exercises 6.36 7y + 2x – 21 + 2 = 0
Find the equations of the straight lines of given 2x + 7y – 19 = 0
gradients cutting the y – axis at the named The equation is 2x + 7y – 19 = 0
points:
Method 2
1. m = 3, (0, 2) 2. m = – , (0, 4) Make y the subject of 2x + 7y – 8 = 0
3. m = 4, (0, 6) 4. m = , (0, - 4) 7y = – 2x + 8
y = – x + compared to y = mx + c, m = –
5. m = -3, (0, -2) 6. m = , (0, 7)

The line parallel to 2x + 7y – 8 = 0 has a gradient

The Equation of a Line through the Point ( ,
) and Parallel to another Line, of – and passes through ( - 1 , 3)
Two or more lines are said to be parallel if they Substitute in y = mx + c
have the same gradient. That is if the gradient of 3=– (- 1) + c
a line is m, then the gradient of the line parallel to 3 × 7 = - 2 (-1) + 7c
it is m. 21 = 2 + 7c

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21 – 2 = 7c 5(y + 3) = - 2(x – 2)
19 = 7c 5y + 15 = - 2x + 4
c= 5y + 2x + 15 – 4 = 0
2x + 5y – 11 = 0
The equation is 2x + 5y – 11 = 0
Put m = – and c = in y = mx + c
y=– x+ Method 2
Make y the subject of 5x – 2y – 11 = 0
The equation is y = – x +
- 2y = - 5x + 11
y= x– compared to y = mx + c, m =
The Equation of a Line which Passes through
the Point ( , ) and Perpendicular to another
Line The line perpendicular to the line 5x – 2y – 11 =
Two lines are said to be perpendicular if the 0, has a gradient of – and passes through (2, -3)
product of their gradient is –1. That is, if the ⇒m = – , x = 2 ,y = -3
gradient of a line is m, then the gradient of the
Substitute in y = mx + c
line perpendicular to it is –
-3=– (2) + c
- 3 × 5 = - 2 (2) + 5c
The equation of the straight line perpendicular to
- 15 = - 4 + 5c
another line of gradient m, passing through the
- 15 + 4 = 5c
point ( , ) is found by making y the subject of
- 11 = 5c
the formula: y – = – ( ) after
c=–
substituting the values of m and ( , )

Worked Examples Put m = – and c = – in y = mx + c

1. Find the equation of the line perpendicular to y=– x–
the line 5x – 2y – 11 = 0, which passes through
The equation is y = – x –
the point (2, - 3).

Solution 2. Find the equation of the perpendicular bisector

In 5x – 2y – 11 = 0 of AB, where A and B are the points
m= (- 4, 8) and (0, – 2).

The line perpendicular to the line 5x – 2y – 11 =

Solution
0, has a gradient of – and passes through (2, - 3) A(- 4, 8) and B(0, – 2)
⇒m = – , , = – 4, = 8, = 0 and = – 2
Substitute in y – = m( ) Perpendicular bisector of AB passes through the
mid – point of AB
y – (- 3) = – (x – 2) (
M= ( )=( ) = (- 2, 3)
y + 3 = – (x – 2)

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The perpendicular bisector passes through the 3. Find the equation of a straight line through P(7,
point (– 2, 3). 5) perpendicular to the straight line AB whose
equation is 3x + 4y -16 = 0
Gradient of AB
– – 4. If the distance of the point (x, y) from the
m= = = =–
– –( origin is always equal to the distance between (3,
Gradient of the perpendicular bisector of AB = -9) and (-1, - 6), find an equation connecting x
and y.
Equation of the perpendicular bisector of AB
which passes through (- 2, 3) Verifying that a Point Belongs to a Line
y– ( ) To verify that a point lies on a line or belongs to a
line:
But m = , = - 2, and =3
I. Substitute the coordinates of the point (x, y)
y– ( ( ) into the equation of the line.
5(y – 3) = ( ) II. If the coordinates satisfy the equation, then the
5y – 15 = 2x + 4 point is on the line.
5y – 2x – 15 – 4 = 0 III. If the coordinates do not satisfy the equation,
5y – 2x – 19 = 0 the point is not on the line .
The equation is – 2x + 5y – 19 = 0
Worked Examples
Exercises 6.37 1. Show whether the point (1, 13) lie on the line
A. Find the equation of the straight line: y = 6x + 7.
1. which passes through (- 3, - 2), perpendicular
to 4x + 3y – 5 = 0 Solution
2. which passes through (4, 0), perpendicular to x Point (x, y) = (1, 13)
+ 7y + 4 = 0 When x = 1,
3. which passes through (3, – 4), perpendicular y = 6x + 7
to the line 5x – 2y = 3 y = 6(1) + 7 = 13
4. which passes through (– 2, 3), parallel to  (1, 13) lie on the line y = 6x + 7
5x – 2y – 1 = 0
2. Show whether the point (13, 30) lies on the line
B. 1. Find the equation of the straight lines y = 2x + 2
through the point (3, –2) which are:
i. parallel to the line2y + 5x = 17 Solution
ii. Perpendicular to the line 2y + 5x = 17 Point (x, y) = (13, 30)
2. Find the equations of the lines passing through When x = 13
the point (4, –2) respectively; y = 2x + 2
i. parallel and y = 2(13) + 2 = 28
ii. perpendicular to the line 2x – 3y – 4 = 0 When x = 13, y 30
 (13, 30) does not lie on y = 2x + 2

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3. Show whether the point (2, 3) belongs to the 2. The coordinates of the points (p, 5) and (-1, q)
line 5x – 2y – 4 = 0 satisfy the equation y = 2x – 1. Calculate the
values of p and q.
Solution
For the point (2, 3), x = 2and y = 3 Point of Intersection of Two Lines
5x – 2y – 4 = 0 To find the point of intersection of two lines,
5(2) – 2(3) – 4 = 0 solve the equations of the lines simultaneously by
0=0 any preferred method unless stated. The solution
L.H.S = R. H. S set is the point of intersection of the lines.
The point (2, 3) belongs to the line 5x – 2y – 4 = 0
Worked Example
4. Given that the point (4, 2) lies on the line y = Find the points of intersection of the lines
2x + c, find c 2x – 3y = 6 and 4x + y = 19

Solution
Solution
y = 2x + c
2x – 3y = 6 ……………(1)
(x,y) = (4, 2)
4x + y = 19…………….(2)
2 = 2(4) + c
eqn (1) × 2
2=8+c
4x – 6y = 12……………(3)
2–8=c
c=-6
eqn (2) – eqn (3);
7y = 7
Exercises 6.38
y=1
A. Which of the point belongs to its line
1. y = 2x + 3, (6, 16) 3. y = 3x – 5, (- 1, -8)
Put y = 1 in eqn (1);
2. 3x + 4y = 1, (-1, 1) 4. 5x – 2y + 7 = 0, (1, -1)
2x – 3(1) = 6
2x = 6 + 3
Challenge Problems
2x = 9
1. A (2, -2) and B (4, 4) are two points
i. Find x=
ii. Find the coordinates of m, the mid – point of x = 4.5
̅̅̅̅  The point of intersection is (4.5, 1)
iii. Find the slope and the equation of the line AB.
iv. The point (–2, r) is on the line AB, find the Exercises 6.39
value of r A. Find the points of intersection of the
following pair of straight lines:
v. The perpendicular bisector of the line segment
1. y = 3x + 2 and 2x + 3y = 17
AB cuts the x – axis at the point p(h, 0) and the y
2. 5x + 3y = 2 and x – y = 6
3. y = 5x + 2 and y = 3x – 1
– axis at the point q(0, k), find the values of h and k.
4. 5x + 7y + 29 = 0 and 11x – 3y – 65 = 0

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Challenge Problems
1. L : 2x – 5y – 9 = 0 and K : 3x – 2y – 8 = 0 2. P = {(x, y) : 4x + 3y = -7}, Q = {(x, y) :5x – 9y
are two lines. L intersects K at the point q. = -13} and R = {(x, y) : 14x – 15y = -33}, x, y ∈
i. Find the coordinates of q R. Prove that P Q⊂ R
ii. Find the equation of the line M such that L is 3. P = {(x, y) : 2x + 3y + 5 = 0} and Q = {(x, y) :
perpendicular and q∈ M 5x – 4y + 1 = 0}, x, y ∈ R. Find P∩Q
iii. Show that the point r (4, -6) is on M

Baffour – Ba Series, Core Maths for Schools and Colleges Page 164
7 PLANE GEOMETRY I Baffour– Ba Series

Lines and Planes 3. A set of points in a line consisting of two

Definitions: distinct end points is called ……
A Point: is an idea associated with position. It is 4. A flat surface which has length and width only
symbolized by a dot (.) and represents specific is called ………
location. It has neither size nor shape 5. A line that starts from a point and end at
infinity is called…….
A Line: is an infinite set of points which extends
indefinitely in two directions The Circle
A circle is a set of points in a plane which are at
the same distance from a fixed point. The fixed
A Line Segment : is a set of points in a line point is called the centre of the circle and the set
consisting of twodistinct ends. It represents a of points forms the circumference of the circle.
collection of points inside the endpoints and it is
named by its end points Parts of a Circle

Endpoint Endpoint Segment

A B Chord

A Ray: is a line segment that has only one

defined end point and one side that extends
endlessly away from the end point. A ray is Diameter
named by its end point and by the other point on
the line.
A B B A Circumference Sector

A Plane: is a flat surface which has length and

Circumference: It is the distance around a circular
width only. A plane therefore has two
region. It is also known as the length or perimeter
dimensions, length and width; no thickness. e.g.
of a circle.
A floor of a football field
Diameter: It is a straight line that divides a circle
into two equal parts
A PLANE Semi - circle: It is half a circle
Chord:It is s a straight line that connects any two
points on a circle.
Exercises 7.1
Arc: It is a portion on the circumference of a
Fill in the blank spaces with the correct response
circle
1. An idea associated with position is
Segment: It is the area bounded by an arc and a
called………
chord.
2. An infinite set of points extending indefinitely
in two dimensions is called …

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Radius : A line drwn from the center of a circle For e.g.
to touch any part of the circumference. The plural a
is radii
Sector: It is area bounded by two radii and an arc
o b
Exercises 7.2
Complete the each with the correct answer I. Place the point “C” of the protractor on
1. Half a circle is called … point “o” of the angle.
2. A straight line drawn from the center of a II. Align the line segment “CB” of the protractor
circle to touch any point on the circumference is with arm “ob” of angle a o b so that CB falls
called … exactly on “ob”. The arm “o a” of angle aob
3. The distance around a circle is called… points to the number
4. Any straight line that passes throught the of degrees the angle measures on the protractor.
center of a circle, touching the circumference at III. In order to determine the size of angles
both ends is called … opening to the right, the inner set of measurement
Angles is used.
An angle is formed when two straight lines meet a
90
at a point. The point where the two straight lines
meet is simply called Vertex.

0,180 180 ,0

C b
(B) Angles opening to the left.

Measurement of Angles
The instrument used to measure angles is called a
protractor. The unit of measure is the degree ( 0 ). I. Place the point “C” of the protractor on
The scale on the protractor is divided into degrees the point O of the angle.
numbered from 00 to 1800 starting from either II. Line up ̅̅̅̅ of the protractor with arm “ ” of
end. 90
angle so that AC falls exactly on arm “
of xoy. At this stage, the of angle xoy will
point to the number of degrees the angle
measures on the protractor.
0 0
180 0, 180, 0
III. In order to determine the size of angles
A C B
opening to the left, use the outer values of
Using the Protractor measure.
(A) Angles opening to the right.

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 o a
90

1800< < 3600 b

0,180 180,0 The values of the type of angles are summarized
C below:
Types of Angles 00< an acute angle < a right angle(900) < a
1. Acute angle: Any angle whose measure straight angle ( 1800 ) < a reflex angle < a
is less than 900. complete turn (3600 )
a
Pair of Angles
1. Complementary Angles: They are any two
angles that sum up to 900.
o b
0 0
0 < θ < 90
a0 a0 + b0 = 900
b0
2. Right Angle: Angle whose measure is
Complementary angles
exactly 900.
For example, in the diagram below,:
a
a
c
o b 600

= 900 300
o b
3. Obtuse Angle: Any angle whose measure is 60 + 30 = 90 , 60 and 300 are complementary
0 0 0 0

greater than 900 but less than 1800 angles

a
2. Supplementary Angles: They are any two
angles that sum up to 1800.
o b
900< < 1800
c0
c0 + d0 = 1800
d0
4. Straight Angle: Any angle whose measure is
Supplementary angles
1800
For example in the diagram below;
a c
o b
= 1800 0
40
0 140
5. Reflex Angle: Any angle whose measure is a b
o
greater than 1800 but less than 3600.

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400 and 1400 are said to be supplementary angles
because 400+ 1400 = 1800

3. Vertically opposite angle

When two lines cross each other, vertically 1800
opposite angles are formed. Vertically opposite
angles are equal Worked Examples
Find the value of in the diagrams below
Consider the diagram below; 1. x + 200 + 450 = 1800 (< s on a straight line)
i. Angle e = Angle e x = 1800 – 200 – 450
f
e e ii. Angle f = Angle f x = 1150
f 0 x 450
20

Worked Examples 2. 4x + x + x = 1800 (< s on a straight line)

The sizes of two angles are (100 – x)0 and (3x + 6x = 1800
50)0. Calculate x if these two angles are
,
supplementary
300
Solution
If (100 – x)0 and (3x + 50)0 are supplementary 3. x – 200 + x = 1800
⇒(100 – x)0 + (3x + 50)0 = 1800 2x = 1800 + 200
1000 – x0 + 3x0 + 500 = 1800 2x = 2000
2x0 + 1500 = 1800 200
,
2x0 = 1800 – 1500
2x0 = 300 x = 1000
x = 150
2. Angles formed in a circle add up to 3600
Exercises 7.4 Consider the figure below:
A. Fill in the spaces correctly:
1. Two straight lines meet at a point to form a
figure that is called … a + b + c = 3600
a b
2. The supplementary angle of 480 is.................. c
3. The complement angle of 650 is ……………
4. Any angle that measures 900 is called……….
5. Two right angles equal to ………………….. Worked Examples
Find the angles marked with letters
Properties of Angles i. x + 900 + 1370 = 3600
1. Angles formed on a straight line sum up to 1800. x = 3600 – 900 – 1370
x
Consider the figure below; 0
x = 133 0
137

Baffour – Ba Series, Core Maths for Schools and Colleges Page 168
ii. x – 100 + x + 400 = 3600
2x = 3600 + 100 – 400 iii. 1800 – 1200 = 600
2x = 3300 x - 100 The supplementary angle of 12x – 9x = 600
x = 1650 x + 400
x0
2. Find the angles marked with letters:

Angles Connected with Parallel Lines 40 a

1. Vertically Opposite Angles 0

When two straight lines intersect, the opposite

angles formed are called vertically opposite Solution
angles. Angles a and 400 are vertically opposite angles.
Therefore 400
Vertically opposite angles are equal to each other.
For example, in the figure below; 3. Find the values of the angles marked with
letters in the diagram below
b
a c
d
1250

i. Angle a = Angle c (Vertically opp. Angles)

ii. Angle b = Angle d (Vertically opp. Angles) Solution
z + 1250 = 1800 (angles on straight line)
Worked Examples z = 1800 – 1250 = 550
1. Given that (12x – 100)0 and (9x + 20)0 are z=x (vertically opposite angles)
0
vertically opposite angles. Calculate: x = 55
i. the value of x.
ii. the value of (12x – 9x)0 y = 1250 (vertically opposite angles)
iii. What is the supplementary angle of (12x – 9x )0
2. Alternate angles
Solution They are angles that are formed at the corners of
i. (12x – 100)0 and (9x + 20)0 are vertically a figure. They are also called Z or N or Σ angles.
opposite angles Alternate angles are equal.
(12x – 100)0 = (9x + 20)0
12x – 9x = 200 + 1000 Consider the diagram below:
3x = 1200
x = 40
e c

ii. 12x – 9x c e
= 12(400) – 9 (400)
= 4800 – 3600 = 1200

Baffour – Ba Series, Core Maths for Schools and Colleges Page 169
i. Angle c = Angle c (Alternate angles) Solution
ii. Angle e = Angle e (Alternate angles) 51o
y1
y
y2
Worked Example o
Find a, b, c and d in the diagram below; 35

y1 and 51o are alternate angles so y1 = 51o

0 y2 and 35o are alternate angles so y2 = 35o
120 c
But y = y1 + y2
b a y = 51o + 35o = 86o

2. In the diagram below, find the value of angle x.

Solution 25o
a = 1200 (Alternate angles)
64o
c + 1200 = 1800 (Straight angles)
1800 –1200 x
o

c = 600
Solution
c = b = 600 (Alternate angles) 25o
y1
y2
Special Alternate Angles
xo
Consider the diagrams below:
y1 + y2 = 64o
x1
But y1 = 25o (Alternate angles)
y1
Y y2 25o + y2 = 64o
x2 y2 = 64o – 25o = 39o
y2 = xo (Alternate angles)
o o
From the diagram above x1 and y1 are alternate x = 39
angles. Therefore, x1 = y1.
Exercises 7.5
Similarly, x2 and y2 are alternate angles. 1. In the figure below, if⃗⃗⃗⃗⃗ is parallel to ⃗⃗⃗⃗⃗ , find
Therefore x2 = y2. the value of angle g
⇒y1 + y2 = x1 + x2 and y1 + y2 = y B
A 50
0

Worked Examples g
Find the value of y in the figure below: 130
0
E D

51o
2. In the figure below, find the value of y
yo
35o

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 Solution
60o
m and n are corresponding angles
y0 m = 300
25o
4. Co-interior Angles
In the diagram below:
3. Corresponding angles
Consider the diagram below: e f

c d c d
a b
d and f are co – interior angles as well as e and c.
c d Co – interior angles are also called supplementary
a b angles because they sum up to 1800.
Angle a corresponds with angle a i. d + f =1800
Angle b corresponds with angle b ii e + c = 1800
Angle c corresponds with angle c
Angle d corresponds with angle d Worked Examples
But corresponding angles are equal 1. In the diagram, PQ is parallel to SR. Find the
⇒a = a , b = b, c = c and d = d value of x
Q R
Worked Examples
1. Find x and y in the diagram below? x + 1300
0
35

x y P S

0
Solution
65
(x + 1300) and 350 are interior opposite angles. (x
Solution + 1300) + 350 = 1800
x = 650 ( Corresponding angles) x + 1300 + 350 = 1800
x + y = 1800 (Angles on a straight line) x = 1800 – 1300 – 350 = 150
650 + y = 1800
y = 1150 2. A straight line intersects three parallel lines as
shown in the diagram below. Find the value of x
2. Find the value of m in the diagram below 0
125

300

x
m

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Solution
Name the angles as shown below;
0
125 x
0
20

y Solution
x Draw a parallel line througg angle x and name

the angles as shown below:

0
y = 125 (Corresponding angles) 900 + 200 + x = 3600
x + y = 1800 (Angles on a straight line) x = 3600 – 900 – 200
x + 1250 = 1800 0 x = 2500
20 x
x = 1800 – 1250
x = 550
0
20

3. In the figure below, find the value of the angle 5. Find the value of angle x.
named y
A x B
E
0
44
0
37 0
130
y
C D F

Solution
Solution x = < BCD (Alternate angles)
0 0
Name the angles as shown below; < BCD + 130 = 180 (Interior opp.<s)
< BCD = 1800 – 1300
44
0 < BCD = 500
0 0
44 37 But x = < BCD = 500
y
Therefore x = 500
x

8. In the diagram below, PQRS is a

0 0 0 parallelogram. If < QPS = 2x, < RSP = 3x and <
x + 44 + 37 = 180
x + 810 = 1800 RST = y, find the value of y
x = 1800 – 810 Q S
x = 990
2x 3x y
x = y (corresponding angles) P S T
But x = 990, ⇒y = 990
Solution
4. Find the value of x in the figure below; 2x and y are corresponding angles

Baffour – Ba Series, Core Maths for Schools and Colleges Page 172
⇒y = 2x 4. a 0
0 20
3x + y = 180 ( < s on a straight line )
3x + 2x = 1800 (But y = 2x) c b
0 0
5x = 180 40
x = 360
5.
0 10x
⇒ y = 2x = 2(36 ) x + 100 x
0
y = 72

9. Find the angle marked . B. 1. In the figure below,< PRS = 1000, < TPV =
F 800 and PV is parallel to QS. Explain why ∆PQR
Not drawn to scale
Ѳ is isosceles
E 40o D C T
0
80
P V
o
110
A B 0
100
Solution S
Q R
Angle ADC = 110o.
(Opposite interior angle of a parallelogram) 2. In the figure below, PQRS is a parallelogram.
Mark in the sizes of the angles x , y and z
Angle EDF = 110o (opposite angle)
S R
θ + 40o + 110o = 180o
y
θ + 150o = 180o o
z
0
θ = 1800 – 1500 = 300 30
0
x
70
52
0

P Q T
Exercises 7.6
A. Find the value of the variables: by variables 3. In the figure below, DE// BC. Mark in the sizes
in the diagrams below: of angles x and y
1. 52
0 A
0 0
70 40
y
0 x
60 D E
2.
350 y 1000
B C F
e
0
82
4. ABCD is a parallelogram in which angle A =
3. C 720 and AB is equal in length to diagonal BD.
x D
Calculate the sizes of all the angle in the figure
y
A B
1180

Baffour – Ba Series, Core Maths for Schools and Colleges Page 173
Triangles
A triangle is a plane figure bounded by three
straight lines. A triangle has three interior angles
that sum up to 1800. The area of a triangle,
A = base × height
In an equilateral triangle, each angle is 600
Types of Triangles
I. Right – angled Triangle V. Isosceles Triangle
It is a triangle in which one angle is a right angle It is a triangle in which two base angles and
or 900 corresponding sides are equal
A a

b c
B C

Note: VI. Scalene Triangle

The symbol represents a right angle or 900 It is a triangle in which all the three sides and
angles are unequal

II. Obtuse – angled Triangle B

It is a triangle in which one angle is an obtuse
angle (greater than 900, but less than 1800)
A A C

Congruent Triangles
They are two or more triangles that have the same
C
B shape or the same size and angles. In other words,
III. Acute - angled Triangle triangles are congruent if:
It is a triangle in which each interior anlge is less I. 3 sides = 3 sides
than 900 II. 2 sides, included angle = 2 sides, included
angle
A
III. 2 angles, one side = 2 angles, corresponding
400 side
Example of a congruent triangle is shown below
800 600
C
B B
E G
IV. Equilateral Triangle
It is a triangle in which all the three sides are A C
equal. F

Baffour – Ba Series, Core Maths for Schools and Colleges Page 174
∆ABC is congruent to ∆ EFG because they have 2. m = 700 ( base angles of an isosceles ∆)
equal corresponding sides n + 700 + 700 = 1800
n
n = 1800 – 700 – 700
Properties of a Triangle n = 400
1. The sum of angles in a triangle is equal to 1800
700 m
c a + b + c = 1800
3. a + 700 + 600 = 1800
b a = 1800 – 700 – 600 600
a
a = 500
700 a b
2. For all Isosceles triangles, the base angles are 0 0
equal. The base angles are the angles that face the b = 70 + 60 ( interior opposite angles)
two equal sides b = 1300

5. 2x + 4x + 3x = 1800
c
9x = 180
x = 200
a b

a = b (base angles of an isosceles ∆) a = 2x + 4x (Interior opposite angles)

a = 2(200) + 4(200) (Put x = 200)
The Exterior Angle Theorem a = 400 + 800
The exterior angle theorem of triangles states that a = 1200
the exterior angle of a triangle is equal to the sum C
of the two interior angles opposite to it (exterior 6. Find the values of angles x, y
angle). y
and z in the figure below;
z
Interior opposite
c angles

d a b e 1350 x
B A
External angle External angle Solution
From the figure above, the theorem is Name the vertices and the angles as shown
summarized as: d = c + b and e = a + c below;

Worked Examples ∆ CDA is an isosceles triangle

Find the angles marked with letters: Therefore n = z (Base <s of a triangle)
1. t + 700 + 500 = 1800 n + n + 1350 = 1800
t = 1800 – 700 – 500 500 2n + 1350 = 1800
t = 600 2n = 1800 – 1350
700 t 2n = 450

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n= C
Some Solved Past Questions
1. In the diagram below, |AE| = |ED| = |DC| = |CE|
n = 22.50 A
y A B
z
⇒z = 22.50
x
E C
m 1350 n x
B D A
0 D
n + x = 180 ( < s on a straight line)
But n = 22.50 Calculate the size of the angle marked x.
22.50 + x = 1800
x = 1800 – 22.50 = 157.50 Solution
Name the angles as shown below
m + 1350 = 1800 y + y + y = 1800 (Equilateral triangle)
m = 1800 – 1350 = 450 3y = 180 o
A m B
y= b
a
∆ CBD is an isosceles triangle
y = 60o e a x
⇒m = y ( Base < s of a triangle) E y y C
y = 450 y

D
7. In the figure below, AC is protruded to D. Find
the value of x and < ACB y + e = 180o (Angles on a straight line)
o o o
B 60 + e = 180
eo = 180o – 60o = 120o
2x+ 70

a + a + eo = 180o (< in an isosceles ∆)

0
x + 170 9x 2a + 120 = 180o (But e =120o )
A C
2a = 180o – 120o
Solution 2a = 60o
x + 170 + 2x + 70 = 9x0 (Exterior < theorem ) a = 30o
170 + 70 = 9x0 – x0 – 2x0
240 = 6x0 y + a + x = 180o (Angles on a straight line)
But y = 60o and a = 30o
x0 = = 40
60o + 30o + x = 180
x = 180o – 60o – 30o = 90o
< ACB + 9x0 = 1800 ( < s on a straight line)
< ACB + 9(40) = 1800 (But x = 40) 2. In the diagram below, find the value of angles
< ACB + 360 = 1800 a, b and c.
< ACB = 1800 – 360 =1440

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 4. Find the value of p
in the diagram below; Po
75O
cO

bc 30o 70o
aO 56O 19O
Solution
Solution p + 30o = 70o(external angle theorem)
bo + 560 + 560 = 180o p = 70o – 30o = 40o
(Base < s of an isosceles triangle )
b = 180o – 56o – 56o = 68o Exercises 7.7
ao + 56o = 180o(< on a straight line)
A. Find the values of the angles marked with
a = 180o – 56o = 124o
variables in the figures below:
c + 56o + 19o = 180o(<s on a straight line)
1. 2.
c = 180o – 56o – 19o = 1050 540
2x
3. Find the value of the angle marked x in the
diagram below:
A 4x 5y
10y p 720
o
55
3.
D C 3x + 600

xo
E 2x
B
(x + 21)0 (x + 17)0
4.
Solution
(3x – 18)0
Name the angles as shown below;
a + a + 55o = 1800 A (x + 7)0
o 0 0
2a + 55 = 180 o (x – 9)0 y
0 0
55
2a = 180 – 55
D a a C 5. C
2a = 125o b
3y0
2a = 125o x
0 o
a = 62.5 B E
7y0 5y0 x0
D
B A
a + b = 180o (angles on a straight line)
62.50 + b = 180o 2x
b = 180o – 62.50 = 117.50 6.

b and x are alternate angles x + 750

⇒b = x. But b = 117.50.
Therefore x = 117.50 1500

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7. Challenge Problems
g
E e
In the diagrams below, determine the value of x
c e 1. C 2.
D
b x
150 a d E
x
A B C D 100
100
B. 1. Find the value of the reflex angle marked y 200
in the diagram below: 700 600
A B
A B
40o
C
3. In the figure below, AB is parallel to DC.
T y A

40o x
D F
E
B
C
2. In the figure below, find the values of angles x 55 0
g
and y
x 600 400
450
D
F
y x y
The Right – Angled Triangle
Any triangle which has one right – angle is called
3. In the figure below ABC is an isosceles a right-angled triangle.This means that in a right-
triangle. Triangle ABD has a right angle at B. < angled triangle, an angle is 900
ADB = 400 , < CBE = 500. Work out for the sizes
of angles x and y D
The Pythagoras Theorem
40
0 (Hypotenuse Rule)
C It states that the square on the hypotenuse of a
E right- angled triangle is equal to the sum of the
50
0 lengths of the squares on the other two sides
x
y
A B This is illustrated in the diagram below:

4. Determine
650 a
the values of e b
angles d, e, f
f a2 = b2 + c2
1320
d
c
This is illustrated in the diagram below;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 178
 4.In the diagram below, Y
what is the length
of XZ?
13cm 5cm
3 5
4 X Z
Solution
From Pythagoras theorem,
132 = /XZ/2 + 52

By Pythagoras theorem, 52 = 32 + 42 /XZ/2 =132 − 52

/XZ/2 = 169 – 25
Worked Examples /XZ/2 = 144
Find the length of the side marked with letters in /XZ/ = √ = 12cm
the triangle below: X
1. 5. XYZ is right angled y
triangle, with length
15cm z
Solution of sides shown below.
2 2 2
a 15 20 Y x Z
a2 225 400 20cm Express z in terms of x and y
a2 = 625,
a=√ = 25cm. Solution
From Pythagoras
2. 522 = 202 2 theorem,
k2 = 522 – 202 20cm 52cm y2 = x2 + z2
k2 = 2704 – 400 z2 = y2 – x2
k2 = 2304 z=√
√
6. In the quadrilateral ABCD below, /AB/ = 3cm.
3. /BC/ = 4cm, /CD/ = 12cm, angle ABC = 900 and
p 37cm
angle ACD = 900. Calculate:
i. the perimeter of ABCD
Solution 21cm ii. the area of ABCD D
12cm
372 = p2 + 212
p2 = 372 – 212 Solution C
P2 = 1369 – 441 i. By Pythagoras theorem,
4cm
P2 = 928 cm /AC/2 = /AB/2 + /BC/2
P= √ cm /AC/ = √ 32 + 42
A 3cm B
P = 30.5cm (3 s. f) /AC/ = √ = 5cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 179
Again by Pythagoras theorem,
/AD/ 2 = /AC/ 2 + /CD/ 2 3. 4. d
But /AC/ = 5cm and /CD/ = 12cm 15cm
45cm
/AD/ 2= 52 + 122
/AD/ 2 = 25 + 144 39cm 27cm
/AD/ 2 = 169
/AD/ = √ = 13cm 5. A

Perimeter of ABCD 17 25
= /AB/ + /BC/ + /CD/ + /DA/
= 3cm + 4cm + 12cm + 13cm
8 D x
= 32cm

ii. Area of ABCD; Pythagorean Triples

= area of ∆ ABC + area of ∆ ACD The Pythagorean triples consist of a set of 3
positive integers that obey the Pythagoras
Area of ∆ ABC = bh theorem. This means that the sum of the squares
of two of them equals the square of the other
= × /AB/ × /BC/
number. For e.g. {5, 12, 13} are Pythagorean
= ×3×4 triples because 52 + 122 = 132. Similarly, {3, 4,
= 6cm2 5} are Pythagorean triples because 32 + 42 = .

Area of triangle ACD; To investigate whether a given set of numbers are

= × /CD/ × /AC/ Pythagorean triples, equate the sum of the squares
of the two smaller integers to the square of the
= × 12 × 5 bigger integer. In other words, equate the sum of
= 30cm2 the squares of the first two integers to the square
of the third integer. For example, {3, 4, 5} is
Area of ABCD; investigated as; 32 + 42 = 52 and not 32 = 42 + 52
= 6cm2 + 30cm2
= 36cm2 The following sets of integers are Pythagorean
triples.
Exercises 7.8 1. {3, 4, 5}, since 52 = 32 + 42
A. Find the unknown lengths: 2. {5, 12, 13}, since 132 = 122 + 52
1. 2. 3. {8, 15, 17}, since 172 = 152 + 82
4. {7, 24, 25}, since 252 = 72 + 242
26cm 5. {6, 8, 10}, since 102 = 82 + 102
a 20cm 10cm
m
b For any two positive integers, a and b, where
16cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 180
a > b, the three sides of the right angled – triangle Therefore the triplets are 5, 12 and 13
can be expressed in terms of a and b to generate
Pythagorean triplets. Exercises 7.9
A. Identify the set of numbers that form the
2ab a2 + b2 sides of a right angled triangle.
1. {9, 12, 15} 2. {8, 15, 17} 3. {7, 24, 25}
a2 - b2 4. {12, 15, 19} 5. {5, 8, 17} 6. {7, 8, 15}
B. 1. A triangle has sides AB, BC and CA
Worked Examples measuring 14, 48 and 50 units.
1. Show that the numbers 9, 12 and 15 represents i. Prove that the triangle is right – angled, and
the sides of a right-angled triangle. calculate its area. Ans: A = 336 sq. units
ii. Calculate the length of the altitude from B to
Solution CA
Let a = 15cm, b = 9cm and c =12 cm
By Pythagoras theorem,
2
2. The points A, B and C have coordinates (-3, 1),
= b2 + c2 (2, -1) and (4, 2) respectively. Show that the
152= 92+ 122 angle ABC is a right angle. Find the coordinates
225 = 81 + 144 of D if :
225 = 225 i. ABCD is a rectangle,
The triangle is a right-angled triangle. ii. ABCD is a parallelogram,
iii. D is the centre of triangle ABC.
2. A triangle has sides 3, 4 and 5 units. Show that
it is a right-angled triangle. 3. Prove that the triangle with sides PQ = 6cm,
PR = 2.5cm and QR = 6.5cn is a right - angled
Solution
Let a = 5, b = 4 and c = 3, 4. Mr. Green tells you that a right angled triangle
By Pythagoras theorem, has a hypotenuse of 13 and a leg of 5. If he asks
a2 = b2 + c2 you to find the other leg of the triangle without
52 = 42 + 32
25 = 16 + 9 using a paper and pencil, what will be your
25 = 25 answer?
⇒The triangle is a right-angled triangle
Generating the Pythagoras Triples
3.Find the Pythagorean triplets, if x = 3 and y = 2. If only one side of a right – angled triangle is
known, then Pythagorean triples can be
Solution generated using the sides:
x2 = 32 = 9 2xy x2 + y2 { a, [ (a2 – 1) ], [ (a2 + 1) ]}
2 2
y =2 =4
x2 - y2 = 9 – 4 = 5 x2 - y2 Pythagorean triples are generated by the formula:
2xy = 2(3) (2) = 12
[ (a2 + 1)]2 = [ (a2 – 1)]2 + a,
x2 + y2 = 9 + 4 = 13

Baffour – Ba Series, Core Maths for Schools and Colleges Page 181
where a is an odd integer greater than one. the length of the other two sides and the area of
the rectangle if the sides are all integers.
Proof
(x + 1)2 – (x – 1)2 = x2 + 2x + 1 – (x2 – 2x + 1) Solution
(x + 1)2 – (x – 1)2 = x2 + 2x + 1 – x2 + 2x - 1)
S R
(x + 1)2 – (x – 1)2 = 4x
(x + 1)2 = (x – 1)2 + 4x 15cm
2 2
(x +1) = (x – 1) + (Divide through by 4)

[ (x +1)]2 = [ ( x - 1)]2 + x P P
Substitute x = a2 to obtain: Applying Pythagorean triples for odd numbers:
2 2 2 2 2
[ (a +1)] = [ (a – 1)] + a ………(1) [ (a2 +1)]2 = [ (a2 – 1)]2 + a2
When a = 15,
This identity can be used to generate Pythagorean [ (152 +1)]2 = [ (152 – 1)]2 + 152
triples a, an odd integer greater than 1. For
example, when a = 3, [ (226)]2 = [ (224)]2 + 152
[ (32 +1)]2 = [ (32 – 1)]2 + a2 1132 = 1122 + 152
⇒length = 112cm, breadth = 15cm and diagonal
[ (10)]2 = [ (8)]2 + a2
= 113cm
52 = 42 + 32
Area = Length × Breadth
Worked Examples = 112 cm × 15 cm
1. The width of a rectangle is 7cm. What are the = 1680 cm2
sizes of the length and diagonals if they are whole
numbers in centimeters? Exercises 7.10
Generate Pythagoras triples using the
Solution following pair of positive integers:
Applying Pythagorean triples for odd numbers: 1. 3 and 2 2. 5 and 2 3. 4 and 1
[ (a2 +1)]2 = [ (a2 – 1)]2 + a2
When a = 7, Challenge Problems
2 2 2
[ (7 +1)] = [ (7 – 1)] + 15 2 2 If three sides of a triangle are x2 – y2, 2xy and
x2 + y2, show that the triangle is a right – angled
[ (50)]2= [ (48)]2 + 152 triangle
252 = 242 + 72
Therefore, the sizes of the length and diagonal are Application of Pythagoras Theorem
24 and 25 respectively Some practical problems as well as real life
situations are solved by application of Pythagoras
2. A Student was asked to prepare a rectangular theorem once a right – angled triangle is formed
model in the form of two joined right angled or produced. For e.g. when a ladder on a vertical
triangles with its shortest length being 15cm. Find wall. The lengths of the involving sides are
Baffour – Ba Series, Core Maths for Schools and Colleges Page 182
calculated by applying the theory of Pythagoras. Solution
This is just a tip of the iceberg. Some other Let the height of the wall be x
applications are discussed below:
By Pythagoras theorem,
172 82 x2
A. Placing a Ladder against a Wall 2
289
When a ladder is placed against a vertical wall, 2
289 17
the foot of the ladder and the ground floor meet at 2 x
225 ,
a right angle. Knowing either how high the ladder
is or the height of the wall or the distance x √ m
between the legs, an unknown side can be x = 15m
calculated using the Pythagoras theorem.
Exercises 7.11
Worked Examples 1. A ladder leans against a vertical wall of height
1. A ladder leans against a vertical wall of length 12m. If the foot of the ladder is 5m away from the
5m. The distance between the foot of the ladder wall, calculate the length of the ladder.
and the wall is 7m. Find the length of the ladder.
2. A ladder leans against a vertical wall of height
16m. If the foot of the ladder is 8m away from the
Solution
wall, calculate the length of the ladder.
By Pythagoras theorem,
2
52 + 72 5
2 m
3. A ladder is 8m long. The foot of the ladder is
2
74, 7 3m away from the base of the wall. How far up
l = √ m = 8.60m ( 3 s. f.) m the wall is the top of the ladder?

2. A ladder leans against a vertical wall of length 4. If a ladder to a slide is 8 feet and the ground
9cm. If the length of the ladder is 12cm, find the from the ladder to the slide is 4 feet, then how far
distance between the foot of the ladder and that of down then how far down will the child slide?
the wall.
5. A 16 ft ladder leans against a wall with its base
Solution 4 ft from the wall. How far off the floor is the top
By Pythagoras theorem, of the ladder?
122 92 f 2
2
Challenge Problems
144 1. A ladder 16m long is placed so that its foot is
144 2 12cm
9cm 3m from a building. How much further must the
2
63 , foot of the ladder be moved from the building in
f √ cm = 7. 94 (3 s. f) order to lower the top of the ladder by 2m?

3. A ladder which is 17m high is placed against a 2. i. The greatest length of an extending ladder is
vertical wall. If the distance separating the foot of 10m. Calculate the greatest distance up a vertical
the ladder and the wall is 8m, what is the height wall the ladder can reach when the foot of the
of the wall? ladder is 6m from the foot of the wall.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 183
ii. When the ladder is adjusted to 8.5m, it reaches Worked Examples
a point 7.5 m above the ground. Calculate the 1. A rectangle has length of 8cm and a breadth of
distance of the foot of the ladder 6cm. How long is its diagonal?
from the foot of the wall.
Solution
Length of the Diagonals of a Plane figure
From Pythagoras theorem,
I. Make a sketch of the figure
d2 = 82 + 62
II. Represent the unknown side (diagonal) by any d2 = 64 + 36
preferred variable d2 = 100 d
6cm
III. Apply Pythagoras theorem to find the length d=√
of the diagonal d = 10cm 8cm

B. The Diagonals of a Rectangle and a Square 2. Find the length of the diagonal of a rectangle
The sides of a rectangle meet at a right angle. A which is 8cm long and 5cm wide
diagonal drawn from one corner to another
divides the rectangle into two equal triangles. Solution
Given the dimensions of the rectangle, the length Let x be the length of the unknown side
of the diagonal can be calculated by application From Pythagoras theorem,
of Pythagoras theorem. Similarly, given the x2 = 52 + 82
length of the diagonal and how long or how wide d2 = 25 + 64 x cm
5cm
the rectangle is, the other or unknown side can be d2 = 89
calculated d = √ = 9.4cm 8cm

From Pythagoras 3. The diagonal of a rectangle is 20cm long. If the

theorem, d length of the rectangle is 17cm, how long is the
d2 = l2 + a2 a
breadth?
l2 = d2 – a2
a2 = d2 – l2 l Solution
Let x represent the breadth of the rectangle
For a square of side a, as shown below; By Pythagoras theorem,
Using Pythagoras theorem: 202 = 172 + x2
I. The length of the x2 = 202 – 172 20cm
d xcm
diagonal, d is a x2 = 111
calculated as: a x=√ = 10.5cm 17cm
d2 = a2 + a2 = 2a2
d=√ ( 4. A square has a side 12cm. What is the length
of the diagonal of the square?
II. The length of a side, a, is calculated as:
( ) Solution
a=√ Let the diagonal of the square be x

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x2 = 122 + 122 ii. the area of the floor. Ans : 60cm2
x2 = 288 12
x=√ x
C. The Diagonals of a Rhombus
12 12 The diagonals of a rhombus bisect each other at
x = 16.97 cm
900. Given the length of the diagonals, the side of
12
Exercises 7.12 the rhombus can be calculated by applying the
1. The length and breadth of a rectangle is 20 cm Pythagoras theorem as shown below;
and 11cm respectively. How long is the diagonal By Pythagoras theorem,
of the rectangle? c2 = a2 + b2 C C
a
a2 = c2 – b2 b b
2. The diagonal of a rectangle is 61cm long. If the
breadth is 11cm, find its length. b2 = c2 – a2
C a C
Perimeter of the rhombus,
3. The diagonal of a rectangle is 12 inches and its P = c + c + c + c = 4c
width is 6 inches. Find its length.
Worked Examples
1. The length of the diagonals of a rhombus are
4. A rectangle which is 8cm by 6cm is divided
10cm and 24cm. Find:
into two right – angled triangles. What are the
i. the side of the rhombus
lengths of the bases and the altitudes of the two
i. the perimeter of the rhombus
different isosceles triangles that can be formed
from these triangles? Solution
Let the shorter diagonal be = 10cm and the
5. Find the area and perimeter of a square whose longer diagonal be CD = 24cm C
diagonal is 12cm long.
12cm
6. The length of a rectangle is 1 cm longer than = × 10cm = 5cm
5cm O 5cm
its width. If the diagonal of the rectangle is 4cm, A B
what are the dimensions of the rectangle? 12cm
= × 24cm = 12cm
Challenge Problems By Pythagoras theorem,
1. In ∆ABC, AB = AC = 12cm and BC = 8cm. = + D
Express the length of the altitude from A to BC as
a surd in its simplest form. = +
2. An equilateral triangle has each side 2a meters =√ = 13cm
long. Find the length of an altitude of the triangle,
and hence find the area of the triangle in terms of a ii. Perimeter of rhombus
P= + + +
3. The sides of a rectangular floor are x m and = = = = 13cm
(x + 7) m. The diagonal is (x + 8) m. Calculate: P = 13 + 13 + 13 + 13 = 52cm
i. the value of x Ans = 5

Baffour – Ba Series, Core Maths for Schools and Colleges Page 185
Alternatively, In general, if the triangle is isosceles, then drop a
P = 4 × 13cm = 52cm perpendicular before you apply the Pythagoras
theorem.
Exercises 7.13
1. The length of the diagonals of a rhombus is Worked Examples
48cm and 14cm. Find the perimeter of the 1. PQR is an equilateral triangle of side 14cm.
rhombus. Find the length of the perpendicular from P to QR
and hence, find the area of the triangle.
2. The sides of a rhombus are 8cm long. One of
its diagonal is 12cm. How long is the other Solution
diagonal? Let the perpendicular from P to QR be h
By Pythagoras theorem,
3. Find the length of a side of a rhombus whose 142 = 72 + h2 P
diagonals are 15cm and 22cm. h2 = 142 – 72
h2 = 147
D. Triangles 14cm
h=√ 14cm
In an equilateral triangle, the lengths of the three h
h = 12.12cm
sides are equal. The diagonal bisects the base at a 7 7
right angle. Given the length of a side, the Q 14cm R
altitude (height) of an equilateral triangle can be
calculated by the use of Pythagoras theorem.
A = bh, but b = 14cm and h = 12cm

By Pythagoras theorem, A= × 14 × 12 = 84cm2

a2 = h2 + b2
a a 2. Calculate the length QR in the triangle PQR
h
P
b b
a
13cm 12cm 15cm
Similarly, in an isosceles triangle, the diagonal
touches the base and divides it into at a right Q N R
angle. Given the height and the two equal sides,
Solution
the third side can be found by applying
/QN/2 = 132 – 122
Pythagoras theorem. This is shown in the diagram
169 – 144 = 25
below;
/QN/ = √ = 5
By Pythagoras theorem, /NR/2 = 152 – 122
a a /NR/ = √ = 9cm
h2 = a2 + b2 h
b2 = h2 – a2
b b
But /QR/ = /QN/ + /NR/
2b = 5 + 9 = 14cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 186
3. A triangle has sides 17cm, 17cm, 16cm. 4. Find the perimeter of the triangle below
Calculate the area of the triangle correct to three significant figures?
S
Solution
By Pythagoras theorem, 14cm
172 = 82 + h2
17cm 17cm
h2 = 172 – 82 h
R 9cm T
2
h = 289 – 64
h=√ = 15cm 8cm 5. Find the altitude of each triangle;
16cm

A = bh, a. b.
5 5 9 9
Substitute b = 16cm and h = 25cm
A = × 16 × 15 = 120cm2 10
6

D. Involving Quadratic Equations

4. The triangle ABC has AB = AC = 7 cm, and BC
Given the value of a side of a right – angled
= 8cm. Find the area of the triangle, giving your
triangle, the next side as a variable and the third
answer to 3 significant figures.
side as an increment or decrement in the variable,
A the actual dimensions (values of the variable) can
Solution
be calculated.
By Pythagoras theorem,
7 I. Apply Pythagoras theorem,
72 = 42 + /AD/2 7
2 2 2 II. Expand the involving brackets,
/AD/ = 7 – 4
III. Equate the equation to zero,
/AD/ = √ B 4 D 4 C IV. Solve the quadratic equation by using the
/AD/ = √ = 5.7446 quadratic method, ignoring negative answers,
V. Substitute the value of the variable to obtained
Area of a triangle = × BC × AD the dimensions of the right – angled triangle.
= × 8 × 5.7 = 23.0cm3
Worked Examples
The length of one leg of a right – angled triangle
Exercises 7.14 is 2cm more than the other. If the length of the
1. The length of the side of an equilateral triangle hypotenuse is 6cm, what are the lengths of the
is 30cm. Find the height of the triangle. two legs?
2. ∆PQR is a triangle with PQ = QR and PR =
6m. If the height of the triangle is 7m, find PQ Solution
3. ∆ LMN has LM = LN, MN = 12cm and area is Draw a sketch of the problem, labeling the known
300cm2. Find the height of the triangle and hence and unknown lengths. If one leg is represented by
find LM. x, the other is represented by x + 2

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Use Pythagoras theorem When x = 8,
to form the equation, The 1st side is 8 – 2 = 6units,
x2 + (x + 2 )2 = 62 The second side is 8units,
x2 + x2 + 4x + 4 – 36 = 0 6 x The third side is : 8 + 2 = 10units
2x2 + 4x – 32 = 0
Exercises 7.15
x+ 2 1. The hypotenuse of a right – angled triangle is
a = 2, b = 4 and c = -32 ( * (2x + 3) cm long, and the other two sides have
lengths x cm and (x + 7) cm. Find x, and calculate
√ √ ( (
x= = = –1 + √ the area of the triangle.
(
⇒x = –1 + √ = 3.123
2. The sides of a right – angled triangle in
ascending order of magnitude are 8cm, (x – 2) cm
If x = 3.123, then x + 2 = 3.123 + 2 = 5.123
and x cm. Find x.
The lengths of the legs are approximately
3.123cm and 5.123cm.
3. The lengths in cm of the sides of a right –
angled triangle are x, (x + 2) and (x + 1), where x
2. The length of the three sides of a right –
> 0, find x.
angled triangle form a set of consecutive even
integers when arranged from least to greatest. If
4. The length of one leg of a right triangle is 1cm
the second largest side has a length of x, form an
more than the other. If the length of the
equation and hence, solve for the three sides.
hypotenuse is 3cm, what are the lengths of the
legs?
Solution
The 3 sides of the right – angled triangle when
5. The length of one leg of a right triangle is 3cm
arranged from the least to the greatest are:
more than the other. If the length of the
(x – 2), x and (x + 2)
hypotenuse is 8m, what are the lengths of the
(x – 2)2 + x2 = (x + 2)2
legs?
(x – 2) (x – 2) + x2 = (x + 2) (x + 2)
x(x – 2) – 2(x – 2) + x2 = x (x + 2) + 2(x + 2)
2 6. Find the lengths of the sides of the right –
x – 2x – 2x + 4 + x2 = x2 + 2x + 2x + 4
2
angled triangle
x – 4x + 4 + x2 = x2 + 4x + 4
2x2 – x2 – 4x – 4x + 4 – 4 = 0 (x + 2)cm (x + 4)cm
x2 (x + 2)
x2 – 8x = 0
x(x – 8) = 0
x cm
⇒x = 0 or x = 8 (x – 2) Solving Other Applications
Worked Examples
When x = 0, the first side is negative. That is: 1. How long must a guywire be to reach from the
0 – 2 = – 2 so ignore the answer x = 0 top of a 30ft pole to a point on the ground 20ft
from the base of the pole?

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Solution From ∆ ABD
Let the length of the guywire be x, ( + 402 = 502
x2 = 302 + 202 ( = 502 – 402
x2 = 900 + 400 ( 900
x2 = 1,300 x 30
h+x=√
x=√ = 36 ft h + x = 30,
20

2. In a right – angled triangle, the hypotenuse is But x = 20.6

39cm and the ratio of the other two sides is h + 20.6 = 30
12: 5. Find the sides. h = 30 – 20.6 = 9.4m
The height of the flagstaff is 9.4m
Solution
Let the sides be (12x) cm, (5x)cm and 39 cm 5. Two points A and C are on the same level
By Pythagoras theorem, ground as the foot of a pole B. The distance
392 = (12x)2 + (5x2) 39cm between A and C is 70m and A and C are on the
12x
1521 = 144 x2 + 25x2 opposite sides of the vertical pole. The distances
1521 = 169x2 from the top of the pole, D, to A and C are 45m
x2 = 9 5x
and 59m respectively. Find :
x = √ = 3cm i. the distance between the foot of the pole, B, and
the point A,
⇒12x = 12(3) = 36cm ii. the height, BD of the pole.
5x = 5(3) = 15cm
The sides of the triangle are 15cm, 36cm, 39cm Solution D
i. From ∆ ABD,
4. A man standing 40m away from a tower x2 + h2 = 45…….(1)
notices that the distances from the top and bottom 85m 59m
h
of a flagstaff on top of the tower are 50m and
45m respectively. Find the height of the flagstaff.
A B C
From ∆ BCD, x 70m - x
2 2
Solution ( + h = 59 ……….(2)
Let h and x be the heights of the flagstaff and
that of the tower respectively as shown below; eqn (2) – eqn (1);
From ∆ ABC, D ( – x2 = 592 – 452
2 2 2
40 + x = 45 h 702 – 140x + x2 – x2 = 1456
x2 = 452 – 402 -140x = 1456 – 702
x2 = 425 C
50m 45m -140x = -3444
x=√ x x = 24.6 m
x = 20.6m The distance between B and A = 24.6m
A 40m B

Baffour – Ba Series, Core Maths for Schools and Colleges Page 189
ii. x2 + h2 = 452 h2 = 532 – (
⇒( + h2 = 452 (But x = 24.6) h=√
2 2
h = 45 – ( h = 39.6m
h=√ The height, BD of the pole is 37.7m
h = 37.7m
The height, BD of the pole is 37.7m 7. Tom and Jerry meet at a corner. Tom turns 900
left and walks 9 paces; Jerry continues straight
6. Two points A and C, are on the same level and walks 12 paces. Find the distance between
ground as the foot of the pole, B. The distance the two of them.
between A and C is 40m and A and C are on the
same sides of the vertical pole. The distances Solution
from the top of the pole, D to A are 53m and 85m Let the distance between Tom and Jerry be x
respectively. Find correct to one decimal place; By Pythagoras theorem,
i. the distance between the foot of the pole, B and
the point A, x2 = 92 + 122
T
ii. the height, BD, of the pole. x2 = 81 + 144
x2 = 225 x
9
Solution x= √ p
Let /BD/ = h and /AB/ = x
x=√ a
From ∆ ABD, c
12 J
x = 15 paces
x2 + h2 = 532……….(1) e
pace
The distance between Tom
s and Jerry is 15 paces
85m 53m h s
Exercises 7.16
1. How long must a guywire be to run from the
C A B
x top of a 20ft to point on the ground 8ft from the
From ∆ BCD,
base of the pole?
( + h2 = 852 ……….(2)
2. If the hypotenus of a right – angled isosceles
eqn (2) – eqn (1) triangle is 6m, what is the length of each of the
( – x2 = 852 – 532 other side‟s?
402 + 80x + x2 – x2 = 4,416
80x = 4,416 – 402 3. Two foggers run 8 miles north and then 5 miles
80x = 2,816 west. What is the shortest distance, to the nearest
x = 35.2 m tenth of a mile they must travel to return to their
The distance between, B and A = 35.2m starting point?

ii. From ∆ ABD 4. A kite is flying so that it is 55 feet high, and its
x2 + h2 = 532 above the point 75feet from the flyer. How long
⇒( + h2 = 532 (But x = 35.2) is the string of the flyer?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 190
5. A right – angled triangle with sides A, B and C II. Each pair of opposite sides is parallel.
with respective sides a, b, c has the following III. It has two diagonals that are not symmetrical.
measurements side A = 3cm, side B = 4cm. What IV. It fits its outline in two ways.
is the length of side C? IV. Area, A = L × B
A= ×
6. A man starting from point A, walks 5km due
east and then 4km due north to point B. Calculate Square: It is a plane figure ( special rectangle)
to one decimal place, the distance from A to B with four sides equal. In the figure below,
direct. = = =
ii. From B he walks 5km north and then 4km west
to C, calculate to one decimal place: Also:
D C
a. the distance from B to C,
b. the distance from A to C.

7. Two points A and C are on the same level

A B
ground as the foot of a pole B. The distance
between A and C is 16m and A and C are on the I. Each interior angle is a right angle.
same side of the vertical pole. The distances from II. Each pair of opposite sides is parallel.
the top of the pole, D, to A and C are 40m and III. It has two diagonals that are symmetrical.
50m respectively. Find : IV. It fits its outline in four ways.
i. the distance between the foot of the pole, B, and V. Area, A = L × L
the point A, A= ×
ii. the height, BD of the pole.
Parallelogram : It is any quadrilateral with both
A Quadrilateral pair of opposite sides equal and parallel. Also:
It is a plane figure bounded by four straight lines.
Examples of quadrilaterals are Rectangles,
Squares, Rhombuses, Kites, Trapeziums and h
Parallelograms. The interior angles of a
quadrilateral sum up to 3600 b
I. Opposite angles are equal.
Types and Properties of Quadrilaterals II. Diagonals bisect each other.
Rectangle : It is a quadrilateral with opposite pair III. It has half turn of symmetry.
of sides equal. Also: IV. It can be formed by two pairs of parallel lines.
D C V. Area = base × perpendicular height
A = bh

Trapezium: It is any quadrilateral with one pair

A B
of opposite sides parallel. Also;
I. Each interior angle is a right angle.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 191
 II. Sum up the angles and equate to 3600.
D C
III. Workout for the value of the variable.
h
Worked Examples
A B 1. In the figure below, 1100
I. It fits it outline in one way. find the value of the
1000 x
II. Area, A = ( h. angle marked y
0
115
C Solution
Rhombus: It is a parallelogram
x + 1100 + 1000 + 1150 = 3600
with four sides equal.
x = 3600 – 1100 – 1000 – 1150
Also:
x = 3600 – 3250 = 350
A B
2. Find the values of the letters in the diagram
below;

D 1200 600
I. It consist of two
congruent isosceles triangle, base to base.
2b a
II. It fits it outline in four ways.
III. Its diagonals are axes of symmetry.
III. Its diagonals bisect at right angles. Solution
For all parallelograms, opposite angles are equal.
IV. Area = × (Product of diagonals)
Therefore, from the diagram, a = 1200 and 2b =
A= ×( 600
2b = 600
b = 300
Kite: It consists of two isosceles triangles with
⇒a = 1200 and b = 300
equal bases. Also;

Exercises 7.17
I. It fits its outline in two ways.
II. One diagonal is an axis of symmetry. A. Find the values of the unknown angles
III. Each pair of adjacent sides is equal. 1. a a
IV. Its diagonals bisect at right angles.
2a 2a
Unknown Angle of a Quadrilateral 2. 680
The sum of interior angles of a quadrilateral is x
3600. 126 0
1060
To calculate the value of an unknown angle in a
3. 1150 x
quadrialateral;
I. Represent the unknown angle by any preferred
650 y
variable. a

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4. 5. A line that divides a figure into two equal halves
x
x is called a line of symmetry. It is also called a
870
0 line of fold.
110 y
940 700
0 Consider the figures below;
70
P Fig. I
A B
0
B. 1. The interior angles of a quadrilateral are 5x ,
3x0, 300 and 7x0. Find the value of x and the
value of each interior angle of the quadrilateral. C
D
M Fig. II
A B
2. A parallelogram has an interior angle of 950.
Find the values of the other three interior angles.
P M
C
3. Find the value of x and the value of each angle D

of a quadrilateral whose values are given as x,

In both figures, PM is the line of symmetry. This
x, x and x. implies that in fig. I, APMD = PBCM and in fig.
II, ABMP = PMDC
4. In the diagram below, ABCD is a kite.
Line of Symmetry of Some Figures:
b
Rectangle, Square, Equilateral triangle, kite…
a 1150 1. Square
L4 L1
L2
300
L3 L3

Calculate the value of the angles a and b. L2 L4

L1
Symmetrical Objects and Line of Symmetry Line of symmetry of a square = 4
Symmetrical objects are objects that can be
2. Rectangle
folded or divided into two equal halves such that L1
one-half fits exactly on the other half. For
example, it is possible to draw a line (mirror) L2 L2
through a rectangle and an isosceles triangle to
divide them into two halves such that one-half fits
L1
exactly on the other half. In these cases, the
Line of Symmetry of a rectangle = 2
rectangle and the isosceles triangle are said to be
Symmetrical.

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3. Equilateral triangle
L1
Number of sides Name
L3 3 Triangle
L2
4 Quadrilateral
5 Pentagon
L2 L3 6 Hexagon
L1
7 Heptagon
Line of symmetry of an equilateral triangle is 3. 8 Octagon
9 Nonagon
4. Kite. 10 Decagon

L1 Sum of Interior Angles of a Polygon

L1
Investigations
Reminder: The sum of interior angles of a
triangle is 1800.
Line of symmetry of a kite = 1
To investigate the sum of interior angles of a
Exercises 7.18 polygon given the number of sides :
1. Name any five figures that are symmetrical I. Draw the polygon.
2. Draw each of the following and indicate the II. Find the total number of triangles in the
lines of symmetry; Polygon by drawing diagonal lines, making
i. Rectangle ii. Rhombus sure that no line crosses another.
iii. Parallelogram iv. Scalene triangle III. Multiply the number of triangles in
v. Circle vi. Isosceles thepolygon by 1800 (the sum of interior angles of
a triangle) to obtain the sum of interior angles, ,
3. Copy and complete the table below: of the polygon.

Object Line of Symmetry 5 sides

4 sides
Square
Rectangle 1
Isosceles triangle 2
1 2 3
Rhombus
2 × 1800 = 3600
Parallelogram
Kite 3 × 1800 = 5400
Circle
Equilateral 7 sides
6 sides
triangle
. 1
Polygons 2 1 3 5
3
A polygon is any plane figure bounded by 2 4
4
three or more straight lines. The following
are names of polygons of three to ten sides. 4 × 1800 = 7200 5 × 1800 =9000

Baffour – Ba Series, Core Maths for Schools and Colleges Page 194
 8 sides 9 sides = 13 × 1800
= 2,3400
1 1
3
3 5
2 6 2 2. 12 sides
7
4 5 4 6
Solution
Sum of interior angles = ( n – 2) 1800
6 × 1800 = 10800 7 × 1800 = 12600
Substitute n = 12
⇒ Sum of interior angles,
Summary = ( 12 – 2) 1800
= 10 × 1800
Number of Number of Sum of Interior = 18000
sides Triangles Angles ( )
3 (3 – 2) = 1 1×1800 = 1800
3. 20 sides
4 (4− 2) = 2 2×1800 = 3600
5 (5 − 2) = 3 3×1800 = 5400
6 (6 – 2) = 4 4×1800 =7200 Solution
7 (7 − 2) = 5 5×1800 = 9400 Sum of interior angles = ( n – 2) 1800
8 (8 − 2) = 6 6×1800=10200 Substitute n = 20
9 (9 − 2) = 7 7×1800 =12600 ⇒ Sum of interior angles
10 (10 − 2) = 8 8×1800 =14400
n (n − 2) (n − 2)×1800 = ( 20 – 2) 1800
= 18 × 1800
From the table, it can be seen that the number = 3,2400
of triangles is two less than the number of sides
of the polygon. The conclusion is that; Exercises 7.19
For any polygon with n sides, the sum of A. Find the sum of interior angles of a polygon
interior angles, = ( 0 with the following sides:
(1) 14 (2) 22 (3) 42
Worked Examples (4) 35 (5) 17 (6) 50
Find the sum of interior angles of a polygon
with thefollowing sides; B. Find the value of the unknown angles;
1. 15 2. 12 3. 20 1.
a
0
60
Solution 0
110
1. 15 sides 0
70

Solution 2. b
Sum of interior angles = ( n – 2) 1800
Substitute n = 15
0 0
⇒ Sum of interior angles, 60 32

= ( 15 – 2) 1800

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3. Exercises 7.20
0
76 A. Find the number of sides of a polygon
125
0
c with the following sums of interior angles:
1. 10800 2. 14400 3. 28800
4. 36000 5. 30600 6. 48600

4. 0 Interior and Exterior Angles of a Polygon

Consider the polygon below;
0
0

0 a b

c d
Finding the Number of Sides of a Polygon
Given the sum of interior of a polygon as;
θ = (n – 2) × 1800, the number of sides, n, of the Angles a and b are pair of inerior angles and
polygon can be found by making n the subject of exterior angles respectively, likewise angles c
the formula as shown: and d.
In θ = (n – 2) × 1800 The sum of the pair of interior and exterior
θ = 1800n – 3600 angles are supplementary (equal to 1800) because
θ + 3600 = 1800 n they are angles formed on a straight line.
n= , where is the sum of interior <s Therefore, from the diagram above;
a + b = 1800 and c + d = 1800

Worked Examples It is conclusive that for all polygons;

1. The sum of interior angles of a polygon is 9000. 1. Interior angle exterior angle 1800
Find the number of sides of the polygon 2. Exterior angle 1800 – interior angle
3. Interior angle 1800 – Exterior angle
Solution
= 9000n = ? Worked Examples
1. What is the value of 1100
n= = = = 7 sides
the angles marked
2. Find the number of sides of a polygon with and in the figure
whose sum of interior angles is 1,9800. below?
0
55

Solution Solution
0 0
= 19800n = ? (
0
– 55 = 1250
0

n= = = = 13 sides 0 0
(
y = 180 – 110 = 700
0 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 196
Exercises 7.21 For a regular polygon, each interior angle
( (
A. List the pair of exterior and interior angles = = = = 1500
in the polygon below;
Method 2
f Because it is a regular polygon, all the 12 sides
e
gh are equal. n =12
Exterior angle, = 300
k
n p
m
Interior angle + exterior angle 1800
B. In the hexagon below, determine the size ⇒Interior angle 1800 – Exterior angle
of the interior and exterior angles. 1800 – 300
= 1500
1120
0
0 125
92 2. A regular polygon has 9 sides. Find:
y i. each interior angle of the polygon,
0
z
86
0 x 75 ii. the Sum of the interior angles of the polygon.

Solution
A Regular Polygon i. For a regular polygon each interior angle,
A regular polygon is a polygon that has allthe (
= , but n = 9
sides and angles equal. Examples are equilateral
(
triangle, rhombus and a square. = = = 1400

For all regular polygons of n sides, it implies that ii. Sum of interior angles,
there are n interior angles and n exterior angles. = (n – 2 ) × 1800 , but n = 9
= (9 – 2 ) × 1800 = 7 × 1800 = 1,2600
If the sum of all interior angles of a regular
polygon is = (n – 2 ) × 1800, then each interior Exercises 7.22
(
angle, = A. Find the value of each interior angle of a
( regular polygon with the following sides;
From = , the number of sides 1. 13 2. 14 3. 17 4. 19 5. 21
n= (for regular polygons)
B. 1. The diagram below shows part of a regular
10 – sided polygon. Workout the size of the angle
Worked Examples
marked x.
1. A regular polygon has 12 sides. What is the
size of each interior angle?
x

Solution
Method 1

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2. The diagram below shows a regular polygon a0 + 880 + 730 + 800 + 360 = 3600
and a regular hexagon. Calculate the size of the a0 = 3600 – 880 – 730 – 800 – 360
angle marked x. a0 = 830

2. The exterior angles of a pentagon are 720,

3x0, 520, 360, and 500 . Determine;
x
i. the value of angle x
Sum of Exterior Angles of a Polygon ii. the value of angle exterior angle 3x
Consider the diagrams below: iii. the value of the interior angle of 3x

a e
a Solution
b
d i. Sum of exterior < s of a polygon = 3600
e 720 + 3x0 + 520 + 360 + 500 = 3600
c b
Fig. I c 3x = 360 – 72 – 52 – 36 – 50
Fig. II
d 3x = 1500
x = 500
When all the exterior angles are cut and moved to
a fixed point, they form a complete ii. The value of angle 3x = 3 × 500 = 150
turn of 3600 as shown below;
iii. Let the interior angle of 3x be y
3x + y = 1800, but 3x = 1500
e a 1500 + y = 1800
b a + b + c + d + e = 3600 y = 1800 – 1500 = 300
d
c
3. The exterior angles of a polygon are 2x0, (x
In general, for every polygon (regular or – 20)0, x0, (3x + 10)0, (x + 15)0 and (2x + 5)0.
irregular), the sum of exterior angles is equal to Find the value of x
3600. ⇒: = 3600
Solution
Worked Examples Exterior angles of a polygon = 3600
1. In the figure below, determine the value of a ⇒2x0 + (x – 20)0 + x0 + (3x + 10)0 + (x + 15)0
a + (2x + 5)0 = 3600
880 2x0 + x0 – 200 + x0 + 3x0 + 100 + x0 + 150 + 2x0
36
0
+ 50 = 3600
0
73 2x0 + x0 + x0 + 3x0 + x0 + 2x0 – 200 + 100 + 150
Not drawn to
80
0 + 50 = 3600
scale
10x0 + 100 = 3600
Solution 10x0 = 3600 – 100
Exterior <s of a polygon sum up to 3600 10x0 = 350 = 350

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Exercises 7.23 i. Find its exterior angle.
A. Find the value of the angles marked x ii. How many sides has the polygon?
1.
Solution
i. Interior angle exterior angle 1800
0
30
50
0 Exterior angle 1800 – interior angle
= But interior angle 1720
Exterior angle 1800 – 1720 80
2.
ii. sum of exterior angles 3600

60
0
But the exterior angle is 80
⇒80

3. 0
Therefore n = = 45 sides
130

3. Each interior angle of a regular polygon is

0
1600, how many sides has it?
+ 40 120
0

0
70 Solution
Interior angle + Exterior angle 1800
Exterior angle 1800 – Interior angle
Finding the Number of Sides, n, of a Regular
But interior angle 1600
Polygon Given its Exterior Angle,
The exterior angle, , of a regular polygon is Exterior angle 1800 – 1600 200
related to the number of sides, n, by the formula: Exterior angle ,
. Therefore, knowing the value of each Substitute exterior angle = 200
exterior angle , the number of sides, n is ⇒200
calculated by the formula; n n= = 18 sides

Worked Examples 4. Find the size of the interior angle of a regular

1. If each exterior angle of a regular polygon is polygon with 5 sides.
300, find the number of sides of the polygon.
Solution
Solution Exterior angle ,

= 300 and n = ?
Substitute n = 5
n= = = 12 sides
Exterior angle 720,
Exterior angle Interior angle 1800
2. A regular polygon has an interior angle of 1720.
⇒Interior angle 1800 – exterior angle

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 1800 – 720 = 1080 = (14)(14) sin 600 = 84.87 cm2

5. Four interior angles of a hexagon are 1300,

Area of the polygon
1600, 1120 and 800. If the remaining angles are
6 × 84.87 = 509.2 cm2
equal, find the size of each of them.
Solved Past Question
Solution Three interior angles of a polygon are 1600
Let each of the remaining angles be x
each. If the other interior angles are 1200 each,
⇒sum of interior angles find the number of sides of the polygon.
= 1300 + 1600 + 1120 + 800 + x + x = 7200
2x = 7200 – 1300 – 1600 – 1120 – 800
Solution
2x = 2380 Let the interior angles of the polygon be n
x = 1190 3(1600) + (n – 3)1200 = (n – 2) 1800
Therefore each remaining angle is 1190 4800 + 1200n – 3600 = 1800 n – 3600
1200n – 1800n = -3600 + 3600 – 4800
6. A regular polygon is inscribed in a circle of -600 n = - 4800
radius 14cm. If each interior angle is 1200,
find: n= = 8 sides
i. the value of each exterior angle,
ii. the area of the polygon. Exercises 7.24
A. 1. The exterior angle of a regular polygon is
Solution 300, find the number of sides of the polygon.
Let the exterior angle be A
Interior angle + Exterior angle = 1800 2. The interior angle of a regular polygon is twice
its exterior angle. Find the number of sides of the
1200 + A = 1800 polygon.
A = 1800 – 1200
14 14 3. A regular polygon has 12 sides. Find:
A = 600 0
60 (i) its exterior and interior angles,
14 (ii) the sum of the interior angles.
ii. From the diagram,
Area of ∆

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8 EQUATIONS AND INEQUALITIES Baffour – Ba Series

Change of subject or Formula

A formula is an equation which expresses one Solution
symbol (called the subject of the formula) in
terms of other symbols. For e.g, v = u + at. ,

Change of subject is the process whereby a

variable is made to stand alone either at the left
side or the right side of an equal sign in a formula 2. Make h the subject of the relation k = mgh
or an equation. For instance, v is said to be the
subject of Solution
k = mgh
The subject of a formula is changed in order to ,
arrange the formula in a more suitable form for
the required calculation h

In dealing with change of subject, we categorize Involving Addition and Subtraction

it into the following types; If it involves addition (+) or subtraction (–), the
1. Involving multiplication sign changes to the opposite sign when
2. Involving addition and subtraction transferred to the other side of the equation. For
3. Involving division or fraction example, to make u the subject of v = u + at,
4. Involving bracket and factorization subtract at from both sides of the equation. That is :
5. Involving squared variables v – at = u + at – at
6. Involving squared root. ⇒ v – at = u or u = v – at

Involving Multiplication Worked Examples

When the equation involves multiplication, divide 1. Make c the subject of y = mx + c
both sides of the equation by the variable or
variables standing with or attached to the one to Solution
be made the subject.For example, to make h the y = mx + c
subject of the relation k = mgh, divide both sides y – mx = c
of the equation by mg to get h standing alone as c = y – mx
shown below;
, 2. Make p the subject of p – 2l = w.
⇒
Solution
p – 2l = w
Worked Examples p = w + 2l
1. Make l the subject of A = lb

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3. Make x the subject of the y = mx + c. 100 × I = 100
100
Solution
y = mx + c = ,
y – c = mx P=
=
x= 3. Make b the subject of the relation,
+c
4. Make t the subject of the relation v = u + at.
Solution
Solution
v = u + at +c
v − u = at -
= ,
=
t=
= (Reciprocate)

Involving Division (Fraction) b=

If it involves division or fraction, multiply
through by the denominator or multiply the 4. Given that = , find the expression
denominator by both sides of the equation and
apply the guidelines under addition, subtraction that is equal to y.
or multiplication, where applicable.
Solution
Worked Examples =
1. Make m the subject of k = mv2 =

Solution = (Reciprocate)
2
y=
2
2× k =2 × mv
2k = mv2 5. Make q the subject of the relation;
= , w=

= Solution
w=
2. Make P the subject of I =
q×w= ×q
Solution qw = n – q
I= qw + q = n (Group like terms)

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q(w +1) = n (Factorization) Solution
(
= R= (a + b) k
(
= 2 × R = 2 × (a + b) k Multiply both sides by 2

w +1= , 2R = (a + b) k.
(
q= = ,
( (
=k
(
Involving Brackets
When the variable to be made the subject is in
bracket, first remove the bracket by expansion, Involving Factorization
and continue the process until that variable stands When the variable being made the subject appears
alone either in front or at the back of the more than once, identify the variables as like
equation. terms and group them at one side of the equation.
Then factorize the group of like terms and follow
Worked Examples the due process.
1. Make a the subject of the relation;
P = 2 (a + b) Worked Examples
1. Make p the subject of mp – a = x – p
Solution
P = 2(a + b) Solution
P = 2a + 2b (Expansion) mp – a = x – p
P − 2b = 2a mp + p = x + a (Group like terms)
p (m +1) = x + a (Factorize like terms)
=
P=
a=
2. Make t the subject of S = ut + t
2. Maked the subject of S = n(a + d )
Solution
Solution
S = ut + t
S = n(a + d )
S = na + nd (Expansion) Multiply through by 2
2 × S = 2 × ut +2 × t
= , (Divide through by n) 2S = 2ut + at
2S = t (2u + a)
d=
(
= ,
3. If R = (a + b) k. Find k in terms of R, a and b =t

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3. In the formula: , express n in terms 1. s = u + t2 (a) 2. 2n + 5 = 7a. (n)
of s, d and a 3. v = u + at (t) 4. v2 = u2 + 2as (s)
5. m = 3(2d + 1) (d) 6. y = 2x + q (x)
Solution
, F. Make the letter in the bracket the subject of
each of the following formulas.
n× × n (Multiply both sides by n) 1. C = 2πr (r) 2. C = 2πrh (r)
nd = s – na 3. T = (R) 4. V = πr2h (h)
nd + na = s
n (d + a) = s Involving Squared Variable and Square Root (√ )
n= 1. If the variable to be made the subject of a
formula is squared, a square root is introduced at
Exercises 8.1 both sides of the equation at the last stage of the
Make the variable in bracket the subject work to make the squared variable a single
A. 1. C = 2πr (r) 2. 2
h( variable.

3. V = IR (R) 4. r2 ( 2. On the other hand, if the subject variable

contains a square root or is in a square root,
B. Express the letters in bracket in terms of square both sides of the equation to eliminate the
the others in each of the following; square root, leaving the variable alone. This is
1. A = 4 (r – y) (y) 3. A = h(a + b) (a) done at the last stage of the process.

2. V = 2a + h (a) 4. S = (1 – d) (d) Worked Examples

Make t the subject of the relation r = √

C. Make the variable in bracket the subject
1. + rh (r)
Solution
2. mt + n = mp + q (m)
3. ax – x + 1 – b = 0 (x) r=√
4. ax = bx + c (x)

= (√ )
D. Change the subject of the following
formulas to the variables in bracket; =
1. t = + . (q) 2. y = 1 + (x)
(9 – t ) = t – 3 (Cross product)
3.l2= (T) 4. P = (m)
9r2 – tr2 = t – 3
5. y = (x) 6. + (u) 9r2 + 3 = t + tr2
9r2 + 3 = t( 1 + r2)
E. Change the subject in each of the following
formula to the variable stated; =t

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Involving Powers or Exponent n, n > 2 Challenge problems
When the variable being made the subject has a
1. Make k the subject of √
power of n, where n > 2, introduce the nth root at
both sides of the equation at the last stage of the ⁄
2. If p = k, express v in terms of p and k.
work to remove the nth power. 3. Given x = y + √ make z the subject of
the formula.
Worked Example (
4. Change the subject of S = to
Change the subject of V = 4π to r
i. a ii. r
Solution 5. Make x the subject of each of the following a.
V = 4π px + qy + r = 0 b. x2 + y2 = r2 c. =
= ( Divide both sides by 4π ) √(
6. If r = , express a in terms of r.
√ =√ ,
Substitution
r= √ It is the act of putting values in place of variables
in a formula. To work out substitution
Exercises 8.2 successfully, make the variable without substitute
A. Make the letter in the bracket the subject of value the subject of the formula and substitute the
the following formula; given values into the new equation.
2
1. v= h (r) 4. s = 4πr2 (r)
Worked Examples
2
2. k = (v) 5. T 2 = (T) 1. If y = 2x + c , find y when x = 6 and c = 8

3. S = (r) 6. 2π fL = (f)
Solution
y = 2x + c
B. Make the letter in the bracket the subject of Substitute x = 6 and c = 8
each formula. ⇒y = 2(6) + 8
√
1. T = (l) 4. v = √ (u) y = 12 + 8 =20

2. S = u + t2 (t) 5. T = 2π√ (l) 2. If q = u t + at, find q when u = 20, t = 10 and

f = 15.
3. V = ( r) 6. T = 2π√ (g)
Solution
C. Solve for the variable in bracket: q = ut + at
1. n = p – 3 √ (t)
By substituting, u = 20, t = 10, and f = 15.
2. M – m √ = r (t)
3. √ =√ (n) q = 20 (10) + (15) (10)
4. √ =√ (t) q = 200 + × 150 = 275

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3. Given that m = 2 and n = , find the values of; Solution
f=
(i) m2 (n – 1) (ii) n2 −
(v +u) f = (v + u)
Solution vf + uf = vu (expansion)
i. Substitute m=2 and n = , in m (n − 1) 2 vu – vf = uf (re-grouping)
v (u – f) = uf (factorization)
⇒ (-2)2( −1) = 4 ( ) = 4( )= 4 ( ) = -1
v=

ii. Substitute m = -2 and n = , in n2 –

Put f = 20 and u = 5 in v=
⇒( ) – = – = + =
(
⇒v = = =

4. Given that C = (F – 32)

i. Find F when C = 40 6. In the relation, = + , if R1 = 1 and R2 =
ii. If C = − 40, find F. 3, find R.

Solution Solution
i. Make F the subject of C = (F−32) Make R the subject first:
by multiplying both sides by 9, = +
⇒9 × C= 9 × (F – 32) = +
9 C = 5 (F – 32)
9C = 5F – 160 (expansion) (Reciprocate)
9C + 160 = 5F
R
= ,
F=
Put R1 = 1 and R2 = 3 in R

When C = 40, ⇒R =
(
F= = 104
7. If = + , find the value R.
(
ii. When C = −40, F =
Solution
F= = = −40
= + = = ,
5. Given that f = find the value of v if
= (Reciprocate)
f = 20 and u = 5.
R= = 3.6

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Exercises 8.3 1. a left hand member (L.H.M),
A. Find the value of the letters in brackets 2. an equal sign,
1. , (p) , m =12, n = 11, t = - 4 3. a right hand member (R.H.M)”

2 3 + 5 = 8
2. = (g), T = 15, l = 45
L.H.M. Equal sign R.H.M
3. (f) v = 4 and u = 12
Variables
4. s = ut + a (a), s = 40, u = 3 and t = 2
A variable is any symbol that is used to represent
5. Given that T = 4 and t = 8, find R a number. Examples of variable are a, b, x, y…
and examples of equations with variables are 3 +
6. Given that = , find k2 when k1= 2 and m = 5 and 2x – 3 = 15

k = 1.
The value(s) of a variable that satisfies or makes
( a mathematical statement true is called the truth
7. From the formula, F =
set or the solution set.
i. find the numerical value of F, when m = 37.6, v
= 0.74, u = 0.59, x = 0.83, g = 32.2
To find the truth set of an equation,
ii. Express v in terms of the other letters
I. Ensure that the variable stands alone either in
front or at the back of the equal sign. Thus, any
B. 1. The perimeter of a regular hexagon is P =
positive or negative integer at the same side of
6x. Change the subject of tox, and hence find x
variable is transposed (transferred) to the opposite
when P = 96.
side of the equal sign to assume the opposite sign
II. Simplify both sides of the equation to obtain
2. The road resistance to a car is R = k . Change
the value of the variable
the subject to v, and find when k = 4 and R =
III. The results may be stated in three different
324.
ways:
a. the solution of the equation is x = a.
3. The rate at which water flows from a valve is v
b. a is the root/truth/solution set of the equation.
= 14√ . Change the subject to h, and find h when
c. {x : x = a} is the truth set of the equation.
v = 28.
General Rules
4. In nuclear physics the formula E = kT occurs. If
I. Remove all brackets by expansion.
E = 4.14 × and T = 300, find k.
II. If there are fractions, multiply the L.C.M. of
the denominators by each term of the equation.
Equations
III. If there is only one fraction, multiply through
An equation is a mathematical statement in which by the denominator of the fraction.
one expression is equal to another. It has two IV. Transpose / group like terms and simplify.
sides separated by an equal sign. It is therefore V. Solve by division, where necessary.
characterized by the following:

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Type 1: Involving Multiplication
Divide the coefficient of the variable by both 2. = -12
sides of the equation to obtain the value of the
Solution
variable.

Worked Examples
= -12 × 3 (Multiply both sides by 3)
1. Solve 5x = 20
4a = -36
Solution a = -9

Type 3: Involving all Procedure (s)

x= 4
Here the principle for addition or subtraction is
applied first, followed by that of division and /or
2. Find the solution set of 3m = 48
multiplication.
Solution
3m = 48 Worked examples
Solve the following equations;
,
1.
m = 16
Solution
Type 2: Involving Division/Fraction
2x + 5 = 15
When the equation involves a fraction:
(Transpose + 5 to become – 5)
I. Identify the denominator of the fraction
II. Multiply the denominator by each term of the
equation. ,
III. Divide each term of the equation by the
coefficient of the variable to obtain the value of
the variable. This step is also called solution by 2. m – 6 = 9
division.
Solution
Worked Examples m–6=9
Solve the following equations:
m=9+6
1. = 21
m = 15
Solution 2 × m = 15 × 2
= 21 3m = 30
4× = 21 × 4 ,
= 84 m = 10
m = 41

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Exercises 8.4 3. 28 – 15a = 84 – 8a
Find the truth set : 4. 4a + 7 + 2a = 11a – 3
1. 50 k + 10 = 160 4. 5(x + 2) = 50 5. 6m + 3 – 8m = -7 – 4m
2. 5y – 3 = 21 6. 2y + 3 = 16 – 2y + 3
3. 20m + 5 = 45 5. 3(3x – 1) – 4 = 11
Fractional Equations
Involving Two or More Variables It is a mathematical equation that involves two or
It is the type of equation which contains two more fractions.
variables of the same kind. For example;
3x – 2 = 2x + 5 Fractional equations are solved as follows:
I. Eliminate all fraction (s) by multiplying each
To solve equations involving two variables of term of the equation by the L.C.M. of the
the same kind: involving fraction(s);
I. Group the like terms or the variable factors at II. Group like terms, if any;
one side of the equation. III. Workout for the value of the involving
II. Simplify both sides of the equation to obtain variable .
the value of the variable.
Worked Examples
Worked Examples Solve the following equations;
Solve the following equations; (1)
1. 3x – 2 = 2x + 5
Solution
Solution
3x – 2 = 2x + 5
3x – 2x = 5 + 2 (Group like terms) Multiply both sides by L.C.M = 9
x=7

2. 8y – 3 = 2y + 15 =6

Solution (2)
8y – 3 = 2y + 15
8y – 2y = 15 + 3 (Group like terms)
6y = 18 Solution
(Divide both sides by 6)
( *

Exercises 8.5
7x = 84
Solve the following equations:
1. 18a = 3a – 75
2. 5x – 9 = 3x + 7

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 (
3. =-2

Solution 6. Solve ( )=
(

Solution
(Remove brackets by expansion)
( )=
12 × – 12 × = 12 × – 12 × ( *
+ =
12 × + 12 × = 12 ×
2x + 6 = 3x – 24
= =3 6 + 24 = 3x – 2x
x = 30
(
4. Solve Some Solved Past Questions
1. Solve −2
Solution
(
Solution
20 × − 20 × 2
( 5x + 4 (3) = 4(3x) – 40
5x + 12 = 12x – 40
5x – 12x = − 40 – 12
-7x = -52
x=

2. Solve the equation: =1

5.
( ( ( Solution
=1
Solution
( ( ( 12 × – = 12 × 1
8x − 4 − 3x + 6 = 12
8x − 3x = 12 + 4 − 6
5x = 10
x= =2

3. Find the truth set of (3y – 1) – (y + 2) =

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Solution 2.
(3y – 1) – (y + 2) =
3. (
(3y – 1) – (y + 2) =
2 (3y – 1) – 3 (y + 2) = 1 4. (
6 y − 2 − 3y – 6 = 1
5.
6y–3y=1+2+6
3y = 9 6. ( ( (
y=3

B. Solve the following equations:

4. Solve 5 (a – 5) – (2a + 6) = 4
1. ( = (

Solution 2. ( – ( =1
5 (a – 5) – (2a + 6) = 4
3. ( –2 ( = –14
2 × 5 (a – 5) – 2 × (2a + 6) = 2 × 4
10 (a – 5) − 2a – 6 = 8 4. ( – ( =
10a – 50 − 2a – 6 = 8
5. ( – ( )=1
10a − 2a = 8 + 50 + 6
8a = 64 6. x ( ) = 10( )
a=8
7 (5x – 4) = x + 1
5. Solve the equation;
8. (x − 3) = (x + 6)
(x + 3) – 2 (x – 5) = 4

Word Problem Involving Equations

Solution
In word problems involving equations, the
(x + 3) – 2 (x – 5) = 4 relationship between certain numbers are stated in
(x + 3) –3 × 2 (x – 5) = 3 × words. It is necessary to state the problem using
mathematical equations for easy solution.
(x + 3) – 3 × 2(x − 5) = 3 ×
x +3 – 6 ( x – 5 ) = 13 Hints on Forming Word Problems
x − 6x = 13 – 3 − 30 a. Let x represent any unknown number
-5x = -20 b. m more than the number = x + m
x=4
c. m less than the number = x – m
Exercises 8.6 d. A number is decreased by m = x – m
A. Solve each of the following: e. A number is increased by m = x + m
1. ( (
f. Half a number = x

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g. Twice a number = 2x
h. Twice the results of m more than a number = =6
2(x + m)
i. Twice the results of m less than a number
3. One – third of a number is decreased by five
= 2(x – m) and the result is sixteen. Write an algebraic
j. The use of the phrases “ the results is”, equation and hence find the number.
“add/sum up to” “is” implies the equal sign
Solution
– 5 = 16
Steps for Solving Word Problems
I. Carefully read the problem. = 16 + 5
II. Choose any preferred variable to represent the 21
number required number.
= 3 × 21
III. Write an equation for the problem.
IV. Solve the equation.
V. Check your results with the words of the
problem. 4. Three times a certain number increased by
seven is 40. Express this in an equation and find
Type 1 : Involving Numbers the number.
Worked Examples
Solution
1. When a certain number is decreased by five,
Let x represents the number
the result is twelve. What is the number?

Solution
I. Let x represents the number
II. x increased by 5 is written as x + 5 ,
III. the results is 12 , implies x + 5 =12 x = 11
IV. Solve x + 5 = 12 to get the value of x
⇒ x + 5 = 12 5. When five times a certain number is increased
x = 12 – 5 = 7 by three, the result is twice the same number
increased by twelve. Find the number.
2. Twice a certain number increased by eight is
twenty. What is the number? Solution
Let x represent the number
Solution
I. Twice a certain mumber is written as 2x
II. plus 8 is written as 2x + 8
III. the results is 20 implies 2x + 8 = 20 ,
x=3

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6. When x is divided by 13, the quotient is 9 and x= = 2.4
the remainder is 3. Find the value of x.
Therefore. the fare for a teacher is Ghȼ2.40
But the fare of a child is half the fare of a teacher
Solution
= (2.4) = 1.2
=9+
Therefore, the fare of a child is Ghȼ1.20
× 13 = 9 × 13 + × 13
x = 117 + 3 = 120 ii. Let the number of students be y
1.2y + 20(2.4) = 240
Some Solved Past Questions 1.2y + 48 = 240
1. When a certain number is subtracted from ten 1.2y = 240 – 48
and the result is multiplied by two, the final result 1.2y = 192
y = 160 students
is four. Find the number.
Therefore 160 students can go on the excursion
Solution
Type 2: Involving Ages
Let x be the number
A. Age Problems Involving a Single person
2 (10 – x) = 4
If the problem involves a single person, the
20 − 2x = 4
solution is similar to the real number problems.
- 2x = 4 – 20
Carefully read the question to determine the
- 2x = - 16
relationship between the numbers.
x=8

Worked Examples
2. A train fair for a school child is half the fare of
1. Five years ago, Esi‟s age was half the age she
a teacher. The total fare for 120 children and 15
will be in ten years. How old is she now?
teachers for an excursion is Ghȼ180.00
i. Find the fare of a child
Solution
ii. How many children will go on the excursion
Let Esi‟s present age = x
with 20 teachers for a total fare of Ghȼ240.00?
Esi‟s age five years ago = x – 5
Esi‟s age in ten years = x + 8
Solution
i. Let the fare for a teacher be x Half her age in ten years = (x + 10)
The fare for a school child = x Five years ago, Esi‟s age was half the age she
will be in ten years;
The fare for 15 teachers will be 15x and that of
120 children = (120) x = 60x
⇒ x – 5 = (x + 10)
2(x – 5) = x + 10 (Solving for x )
The total fare for 120 children and 15 teachers is
2x – 10 = x + 10
Ghȼ180
2x – x = 10 + 10
⇒ 60x + 15x = 180
x = 20
75 x = 180
Esi is 20 years now.

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A. Age Problems ⇒5 (2x) + 2 + 2x + 2 + x + 2 = 58
If the problem involves the ages of two or more 10x + 2 + 2x + 2 + x + 2 = 58
people, then using a table will be a good idea. 13x = 58 – 2 – 2 – 2
This helps to organize the information to write 13x = 52
the equations. x=4
John is 2(x) = 2(4) = 8 years old.
Worked Examples
1. John is twice as old as his friend Peter. Peter is Solved Past Questions
5 years older Alice. In 5 years, John will be three 1. The sum of the ages of two brothers Kofi and
times as old as Alice. How old is Peter now? Kwaku is 35. Kofi‟s age is two thirds of Kwaku‟s
age. Find their ages.
Solution
Let x be Peter‟s age now. Solution
Kwaku‟s age = x years,
Age now Age in 5 years Kofi = years
John 2x 2x + 5
Sum of their ages = 35
Peter x x+5
Alice x–5 x–5+5
x+ = 35
In five years , John will be three times as old as
3×x+3× = 3 × 35
Alice;
3x + 2x = 105
⇒2x + 5 = 3(x – 5 + 5)
5x = 105
2x + 5 = 3x
x = 21
2x – 3x = -5
-x = -5
Kwaku‟s age = 21 years
x=5
Peter is now 5 years old. Kofi‟s age = × 21 = 14years

2. John‟s father is five times older than John and 2. Kojo is n years old now;
John is twice as old as his sister Alice. In two i. How old was he 5 years ago?
years time, the sum of their ages will be 58. How ii. How old will he be 10 years from now.
old is John now? iii. If his age in 10 years ago will be four times
his age 5 years ago, how old is he now?
Solution
Age now Age in two years Solution
Father 5(2x) 5(2x) + 2 Let Kofi‟s age be n.
John 2x 2x + 2 Then 5 years ago, his age was (n – 5) years.
Alice x x+2 10 years time, his age will be (n +10) years.
n + 10 = 4 (n – 5)
In two years time, the sum of their ages will be n + 10 = 4n – 20
58.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 214
n – 4n = - 20 −10 2. When a certain number is increased by five,
-3n = -30 the result is six times the same number decreased
n = 10 years by ten. What is the number?

3. Abanga‟s age in the next 10 years will be four 3. If of a numbers is 4 more than the number,
times his age 5 years ago. How old is Abanga
what is the number? Ans 40
now?
4. I am thinking of a certain number. If I
Solution
multiply by 6, add -18 to the product and multiply
Let x be Abanga‟s age
by one – third of the result, I get -2. What
x +10 = 4 ( x – 5 )
number am I thinking of?
x + 10 = 4x – 20
x - 4x = −20 −10
5. One number is greater than another by fifteen.
−3x = −30
If five times the larger number minus twice the
x = 10 years
smaller number is three, what are the numbers?

4. Awa is m years old now and Fatou is y years

6. A certain rational number is halved, and 23 is
older than Awa. If (x – 5) years ago, Fatou was
added to the results. The same number is doubled,
twice as old as Awa, express x in terms of y and m.
and 21 is added to the results. If the two numbers
obtained are equal, find the rational number.
Solution
Awa‟s age = m
7. A boy was asked to subtract seven from a
Fatuo‟s age = m + y
certain number and divide the results by five.
(x - 5) years ago
Instead, he subtracted five and divided the results
Awa‟s age was = m – (x – 5) by seven. His answer was four less than it should
=m–x+5 have been. What was the given number?
Fatuo‟s age = m + y – (x – 5)
=m+y–x+5 C. 1. A father is now twice as old as his son. If
Fatou was twice as old as Awa, the sum of their ages ten years ago was fifty –
m + y – x + 5 = 2(m – x + 5) two, find their present ages.
m + y – x + 5 = 2m – 2x + 10
- x + 2x = 2m – m – y + 10 – 5 2. A boy is three years younger than his sister. If
x = m – y + 10 his age three years ago was two – thirds of her
age at that time, what are their present ages?
Exercises 8.7
A. Write an equation and solve it. 3. A man is x years old and his son is 30 years
1. Six times a certain number decreased by two is younger. In 2 years, the father‟s age will be twice
four times the same number increased by twelve. the son‟s age. Form an equation, and hence find
Find the number. the present ages of the father and son.

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4. Ben is eight years older than Sarah. Ten years 2. Adam is about to embark on a journey on a
ago, Ben was twice as old as Sarah. Find the narrow country lane that covers 32km and
present ages of Ben and Sarah. Ans 10, 18 decides to go x km/h. On second thoughts, he
calculates that if he increases the speed by 4km/h,
5. Mary is three times older than her son. In 12 his journey time can be cut down by 4 hours.
years, Mary‟s age will be one year less than twice find x. Ans 4km/h
her sons age. How old is each now? Ans 11, 33
3. A butcher bought some pheasants for
6. Sally is three times as old as John. Eight years Ghȼ100.00. Had each cost Gh ȼ1 less, he would
from now, Sally will be twice as old as John. have bought 5 more. How many pheasants did he
How old is John? Ans 8, 24 buy? Ans: 20

7. Kim is 6 years more than twice Tims age. Two 4. A group of student went to a restaurant for a
years ago, Kim was three times as old as Tim. meal. When the bill of Ghȼ175.00 was brought
How old was Kim two years ago? 10, 24 by a waiter, two of the cheeky ones from the
8. Leah is two less than three times Rahel‟s age. group just sneaked off before the bill was paid,
Three years from now, Leah will be seven more which resulted in the payment of extra Ghȼ10.00
than twice Rahels‟ age . How old will Rahel be in by each remaining student. How many students
three years from now? were in the group as well? Ans 7

9. Becca is twice as old as Susan and George is 9 5. The reciprocal of the sum of reciprocals of two
years older than Susan. Three years ago, Becca numbers is 6. The sum of the numbers is 25. Find
was 9 less than three times Susan‟s age. How old the numbers. Ans 10, 15
is George now?
6. Andy has more money than Joe. If Andy gave
10. Lauren is three less than twice Andy‟s age. Joe Ghȼ20.00, they would have the same amount.
Four years from now, Sam will be two more than If Joe gave Andy Ghȼ22.00, Andy would then
twice Andy‟s age. Five years ago, Sam was three have twice as much as Joe. How much does each
times Andy‟s age. How old was Lauren five years one actually have? Ans 106, 146
ago?
7. Noah wants to share a certain amount of
Challenge Problems money with 10 people. However, at te last
1. A man has a daughter and a son. The son is miniute, he is thinking about decreasing the
three years older than the daughter. In one year, amount by 20 so he can keep 20 for himself and
the man will be six times as old as the daughter is share the money with only 5 people. How much
now. In ten years, the man would be fourteen money is Noah trying to share if each person still
years older than the combined ages of his gets the same amount. Ans 44
children at that time. What is the present age of
the man. Ans: 7, 9, 10 Consecutive Numbers
Consecutive numbers are two or more numbers

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that follow each other. For example, 1, 2, 3 or 9, x + x + 2 + x + 4 = 18
10, 11 etc. For all consecutive numbers, if the 3x + 6 = 18
first number is represented by the variable x, then 3x = 18 − 6
second number is x + 1, the third number is x + 2, 3x = 12
and the fourth number is x + 3, in that order. x=4
4, 4 + 2 and 4 + 4 = 4, 6, 8
For the sum of consecutive odd or even numbers, The 3 consecutive even numbers are 4, 6 and 8.
the first number is x, the second number is x + 2,
the third number is x + 4, the fourth is x + 6, in 3. The sum of three consecutive odd numbers is
that order. 27. What are the numbers?

Worked Examples Solution

1. The sum of three consecutive numbers is 18. Let the first odd number be x.
Find the numbers. ⇒second odd number will be x + 2,
⇒ the third odd number will be x + 4,
Solution The sum of the three even numbers is 27
Let the first number = x ⇒ x + (x + 2) + (x + 4) = 27
⇒ the second number = x + 1, x + x+ 2 + x + 4 = 27
⇒ the third number = x + 2 3x + 6 = 27
The sum of the three numbers is 18 3x = 27 − 6
⇒ x + (x + 1) + (x + 2) = 18 3x = 21
x + x + 1 + x + 2 = 18 x=7
3x + 3 = 18 The three consecutive odd numbers are: 7, 9, 11
3x = 18 − 3
3x = 15 4. Five times the smallest of three consecutive
x=5 odd integers is seven more than twice the largest.
⇒the first number, x = 5 Find the largest integer.
⇒ the second number, x + 1 = 5 + 1 = 6
Solution
⇒third number, x + 2 = 5 + 2 = 7 Let the three consecutive odd integers be:
The three consecutive numbers are 5, 6, 7. x, x + 2, and x + 4 respectively.
Five times the smallest is seven more than twice
2. The sum of three consecutive even numbers is the largest;
18. Find the numbers. ⇒ 5(x) = 7 + 2 (x + 4)
5x = 7 + 2x + 8
Solution 5x – 2x = 7 + 8
Let the first even number = x 3x = 15
⇒the second even number = x + 2, x=5
⇒ the third even number = x + 4 The three consecutive odd numbers are:
The sum of the three even numbers is 18 5, 5 + 2, and 5 + 4 = 5, 7, 9
x + (x + 2) + (x + 4) = 18 The largest integer is 9.

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Involving Product 14 and 16 or -16 and - 14.
Worked Examples
1. Find two consecutive odd integers whose Solved Past Questions
product is 99. 1. Find three consecutive odd integers such that
the sum of the last two is fifteen less than five
Solution times the first.
Let the two consecutive odd integers be:
x, and x + 2 respectively. Solution
x (x + 2 ) = 99 Let the three consecutive odd integers be x,
x2 + 2x = 99 x + 2 and x + 4,
x2 + 2x – 99 = 0 The sum of the last two is fifteen less than five
(x – 9) (x + 11) = 0 (by factorization) times the first;
(x – 9) = 0 or (x + 11) = 0 ⇒x + 2 + x + 4 = 5x – 15
x = 9 or x = -11 2x + 6 = 5x – 15
6 + 15 = 5x – 2x
When x = 9, 21 = 3x
The 1st odd integer is 9 and the 2nd is (9 + 2) = 11.
x= =7
The consecutive odd integers are:
When x = - 11,
7, 7 + 2, 7 + 4 = 7, 9, 11
The 1st odd integer is -11 and the 2nd is -9.
Exercises 8.8
Thus, the two consecutive odd consecutive
1. Five times the second of three consecutive
intgers are 9 and 11 or -11 and -9.
even intergers is six more than twice the sum of
2. Find two consecutive even integers whose the first and third integers. Find the middle even
product is 224. integers.

Solution 2. Three times the second of three consecutive

Let the two consecutive even integers be: even integers is twelve less than twice the sum of
x and x + 2 respectively. the first and third integers. Find the largest even
x (x + 2 ) = 224 integer.
x2 + 2x = 224
x2 + 2x – 224 = 0 3. Find two consecutive integers such that five
(x – 14) (x + 16) = 0 times the first equals ten more than three times
(x – 14) = 0 or (x + 16) = 0 the second.
x = 14 or x = -16 4. Four times the smallest of three consecutive
integers is three more than three times the largest.
When x = 14, the second is (14 + 2) = 16. Find the middle integer.
When x = - 16, the second is (-16 + 2) = - 14.
5. Five times the second of three consecutive odd
Thus, the two consecutive odd integers are: integers is twelve less than twice the sum of the

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first and third integers. Find the largest odd 9. When the sum of four consecutive even
integer. numbers is divided by 7, the result is 4. Find the
integers.
6. A man has four sons each of whom (except the
youngest) is 2 years older than his next younger 10. Find four consecutive multiples of five whose
brother. If the sum of the boys ages is 52 years, sum is 90.
how old is the eldest son?
C. Write an equation for each and solve
7. Find five consecutive even integers if the sum 1. Find three consecutive multiples of 3 if the
of the first and fifth is two less than three times sum of the first and third is 12.
the fourth.
2. Find three consecutive integers if twice the
B. Write an equation for each and solve middle integer is equal to the sum of the first and
1. Find four consecutive odd integers if the sum third.
of the first and fourth is three less than three
times the second. 3. Find four consecutive integers such that the
sum of the two largest subtracted from three
2. Find four consecutive integers if the sum of the times the sum of the two smallest is 70.
second and fourth is 48.
4. Find four consecutive odd integers such that
3. Find three consecutive integers such that the the sum of the two smallest added to four times
sum of the first and third is – 34. the smallest is 92.

4. Find two consecutive odd integers if twice the 5. Half of the smaller of two consecutive even
larger, increased by the smaller, equals 85. integers is equal to two more than the larger
integer. Find the numbers
5. Find three consecutive even integers if their
sum, decreased by the third equals – 22. 6. The sum of reciprocal of two consecutive
integers is . Find the integers. Ans: 6, 7
6. The ages in years of three brothers are
consecutive. The sum of their ages is 39 decreasd Inequalities
by the age of the youngest. What are their ages?
An inequality is a mathematical statement formed
7. The sum of three consecutive even integers is
by relating two expressions to each other by using
50 more than the third integer. Find the three
any of the following signs or symbols;
integers
< less than
8. The sum of three consecutive even integers is > greater than
equal to 9 less than 4 times the least of the less than or equal to
integers. Find these integers. greater than or equal to
For example, 3x > 9, 6 + x < 20 , 10, etc

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Solving Inequalities 4. Solve + 4 > 17
All the steps involved in solving equations are
applicable in solving inequalities except that Solution
when we multiply or divide both side of an + 4 > 17
inequality by a negative number, the direction of
the inequality sign reverses. That is; > 17 – 4
changes to and vice – versa whilst ≤ > 13
changes to and vice – versa. × – 2 > 13 × – 2
x < – 26 (sign reverses)
Worked Examples
1.
5. Solve – ( 3x – 1

Solution
Solution
– ( 3x – 1
– – 3x –
12 × – 12 × – 12 × 12 × 3x – 12 ×
4x – 3x– 6 36x – 16
4x – 3x – 36x –16 + 6
, –35x –10
x (division by a negative number)

x
2. Solve
Exercises 8.9
Solution
A. Determine the solution set
1. 3x – 13 > 26 2. 2x – 7 < 32 – x
3. +9 −5 4. 14 8 – 2x
5. 14 > 8 − 2x 6. 8 − 7x < 11

(sign reverses)
B. Determine the solution set
1) - (2 + x) < 3 + (-7) 2) 3x > (8 + x) – 2
3. Solve
3) – x x+ 4) x +
Solution
5) − x>

6. If x = {1, 3, 5, 7, 9, 11, 13, 15}, find the truth

, set of x – 3 > 10.

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Inequalities on a Number Line 2. Identify the inequality shown in the number
The truth set of inequalities can be represented on line below;
a number line by the use of a directed shaded
circle or directed none shaded circle.
-3 -2 -1 0 1 2 3 4
For all inequality signs:
The shaded circle is used for signs and Solution
the none shaded circle is used for ˂ . Because the circle is shaded and directed to the
The circle, either shaded or non-shaded is made left, it represents the inequality symbol <. The
to stand on top of the number that represents the shaded circle stands on 2 on the number line.
truth set. Therefore x < 2
I. If the sign is the arrow is directed to
the right. 3. What inequality is represented on the number
II. If the sign is the arrow is directed to line below?
the left.
III. If the condition is given on the set of
integers (Z), maintain the direction of the arrow -3 -2 -1 0 1 2 3 4
and put a dot on the integers representing the
solution set as shown in example 4 below. Solution
None shaded circle directed to the right
Worked Examples represents >. Since the none shaded circle stands
Solve the inequalities and represent the answer on on -3, the inequality is x > -3
a number line.
1. 4. Solve 5 – 2x x + 2, where x is an integer and
illustrate your answer on a number line.
Solution
Solution
5 – 2x x + 2
5 – 2 x + 2x
5 – 2 3x
Stand on the number that represents the truth set 3 3x
using non – shaded circle because the sign is 1 x or x 1
(
  
-3 -2 -1 0 1 2 3 4
-3 -2 -1 0 1 2 3 4 5. Determine the inequality on the number line
Arrow the non-shaded circle to the left and write below;
the truth set on it.

-3 -2 -1 0 1 2 3 4

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Solution 5. ≤ +x 6. (x– 9) > x + 3
The non shaded circle on -1, directed to the right
represent: x > −1 or −1 <x
Word Problems (Linear Inequalities)
The non shaded circle is on 3, directed to the left
Very often, problems involving linear inequalities
so the truth set is x < 3. The two inequalities put
are given in words. It is advisable to construct an
together = −1 <x < 3
open sentence and then find the truth set.

Exercises 8.10
Steps
A. Identify the inequality on the number lines
1. Read the problem carefully.
2. Assign a variable to represent the unknown.
1.
3. Determine which inequality is required
-3 -2 -1 0 1 2 4 4. Wite an inequality based on the given
3
information.
2.
5. Solve the inequality to find the answer.
-3 -1 0 6. Check your answer.
-2 1 2 3 4
3.
Worked Examples
-3 -2 -1 0 1 2 3 4 1. A man, 40 years old is more than twice as old
as his son. How old is the son?
4.
-6 -4 -2 0 2 4 6 8
Solution
5.
Let the son‟s age be x
2 3 4 5 6 7 8 9 ⇒40 > 2x
< ,
6.
-3 -1 0 x < 20
-2 1 2 3 4
Therefore, the son is less than 20 years.
7.
-3 -2 -1 0 1 2 3 4 2. Two sides of a triangle have lengths 6cm and
8cm. What is the length of the third side?
B. List the members of the following sets
1. {2 < x < 10} 2. -10 < x < 5 Solution
3. 21 < y < 27 4. 3 < x ˂ 5 6+8>x
5. 2<x<5 6. -2 ≤ x < 2 14 > x
8cm
Also, 6 + x > 8 x
C. Solve the following and represent your
answer on a number line. x>8–6
1. (x – 1) + 2 ≤ 2.
(
˂1 x>2 6cm

3. (2x + 5) ˂ 4. (x – 1) − (x – 3) ≤ 1 Again, 8 + x > 6

Baffour – Ba Series, Core Maths for Schools and Colleges Page 222
x>6–8 Or more than 6 days;
x > −2, Car = 50 + 30d = 50 + 30(7) = Ghȼ260.00
But x > −2 satisfies the first two inequalities. Van = 80 + 25d = 80 + 25(7) = Ghȼ255.00
Thus, 2 < x < 14, meaning the third side has a It is cheaper to rent the van for more than 6 days.
length between 2cm and 14cm.
Exercises 8.11
3. Every number of set of integers is greater than Write an open sentence, and solve
40 and is less than 44. What are the members of 1. Two sides of a triangle have lengths 10 cm and
the set? 14 cm. What is the length of the third side?

Solution 2. The smallest number in a set of integers is 572

Let x be a member of the set, then x > 40 and x < and the largest is 999. What are the possible
44. The two can be put together as; numbers of the set?
40 < x < 44.Therefore, the truth set of this
inequality is {41, 42, 43} 3. Your score in the 4 mathematics test were 71,
74, 75 and 86. What is the lowest score you need
4. You want to rent a car for a rental cost of in your next test in order to have an average score
Ghȼ50.00 plus Ghȼ30.00 per day, or a van for of at least 80.
Ghȼ80.00 plus Ghȼ25.00 per day. For how many
days at most is it cheaper to rent a car? 4. The sum of two consecutive integers is less
than 79. Find the integer with the greatest sum.
Solution
Let d = days required, Compound Inequalities
Car = 50 + 30d, Van = 80 + 25d When two simple inequalities are connected by the
word “and” or the word “or” a compound
At most means “not more than, so ≤ is the
inequality is formed. For example, x < 6 and
sign. The inequality is ;
x < 9 , 2x – 9 ≤ 5 or – 4 x ≥ - 12
50 + 30d ≤ 80 + 25d
5d ≤ 30 Compound Inequalities Connected by “or”
d≤6 A compound inequality connected by the word
“or” is true if one or the other or both of the
Check simple inequalities are true. It is false only if both
Car = 50 + 30d = 50 + 30(6) = Ghȼ230.00 simple inequalities are false.
Van = 80 + 25d = 80 + 25(6) = Ghȼ230.00 To determine whether such statement is true or
The rental cost are the same for 6 days. false;
a. check if the first statement is true or false
For less than 6days,
b. check if the second statement is true or false.
Car = 50 + 30d = 50 + 30(5) = Ghȼ200.00
Van = 80 + 25d = 80 + 25(5) = Ghȼ205.00
It is cheaper to rent the car for less than 6 days. Draw conclusions base on the following:
1. if a is true and b is true, the inequality is true;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 223
2. if a is true and b is false, the inequality is true; 2. if a is true and b is false, the inequality is false;
3. if a is false and b is true, the inequality is true; 3. if a is false and b is true, the inequality is false;
4. if a is false and b is false, the inequality is 4. if a is false and b is false, the inequality is
false. false.

Worked Examples Worked Examples

Determine whether each compound inequalities is Determine whether each compound inequality is
true or false. true or false;
1. 2 < 3 or 2 > 7 2. 4 < 3 or 4 ≤ 7 1. 5 > 2 and 5 < 7 2. 6 < 5 and 6 > 2

Solution Solution
1. 2 < 3 or 2 > 7 1.5 > 2 and 5 < 7
Let a represents 2 < 3 and b represents 2 > 7. Let a represents 5 > 2 and b represents 5 < 7.
2 < 3 is true, 2 > 7 is false. 5 > 2 is true, 5 < 7 is true.
(a is true, b is false) (a is true, b is true)
The statement 2 < 3 or 2 > 7 is true. The statement 5 > 2 and 5 < 7 is true.

2. 4 < 3 or 4 7 2. 6 < 5 and 6 > 2

Let a represents 4 < 3 and b represents 4 7. Let a represents 6 < 5 and b represents 6 > 2.
4 < 3 is false, 4 7 is false. 6 < 5 is false, 6 > 2 is true.
(a is false, b is false) (a is false, b is true)
The statement 4 < 3 or 4 7 is false. The statement 6 < 5 and 6 > 2 is false.

Exercises 8.12 Exercises 8.13

Show whether the following is true or false; Determine whether each is true or false;
1. 3 < 5 or 3 < 10 2. 4 < 8 or 4 > 2 1. – 6 < 5 and – 6 > - 3 2. 1 < 5 and 1 > - 3
3. 4 - 4 or 0 0 4. 6 < 5 or – 4 > -3 3. 4 ≤ 4 and – 4 > - 3 4. 3 < 5 and 3 ≤ 10

Compound Inequalities Connected By “and” Graphing Compound Inequalities

A compound inequality connected by the word a. Using the connective “or”
“and” is true if and only if both simple The solution set to the compound inequality using
inequalities are true. the connective “or” is the union of the solution
sets to each of the simple inequalities.
To determine whether such statement is true or
false; If A and B are the set of numbers, then the union
a. check if the first statement is true or false of A and B is the set of all numbers that are in
b. check if the second statement is true or false. either A or B, denoted as A∪B.

Draw conclusions base on the following: To represent such inequalities on a number line,
1. if a is true and b is true, the inequality is true; graph each inequality separately on the same
number line.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 224
Worked Examples If A and B are the set of numbers, then the
1. Illustrate the solution set of the compound intersection of A and B is the set of all numbers
inequalities x > 4 or x < - 1 on a number line. that are in both A and B, denoted as A B.

Solution To represent such inequalities on a number line;

The union of these two sets is the set of numbers I. graph each inequality separately on the same
less than – 1 (excluding – 1) and set number line,
of numbers greater than 4 (excluding 4) II. identify the overlapping interval as the solution
set of the compound inequality.

-2 0 1 2 3 4 5
Worked Examples
-1
1. Graph the solution set to the compound
2. Show the solution set of the compound inequality x > 2 and x < 5 and list the intersection.
inequality x > 3 or x > 1 on a number line.
Solution
Solution
The union of these two sets is the set of numbers
-1 0 1 2 3 4 5 6
greater than 1.
The intersection of these two solution set is the
portion of the number line that is shaded, just
-2 -1 0 1 2 3 4 5 the part between 2 and 5, excluding 2 and 5. That is
{x : x = 3, 4}
3. Sketch the graph of the solution set of the
compound inequality x ≤ 0 or x > 3 on a number
2. Illustrate the solution set of the compound
line.
inequalities x ≤ 4 and x ≥ 0 on a number line.
Solution
Solution
The union of these two sets is the set of numbers
 -
less than or equal to 0 ( incuding 0) 1  -
and numbers greater than 3 (excluding 3). -2 -1 0 1 2 3 4 15

 - The intersection of these two solution set is the

1 portion of the number line that is shaded, just the
-2 -1 0 1 2 3 4 5 part from 0 to 4, inxcluding 0 and 4. That is {x : x =
0,1, 2, 3, 4}
b. Using the connective “and”
The solution set to the compound inequality using 3. Illustrate the solution set of the compound
the connective “and” is the intersection of the inequalities x ≤ 2 and x > -1 on a number line.
solution sets to each of the simple inequalities.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 225
Solution
Compound inequalities are solved by following the
 - steps below:
1
-2 -1 0 1 2 3 4 5
The intersection of these two solution set is the Method 1
portion of the number line that is shaded, just the I. Form two inequalities such that each one
part from 0 to 2, excluding -1 but including 2. contains one of the given inequality.
That is {x : x = 0,1, 2} II. Solve each inequalities separately to obtain the
truth set of each.
Exercises 8.14 III. Put the solution sets together as one
Illustrate the following on a number line and inequality, using set builder notation.
list the members in each case:
A. 1. x > -1 or x < 3 4. x < 7 or x > 0 Method 2: (solving the original equation)
2. x ≤ 6 or x > - 2 5. x ≤ 6 or x > 9 I. Group like terms, by transposition such that
3. x ≥ 2 or x ≥ 5 6. x > 3 or x < -3 the variable factor remains at the middle.
II. Divide through by the coefficient of x.
B. 1. x > -1 and x < 4 4. x ≤ 6 and x > 9 III. Obtain the solution set in the interval
2. x ≤ 3 and x ≤ 0 5. x ≥ 4 and x ≤ - 4 notation.
3. x > - 2 and x ≤ 4 6. x ≥ 6 and x ≤ 1  Use “[” and “]” if reading from the left,
the first inequality is ≤ and the second is
C. 1. {x : x > 1} {x : x < 4} ≤.
2. {x : x > -3} {x : x < 3}  Use “(” and “)” if reading from the left,
the first inequality is < and the second is <
3. {x : x < 2} {x : x > 5}
2 .
4. {x : x > 9}  Use “[” and “)” if reading from the left,
5. {x : x2 < 2} ∪ {x : x > 5} the first inequality is ≤ and the second is <
or vice versa
Solving Compound Inequalities
An inequality may be read from left to right or Worked Examples
from right to left. For example, 2 < x is read in Find the solution set of 5 < 3x – 1 < 17 and
the usual way as “2 is less than x”. Reading from represent your answer ona number line.
the right, we say “x is greater than 2” . The
meaning is clearer when the variable is read first. Solution
Method 1
For compound inequality, another notation is 5 < 3x – 1 < 17
used. For example, x > 3 and x < 6 is commonly ⇒5 < 3x – 1 and 3x – 1 < 17
written as 3 < x < 6. This is read from the left to 5 + 1 < 3x and 3x < 17 + 1
right as “3 less than x is less than 6.” Reading the 6 < 3x and 3x < 18
variable first, we say “x is greater than 3 and less <x and x<
than 6.” So x is between 3 and 6, and reading x
2<x and x<6
makes it more understandable.

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The solution set is { x : 2 < x < 6} 1 ≤ 2x < 10
2 x<6 ≤ <
≤x<5
3
The solution set is [ , 5 )
Method 2
Solving in the original form;
5 < 3x – 1 < 17 [ )
5 + 1< 3x < 17 + 1 (Add 1 to both sides) -1 0 1 2 3 4 5
6 < 3x <18
Exercises 8.15
< < (Divide throught by 3) A. Solve and graph each inequality.
2<x<6 1. 5 < 2x – 3 < 1 2. – 1 < 5 – 3x ≤ 14
The solution set is (2, 6) 3. -1 < 3 – 2x < 9 4. -2 < 3x + 1< 10
5. – 1 ≤ 3 – 2x < 11 6. 3 ≤ 3 – 5(x – 3 ) ≤ 8
( )
3
B. Solve and grapheach inequality.
2. Solve the inequality – 2 ≤ 2x – 3 < 7 and graph 1. 2 ≤ 4 – ( x – 8 ) ≤ 10 2. -3 < <7
the solution set.
3. 0 ≤ <7 4. -2 < <7
Solution 5. -3 < ≤5
– 2 ≤ 2x – 3 < 7
– 2 + 3 ≤2x – 3 + 3 < 7 + 3

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9 BEARINGS AND VECTORS Baffour – Ba Series

Meaning of Bearings Find the bearings of M, N, P, Q, R, S, T, U and

It is the measurement of direction using angles. W from the center O in the diagram below;
This is done with the help ofthe cardinal points; N
U
North, South, East and West. M
0
50
0
There are two ways of measuring the bearing of a 20
W 15
0
20
0
E
point from another point (o). 0
67 0
25 P
1. Using the North-pole and measuring in the T
clock wise direction. R Q
2. Using the North or South Pole only and
S
measuring eastward or westward. NOT DRAWN TO SCALE
Solution
Measuring in the Clockwise Direction from the The Bearings from the center, O, are:
North - pole M = 0700
By this method, measurement is taken in the P = 900+ 200 = 1100
clockwise direction from the North Pole. The Q = 900 + 200 + 450 = 1550
bearing is then written in 3 – digits. Thus, angles R = 900 + 900 + 670 = 2470
less than 100 are preceded by zero(s). S = 900 + 900 = 1800
0
360 N 000
0 T = 900 + 900 + 670 + 80 = 2550
D A U = 900+ 900 + 900 + 400 = 3100
W = 900 + 900 + 900= 2700
0
60
0
30
W 2700 0
25 500 E 0900 2. Represent the following bearings in a diagram:
0
F
30 i. V = 1450 ii. K = 0750 iii. B = 2400
C B iv. M = 1150 v. I = 2040 vi. A = 3300
S
180
0 Solutions

From the diagram, the bearings of the point A, B, i. ii. 0

75
o o
C, D and F from the center, O, is 55
0 K
A = 0600 B = 900 + 500 = 1400
V
C = 900 + 900 + 300 = 2100
iii. iv.
D = 900 + 900 + 900 + 300 = 3000
F = 900 + 900 + 300 + 350 = 2450 o
o 0
25
0
60
Worked Examples M
B

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v. vi. Consider the diagram below:
A N
D
o 60
0 A
o 0
0 60
0 70
24 W 25
0
E
0
I 35
C
Exercises 9.1 B
S
A. Find the bearings of A, B, C, D, E and F in
The bearings of A, B, C and D from the center,
the diagram below;
O, are found by finding the acute angle each of
N them makes with either the North – pole or South
A B
– pole.
F 50 C 1. Since A makes an angle of 600 with the North
0 400 0 and lies between the North and East, the bearing
15 33
W 0
E of A is N 600 E.
64
280 2. Since B makes an angle of 350 with the South
D Pole and lies between South and East, the bearing
E of B is S 350 E.
S
3. Similarly, the bearing of C is found by
B. Represent each bearings on a separate
calculating the angle C makes with the South.
diagram;
That is 900 – 250 = 650. Because C lies between
1. A = 2130 2. D = 2030 3. B = 1220
South and West, its bearing is on S 650 W.
4. E = 2900 5. C = 0480 6. F = 1550
4. Likewise, the bearing of D is found by finding
Using the North and South Poles and the angle it forms with the North Pole. That is 900
Measuring Towards the East or West − 700 = 200. Because D is located between North
and West, it is said to be on a bearing of N 200 W.
STEPS:
Here the acute angle formed with the North or
Worked Examples
South Pole is measured in the Eastward or
A. Find the bearings of M, P, U, Q, R, S, and T
Westward direction. Thus, measurement is taken
from the center O, in the diagram below;
in: North – West (N – W), North-East (N – E)
and South - East (S− E), South - West (S − W) N
M
but not the other way round. The bearing is
written by either the North (N) pole or South (S) U 28 0
0 P
0 33
pole first, followed by the acute angle formed 58
W 0 0 E
with it and finally, the East (E) or South (S) 35 46
0
indications. 10
R Q
T
S

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The bearings of M, P, U, Q, R, S, T, from the N
center O are:
D
M = N 280 E, P = N 610 E, U = N 320 W
A
Q = S 440 E, T = S 100 W, R = S 550 W 450 650
W E
B. Show the following bearings from thecenter O, 0
50
on a diagram and write the alternative bearings of 250
B
each; C
1. G = 2800, 2. H =2050, S
3. K = 2480, 4. L = 0550,
Solution
Solutions Bearings of A, B, C and D from the center,O, is:
N
1. G The angle G makes with A = North 650 East = N 650 E
0
the North-pole; B = South 500 East = S 500E
10
W E = 900 −100 C = South 250 West = S 250W
= 800. D = North 450 West = N 450W
S
The alternate bearing of G = N 800 W. 4. Find the bearings of J, K, L and M from the
center.
2. N The angle H makes
with the South – pole N
0
M
W E = 25 . J
0
25 The alternate bearing of H
H = S 250 W 70
0
68
0
S W E
3. 0
33 65
0

The angle, k makes with E

N
the South – pole L K
0
W E = 68 .
68
0
The alternate bearing of k S
K = S 680 W Solution
S
Bearing of J from O = (900– 680) = 220 from the
4. N
The angle L makes with North- pole towards east = N 220 E
the North-pole = 550. Bearing of K from O = (900 – 650) = 250 with the
0
L
55 The alternate bearing of L
W south-pole towards east = S 250 E
E = N 550 E
Bearing of L from O = (900 – 330) = 570 with the
S south-pole towards west = S 570 W
Bearing of M from O = (900 – 700) = 200 with the
C. In the diagram below, determine the bearings north-pole towards west = N 200 W
of A, B, C and D from the center,

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Some Solved Past Questions Exercises 9.2
1) The bearing of town A from town B is 1260. A. Write the alternate bearings of the
Show this in a diagram. following;
1) S330 W 2) N250 W 3) N 600 W
Solution 4) N150 E 5) S730 E 6) S 440 W
N
B B. Write the alternate bearings:
W E 0 0 0 1) 2300 2) 0380 3) 1550
360 (90 – 36 ) = 54 with
South Pole towards East, 4) 1050 5) 3030 6) 3550
⇒126 0 = S 540 E
S A
Distance Bearing
2. The bearing of Asuofua from Kejetia is 3160. When bearing is measured in the form (akm, ),
Represent this in a diagram and give an where a is the distance and b0 is the direction, it is
alternative name for this bearing. called distance bearing. It is written with
distance, followed by the direction, which is in
Solution the form (distance, direction). For example, in the
N (900 – 460) = 440 with diagram below, N
A
North Pole towards West, W A E
0
45
46
0
K ⇒3160 = N 440 W
E 5km
W
B
S S
3. Identify the bearing represented in the diagram The distance bearing of AB = (5km, 1350)
below:
N Worked Examples
Solution W
O
0 E 1. A ship sails from port R on a bearing 0650 to
66
Bearing of M from O port K in a distance of 5km. Show this in a
= 900 + 660 diagram and write the distance bearing of RK
= 1560 S M
Solution
0 N
4. Tepa is located S30 W of Abuakwa. Represent
this in a diagram and find the bearing of Tepa K The distance bearing of
0
from Abuakwa.
65 5km RK= (5km, 0650)
W R E

Solution
N S
Bearing of Tepa from
A 2. A village R is 8km from another village P on a
W E Sunyani;
30
0
= 900 + 900 + 300 bearing of 1550 Show this in a diagram and write
= 2100 RP in distance bearing.
Tepa
S

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Solution N The distance bearing of RP same diagram and alternating the angle 200.
= (8km, 1550) Measurement is then taken from the North Pole in
P the clockwise direction.
W 0
E
65 N
8km
R A
S W E
0
N 20
Exercises 9.3 Bearing of
A.1. Represent the following bearings on the 0
B from A
20 S
same diagram; W
B
E The back bearing;
i. A = S200 W, ii. B = S200 E, Bearing of A
= 900 + 900 + 200
iii. C = N390 W iv. D = S 530 W from B S
= 2000
B. Write the bearings from o;
N
Worked Examples
1. N 2.
R 1.The bearing of P from Q is 0600. Find the
6km
0
bearing of Q from P.
W o 45 o
150 E W E
7km Solution N Q from P
S
T S P
P from Q W 0
E
60
3. 4. N
N N
Q 60
0
S
V W E
11km The bearing of Q from P;
o
W
o 250 W E = 900 + 900 + 600
E 3km 700
S = 2400
U
S 2. The bearing of a point M from T is 2130. Find
S
Back Bearings or Opposite Bearings. the bearing of T from M.
The back bearing of say A from B is the bearing
of B from A. For example, the bearing of A 200 Solution
N
from B is represented in the diagram below; M from T
T
W E
N 0
A T from M 33

200 N
W B E 0 S
M 33
W E The bearing of T from M
= 0330.
S
S
3. The bearing of D from F is 3200. Find the
The back bearing which is the bearing of B from bearing of F from D.
A is found by using A as the turning point on the

Baffour – Ba Series, Core Maths for Schools and Colleges Page 232
 F from D B.What is the back bearing of the following?
N
1. Bole from Bamboi = S630E
2. Konongo fromOdumase = 2200
F 3. Keta from Krachie = 3150
W E
50
0
4. Secondi from Takoradi = S270W.

S N D from F C. Study the diagrams below carefully and

rom D 0
W
50 D E use them to answer the questions that follow;
i. P ii. A
S 0
20
0
42
Solution
Bearing of F from D Q B
= 900 + 500 iii. N iv.
L
= 1400 35
0

0
NOTE T 57
I. For all bearing less than or equal to 1800, the K
back bearing is determine by adding 1800 to the i. The back bearing of Q from P is…
given bearing (angle). That is : θ + 1800, where 0 ii. The back bearing of B from A is…
≤ θ ≤ 1800. For example, if the bearing of P from iii. The back bearing of T from N is…
Q is 0600, the bearing of Q from P is: 1800 + 600 iv. The back bearing of L from K is……
= 2400
Other Applications
II. For all bearings greater than 1800, the back-
Bearings, Pythagoras and Trigonometry;
bearing is determined by subtracting 1800 from
To solve problems involving bearings,
the given bearing (angle). That is:
Pythagoras and trigonometry;
θ – 1800, where 1800 < θ < 3600. For e.g., if the
I. Make a sketch of the problem.
bearing of A from B is 2050, then the bearing of
II. If one of the angles of the triangle formed is a
B from A= 2050 – 1800 = 0250.
right angle ( , use Pythagoras theorem to find
the length of the third side given two sides.
Exercises 9.4
III. Knowing the three sides of the triangle,
A. 1. The bearing of P from Q is 0450. What is
calculate the values of the other two angles in the
the bearing of Q from P?
triangle using any of the trigonometric ratios:
2. The bearing of town M from town K is 3250.
SOH, CAH, TOA
Find the bearing of the town K from town M.
IV. To find the bearing of say A from B, is to find
3. What is the bearing of R from T if the bearing
the total angle turned through in the clockwise
of T from R is 3000?
direction from the north pole of B to the direction
4. The bearing of Asuofua from Kokoben is 1250,
of A.
what is the bearing of Kokoben from Asuofua?

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Type 1 (Worked Examples) d = 26 km
1. Ata‟s house is 5 km due east of Bortey‟s house.
If Cudjoe‟s house is 10 km due south of Bortey ii. 16 km
B o
house, find, correct to 1decimal place, the bearing From the ∆
and distance of Ata‟s house from cudjoe‟s house. tan θ = 20 km
d
Solution θ= ( ) θ
0
Let A, B, and C represent the positions of Ata‟s, θ = 39 A
Bortey‟s and Cudjoe‟s houses respectively. + θ + 900 = 1800
Let (r, θ) be the position of Ata‟s house from + 390 + 900 = 1800
Cudjoes‟s. N
= 1800 – 900 – 390
= 510
From ∆ABC, B 5 km
A Bearing of B from A;
tan θ =
10km r = 900 + 900 + 900 +
θ= ( ) θ = 900 + 900 + 900 + 510
0 C E
θ = 26.57 = 3210

By Pythagoras theorem, 3. Salifu walks 500m due north, then 250m due
r2 = 102 + 52 east and finally 500m on a bearing of 0550,
r2 = 100 + 25 i. Sketch a diagram to illustrate Salifu‟s
r2 = 125 movement;
r=√ = 11.2 ii. Calculate to the nearest whole number, how far
The bearing of Ata‟s house from Cudjoes‟s north Salifu has moved from the starting point;
house = (11.2km, 0270) iii. Calculate to the nearest whole number, how
far east he has moved from the starting point;
2. Two buses A and B are at a station. If bus A iv. Calculate the bearing of Salifu‟s final position
moves 20km due south and bus B moves 16km from the starting point.
due west,
i. how far apart are the two buses, Solution
ii. what is the bearing of bus B from bus A. i.
500m
Solution 0
55 y
i. Let the position of the station be o and the 250m 0
35
distance between the buses be d x
B 16 km o
500 m

500m

2 2 2
d = 16 + 20
20 km
d2 = 256 + 400 θ
d
d2 = 656
d=√ A

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ii. How far north salifu has moved from the a. north of its initial position,
starting point = y + 500 b. west of its initial position. a = 8.7km, b = 5km
But from the diagram, 4. An aeroplane flies 150 km due south then
sin 350 = 150km on a bearing of 0450. With respet to its
y = 500 sin 350 initial position, find:
y = 287m a. how far south the plane is;
b. how far east the plane is;
287m + 500m = 787m c. how far awy and its bearing.
Ans: a. 44km, b. 106km, c. (115km, 1150)
iii. how far east he has moved from the starting
point= x + 250 Type 2
But from the diagram, (Two or More Bearings on the Same Diagram)
cos 350 = Worked Examples
0
x = 500 cos 35 1. An aircraft flew from port A on a bearing of
x = 410m 1300 to another port B 80km apart. It then flew
410m + 250m = 660m from port B on a bearing of 0400 to port C, a
distance of 120 km. Calculate:
iv. Let the bearing of salifu‟s final position from i. the distance from port A to port C, to the
the starting point be θ nearest kilometer
tan θ = ii. the bearing of port A from port C, to the
nearest degree
θ= ( )= 0400 (Nearest whole number)
Solution
Trial Test
1. A man walks 500m due north, and then 350m θ C
due west. What is his bearing from the starting A
points? A = 3250
120km
80km
2. Baru stands at the center of a football field. He
first takes 10 steps north, then 5 steps east and B
finally 10 steps on a bearing of 0450
a. how far east is Baru‟s final position from the
From the diagram,
center? A = 12 steps
ABC is a right – angled triangle
b. how far north is Baru‟s final position from the
By Pythagoras theorem,
center? A = 17 steps
= +
c. Find the bearing and distance of Baru‟s final
= +
position from the center.A= (21s, 0350)
= 6400 + 14400
3. A car is driven 10km on a bearing of 3000 find = 20,800
how far the cars final position is: =√

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 = 144km (Nearest kilometer) α= – 610 = 290
ii. From the diagram,
tan θ =  The bearing of school Q from R,
= + +
θ= ( ) = ( Nearest degree) = 90 + 90 + 290
0 0

 The bearing of port A from port B, = 2090

R
= + + + = 11km
iii. a. From the diagram,
2. Three schools, P, Q and R are situated as H = 11km and
follows: Q is on a bearing of 5km, 1430 from P A = TR, P
and R is on the bearing 11 km, 0530from P. Find: T
5km
i. the distance between school Q and School R, to CAH is applicable
the nearest kilometer, cos 240 = Q
ii. the bearing of school Q from school R to the
nearest degree, cos 240=
iii. T is a school situated in – between Q and R,
such that PT is perpendicular to QR; = 11 cos = 10km
a. find the distance of T from R,  The distance of school T from school R is10km
b. how far is school T from school P, to the
nearest kilometer? b. By Pythagoras theorem,
= +
Solution – =
i. From the diagram, – =
PQR is a right – angled triangle = 121 – 100
By Pythagoras theorem, = 21
= + R
θα
=√
= + 11km = 5km
= 25 + 121  School T is 5km away from school B
= 146 P
=√ 3. Two cars A and B moved from a station at the
= 12km same time. Car A moved on the bearing
5km
from the station and Car B moved on a bearing
( Nearest kilometer) Q from the station. If 3 hours later, the
distance of A from the station was 16km and the
ii. From the diagram, distance of B from the station was 12km, find;
tan θ = i. the distance between the two cars after 3 hours
θ= ( )= 240 (to the nearest degree) ii. the bearing of car A from car B after 3 hours,
to the nearest degree
+

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Solution 2500 =
From the diagram,∆ AOB is a right – angled x2 = 250
triangle x=√ = 16km
B
By Pythagoras theorem, θ
12km
= + = 3x, but x = 16
= + O = 3 × 16 = 48km
= 144 + 256 45 0 The distance from F to G is 48km.
= 400
16km ii. From the diagram,
=√
= 20km tan θ = =
A
The distance between the two cars is 20km θ= ( )= 720

ii. From the diagram,

Bearing of A from B = 900 + 450 + θ The bearing of G from E,
= 900 + 900 + 440 + θ
tan θ = ( )
= 900 + 900 + 440 + 720 = 2960
θ= ( ) = 530
Exercises 9.5
Bearing of A from B; 1. With respect to a fixed point, O, a point P is at
= 900 + 450 + θ (12km, 0350) and another point Q is at
= 900 + 450 + 530 (16km,1250). Find the bearing and the distance of
= 1880 P from Q.

4. A ship sailed from port E to port F, x km away 2. Two buses start off from a ststion, O. The first
on a bearing of . From there, it sailed to port bus ends its journey at station P(39km, 0500)
G, three times the distance from E to F on a from O. If the final stop of the second bus is
bearing of . If the distance from E to G is Q(25km, 3200) from O, find the position of Q
50km, find to the nearest whole number; from P.
i. the distance from F to G
ii. the bearing of G from E 3. A ship sailed from port P on a bearing of 1130
to port Q, 32km apart. It then sailed from port Q
G
Solution 50km to another port R on a bearing of 0230, a distance
From the diagram, of 54km. Calculate:
E
∆ EFG is a right – θ i. the distance from port P to port R, to the nearest
angled triangle, kilometer,
3x ii. the bearing of port P from port R, to the
By Pythagoras theorem,
km x
= + km nearest degree,
= +( F
4. The location of three schools A, B and C are as
2500 = +
follows: B is on a bearing of 1500 from A and C

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is on bearing of 2400 from A.If A and B are 13km Representation of Vectors
apart and A and C are 19km apart, calculate; A free vector is a vector that does not start from
i. how far B is located from C, to the nearest the origin as shown below;
kilometer.; y
B
ii. the bearing of C from B, to the nearest degree.

5. Two aircrafts, P and Q flew from the same port A

at the same time. Craft P flew at 26km/h on a x
O
bearing of 3450 and craft Q flew at 41km/h on a
bearing of 0750. Find after three hours; Free vectors are represented by a pair of capital
i. the distance between the two crafts, to the letters with an arrow over their top.
nearest kilometer. The arrow indicates the direction of movement.
ii. the bearing of P from Q, to the nearest degree. Thus, movement from A to B is written as ⃗⃗⃗⃗⃗
and movement from B to A is written as ⃗⃗⃗⃗⃗ or
6. a. Bebu village is x km from Kejetia on a ⃖⃗⃗⃗⃗⃗ . These are called free vectors. For example,
bearing 3360. Adumasa village is 4x km from
Bebu on a bearing 2460. If the distance from if ⃗⃗⃗⃗⃗ = ( ) then x - component is on top andy -
Kejetia to Adumasa is 120km, calculate; component is at the bottom.
i. the distance from Bebu to Adumasa, to the
nearest kilometer. Position Vectors
ii. the bearing of Adumasa from Kejetia, to the It is a vector that has the origin as a starting point.
nearest degree. y
b. Senfi is a village located on Kejetia –
P
Adumasa road, which is perpendicular to Bebu.
Find the distance of Senfi from Adumasa to the
nearest kilometer. x
O

Vectors The vector ⃗⃗⃗⃗⃗ indicates a movement from the

A vector is any quantity that has magnitude and origin O, to a pointP. p(small letter) is called the
direction. This means that its measurement position vector of the point OP.Thus, if A = (x, y),
requires two or more numbers. Examples of then ⃗⃗⃗⃗⃗ = ( ).
vector quantities are weight, Velocity and
displacement.
Similarly, if B is the point (h, k), then ⃗⃗⃗⃗⃗
represents the position vector b of B, and b = ( )
On the other hand, any quantity that has
magnitude only is called a scalar quantity. Column Vector
Examples of scalar quantities are mass, speed It is a vector written in vertical form ( ). For
and distance etc. example, ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( )

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Inverse or Negative Vector Perpendicular Vectors
If ⃗⃗⃗⃗⃗ = ( ), then the inverse of ⃗⃗⃗⃗⃗ , written as Two vectors a and b are perpendicular if;
1. The product of their gradients is – 1.
⃗⃗⃗⃗⃗ = ( )
2. a is the rotation of b or vice – versa through an
angle of 900 or 2700clockwise or anti –
Worked Examples clockwise about the origin.
1. If ⃗⃗⃗⃗⃗⃗⃗ = ( ). Find the inverse of ⃗⃗⃗⃗⃗⃗⃗ . 3. If their dot product is zero i. e. a × b = 0

Solution Worked Examples

⃗⃗⃗⃗⃗⃗⃗ = − ( ) ( ). Find a vector that is perpendicular to a = (8, - 4)

Solution
2. If ⃗⃗⃗⃗⃗ = ( ) find ⃗⃗⃗⃗⃗ .
Method 1
Let the vector perpendicular to a = (8, - 4) be b =
Solution
(x, y ).
⃗⃗⃗⃗⃗ = − ( )= ( )
Then a × b = 0
⇒8x + (- 4y) = 0
Exercises 9.6
⇒8x – 4y = 0
Find inverse of the following vectors.
(1) ⃗⃗⃗⃗⃗ = ( ) (2) ⃗⃗⃗⃗⃗ = ( ) When x =1 and y = 2
(3) ⃗⃗⃗⃗⃗⃗⃗ = ( ) (4) ⃗⃗⃗⃗⃗ = ( ) 8(1) – 4(2) = 0
Therefore b = (1, 2) is perpendicular to a
Equal Vectors
Two or more vectors are said to be equal if their x When x = 2 and y = 4
and y components are the same. For example, if 8(2) – 4(4) = 0
⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗⃗⃗ = ( ), then ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗ Therefore b = (2, 4) is perpendicular to a.

Worked Examples When x = -1 and y = -2

1. If ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ), what is the 8(-1) – 4(-2) = 0
relationship between the two vectors? Therefore b = (-1, -2) is perpendicular to a.
The vectors perpendicular to a = (8, - 4) are
Solution (1, 2), (2, 4) (-1, -2) .
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ = ( ). Therefore, ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ are equal
vectors. Method 2
a = (8, - 4)
Zero Vector Let the vector perpendicular to a = (8, - 4) be
It is a vector that has no component. This b = (x, y).
means that both x and y components are zero. This means that = -1
For example, ⃗⃗⃗⃗⃗ = ( ) But = and =

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 × = -1 -4×-2=8
× 16 = - 8
× = -1
8≠-8
⇒ = so x = 1 and y = 2
The vectors are not parallel
The vector perpendicular to a = (8, - 4) is
b = (1, 2)
2. m = ( ) n=(

Exercises 9.7
Find at least one vector that is perpendicular Solution
to each of the following: For two vectors to be parallel, their cross product
1. a = (12, 3) 2. b = (2, -5) 3. a = (-6, -7) is equal
⁄
4. a = (- 4, 10) 5. a = (-3, -5) 6. a = (8, 2) ( ) =( )
4×2=8
Parallel Vectors
× 16 = 8
Two or more vectors are parallel if they are
multiples of each other. For eg, if a = (2, 4),: 8≠8
⇒-1a = -1(-2, -4) The vectors are parallel
a = (2, 4) = (1, 2)
Exercises 9.8
2a = 2(2, 4) = (4, 8)
A. Given that a = (6, -9), b = (10, 7), c = (- 4, -12)
The vectors (2, 4), (-2, -4), (1, 2), (4, 8) are said
and d = (8, 1), e = (- 6, 9) and f = (2, 6)
to be parallel.
i. Name two vectors that are parallel.
ii. Find any three vectors parallel to vectors a, b
Worked Example
and c.
State whether the given pair of vectors are
parallel or not.
B. 1. A is the point (4, –1), B is (11, 5) and C is
1. m = ( ) n=(
(6, 2). Find the coordinates of D, such that: i. ⃗⃗⃗⃗⃗
= 2⃗⃗⃗⃗⃗ ii. ⃗⃗⃗⃗⃗ = 2⃗⃗⃗⃗⃗
Solution
2. In the figure below, BQ and RP are parallel;
For two vectors x and y to be parallel, x = ky
QP and BR are parallel. ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ =
where k is a scalar.
For y coordinate, - 4 × - 4 = 16 T 2⃗⃗⃗⃗⃗ . ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ represent vector a and b
respectively. Q
For x coordinate, ×-4=2≠-2 F
C
Not parallel vectors b
P
A a B
Method 2
For two vectors to be parallel, their cross product R
is equal. i. Express in terms of a and b, the vector
⁄
( ) =( ) represented by:

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 a. ⃗⃗⃗⃗⃗ b. ⃗⃗⃗⃗⃗ c. ⃗⃗⃗⃗⃗ d. ⃗⃗⃗⃗⃗ Solution
ii. Why are C, B, R collinear? |⃗⃗⃗⃗⃗ | = √(
= √ = √ = 4 units
2. In the figure below, M is the midpoint of AB,
5. If x = ( ) and y = ( ), find correct to one
and MN//BC. ⃗⃗⃗⃗⃗ represents vector c, and ⃗⃗⃗⃗⃗
represents vector b decimal place /x + y/.
A
Solution
= ( ) and y = ( ),
M N x+y=( )+( )=( )

B C
/ x + y/ = √ (
i. Show that ⃗⃗⃗⃗⃗⃗⃗ represents ( = √ = √ = 7.1(1 d. p)
ii. What can you say about N?
Exercises 9.9
Magnitude or Length of a Vector Find the length of the following vectors;
If ⃗⃗⃗⃗⃗ = ( ) then the magnitude or length of ⃗⃗⃗⃗⃗ 1. ⃗⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗⃗⃗ = ( ) 3. ⃗⃗⃗⃗⃗ = ( )
is written as: /⃗⃗⃗⃗⃗⃗ / = √ 4. ⃗⃗⃗⃗⃗ = ( ) 5. ⃗⃗⃗⃗ = ( ) 6. ⃗⃗⃗⃗⃗ = ( )

Worked Examples Challenge Problem

1. If ⃗⃗⃗⃗⃗⃗⃗ =( ), find the length of ⃗⃗⃗⃗⃗⃗⃗⃗ . ABCD is a square of side 1 unit. ⃗⃗⃗⃗⃗ represents
vector u, ⃗⃗⃗⃗⃗ represents v. Name line segments
Solution representing – u, – v, u + v and u – v . Calculate
| ⃗⃗⃗⃗⃗⃗⃗ | = √ + and
=√ =√ = 13units
Scalar Multiplication
2. Find |⃗⃗⃗⃗⃗ | if ⃗⃗⃗⃗⃗ = ( ) It is the act of multiplying a number by the x and
y components of a vector.
Solution If a = ( ) then k × a = ( ) also written aska =
|⃗⃗⃗⃗⃗ | = √ =√ = √ units
( ).k is called a scalar and kais called a
3. Find the length of ⃗⃗⃗⃗⃗ if ⃗⃗⃗⃗⃗ = ( ) scalarmultiplication.

Solution Worked Examples

|⃗⃗⃗⃗⃗ | = √ 1. Given that a = (3, 2) and b = (-4, 3). Find:
=√ =√ = 5 units (i) 3a (ii) -5b

4. What is the magnitude of ⃗⃗⃗⃗⃗ = ( )? Solution

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(i) 3a = ( )= ( ) Solution
6 (r + 2s)
(ii) -5b = ( ) =( )
= 6[( )+ 2( )]
= 6[( ) + ( ) ]
2. Given that c = (4, 7) and d = (3, 6) Find:
a. 2c b. 3d c. 2d d. - 4d = 6[( )]
= 6( ) = ( )=( )
Solutions
(a) 2c = ( ) = ( ) 6. Given that u = ( ) and v = ( ), find:
(b) 3d = ( ) = ( ) (u + v).

(c) 2d = ( ) = ( )
Solution
(d) - 4d = ( ) = ( ) (u + v)
= [ ( ) + ( )]
Exercises 9.10 ⁄
= [( )+ (⁄)
If p = ( ), t = ( ) r = ( ) and u = ( ), find:
= [ ( ) + ( )]
1) −2r 2) − p 3) −3t 4)
= [( )]
Addition of Vectors ⁄ ⁄
= [( )]= ( * = ( )
Two or more vectors can be added by adding ⁄
their respective x and y components. That is, if
A = (x1, y1) and B = (x2, y2), then : Exercises 9.11
⇒⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = a + b = ( ) Given that a = ( ), b = ( ), c = ( ), find:
1. +c 2. 2a + 3b 3. c + b
Worked Examples 4. c + 4b 5. 4a + 2c 6. b + a
(1) If m = ( ) and n = ( ). find m + n.
Subtraction of Vectors
Solution The difference between two or more vectors
m+n=( ) ( )= ( )=( ) is found by subtracting the respective x and y
components of the vector. That is: if A = (x1, y1)
(2) Find the sum of s and t, if s = ( ) and and B = (x2, y2) then:
⃗⃗⃗⃗⃗ – ⃗⃗⃗⃗⃗ = a – b = ( ) − ( ) = ( )
t=( )

Solution Worked Examples

s+t= ( )+ ( )=( )=( ) (1) If a = ( ) and b = ( ), find a – b and b – a.

Solution
5. If r = ( ) and s = ( ) calculate, 6(r + 2s)

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a–b=( )−( )=( )=( ) Relating Free Vectors and Position Vectors
⃗⃗⃗⃗⃗ = ( ) − ( )
4 B
b–a=( )−( )=( )=( ) =( )
y
2. If x = ( ) and y = ( ) find 3x – y. A =( )
1
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
Solution
1 x 5
x= ( ) and y = ( ) ⃗⃗⃗⃗⃗ = b – a
3x – y = 3( ) – ( ) For all free vectors such as ⃗⃗⃗⃗⃗ , ⃗⃗⃗⃗⃗ , ⃗⃗⃗⃗⃗ , ⃗⃗⃗⃗⃗ etc,
=( )–( )=( )=( ) the following generalizations can be made:
(i) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗ (ii) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
3. If r = ( ) and s = ( ), calculate 2r – 3s.
(iii) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗ (iv) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
Solution
r = ( ) and s = ( ) Worked Examples
Given that a = (3, 6), b = (-7, 2), c = (2, 5)
2r – 3s = 2( ) – 3( )
and d = ( ). Find:
=( )–( )= ( )= ( )
(1) ⃗⃗⃗⃗⃗ (2) ⃗⃗⃗⃗⃗ (3) ⃗⃗⃗⃗⃗
(4) ⃗⃗⃗⃗⃗ (5) ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
4. Simplify; ( ) + ( ) + ( )

Solution
Solution
(1) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
( )+ ( )+ ( )
= ( ) – ( )= ( )= ( )
⁄ ⁄ ⁄
=( *+( *+( *
⁄ ⁄ ⁄

=( )+( )+ ( )=( ) (2) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗

= ( ) − ( )=( )=( )
Exercises 9.12
1. If u = ( ) and v = ( ), find 2u + 3v. (3) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
=( )–( ) =( )=( )
2. If a = ( ) and b = ( ), find 2a – b.

3. If r = ( ) and s = ( ), find r + 2s (4) ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗

4. Find p + 2q + r, if p = ( ), q = ( ) and =( )–( )=( )=( )

r = ( ),
(5) ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ − ⃗⃗⃗⃗⃗
5. Given the vectors r = ( ), p = ( ) and = ( ) −( ) + ( ) − ( )
q = 2p − r, find q. =( )+( )= ( )

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Exercises 9.13 Solution
1. Find ⃗⃗⃗⃗⃗ for the points P(-2, 3), Q(4, 7) Method 1
A (3, 4) and B = (7, 12)
2. A is (2, 5), B is (6, 12), C is (-3, 10) and D is M=[ ( , ( ]
(8, 19). Convert in component forms the
M=[ ( , ( ]
vectors:⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗
M=[ ( , ( ]
3. Given P (-7, 1), Q (-3, 6), R (7, 2) and S (3, -3) M=(
i. Find ⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ . Therefore the mid – point is (5, 8)
ii. Calculate the lengths of ⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ .
Method 2
A (3, 4) and B = (7, 12)
4. The point A is (2, 3) and ⃗⃗⃗⃗⃗ is( ). What are
M = (a + b)
the co-ordinates of B?
⁄
M = [( ) + ( )] = ( ) = ( *= ( )
⁄
5. If P, Q and R are the points (2, 1), (-2, 3) and
(1, 4) respectively, find the vector:
Exercises 9.14
⃗⃗⃗⃗⃗ – ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
1. Given that M = (6, 5), N = (14, 11) and
P = (2, 3) ind C, the mid points of:
Mid – Point of a Vector
i. ⃗⃗⃗⃗⃗⃗⃗ ii.⃗⃗⃗⃗⃗⃗⃗⃗ iii. ⃗⃗⃗⃗⃗⃗
If M is the midpoint of AB, then;
m= ( a + b). If A is the point (x1, y1) and B is 2. O is the origin, A is (3, –1) and B is (5, 5).
the point (x2,y2) in the OXY plane, then the Given that ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ ,
midpoint, m, of AB follows that MA = MB. i. find the coordinates of C.
ii. find also the midpoint of AC and the length of
A B (x2,y2)
(x1,y1) M AC.
)
)
M=[ ( , ( ] The Triangular Law of Vectors
Consider the diagram below:
C
In summary, use any of the following methods to 5
find the midpoint of a line. 4
3
1. M = [ ( , ( ]
1 A B
2. M = (a + b) 2
1 2 3 4 5

Worked Examples From the diagram, ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗

+ ⃗⃗⃗⃗⃗ . ⃗⃗⃗⃗⃗ is
If A (3, 4) and B (7, 12) are points in the OXY said to be the resultant vector of ⃗⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗ .
plane, find, C, the midpoint of AB. ⇒⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗

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Similarly, if ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ then, ⃗⃗⃗⃗⃗ is the But ⃗⃗⃗⃗⃗ = r – q, by substitution
resultant vector of ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ . ( )=r–( )
( )+( )=r
Worked Examples
( )=r
1. Given that ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ) find the
( ) =r
resultant vector.
⃗⃗⃗⃗⃗ = t – p, by substitution
Solution
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ ( )=t–( )
= ( ) + ( )=( ) ( )+( )=t
( )=t
2. If ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ). Find ⃗⃗⃗⃗⃗ . ( ) =t
⃗⃗⃗⃗⃗ = r – t, but r = ( ) and t = ( )
Solution ⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
= ( )+ ( )=( ) 6. ABCD is a parallelogram, with vertices
A(x, y), B(5, 7), C (4, 3) and D (1, 2).
3. Find ⃗⃗⃗⃗⃗⃗ if ⃗⃗⃗⃗⃗⃗⃗ ( ) and ⃗⃗⃗⃗⃗⃗ = ( ) i. Find ⃗⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ and hence find the values of x
and y.
Solution ii. Calculate the magnitude of ⃗⃗⃗⃗⃗ .
⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ = ( ) + ( )= ( ) Solution
i. ⃗⃗⃗⃗⃗⃗ = b – a
4. If ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ), find ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )

Solution
⃗⃗⃗⃗⃗⃗⃗ = c – d
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )=( )
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ – ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )=( )
–
For parallelogram ABCD, ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗

5. P, Q, R and T are points in the Cartesian plane. ( ) =( )

The coordinates of P and Q are (4, 1) and (-3, 2) ⇒ 5 – x = 3 and 7 – y = 1
respectively. ⃗⃗⃗⃗⃗ = ( ), ⃗⃗⃗⃗⃗ = ( ). Find ⃗⃗⃗⃗⃗ 5 – 3 = x and 7 – 1 = y
x = 2 and y = 6
Solution
ii. A (2, 6) and C (4, 3)
P(4, 1), Q(-3, 2), ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( )
⃗⃗⃗⃗⃗ = c – a
⃗⃗⃗⃗⃗ = r – t

Baffour – Ba Series, Core Maths for Schools and Colleges Page 245
⃗⃗⃗⃗⃗ = ( ) – ( ) = ( ) Solution
i. P(6, 4), Q(-2, -2) and R(4, -6)
/⃗⃗⃗⃗⃗ / = √ = √ =√ = 3.6 units M = ( p + q)
M= [( )+ ( )]
7. The vertices of a triangle ABC are A(1, -3), ⁄
B(7, 5) and C(-3, 5) M= ( )= ( *=( )
⁄
i. Express ⃗⃗⃗⃗⃗ , ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ as column vectors.
ii. Show that the triangle ABC is isosceles. S = ( p + r)
iii. Find the coordinate of the midpoint of ⃗⃗⃗⃗⃗
S= [( ) + ( )]
iv. Find the equation of ⃗⃗⃗⃗⃗
⁄
S= ( )=( *=( )
⁄
Solution
i. A(1, -3), a = ( ), B(7, 5), b = ( )and C(-3, 5),
ii. ⃗⃗⃗⃗⃗⃗⃗ = r – q
c=( ) ⃗⃗⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )
⃗⃗⃗⃗⃗ = b – a = ( ) – ( ) = ( )
⃗⃗⃗⃗⃗ = c – b = ( ) – ( ) = ( ) ⃗⃗⃗⃗⃗⃗⃗ = s – m
⃗⃗⃗⃗⃗ = c – a = ( ) – ( ) = ( ) ⃗⃗⃗⃗⃗⃗⃗ = ( ) – ( ) = ( )

ii. To show that triangle ABC is isosceles find; iii. ⃗⃗⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗⃗⃗ = ( )
⃗⃗⃗⃗⃗⃗⃗⃗ , /⃗⃗⃗⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ 2⃗⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗⃗
/⃗⃗⃗⃗⃗ / = √ = √ = 10 units ⃗⃗⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗⃗ are parallel vectors

/⃗⃗⃗⃗⃗ / = √ = √ = 10 units iv. Equation of ̅̅̅̅

M = ( ) and S = ( ).
/⃗⃗⃗⃗⃗ / = √( ( = √ = 8.9 units Gradient of ̅̅̅̅ = =–
Since /⃗⃗⃗⃗⃗ / = /⃗⃗⃗⃗⃗ /, ∆ABC is isosceles y – y1 = m(x – x1)
(x1, y1) = (2, 1)
Some Solved Past Questions
y–1=– (x – 2)
1. P(6, 4), Q(-2, -2) and R(4, -6) are the vertices
of triangle PQR. 3(y – 1) = -2 (x – 2)
i. Determine the coordinates of M and S, the 3y – 3 = -2x + 4
mid points of ̅̅̅̅ and ̅̅̅̅ respectively. 2x + 3y –3 – 4 = 0
2x + 3y – 7 = 0
ii. Find ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗ .
The equation of ̅̅̅̅ is 2x + 3y – 7 = 0
iii. State the relationship between ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗ .
iv. Find the equation of ̅̅̅̅.
2. In ∆ABC, ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ). If M is

Baffour – Ba Series, Core Maths for Schools and Colleges Page 246
the mid point of ⃗⃗⃗⃗⃗ , express ⃗⃗⃗⃗⃗⃗ as a column
vector. 4. A and B are the points (–1, 1) and (3, 4)
respectively,
Solution i. write down the components of the position
⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ) vectors a and b .
C ii. write down the components of the vector
From triangular law,
represented by ⃗⃗⃗⃗⃗ .
⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
iii. calculate the magnitude of b – a
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ – ⃗⃗⃗⃗⃗
B
⃗⃗⃗⃗⃗ = ( ) – ( ) A 5. A (4, 7) is the vertex of triangle ABC.
⃗⃗⃗⃗⃗ = ( )=( ) ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ).
a. find the co-ordinates of B and C.
If M is the mid point of ⃗⃗⃗⃗⃗ , b. if M is the midpoint of the line ⃗⃗⃗⃗⃗ find ⃗⃗⃗⃗⃗⃗ .
M = ⃗⃗⃗⃗⃗
C 6. A (-2, 3), B (2, -1), C (5, 0) and D (x, y) are
⁄
M= ( )= ( *=( ) the vertices of the parallelogram ABCD.
⁄
a. Find ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗⃗ Hence find the coordinates D.
From ∆MBC, b. Calculate, correct to one decimal place,
A M B
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ .

( ) + ( ) = ⃗⃗⃗⃗⃗⃗
7. The points P, Q, R, and S are vertices of
⃗⃗⃗⃗⃗⃗ = ( )
parallelogram in the Cartesian plane. The co-
ordinates of P and R are (-8, 2) and (5, -2)
Exercises 9.15 respectively and ⃗⃗⃗⃗⃗ ( ). Find:
1. Given that ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ) i. the co-ordinates of Q and S,
Find: (i) ⃗⃗⃗⃗⃗ (ii) ⃗⃗⃗⃗⃗ (iii) | ⃗⃗⃗⃗⃗ ii. the magnitude of PR.

8. a. A, B, C and D are four points such that

2. If A = (-6, -9), B = (8, 5) and C = (0, 3),
A (-3, 2), C (6, 3), ⃗⃗⃗⃗⃗ ( ) and ⃗⃗⃗⃗⃗ ( ).
find ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ .
Calculate:
3. O is the origin and A and B are the points (4, 5) i. the coordinates of B and D,
and (–1, 3) respectively. ii. the vectors ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗
i. Express 2a – 3b in component form. b. what is the relationship between ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗
ii. What are the components of ⃗⃗⃗⃗⃗ ?
iii. What are the components of ⃗⃗⃗⃗⃗ , where P is 9. The vertices of a triangle are P ( 1,-3), Q( 7, 5)
R (-3, 5).
the midpoint of ⃗⃗⃗⃗⃗ ?
iv. Calculate the length of AB, leaving your i. Express ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗ as column vectors.
answer in surd form. ii. Show that triangle PQR is isosceles.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 247
10. PQRS is quadrilateral with P (2, 2), S (4, 4) 2. If M = ( ), N = ( ) and R= ( ), find the
and R (6, 4). If ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ , find the coordinates of Q. values of a and b if M – N = R.

Challenge Problem Solution

1. P(-1, -2), Q(5, k), R(8, 2) and S(h, 1) are the M–N=R
four vertices of the parallelogram PQRS. Find the ( )–( ) = ( )
values of h and k. −( )=( )−( )
−( ) =( )
2. P(-1, 5), Q(-2, 1), R(3, -2) and S(a, b)are the
four vertices of a parallelogram, with PQ parallel −( )=( )
to RS. Find two pairs of coordinates of the point S. ( )=−( )
( ) ( ),
3. A rhombus ABCD has A at (0, 2) and B at (5,
a = -3 and b = -10
3). If the diagonals intersect at (0, 3), find the co-
ordinates of C and D.
3. Given that ( ) = ( ) find the values of x

4. In ∆ABC, ⃗⃗⃗⃗⃗ = ( ) and ⃗⃗⃗⃗⃗ = ( ). If P is the and y.

mid point of ⃗⃗⃗⃗⃗ , express ⃗⃗⃗⃗⃗ as a column vector. Solution

Ans ( )
( )= ( )

Vector Equality ( )= ( )
Two or more vectors are said to be equal if the x
( )=( )
and y components are of the same value. That is
x = 7 and y = 1
if A = ( ) and B = ( ), then A and B are equal
vectors written as A = B. 4. If ( ) + ( ) = ( ). Find m and n.

Worked Examples Solution

1. Given that A = ( ), B = ( ) and C =( ), find ( )+( )=( )
the values of x and y if A + B = C. ( )=( )
( )=( )
Solution
A + B = C
⇒13 + m = 20
( )+ ( ) = ( ) m = 20 –13 and n = 14
( )=( ) m = 7 and n = 14

( )= ( ) Solved Past Question

( )= ( ) The vectors a = ( ), b = ( ) and c = ( )are in
the same plane. If 3a – 2b = c, find:
Baffour – Ba Series, Core Maths for Schools and Colleges Page 248
i. the vector b 5. Find the values of x and y in the equation,
ii. and express your answer in the form ( ) ( ) ( )
√ where p and q are integers and d = b – c.
6. If p ( ) and = ( ),
Solution
i. calculate q – p , correct to one d. p,
i. a = ( ), b = ( ), c = ( ) ii. find the vector r such that p + r =2q.
3( ) – 2( ) = ( )
The Values of Scalars in Equal Vectors
( )–( )=( )
Given a vector equation with two different
( )–( )=( ) scalars, such as m( ) + n( ) = ( ), where m and
( )=( ) n are scalars, the values of the scalars (m and n)
are found as follows:
⇒ 2x = -6 and 2y = 10 I. Expand the given equation
II. Equate the x component of the L.H.S of the
x= and y =
equation to the x – component of R.H.S. and
x = -3 and y = 5 name it as eqn (1)
b=( )=( ) III. Similarly, equate the y component of the
L.H.S of the equation to the y – component of
ii. d = b – c R.H.S. and name it as eqn (2)
d=( )–( )=( ) IV. Solve the pair of equations obtained
simultaneously to obtain the values of the scalars

| d | = √( =√ = 3 √ units
Worked Examples
If a = ( ), b = ( ) and c = ( ), find the scalars
Exercises 9.16
m and n such that c = ma + nb, where m and n are
A. 1. If A = ( ) and B = ( ), find the scalars.
values of x and y, if A is equal to B.
Solution
2. If the vectors ( ) =( ). Find the values a =( ), b =( ) and c = ( )
of V and U. c = m a +n b
( ) = m ( )+ n ( )
3. Find the values of a and b, if A = ( ) and B ( ) =( )+ ( )
= ( ) are equal.
⇒ – 4 = m + 2n…………….(1)
4. Find x and y from the vector equation –11 = -m + 3n……………(2)
( )+( )=( )
eqn (1) + eqn(2)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 249
-15 = 5n Comparing the two planes, it can be seen that the
n = -3 North – pole represents y – axis positive (+ ve),
the South Pole represents y – axis negative (–)
Put n = -3 in eqn (1) whilst the West represents x – axis negative (–)
- 4 = m + 2(-3) and the East side represents x – axis positive (+).
-4=m–6 That is; (N = + y), (S = - y), (W = - x) and (E = + x).
-4+6=m
Therefore in the Cartesian form, the following
2=m
m=2 generalizations can be made:
(m, n) = (2, -3) I. If ⃗⃗⃗⃗⃗ = ( ), then ⃗⃗⃗⃗⃗ can be written in
Cartesian form as⃗⃗⃗⃗⃗ = ( ),
Exercises 9.17
1.If a = ( ), b = ( ) and c = ( ), find scalars II. If ⃗⃗⃗⃗⃗ = ( ), then ⃗⃗⃗⃗⃗ can be written in
p and q such that c = pa + qb, Cartesian form as ⃗⃗⃗⃗⃗ = ( ),
iii. If ⃗⃗⃗⃗⃗ = ( ), then ⃗⃗⃗⃗⃗ can be written as
2. If a = ( ), b = ( ) and c = ( ), find:
i. m and n such that c = ma + nb, where m Cartesian form⃗⃗⃗⃗⃗ = ( ),
and n are scalars. iv. If ⃗⃗⃗⃗⃗ = ( ) then ⃗⃗⃗⃗⃗ can be written as
ii. if d = c – 2a
Cartesian form as ⃗⃗⃗⃗⃗ = ( ),
3. The vectorsp = ( ), q = ( ) and r = ( q – p) Note: Negative and positive signs are not used to
i. find the vector r. represent vectors in the Cartesian form. For
ii. if mp + nq = ( ), find m and n, where m and n example, ⃗⃗⃗⃗⃗ ( ) ( )
are scalars.
Worked Examples
4. If p = ( ) q=( ) and r= ( ), find: Write the following vectors in Cartesian form.
a. m and n, such that r = mp + nq, where m and n 1. ⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗ =( )
are scalars.
b. find│g│, if g =3q + r 3. ⃗⃗⃗⃗⃗⃗ =( ) 4. ⃗⃗⃗⃗⃗ =( )

Cartesian Form of a Vector Solutions

Comparing the cardinal points to the OXY, plane, 1. ⃗⃗⃗⃗ =( ) is represented in the x – y plane
N y as ( ), which is also equivalent to ( ) in the
+ ve Cartesian form. Therefore:
W 1. ⃗⃗⃗⃗ = ( ) ( ) 2. ⃗⃗⃗⃗⃗ = ( ) = ( )
E x x
-ve + ve
-ve 3. ⃗⃗⃗⃗⃗⃗ = ( ) = ( ) 4. ⃗⃗⃗⃗⃗ = ( ) = ( )
S y

Baffour – Ba Series, Core Maths for Schools and Colleges Page 250
Exercises 9.18 Let be the angle the vector makes with the x –
A. Express the vectors in Cartesian form: axis y
A

1. ⃗⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗ = ( ) 1. a = ( )
8
tan θ =
3.⃗⃗⃗⃗⃗⃗⃗ = ( ) 4. ⃗⃗⃗⃗⃗ = ( )
O 5 x
0
θ ( ) = 58 y
B. Express the vectors in the form ( )
where a and b are rational numbers B
2. b = ( )
1. ( ) 2. ( ) 3. ( )
2
4. ( ) 5. ( ) 6. ( ) Tan θ = x O
7
θ= ( )
Angle of a Position Vector with the x - axis
θ = 160 O 4
To find the angle a position vector makes with x
the x – axis:
3.c = ( ) 9
I. Write the vector in the form ⃗⃗⃗⃗⃗ = a = ( ) tan θ =
II. Make a sketch of the vector in the OXY plane C
y
as shown below: ( ) = 660
y
4. d = ( ) x 3 O
A(x, y)
tan =
11
y
( ) = 750
O x
x
y
III. Calculate the angle (the angle the vector Exercises 9.19
makes with the x – axis, using the ratio: Calculate the angle each of the following
vectors makes with the x – axis:
a. tan = = , b. θ = ( )
1. a = ( ) 2. b = ( ) ( )

IV. Avoid negative signs in the calculations ( ) 5. v = ( ) ( )

because they indicate the quadrant within which
the vector falls Magnitude – Bearing Vectors as Column or
Component Vectors [a units, b0 to ( )]
Worked Examples A vector written in the form (a units, b0) is
Find the angle the following vectors make with called a magnitude – bearing vector because a is
the x – axis to the nearest degree; the magnitude of the vector and b0 is the bearing
1. a = ( ) 2. b = ( ) 3. c = ( ) 4. d = ( ) of the vector. Such vectors can be written in the
Solution form ( ), called column vector.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 251
Method 1 4. If < ( falls in the
Reminder : SOH, CAH, TOA second quadrant.
Given ⃗⃗⃗⃗⃗ = ( to be written in the column  The angle the vector makes with the x – axis,
or component form ( ) : θ= ⇒( =( )
I. Make a sketch of the vector in the OXY
5. When the bearing is given in cardinal point
plane as shown below;
such as N E, N W, S W and S E, the
y
angle between the points, , is always formed
C(x, y) with the vertical axis (y – axis). The angle formed
a with the x – axis, =
y
θ
x
O x Summary y

II. Let the angle the vector makes with the x –

( * ( *
axis be θ. From the diagram,
cos θ= a a
x = a cos θ (
x
sin θ =
a a
y = a sin θ (
( ) = ( ) ( * ( *

Note:
1. If < then (a, b0) falls in the first Worked Examples
quadrant. Express the following as column vectors:
 The angle the vector makes with the x – axis. 1. ⃗⃗⃗⃗⃗ = (5km, 0250) 2. ⃗⃗⃗⃗⃗ = (7km, 1260)
= ⇒( =( ) 3. ⃗⃗⃗⃗⃗ = 10km, 2370) 4. ⃗⃗⃗⃗⃗⃗ = 3km, 3100)
5. ⃗⃗⃗⃗⃗ = (
2. If < ( falls in the
fourth quadrant. Solution
 The angle the vector makes with the x – axis, 1. ⃗⃗⃗⃗⃗ = ( lies in the first quadrant.
θ= ⇒( =( ) Let a = 5km and b =
The angle the vector makes with the x – axis, θ =
A
3. If < ( falls in
0
the third quadrant. θ= = 65 0
25
 The angle the vector makes with the x – axis, 0 θ
θ= ⇒( =( ) ⃗⃗⃗⃗⃗ = (5km, 0250 )
Required angle
=( )

Baffour – Ba Series, Core Maths for Schools and Colleges Page 252
=( )=( ) θ=
Required angle
θ=
θ= 0
2. ⃗⃗⃗⃗⃗ = ( lies in the fourth θ
230
quadrant. Let a = 7km and b =
⃗⃗⃗⃗⃗ = (
The angle the vector makes with the x – axis,
E
= =( )= ( ) =( )
Required angle
θ=
θ= 0 Method 2
360 This method also makes use of the fact that given
⃗⃗⃗⃗⃗ = ( ⃗⃗⃗⃗⃗ = ( = ( ), where a is the
B
=( )=( )=( ) magnitude of the vector and θ is the angle the
vector makes with the x – axis in the anti –
clockwise direction. Simply put;
3.⃗⃗⃗⃗⃗ = ( lies in the third quadrant.
I. For all given bearings less than 900,
Let a = 10km and b =
i.e 900 < b < 900 , θ = 900 – b0
The angle the vector makes with the x – axis, θ =
II. For all given bearings greater than 900,
Required angle i.e 900 < b < 3600 , θ = 4500 – b0
θ=
θ= 0
θ With this method, the quadrant where the vector
570
lies is not necessary. It shows up after the vector
⃗⃗⃗⃗⃗ = ( C
is expressed in the form:( )
=( )= ( )=( )
Note:
4. ⃗⃗⃗⃗⃗⃗ = ( lies in the second When the bearing is given in the form of cardinal
points such as:
quadrant. Let a = 3km and b =
a. N E, then b = <
The angle the vector makes with the x – axis,
⇒θ =
θ=
b. N W, then b = ( – ) + 2700 > 900
θ= =
⇒θ = –( –
D c. S W, then b = + >
⃗⃗⃗⃗⃗⃗ = (
⇒θ = –( +
=( ) 400 0

=( )= ( ) d. S E, then b = ( – >
Required angle ⇒θ = –( –

5. ⃗⃗⃗⃗⃗ = ( lies in the third Worked Examples

quadrant. Let a = 6km and b = Express the following as column vectors:
The angle the vector makes with the x – axis, 1. ⃗⃗⃗⃗⃗ = (

Baffour – Ba Series, Core Maths for Schools and Colleges Page 253
2. ⃗⃗⃗⃗⃗ = ( =( )=( )
3. ⃗⃗⃗⃗⃗ = (
4. ⃗⃗⃗⃗⃗⃗ = ( 5. ⃗⃗⃗⃗⃗ = (
5. ⃗⃗⃗⃗⃗ = ( Given S W, b = + >
θ= –( +
Solution For S W, b = 1800+ 230
1. ⃗⃗⃗⃗⃗ = ( = 1800 >
Let a = 5km and b = < θ= –
Θ=
θ= = 650 ⃗⃗⃗⃗⃗ = (
=( )= ( )
⃗⃗⃗⃗⃗ = (
=( )= ( ) Some Solved Past Questions
1. A(9, 5) and B(3, 11) are points in the OXY
2.⃗⃗⃗⃗⃗ = ( plane. If C is the midpoint of AB, find:
Let a = 7km and b = > i. ⃗⃗⃗⃗⃗ ,
θ= ii. the value of the acute angle between OC and
θ= = 3240 the x – axis, correct to the nearest degree.

⃗⃗⃗⃗⃗ = ( Solution
i. A(9, 5) and B(3, 11)
=( )= ( )
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = ( + b)
3. ⃗⃗⃗⃗⃗ = (
⃗⃗⃗⃗⃗ = [( )+ ( )]
Let a = 10km and b = >
⁄
θ= ⃗⃗⃗⃗⃗ = ( )= ( *= ( )
⁄
θ= =

ii. Let θ be the value of the acute angle between

⃗⃗⃗⃗⃗ = (
OC and the x – axis,
=( )= ( ) OC= ( )
tan θ =
4. ⃗⃗⃗⃗⃗⃗ = (
θ= ( ) = 530
Let a = 3km and b = 3100 > 900
θ=
2. M(9, 7) and N (7, 23) are points in the OXY
θ= =
plane.
a. Find the coordinates of the point R such
⃗⃗⃗⃗⃗⃗ = (

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that ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗⃗ I. Find the magnitude of ⃗⃗⃗⃗⃗ by using the
b. Calculate ; formula; / ⃗⃗⃗⃗⃗ = √
i. / ⃗⃗⃗⃗⃗ / II. Find the angle (acute), θ, the vector makes
ii. correct to the nearest degree, the angle that ⃗⃗⃗⃗⃗ with the x – axis by using the formula;
makes with the x - axis. tan θ = , ⇒θ = ( )
III. Identify the quadrant where the vector lies.
Solution Thus:
a. ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗⃗ a. If ( )is in the first quadrant, the bearing
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ + (⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗ ) –θ
⃗⃗⃗⃗⃗ = ( ) + [( ) ( )] b. If ( )is in the second quadrant, the bearing
⃗⃗⃗⃗⃗ = ( ) + ( ) +θ

⃗⃗⃗⃗⃗ = ( ) + ( ⁄ c. If ( )is in the third quadrant, the bearing

*=( )+( )=( )
⁄
–θ
d. If ( )is in the fourth quadrant, the bearing
b. i. / ⃗⃗⃗⃗⃗ / = √ = √ = 17 units
ii. Let θ be the angle ⃗⃗⃗⃗⃗ makes with the x axis; +θ
tan θ =
Worked Examples
θ= ( ) = 620
Express in magnitude – bearing form;
1. ⃗⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗ = ( )
Exercises 9.20
A. Express the following as column vectors: 3. ⃗⃗⃗⃗⃗ = ( ) 4. ⃗⃗⃗⃗⃗⃗ = ( )
1. ⃗⃗⃗⃗⃗⃗ = (9km, 0530) 2. ⃗⃗⃗⃗⃗ = (10km, 1060)
Solution
3. ⃗⃗⃗⃗⃗ = (14km, 2170) 4. ⃗⃗⃗⃗⃗ = (6km, 3320) 1. ⃗⃗⃗⃗⃗ = ( )
/ ⃗⃗⃗⃗⃗ = √ =√ = 6 units
B. Express as vectors of the form ( )
1. ⃗⃗⃗⃗⃗ = (5km, S440W) tan θ =
2. ⃗⃗⃗⃗⃗ = (12km, S600E) θ= ( )
3. ⃗⃗⃗⃗⃗⃗⃗ = ( θ= (nearest degree)
4. ⃗⃗⃗⃗⃗⃗ = ( ⃗⃗⃗⃗⃗ = ( )lies in the first quadrant. Therefore the
bearing, –θ
Column Vectors as Magnitude –Bearing But θ = , A
0
Vectors ( ) ⇒ – 59 15
O
Given the vector⃗⃗⃗⃗⃗ = ( ), it can be written in the
0
b =
form ( ) called magnitude – bearing ⃗⃗⃗⃗⃗ = (15 units, )
vecto rby going through the following steps:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 255
2. ⃗⃗⃗⃗⃗ = ( ) Exercises 9.21
A. Write as magnitude – bearing vector:
/ ⃗⃗⃗⃗⃗ = √ =√ = 8 units
1. ⃗⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗⃗ = ( ) 3. ⃗⃗⃗⃗⃗ = ( )
tan θ = 4. ⃗⃗⃗⃗⃗ = ( ) 5. ⃗⃗⃗⃗⃗ = ( ) 6. ⃗⃗⃗⃗⃗⃗ = ( )

θ= ( )= (nearest degree)
B. Express in the form ( ):

⃗⃗⃗⃗⃗ = ( ) lies in the second quadrant. 1. ⃗⃗⃗⃗⃗⃗ = ( ) 2. ⃗⃗⃗⃗⃗⃗ = ( ) 3. ⃗⃗⃗⃗⃗ = ( )

Therefore the bearing, +θ 4. ⃗⃗⃗⃗⃗ = ( ) 5. ⃗⃗⃗⃗⃗ = ( ) 6. ⃗⃗⃗⃗⃗⃗ = ( )
But θ = 300, B
8
⇒b0 = 2700 + 300 = 3000
Application
⃗⃗⃗⃗⃗ = (8 units, 3000) 30
0
O
UsingVector Approach to Solve Bearings
I. Make a sketch or accurate drawing of the
3. ⃗⃗⃗⃗⃗ = ( ) diagram representing the problem.
/ ⃗⃗⃗⃗⃗ = √ =√ = 10 units II. Apply the triangular law of vectors which
states that ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ , for a right triangle with
θ= ( )= (nearest degree) vertices, A, B and C.
III. Express the bearing of A from B as a vector
and the bearing of C from B as a vector and
⃗⃗⃗⃗⃗ = ( ) lies in the third quadrant. Therefore the
substitute in ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ to get the component
bearing, –θ
But θ = , O vector ⃗⃗⃗⃗⃗ = ( ).
⇒ – = 0
IV. To find the distance between A and C is
37
⃗⃗⃗⃗⃗ = (10 units, ) 10 to find the magnitude of ⃗⃗⃗⃗⃗ , such that:
C / ⃗⃗⃗⃗⃗ / = √
4.⃗⃗⃗⃗⃗⃗ = ( ) V. To find the bearing of A from C is to find the
/ ⃗⃗⃗⃗⃗⃗ = √ =√ = 13 units angle θ,⃗⃗⃗⃗⃗ makes with the x – axis, using the
relation, θ = tan-1( )
θ= ( )= (nearest degree)
VI. To obtain the bearing of A from C, take
⃗⃗⃗⃗⃗⃗ = ( ) lies in the fourth quadrant. measurement from the north pole of C to the line
A (direction of A).

Therefore the bearing, Worked Examples

+θ O
0 1. An aircraft flew from port A on a bearing of
39
But θ = , to another port B 80km apart. It then flew
⇒ + = 13 from port B on a bearing of to port C, a
D
⃗⃗⃗⃗⃗⃗ = (13 units, 1290) distance of 120 km. Calculate:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 256
i. the distance from port A to port C, to the 2. Three schools, P, Q and R are situated as
nearest kilometer; follows: Q is on a bearing of 5km, 1430 from P
ii. the bearing of port A from port C, to the and R is on the bearing 11 km, 0530from P.
nearest degree. i. Find the distance between school Q and School
R, to the nearest kilometer,
Solution ii. Find the bearing of school Q from school R to
the nearest degree.
θ C
R
A Solution θ
i. 11km
120km
80km
P
B
From the diagram and the triangular law of 5km From the triangular law,
vectors,⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
Resolving along the x – axis, Q
⃗⃗⃗⃗⃗ = ( )+( ) Resolving along the x – axis,
⃗⃗⃗⃗⃗ = ( )+( )=( ) ⃗⃗⃗⃗⃗ = ( )+ ( )
⃗⃗⃗⃗⃗ = ( )+( )=( )
/ ⃗⃗⃗⃗⃗ / = √( (
=√ = 144km (Nearest km)
/ ⃗⃗⃗⃗⃗ / = √( (
ii. From the diagram, bearing of A from C / ⃗⃗⃗⃗⃗ / = √ = 12km (Nearest kilometer)
= 900+ 900 + (900 – θ0)
ii. From the diagram, bearing of Q from R
Angle ⃗⃗⃗⃗⃗ makes with the x – axis, = 900 + 900 + ( 900 – θ0 )
tan θ =
Angle ⃗⃗⃗⃗⃗ makes with the x – axis,
tan θ = ( )
tan θ = = ( )
-1
θ = tan ( )
0
θ = tan-1( ) = 610 at the x – axis
θ = 16 (nearest degree) at the x – axis
Bearing of Q from R;
Bearing of A from C;
= 900 + 900 + ( 900 – θ0 ) C = 900 + 900 + ( 900 – θ0 ) 0
R
0 0 0 0 0 61 0
= 900 + 900 + ( 900 – 160 )
16 0
74
= 90 + 90 + ( 90 – 61 ) 29
= 2540 = 2090 Q
A
⃗⃗⃗⃗⃗ = (144km, 2540)compare to solution at pg 274 ⃗⃗⃗⃗⃗ = (12km, 2090)compare to solution at pg 274/5

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3. The bearing of P from X, 10km away is 0250. ⃗⃗⃗⃗⃗ =( )+( )
Another point Q is 6km from X and on a bearing
of 1620. Calculate: ⃗⃗⃗⃗⃗ = ( )= ( )
i. the distance PQ; ii. the bearing of P from Q.
/ ⃗⃗⃗⃗⃗ / = √( (
P
Solution / ⃗⃗⃗⃗⃗ / = √ = 114.3390 km
i. From the diagram, The distance from the finishing point is
⃗⃗⃗⃗⃗ =⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ 10km 114km (to the nearest kilometer)
⃗⃗⃗⃗⃗ = – ( The angle ⃗⃗⃗⃗⃗ makes with the x - axis
) +( )
0
θ = tan-1( )
⃗⃗⃗⃗⃗ = ( ) X 65
72
0 θ = 690 (nearest degree) at the x – axis
⃗⃗⃗⃗⃗ = ( ) 6km 180 K
Bearing of K from F;

114km
= 900 – 690 = 210
/ ⃗⃗⃗⃗⃗ / = √( ( Q
⃗⃗⃗⃗⃗ = (114km, 0210) 68
0
/ ⃗⃗⃗⃗⃗ / = √ = 15km (Nearest kilometer) F
Exercises 9.22 5
ii.From the diagram, bearing of P from Q
Use the vector method in each case.
= 900 – θ0
1. Two boats A and B leave a port at the same
time. A travels 15km on a bearing of 0200 while B
Angle ⃗⃗⃗⃗⃗ makes with the x – axis,
travels 14km on a bearing of 2900. Calculate,
tan θ = ( ) correct to two decimal places, the
θ = tan-1( ) = 810 at the x – axis a. distance between A and B,
P b. bearing of A from B.
Bearing of P from Q;
15km

0 2. From Bamako, Enugu is 1800km away on a

= 900 – θ0 9
0 bearing of 1660 and Monrovia is 900 km away on
= 900 – 810 = 0090 Q 81
a bearing of 2060. Calculate :
⃗⃗⃗⃗⃗ = (15km, 0090) i. the distance between Enugu and Monrovia,
K ii. the bearing of Enugu from Monrovia.
680
5. From Kwadaso, I travelled 100km
on a bearing of 1580, and then 3. The bearing of Q from P is 1500 and the
80km on a bearing of 2600. 100km bearing of P from R is 0150. If Q and R are 24km
Find the distance and and 32km respectively from P:
bearing from my finishing point.
i. represent this information on a diagram;
680 o ii. calculate the distance between Q and R, correct
100
Solution to two decimal places;
F 80km iii. find the bearing of R from Q, correct to the
From the diagram,
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ nearest degree.

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10 STATISTICS I Baffour- Ba Series

Idea of Statistics instance, a tree of height 9.8m might actually be

It is a branch of Mathematics which deals with 9.812. Age is also another example of continuous
the collection and study of numbers in order to data. Although the values are usuallytaken as
get facts or information. whole numbers like 15, 18, 26 years, they can
Data: It is information collected from various assume intermediate values like 15 or 18 years,
sources through observation or from results of an
2 months…..
experiment. These are raw data, array and
grouped data.
Exercises 10.1
Determine whether the following data is
Types of Data
continuous or discrete.
Raw Data: It is a data which is in the form in
1. The number of children in the various families
which it was collected from source.
in the community .
2. The monthly income of workers in a factory.
Array: It refers to data which has been arranged
3. The marks obtained by a student in an
in ascending or descending order of magnitude.
examination.
4. The distance travelled by a day student to
Grouped Data: It is a data which has been
school each school day.
arranged into classes like 0 – 9, 10 – 19 etc.
5. The amount of pocket money given to students.

Discrete data: It is data that can be counted, for

Sources and Collection of Data
example the number of people, cattle, sheep,
A data or information can be collected from many
table, chair etc. The count of discrete data is
sources such as schools, hospitals and business.
whole numbers, since we cannot get half a
person, or 2 sheep on a farm. Similarly, the In schools, records such as attendance, personal
number which appears on a die when thrown can records, results of an examination and others are
only be 1, 2, 3, 4, 5 or 6. Other examples of kept. At the hospitals, records of attendance,
discrete data are vehicles passing a station, and diseases treated number of patients treated etc are
the grade obtained by a student in a class. also collected and kept.

Continuous data: It is a data which result from In order to collect data, a survey must be
measurement, for example, the height of a person, conducted. The result of a survey is called a raw
the height of a tree or the, the time taken to data. For example, 11 students in Asuofua D/A
complete an activity. The count of continuous J.H.S. one were asked to mention the days of the
data is not a whole number but rounded off. For week in which they were born and the following

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results were obtained; Thursday, Saturday set of data.The mode is not a calculated value but
Thursday, Wednesday, Tuesday, Friday, a sight value which must be carefully identified
Wednesday, Monday, Tuesday, Wednesday, by sight.
Sunday,
Worked Example
When data are collected from a source as above, The ages of 10 K.G pupils are recorded as; 4, 4,
it is called a Raw Data. 5, 5, 6, 5, 6, 6, 6, 6. Find the modal age.

Raw data is not in organized form. There is Solution

therefore the need to process raw data in an The mode or the modal age is 6 because it‟s the
organized form called information to enhance number that appears most.
understanding. From the above example,
information can be obtained in a tabular form as Median
follows: The median of a set of numbers is the middle
value or the arithmetic mean of the two middle
Day Tally No. of pupils values when they are ordered in ascending or
Sunday / 1 descending order of magnitude.
Monday // 1
Tuesday /// 2 Steps in finding the median of a raw data with
Wednesday /// 3 “n” numbers;
Thursday //// 2 Method I
Friday /// 1 I. Re – arrange the numbers in either ascending or
Saturday // 1 descending order.
II. Count to locate the middle value as the
In this case, one can easily know the number of median.
student born on each day of the week. III. If two numbers, a and b, appear as the middle
values, find the sum of the two and divide by 2 to
Measures of Central Tendencies
obtain the median. That is median = .
(Mode, Median and Mean)
The mode, median, and mean arethe three most
common examples of measures of central Method II
tendencies. They are also called averages. They I. Arrange the numbers in order of magnitude and
refer to a single digit which enables one to assess count to ensure that they are up to “n” numbers.
the position in which a group is with respect to II. If n is odd, then the median is the middle term.
others. That is; ( term or position
III. If n is even, then the median is the arithmetic
Mode mean of the two middle terms. That is:
The mode is the most commonly occurring ( and ( terms or positions
number or the number that repeats itself most in a

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Worked Examples Total number of entries, n = 10 (even) Median
1. Find the median of the following numbers: = ( and ( terms
47, 30, 56, 31, 55, 43 and 44
( = 5th and ( + 1 = 6th
Solution
Method I The median positions are 5th and 6th
Re-arrange the numbers in ascending order to get: 4, 4, 4, 5, 5, 5, 6, 6, 7, 7
30, 31, 43, 44, 47, 55, 56,
The middle number is 44. ⇒median = 44 6th Position number
th
5 Position number
Method II Median = =5
Re-arrange the numbers in ascending order to get:
30, 31, 43, 44, 47, 55, 56,
Mean
Total number of observations is 7
The mean is the ratio of the sum of all the values
Therefore, n = 7 and n is odd
in a raw data to the number of entries.
Median = ( term or position
= ( term or position The mean is denoted by ̅.Thus, given the set of
values: x1, x2, x3, x4, ..xn, the mean,
= ( = 4th position ∑
( ̅) = =
30, 31,43, 44, 47, 55, 56, Where n is the position of the last number of the
entries.
4th Position number
Worked Examples
Therefore, the median is 44.
1. Find the mode, median and mean of the
following numbers: 4, 24, 10, 17, 19, 21, 10
2. Find the median of the following numbers; 4,
5, 4, 6, 7, 5, 7, 4, 6, 5 Solution
Mode = 10 (Most frequently number)
Solution
Method I Median: Ordering the numbers in ascending
Re-arrange the numbers in ascending order to get: order, 4, 10, 10, 17, 19, 21, 24
4, 4, 4, 5, 5, 5, 6, 6, 7, 7 The middle number is 17. ⇒median = 17
The middle numbers are 5 and 5. Therefore,
median = =5 ̅= = = 15

Method II 2. The masses of 12 boxes measured to the

Re-arrange the numbers in ascending order to get: nearest kg are: 26, 23, 27, 28, 29, 22, 23, 27, 20,
4, 4, 4, 5, 5, 5, 6, 6, 7, 7 20, 24, and 25. Find the mode, median and mean

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Solution 3. The scores of 10 students in an examination are
Mode = 20, 23 and 27 given as follows; 45, 12, 75, 81, 54, 51, 24, 67,
19 and 39. What is the median and mean score?
In ascending order: 20, 20, 22, 23, 23, 24, 25, 26,
27, 27, 28, 29 4. The price of 12 commodities in cedis was
The middle numbers are 24 and 25. recorded as: 5, 2, 5, 3, 3, 5, 3, 2, 4, 3, 3, 2.
Median = = = 24.5 i. What the modal price?
ii. Find the median price.
Mean ( ̅ ) = iii. Determine the mean price.

Where n = number of entries = 11

5. The ages in years of 8 boys are: 14, 14, 15, 15,
̅= = = 5.36(2 d.p.) 12, 11 13, 10.What is the average age?

3. Find the mean, median and modal heights of Calculations Involving Mean
the following distribution: 170cm, 171cm, I. Represent any unknown number by any
173cm, 174cm, 177cm, 177, 184cm, 186cm, preferred variable.
II. Identify the total number of entries.
Solution III. Find the sum or total of the given data.
i. Mean ( ̅ ) = III. Identify the value of the mean, if given.
Where n = number of entries = 8 IV. Use the formula below and workout the value
̅= = 176.50 cm of the variable.
Mean =
ii. To find the median, arrange the numbers in
ascending order to obtain: 170cm, 171cm, Worked Examples
173cm, 174cm, 177cm, 177, 184cm, 186cm. 1. The mean of the numbers 4, 3, 3, and x is 5,
The two middle values are 174 and 177. Find the value of x.

Median = = 175. 50 cm Solution

̅= =5
iii. The modal height = 177cm ⇒ =4×5
10 + x = 20
Exercises 10.2 x = 20 – 10
1. The following temperature in °C was recorded
x = 10
in 10 cities in Europe; - 4, 5, 2, 0, - 6, - 4, 3, - 6,
- 4, -7. Find the modal temperature.
2. The average of the numbers 4, 10, 24, x and 16
is 13. Find the value of x.
2. The marks scored by 8 pupils in a science test
are 3, 7, 8, 8, 5, 8, 4, 8.What is the median mark?
Solution
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 ̅= = 13 Correct average age = = 66 years
⇒ = 13 × 5
54 + x = 65 5. The ages of 5 women are 48, y, 52, 50 and (2y
x = 65 – 54 = 11 – 5). If their average age is 47 years, find the age
of the oldest woman.
3. The average age of a class of 30 students is 18
years and the average age of another class of 45 Solution
students is 20 years. Find the average age of the ̅= = 47
students of the two classes.
= 47
Solution 145 + 3y = 5 × 47
If the average age of 30 students is 18 years 145 + 3y = 235
⇒the total age of 30 students = 30 × 18 3y = 235 – 145
= 540 years 3y = 90
y = 30
If the average age of 45 students is 20 years ⇒The ages of the women are 48, 30, 52, 50 and 55
⇒the total age of 45 students = 45 × 20 The oldest woman is 55 years old.
= 900 years
Exercises 10.3
Total number of students in the combined class 1. The following are the ages in years of members
= 30 + 45 = 75 students of a group:8, 11, 8, 10, 6, 7, 3x, 11, 11. If the
mean age is 9 years, find the value of x.
Total age of 75 students,
= (540 + 900) years = 1,440 years 2. The mean of 10 positive numbers is 16. When
another number is added, the mean becomes 18.
Average age for the two classes; Find the number added.
= = 19.2 years
3. The mean age of 4 men is 19 years, 11 month.
When a fifth man joins them, the mean age of all
4. The average test mark for a class of 20 students
the 5 men is 20 years 7 months. How old is the
is 65. Kofi‟s mark was recorded as 35 instead of
fifth man?
55. What is the correct average mark for the test?
4. The mean of 5 numbers x, 2, 3, 5, and 9 is 3,
Solution
find the value of x .
Total marks when 35 was recorded.
= 20 × 65 = 1300
5. 40% of the employees of a factory are workers.
All the remaining employees are executives. The
The difference between the correct and incorrect
annual income of each worker is Gh¢390.00 and
mark = 55 – 35 = 20

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the annual income of each executive is drawn to across the first 4 strokes. Thus //// for 5
Gh¢420.00. What is the average annual income and //////// // for 12.
of all the employees in the factory? Column 3 (frequency): is made up of the digital
representation of the tally.
6. A year ago, the average annual income of Jack Column 4 (fx): consist of the product of each
and Jill was Gh¢3,800.00, the average annual event and its corresponding frequency.
income of Jill and Jess was Gh¢4,800.00 and the
average annual income of Jess and Jack was ∑ is the sum of all the frequencies.
Gh¢5,800.00. What was the average annual ∑ is the sum of the product of frequencies and
income of the three people? their corresponding events.

7. The average of all scores in a certain algebra Worked Examples

test was 90. If the average of the 8 male students 1. The ages of 20 school children were recorded
score was 87 and the average of the female as follows;
students score was 92, how many females took 13 9 15 17 13 17 15 11 9 9
the test? 9 11 9 11 15 11 15 11 11 11
Make a frequency table for the data.
Frequency and Frequency Diagrams
The number of times an event scores in a given Solution
data is called its frequency. Age (x) Tally Frequency (f) fx
9 //// 5 45
Frequency Diagrams for Ungrouped Data 11 //// // 7 77
It is a table that shows the event and the number 13 // 2 26
of times each event occurs. It is usually divided 15 //// 4 60
17 // 2 34
into four (4) sections as shown below:
∑ = 20 ∑ =242

Range(x) Tally f fx
2. The table below gives family sizes recorded by
30 respondents. Construct a frequency table for
∑ ∑fx =
the data.
Column 1(x) : is made up of the events usually 5 4 3 4 4 5 4 4 4 3 3 3 5 4 3
arranged downward in increasing order. It is also 6 3 2 4 5 3 3 4 3 3 6 5 2 5 5
called the range.
Column 2 (Tally): consist of slight tilted strokes Solution
like /, that represents the number of
times an event occurs in a data. That is / for 1, // Sizes Tally Frequency fx
for 2, /// for 3 and //// for 4. However, at every (x) (f)
2 // 2 4
fifth successive occurrence, a horizontal line is
3 //////// 10 30
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 4 //// //// 9 36 column3 of the table. Note that the mode is not
5 //// // 7 35 the highest frequency (in column 3) but its
6 // 2 12 corresponding value in column 1.
∑f = 30 ∑fx = 117
From the table below;
Exercises 10.4
1. The scores obtained by 35 students in a Ages (x) A B C D E
Mathematics test are given below; Frequency (f) 3 1 2 5 3
17 16 16 16 17 18 16 18 18 17 18
17 19 18 19 19 18 18 17 20 19 20 The mode is D because it has the highest
20 19 17 17 17 20 16 20 19 17 17 frequency of 5, but the mode is not 5 because it is
16 19 the highest frequency. The mode is not selected
Construct a frequency table for the data. from column 3 but from column 1 of the
frequency distribution table.
2. Dinah tossed two dice 40 times and recorded
the total scores. The results are presented below: II. Mean
8 4 8 4 10 5 10 6 3 10 6 8 6 3 6 On the frequency distribution table, the mean ( ̅)
5 3 7 4 3 7 4 11 7 4 11 9 3 11 5 ∑
is calculated by the formula;( ̅) = ∑
5 10 3 3 11 3 7 3 4 3
Construct a frequency table for the data The respective values are then substituted in the
formula to obtain the mean.Care must therefore
3. A survey was carried out to determine the ages be taken to compute the summation of “f” and
of 40 children, the results are as follows: “fx” in order to obtain accurate values of ∑
16 14 13 13 15 11 13 11 12 13 and ∑ .
14 12 16 14 14 13 13 12 13 13
11 12 13 13 13 15 12 12 11 12 The value of the mean is usually corrected to 2
15 13 12 12 13 12 11 16 13 14 decimal places.
Construct a frequency diagram for the data.
Worked Example
Note:the frequency table can be drawn without Copy and complete the table below and use it to
the column 4 or fx. The column 4 becomes find the mean to 2 decimal places.
necessary when the questions demand a
calculation of the mean. x Tally Frequency(f) fx
1 // 2
Mode, Median and Mean on an Ungrouped 2 /// 3
Frequency Table 3 //// / 6
I. Mode 4 4
∑f = ∑fx=
The mode in the frequency table is any number at
column1 that has the highest frequency in
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Solution Note that to locate the middle of anything is to
divide by 2.
x Tally Frequency(f) fx If ∑ is odd, determine the median by adding 1
1 // 2 2 to ∑ to make it even and the sum divided by 2
2 /// 3 6 to give the position of the median. That is:
3 //// / 6 18 ∑
4 / 1 4 Median position = ( )th
∑f = 12 ∑fx = 30
By adding frequencies from the top to the bottom
∑ or the other way round or counting the strokes of
Mean ( ̅) = ∑
= = 2.50
the tally, the number which corresponds to the
∑
2. The table below shows the frequency median position, ( )th is the median.
distribution of the sizes of shoes produced by
a factory If ∑ is even, then two median positions which
∑ ∑
Size 5 6 7 8 9 10 11 are ( )th and ( )th + 1 are used to determine
Frequency 4 6 9 15 14 7 5 the two middle numbers, the sum of which is
divided by 2 to get the median value.
Calculate, correct to the nearest whole number,
the mean size of the shoes Worked Examples
1. The height in centimeters of some school
Solution children were recorded as: 165, 165, 155, 174,
∑
Mean ( ̅ = ∑ 159, 169, 155, 150, 155, 154
∑ i. Draw a frequency table for the distribution.
(5 × 4) + (6 × 6) + (7 × 9) + (8 × 15) + (9
ii. Use your table to find the mode and the
× 14) + (10 × 7) + (11 × 5) median of the distribution.
= 20 + 36 + 63 + 120 + 126 + 70 + 55 = 490
Solution
i.
∑ = 4 + 6 + 9 + 15 + 14 + 7 + 5 = 60
Height Tally Frequency(f)
∑ (cm)
(̅ = = = 8 (Nearest whole number)
∑ 150 / 1
iii. Median 154 / 1
To find the median from a frequency distribution 155 /// 3
table 159 / 1
165 // 2
I. Find the total frequency i.e. ∑ .
169 / 1
II. Identify whether ∑ an even number or an 174 / 1
odd number. ∑ = 10

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Mode is the highest frequency = 155cm i. The modal mark is the mark that has the highest
frequency = 6
Median: since ∑ = 10 (even number) ii. The total number of candidates who took the
∑ ∑
So ( )th and ( ) +1 test = 2 + 3 + 5 + 7 + 8 + 13 + 7 + 5 = 50
∑
th th
iii. Mean (̅) = ∑
= = 5.6
= ( ) and ( ) + 1
Median :∑ = 50 (an even number)
= 5th and 6th positions
∑ ∑
By adding frequencies from the top, the 5 th ( )th and ( )th + 1
position = 155 and the 6th position = 159
= ( )th and ( )th + 1
Median = = = 157cm.
= 25th and 26th position
= 5 and 6
2. The table below shows the marks scored out of
10 by some candidates in a test. Median = = = 5.5

Marks 1 2 3 4 5 6 7 8 3. The table below shows the distribution of ages

Frequency 2 3 5 7 8 13 7 5 in years of children who were treated in a clinic.

From the table, find: Age (yrs) 1 2 3 4 5

i. the modal mark. Frequency(f) 8 5 2 4 6
ii. how many candidates took the test?
iii. calculate the mean and median mark for the
test. Find the mean, mode and median.

Solution Solution

Ages Tally Frequency fx

Marks(x) Tally Frequency(f) fx
x (f)
1 // 2 2
1 //// /// 8 6
2 /// 3 6
2 //// 5 8
3 //// 5 15
3 // 2 6
4 //// // 7 28
4 //// 4 12
5 //// /// 8 40
5 //// // 6 25
6 //////// 13 78
∑f = 25 ∑fx =57
///
7 //// // 7 49
∑
8 //// 5 40 i. Mean ( ̅) = = = 2.28
∑
∑f = 50 ∑fx= 258
ii. The modal age is the age with the highest
frequency = 1

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iii. Median: ∑ = 25 (odd number) Exercises 10.5
∑ th th th th 1. The following is a record of scores obtained by
( ) =( ) = ( ) = 13 position.
th 30 J.H.S. 2 pupils in a test marked out of 5:
But 13 position = 2
5 3 2 4 5 2 4 3 1 3 3 4 2 3 4
The median = 2
5 3 4 3 2 4 3 1 2 3 3 2 4 2 1
a. Construct a frequency table for the data.
4. The table below shows the distribution of
b. Find the median marks.
marks scored by a group of students.
c. Calculate the mean of the distribution.
Marks 1 2 3 4 5 6 7 8 9 10
2. The table below gives the ages of the members
No. of
of a juvenile club.
student 2 2 3 3 k 2 2 0 1 1
Ages (yrs) 8 9 10 11
If the mean of the distribution is 4.6, find the
Frequency 5 10 6 8
value of k.
i. How many people are in the club?
Solution
ii. Determine the median and mean ages.
Marks Frequency (f) fx iii. What is the modal age?
1 2 2
2 2 4 3. The table below shows the distribution of 23
3 3 9 students visiting a clinic on a particular day.
4 3 12
5 k 5k Ages (x) 15 16 17 18 19 20
6 2 12 No. of stud 2 3 7 5 2 4
7 2 14
8 0 0
i. What is the modal age of the students?
9 1 9
10 1 10 ii. Find the mean and median age.
∑ = ∑ =
16 + k 72 + 5k 4. The score obtained by 35 students in a
mathematics test are given below:
∑ 12 26 15 14 14 12 16 26 12 16
Mean ( ̅ ) = ∑
= 4.6
15 15 14 26 26 15 14 19 12 19
⇒ = 4.6 26 19 19 19 14 15 26 12 14 15
72 + 5k = 4.6(16 + k) 26 12 15 26 12
72 + 5k = 73.6 + 4.6k i. Construct a frequency table for the data.
5k – 4.6k = 73.6 – 72 ii. Use your table to calculate the mean and
0.4k = 1.6 median.
k=4 iii. What is the mode of the distribution?

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Relative Frequency
The relative frequency of an event is the SCORE ON DIE FREQUENCY
probability of an event on a frequency table. To 1 5
be precise, it is the frequency of an event 2 6
compared to the total frequency. 3 11
The relative frequency of an event, 4 8
5 6
Rf = 6 4

Calculate the relative frequency of the event that

Worked Examples
the score is an even number.
1. A survey was conducted to determine the
number of children per family at Pasro D/A
Solution
J.H.S. 1. The results were recorded in the table
Total frequency
below:
= 5 + 6 + 11+ 8 + 6 + 4
= 40
No. of Chn/family No. of family
2 2 Even numbers on the die are 2, 4, 6.
3 10 Frequency of 2 + frequency of 4 + frequency of 6
4 9 = 6 + 8 + 4 = 18
5 7
6 2 Rf =

Find the relative frequency of;

Exercises 10.6
a. 3 children per family,
1. If a number is selected at random from the
b. 5 children per family.
table below, what is the probability that the
Solution number is 5?
a. From the table, the total number of families
= 2 + 10 + 9 + 7 + 2 = 30, Number 1 3 5 7 9
The number of families with 3 children per Frequency 25 15 8 10 2
family = 10
2. The frequency distribution table below shows
Rf =
the ages of some nurses:

b. Total number of families = 30

Age (yrs) 24 29 34 39 44
Number of families with 5 children per family = 7
Frequency 7 8 5 6 4
Rf = =
2. A fair die was tossed 40 times. The outcomes If a person is selected at random from the nurses,
are given in the table below; find the probability that the person chosen is;

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a. 34 years b. more than 34 years c= × 360

3. The frequency table below shows the age iv. Use the values of the angles calculated to draw
distribution of 20 pupils in a class. the sectors in the circle. Make sure that all the
angles calculated sum up to 3600.
Age (yrs) 9 10 11 12
Frequency 2 8 6 4 Drawing a Pie – Chart
1. Draw a sizeable circle using a compass and a
What is the probability that a pupil selected at well sharpened pencil.
random is:
a. 9 years old b. less than 12 yrs old.

Graphical Representation of Data

Data collected can be represented in various
forms such as; 2. Draw a line from the center of the circle to
i. Pie chart ii. Bar chart touch the circumference as a starting line.
iii. Histogram iv. Cumulative frequency curve
v. Leaf and stem plot

Pie – Chart
It is the diagrammatic representation of data or
3. Place the protractor on the radius (straight line)
information by using degrees or sectors of a
such that the base line and the center of the
circle. Pie charts are also called circle graph.
protractor coincide exactly with the radius and
the center of the circle respectively.
Calculating the Angles of a Pie Chart
Note:
To get the angles of a pie-chart:
All angles can be drawn in one direction, either
I. Remember the fact that the sum of angles in a
clockwise or anticlockwise. Continue the process
circle is 360
by using the second radius of the central angle
II. Find the sum of all the items under
drawn, to draw subsequent angles until all
consideration.
central angles are drawn.
III. Calculate the angle of each item by using the
Write the name and angle of each sector in that
formula;
sector.

An angle = 360 Worked Examples

For e.g. given the items a, b, and c 1. The table below shows the distribution of
Angle a = × 360 pupils in a J.H.S. 1 who speak some of the
Ghanaian languages.
b= × 360

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 Ghanaian No. of students who Item Calculation Angles
language speak the Language A × 360° 42°
Nzema 5
B × 360° 90°
Ga 20
Twi 30 C × 360° 156°
Ewe 25 D 72°
Fante 10 × 360°

i. Draw a pie chart for the distribution.

ii. What is the modal language?

Solution 156°
i. Total number of students who speak the D 72° C
language = 5 + 20 + 30 + 25 + 10 = 90.

Nzema = × 360°=20°
Ga = × 360° = 80° Nzema 3. The table below shows the performance of
Fante 0
Twi = × 360° = 120° 400 20
Solomon in his final examination.
0 1000 Ewe
Ewe = × 360° = 100° Ga 80 Subject Score
1200 English 54%
Fante = × 360° = 40°
Twi Mathematics 36%
Ga 68%
Science 50%
Not drawn to scale
Social studies 32%

i. The modal language is the language with the Draw a pie chart to represent this information,
highest frequency = Twi.
2. The following table shows the distribution of Solution
grades obtained by 120 students in an Total score = 54 + 36 + 68 + 50 + 32 = 240
examination.
Grade A B C D Item Calculation Angles
No. of students 14 30 52 24 English × 360° 81°
Maths × 360° 54°
i. Draw a pie chart for the distribution.
Ga × 360° 102°
Solution Science × 360° 75°
Total number of students
Soc studies × 360° 48°
=14 + 30 + 52 + 24 =120

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 Taxes Gh¢16
Miscellaneous Gh¢22

Ga i. Draw a pie chart for this information.

102° ii. What item received the highest expenditure?
iii. What fraction of the amount is spent on
81°
clothing?
English

3. The table below shows the number of students

who offered certain subjects in a
school.
Exercises 10.7
1. The Smith family has an annual income of Subject No. of students
Gh¢15,000.00. They planned to budget it in the
Mathematics 45
following way:
Physics 39
Chemistry 28
Expenses Amount(%)
Biology 14
Food 10%
Economics 36
Shelter 25%
History 18
Clothing 20%
Advancement 15%
Draw a pie chart for the above information
Operating expenses 18%
Savings 12% 4. In an election, the number of votes won by
five political parties A, B, C, D and E in a village
i. Make a pie chart to show the distribution of the were recorded as follows:
money.
ii. What fraction of the money is spent on Party A B C D E
savings? No. of votes 14 11 19 52 24

2. Mr. Green‟s income is Gh¢180.00 per week. i. Draw a pie chart to illustrate the information.
He spends his income as follows: ii. What percentage of the total votes did the
winner obtain?

Rent Gh¢36 Calculations Involving Pie - Charts

Food Gh¢66 Sometimes a central angle of a pie chart could be
Clothing Gh¢10 represented by a variable. To determine the value
Electricity Gh¢18 of the variable, find the sum of all the central
Medicals Gh¢12

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angles and equate it to 360°.For example, in the Savings;
diagram below; If 25° = Gh¢10, then 165° = = Gh¢66.00.
x° + 115° + 72° + 23° = 360°
The amount spent on savings is Gh¢66.00,
x° + 210° = 360° Others Food
x° = 360° 210° x0 1150 Total amount;
x = 150°
230 Savings If 25° = Gh¢10, then 360° = = Gh¢144.00
720 The total amount spent is Gh¢144.00,

Clothing Alternatively;
Applying Method of Direct Proportion Total amount
Also the value of an item could be given to = 10 + 20 + 48 + 66
determine the value of other items on pie a chart. = Gh¢144.00
Similarly, the value of the total item can be given
to find the value of other items. Such problems Method 2
are solved by method of direct of proportion. For Using the given value as a standard, express each
example, in the pie chart below; item as a proportion of the angle of Electricity
and the given value.
⇒Electricity : Rent
25° : 50° = 10 : x
Food =
Saving
120°
165°
= = Gh¢20.00

Amount spent on Rent is Gh¢20.00

If the amount spent on electricity is Gh¢10.00, Electricity: Food

the values of all the other items can be calculated 25° : 120° = 10 : x
as follows;
=
Method 1 = = Gh¢48.00
Rent; The amount spent on food is Gh¢48.00,
If 25° = Gh¢10, then 50° = = Gh¢20.00
The amount spent on rent is Gh¢20.00 Electricity: Savings
25° : 165° = 10 : x
Food; =
If 25° = Gh¢10, then 120° = = Gh¢48.00 = = Gh¢66.00
The amount spent on food is Gh¢48.00, The amount spent on savings is Gh¢66.00,

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Electricity: Total =
25° : 360° = 10 : x
= 2. The pie chart below shows the performance of
= Serwaa in her final examination.

x = Gh¢144.00 (Total Amount)

French
x° Social
Worked Examples Studies
Maths 25°
1. The pie chart below shows the monthly
expenditure of Mr. Awuah whose monthly
income is Gh¢180.00

x
Clothing Rent i. What is the angle for French?
500 ii. If Serwaa scored 60% in Social Studies, what
Fuel was her score in Economics?
Food
600
1200
Solution
Let x represent the angle for French;
x + 55° + 90° + 120° + 25° = 360°
i. What is the size of the angle representing
x + 290° = 360°
Savings?
x = 360° − 290° = 70°
ii. How much does Mr. Awuah spend on Rent?
iii. What fraction of Mr. Awuah‟s income is spent ii. If 120° = 60%
on Food?
Then 90° = × 60% = 45%.
Solution
i. Let x represent angle of Savings Exercises 10.8
= x + 50° + 60° + 90° + 120° = 360° 1. The pie chart below shows the program
⇒x + 320° = 360° analysis of a T.V station. The station telecast 6
hours each day.
x = 360° – 320°= 40°

ii. Mr. Awuah‟s monthly income is Gh¢180.00, Sports

then amount spent on rent; 60°
360° = Gh¢180 60° Drama
90° = = Gh¢45.00
75°
Documentary

iii. Fraction spent on Food;

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i. What is the size of the angle marked „variety‟? 4. The following table gives a data from a
ii. How many hours in a day do the station newspaper survey in Kumasi. 400 people were
telecast news? asked to choose anyone of the newspaper they
prefer to read.
2. The pie chart below shows the monthly
expenditure of Mr. Brown who receives
Gh¢5,400.00 a month. Newspaper No. of people
Graphic 150
Ghanaian Times 100
Rent
Mirror 75
x° Food
Clothing Daily Guide x
105°
65° Free Press 25
Savings
84° i. Find the value of x.
ii. Draw a pie chart for the distribution.

i. What is the angle for rent?

ii. Calculate the amount Mr. Brown put in his 5. The pie chart below shows how Mr. Brown
saving account. spent his salary in the month of January, 2010.
iii. What fraction of his income was spent on He spent Gh¢180.00 on rent.
rent?
y°
3. The pie chart below represents the number of Food
Clothing
different cars that were parked at a car park. 108°
60°
Rent
72°

Toyota Hyundai
x° 100°
i. Find angle y representing savings for the
month.
ii. What‟s the man‟s salary for the month?
iii. By how much does expenditure on others
exceed that on clothing.
If 9 Opel cars were parked,
i. Find the angle representing Toyota. Bar Chart / Graph
ii. How many Toyota cars were parked at the Any graph in which bars of various lengths are
park? used to represent different quantities is called a
iii. How many cars were parked at the park? Bar Graph or a Bar Chart.
iv. What percentage of the cars parked is Nissan?

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Drawing a Bar Chart VI. The length of each bar must correspond to its
The following points are helpful in drawing the frequency.
bar chart: VII. The title of the chart must be written
I. The bars must be of equal size.
II. The bars must be separated by blank spaces of Worked Examples
the same size (equal spacing). 1. The table below shows the scores obtained by
III. The x- axis and y- axis must be divided into 20 pupils in a test
equal segments and labeled to a convenient scale.
IV. The frequencies are always shown on the y – Scores 4 5 6 7 8
axis whilst the other component being marks, Frequency 3 2 5 4 6
ages, scores, height or whatever are shown on
the x- axis. i. Draw a bar chart for the distribution
V. Present the values on the x-axis in such a way
that it can use more than half of the graph sheet. Solution

6 Scores obtained by 20 Pupils in a test

Frequency

4 5 6 7 8 Scores
Soft drink No. of pupils
3. The following table is a result of a survey Pepsi 5
conducted in a class of a J.H.S to find the favorite Coke 6
soft drink of each pupil in the class. Pee cola 8
Fanta 3

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 Muscatella 5
Mirinda 4 Draw a bar chart showing this information, using
Club 6 a scale of 2cm to 1 unit on the vertical axis.
Sprite 3

Solution

10
A Bar Chart Showing the Favorite Drink of Pupils in a Class
No. of pupils

4
Muscatella

2
Pee Cola

Mirinda

Sprite
Club
Fanta
Pepsi

Coke

Soft Drinks

Exercises 10.9 2. A group of 20 students took an examination in

1. The ages of 30 students in a class is shown Twi and the following results were recorded (out
below: of 10 marks).
15 13 12 14 13 12 15 15 15 14 13 12 12 6 5 4 6 7 5 7 4 6 5
14 14 15 11 12 11 13 15 15 16 13 15 16 13 10 8 8 9 9 9 9 6 6 9
12 11 12. i. Construct a frequency distribution table for this
set of data.
i. Construct a frequency table for the data.
ii. Use your table to determine the median and
ii. Calculate from the table, the mode, median and
mean.
mean.
iii. Draw a bar chart for the information.
iii. Draw abar chart for the data.

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3. A die was tossed 50 times and the following
results were recorded: marks 1 2 3 4 5 6 7 8 9 10
3 2 6 4 6 2 5 1 4 5 2 5 4 1 2 6 Freq 2 4 5 6 10 1 2 3 1 1
1 3 2 6 3 1 3 5 6 5 2 4 6 4 2 1
1 6 1 2 4 2 5 6 1 6 6 2 1 34 2 Use the distribution to draw a bar chart.
5 1
i. Construct a frequency table distribution table
Extracting Information from a Bar Chart
for the data.
In some situations, the bar chart may be drawn
ii. Determine the mode, median and mean from
for students to answer some questions. It is quite
the table.
necessary to note the following:
iii. Use the information to draw a bar chart.
i. Consider the height of the tallest or the longest
bar (Mode).
4. Asibey performs an experiment with two dice,
ii. Consider the height of each bar
the numbers that showed up were added to obtain
iii. Add up the frequencies of all the bars
results shown in the table below:
iv. Find the differences between the frequencies
of all the bars
Score 2 3 4 5 6 7 8
Frequency 2 2 5 1 1 3 4
Worked Examples
Using this information, draw a bar chart. 1. The bar chart below shows the mark
distribution in a test.
5. The frequency table below is the mark i. What is the modal mark?
distribution of a class of 35 pupils in a ii. How many pupils took the test?
Mathematics examination.

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Solution

Frequency
8

4 5 6 7 8 Marks
= 4 + 5 + 7 + 6 + 3 = 25.
Solution
i. The modal mark is the mark with the longest 2. The bar chart below is for the distribution of
(or highest) frequency (or bar) = 6 marks in a class test.
To find the total number of pupils who took the
test, add up the frequencies of all the bars
Frequency

2 3 4 5 6 7 8 9 10 Marks

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i. Write down the frequency table for the 6 //// 5 30
distribution. 7 //// 4 28
ii. Use the table to find the mean mark. If the pass 9 // 2 18
mark is 4, how many pupils failed the test? 10 /// 3 30
Σf = 25 Σfx
Solution =150
a. i
Marks(x) Tally Frequency(f) fx
ii. The mean ( ̅) = = =6
2 /// 3 6
The number of people who failed the test if the
3 // 2 6
pass mark is 4 = 3 + 2 = 5
4 / 1 4
5 //// 4 20

Exercise 10.10
1. The bar chart below shows the monthly rainfall distribution (in mm) for some selected regions in
Ghana
Rainfall

1000

800

600

400

200
Western
B. Ahafo
G. Accra

Northern
Ashanti

Regions
Eastern
Central

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Use the bar chart to answer the following iii. How much more rainfall was recorded in the
questions: Northern region than Ashanti.
i. Which region recorded the highest amount of iv. What was the total rainfall recorded by the
rainfall entire regions in the chart?
ii. What was the least amount of rainfall
recorded?

2. The bar chart below shows the sizes of shoes sold in a shop

20
Frequency

15

10

41 42 43 44 45
Shoe Size

From the bar chart; a grouped frequency table is obtained. The groups
i. Which shoe size is the most common? into which the entries are put are referred to as
ii. Which shoe size is the least common? classes
iii. How many more size 43 shoes were sold than
the size 41? The grouped frequency table is usually used for
iv. How many shoes were sold in all? data‟s with many entries. Below is an example of
a grouped frequency table
Grouped Frequency Table
When the values of the entries are put into well Class Marks Frequency
organized groups and presented in a tabular form, (x) (f)
25 – 29 3

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 30 – 34 4 For grouped frequency distribution, each class
35 – 39 7 boundary is usually half way between the upper
40 – 44 6 limit of one class and the lower limit of the next
45 – 49 2 class. That is:
= (upper limit of one class + lower limit of the next class)
On the above table, column 1 represents the
group put into the classes: 25 – 29, 30 – 34, For instance, to determine the class boundary of
35 – 39, 40 – 44, 45 – 49 and column 2 represents the class 25 – 29 and the next class 30 – 34,
the respective frequencies. If the class 25 – 29 has I. Identify the upper class limit of 25 – 29 as 29
a frequency of 3, it means that there are three and the lower class limit of 30 – 34 as 34
different or same numbers within 25 and 29 II. Find the sum of the limits identified in I as 29
+ 30 = 59
Class Limits III. Find half of the sum obtained in II as:
Consider the table below; ( = 29.5
IV. Recognize 29.5 as the upper class boundary
of the class 24 – 29 and the lower class boundary
Class Marks Frequency of the next class 30 – 34
(x) (f)
25 – 29 3 Steps in Finding the Class Boundary of a
30 – 34 4 Grouped Data
35 – 39 7 To find the class boundary of a grouped data,
40 – 44 6 follow the steps below;
45 – 49 2 I. Find the difference between the upper class
limit of a particular class interval and the lower
It is seen that the classes contain a range of values class limit of the next class and divide by 2. For
i. e. 25 – 29, 30 – 34, 35 – 39, 40 – 44, 45 – 49. example, the class boundary of 25 – 29, 30 –
This means that each class has a smallest value 34… is determined as (30 – 29) = 0.5
and a largest value. These two values are called
II. Subtract the answer obtained from the lower
Class limits. Thus, we have the lower class limit
class limit of each class and add the answer to the
representing the smallest number and the upper
upper class limit of each class to obtain the class
class limit representing the largest number. For
boundaries.
example in the class 25 – 29, the lower class limit
is 25 and the upper class limit is 29
Worked Examples
Lower class limit 25 – 29 upper class limit Copy and complete the table below;

Class Boundaries Marks Class

It is the point where one class separates itself (x) Boundaries
from the other. It is also explained as the smallest 5–9
and largest values an item in that class can take. 10 – 14

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 15 – 19 It is the average of the lower and upper limits of a
20 – 24 class. This is achieved by adding the lower and
25 – 29 upper class limits and dividing the sum by two to
get the mid – point of the class. The midpoint is
Solution often called class mark. For example, the class
Method 1 mark of the class 5 – 9 is determine as : = =7
( = 9.5 ( = 14.5
Worked Example
( 19.5 ( = 24.5
Copy and complete the table below:

Marks Class Marks (x) Class Mark

(x) Boundaries 5–9
5–9 4.5 – 9.5 10 – 14
10 – 14 9.5 – 14.5 15 – 19
15 – 19 14.5 – 19.5 20 – 24
20 – 24 19.5 – 24.5 25 – 29
25 – 29 24.5 – 29.5
Solution
Method 2 (Calculations are not necessary)
Consider 5 – 9 and 10 – 14
( – ) = 0.5 Marks (x) Class Mark
Subtract 0.5 obtained from the lower class limit 5–9 7
of each class and add 0.5 to the upper class limit 10 – 14 12
of each class 15 – 19 17
20 – 24 22
Marks Class 25 – 29 27
(x) Boundaries
5–9 4.5 – 9.5 Class Size/Width
10 – 14 9.5 – 14.5 The class width or class size or class interval is
15 – 19 14.5 – 19.5 the difference between the upper and lower class
20 – 24 19.5 – 24.5 boundaries of aclass. Forinstance, the class width
25 – 29 24.5 – 29.5 or class size of a class with boundaries 4.5 and
9.5 = 9.5 – 4.5 = 5
Class Mid – point/Class Mark

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 Format of a Grouped Frequency Table

Intervals Class Boundary Class Marks (x) Frequency (f) fx

∑ = ∑ =

Worked Examples
Below is a record of the marks (out of 50) obtained by 22 students in a test.
14 12 8 19 13 27 22 21 32 30 35 39 36 37 42 45 40 41 44 46 44 46
Prepare a group frequency table for the data using the intervals 1 – 10 , 11 – 20 …
Solution

Marks Class Boundaries Class Marks Frequency (f) fx

1 - 10 0.5 – 10.5 5.5 1 5.5
11 – 20 10.5 – 20.5 15.5 4 62
21 – 30 20.5 – 30.5 25.5 4 102
31 – 40 30.5 – 40.5 35.5 6 213
41 – 50 40.5 – 50.5 45.5 7 318.5
∑ = 22 ∑ 701

Exercises 10.11 44 32 40 48 28
1. The following is a record of scores out of 100 Draw a grouped frequency table for the intervals
obtained by 40 candidates in a mathematics 10 – 14, 15 – 19, …
examination
50 70 66 52 56 68 64 52 68 56 3. The marks scored by a group of students in a
56 50 50 54 64 50 58 58 60 56 test are given below;
74 50 72 51 59 54 52 60 66 58
14 62 51 37 70 42 12 52 31 53 45 70 72
52 56 53 58 68 52 64 62 52 56
54 76 27 21 58 57 22 68 13 56 75 54 40
Draw a grouped frequency table for the intervals
59 63 19 59 70 51 41 55 56 36 56 73 57
50 – 54, 55 – 59, 60 – 64 …
50 . Group the data in tens and construct a
2. The following is a record of scores out of 100 frequency table
obtained by 35 candidates in a mathematics
examination Mean of a Grouped Data
36 18 44 30 40 10 36 20 34 46 To calculate the mean from a grouped frequency
12 44 34 30 48 18 20 42 16 44 table, make sure the table is of the format shown
32 40 48 28 30 48 18 20 42 16 below:

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 Class Class 25 – 29 3
Interval Mark(x) f fx 30 – 34 7
35 – 39 26
40 – 44 29
∑
45 – 49 25
∑ =
50 – 54 6
55 – 59 2
II. Carefully complete the table and accurately
compute the values of ∑ and Find from the table:
∑ . i. the mean ii. the mode
III. Substitute the values of iii. the modal class iv. the median class
∑ and ∑ into the formula below to obtain the
∑
mean value of the data : Mean ( ̅ ) = ∑ Solution

Mode of a Grouped Data Masses Class

(kg) Marks(x) f fx
For grouped data, the mid – point of the class
20 – 24 22 2 44
with the highest frequency may be taken as the
25 – 29 27 3 81
crude estimate of the mode. The class in which
30 – 34 32 7 224
the mode occurs is called the modal class
35 – 39 37 26 962
40 – 44 42 29 1218
Median of a Grouped Data 45 – 49 47 25 1175
For a grouped data, it is a bit complicated to 50 – 54 52 6 312
determine the median but not the class within 55 – 59 57 2 114
which the median value falls. These are few ∑ = ∑ =
hints: 100 4130
I. Find half of the total frequency and identify the
value. ∑
Mean ( ̅ ) = ∑
= = 41.30 kg
II. Count from the top of the table to obtain a sum
ii. The mode is the midpoint of the highest
of frequencies that equals the value obtained.
frequency = 42
III. Identify the class within which the values
falls as the median class. iii. From the table, the highest frequency is 29
A good estimate of the median is best obtained and falls within the class 40 – 44. Therefore, the
from a cumulative frequency curve modal class is 40 – 44

Worked Examples iv. The median class

1.The table below shows the distribution of the
= ×∑ = × 100 = 50
masses of parcels in a local post office
Adding frequencies from the top, 50 can be
Masses (Kg) Frequency located in the class 40 – 44. Therefore, the
20 – 24 2 median class = 40 – 44

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2. The marks obtained by 40 students in an ∑
b. i. ̅ = ∑
= = 78.88
examination are as follows;
63 76 87 61 78 85 77 87 74 77 ii. P(at least 75) = =
80 77 74 88 72 78 79 89 85 90
77 70 81 69 75 78 73 86 83 91 Exercises 10.12
69 96 65 88 84 74 84 81 83 75 1. The table below shows the distribution of
Copy and complete the table below using the data marks scored by a group of 30 students. Find: the
above mean. Ans = 27.4

Class Marks 10 – 15 – 20 – 25 – 30 – 35 –
Class Midpoint 14 19 24 29 34 39
Boundaries (f) (x) fx No of
59.5 – 64.5 62 Students 1 3 4 6 7 4
64.5 – 69.5 67
69.5 – 74.5
74.5 – 79.5 2. The marks obtained by candidates in a
79.5 – 84.5 mathematics examination were first grouped
84.5 – 89.5 0 – 9, 10 – 19, 20 – 29, 30 – 39 and so on.
89.5 – 94.5
The midpoint of the mark of each group was
94.5 – 99.5 97 97
Total ∑ = 40 ∑ = taken as the mark representing the group. The
following table gives the distribution of the
∑ marks;
b. i. Using the relation ̅ = ∑
, or otherwise, find
the mean ̅ Midpoint 4 .5 14.5 24.5 34.5 44.5
ii. Calculate the probability that a student chosen Freq. 18 19 x 12 9
at random obtained at least 75 marks Midpoint 54.5 64.5 74.5 84.5
Freq. 5 2 2 1
Solution
a. If the mean mark for the candidate was found to
Class be 26.06;
Class Class Midpt i. determine the value of x
Boundaries (f) (x) fx ii. draw a histogram for the distribution
60 - 64 59.5 – 64.5 2 62 124 iii. find the probability that a candidate chosen at
65 – 69 64.5 – 69.5 3 67 201 random obtained 55 marks or more
70 – 74 69.5 – 74.5 6 72 432
75 – 79 74.5 – 79.5 11 77 847
The Assumed Mean
80 – 84 79.5 – 84.5 7 82 574
85 – 89 84.5 – 89.5 8 87 696 The assumed mean as the name suggests, is an
90 – 94 89.5 – 94.5 2 92 184 assumption or a guess of the mean. The assumed
95 – 99 94.5 – 99.5 1 97 97 mean is usually denoted by the letter “A”. It
∑ = 40 ∑ =3155 doesn‟t need to be correct or even closer to the
actual mean and the choice of the assumed mean
is usually at one‟s own discretion (usually any of
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the class marks estimated to be closer to the
mean) except where the question explicitly states Class x d= x–A f fd
a certain assumed mean value. 0–4 2 -15 45 -675
5–9 7 -10 58 -580
The assumed mean is used to calculate: the actual 10 – 14 12 -5 27 -135
mean, the variance and the standard deviation. 15 – 19 17 0 30 0
20 – 24 22 5 19 95
Given the assumed mean, the actual mean 25 – 29 27 10 11 110
∑ 30 – 34 32 15 8 120
is calculated by the formula: ̅ = A + ∑ 35 – 39 37 20 2 40
∑ ∑ =
These values are obtained by completing the -1025
frequency table such as the one below:
∑
̅ =A+ ∑
class x d=x-A Freq. fd
(f) But A = 17, ∑ = 200 and ∑ = - 1025
̅ = 17 +
∑ ∑
̅ = 17 – 5.125 = 11.88
Worked Examples
2. The table below shows the marks obtained by
1. The student body of a certain school were
students in an examination
polled to find out what their hobbies were. The
number of hobbies of each student was recorded
Marks 0 – 9 10 – 20 – 30 – 40 –
and the data obtained were grouped into classes
19 29 39 49
as shown on the table below.
Freq 2 3 15 10 10

Number of hobbies Frequency

Assuming the mean is 15, calculate the actual
0–4 45
mean mark
5–9 58
10 – 14 27
15 – 19 30 Solution
20 – 24 19 A = 15
25 – 29 11 class x d= x–A f fd
30 – 34 8 0–9 4.5 -10.5 2 -21
35 – 39 2 10 – 19 14.5 -0.5 3 -1.5
20 – 29 24.5 9.5 15 142.5
Using an assumed mean of 17, find the mean of 30 – 39 34.5 19.5 10 195
the number of hobbies of the students in the 40 – 49 44.5 29.5 10 295
school. ∑ ∑ =
= 610
Solution 40
Assumed mean, A = 17
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 ̅ =A+
∑ 30 – 39 13
∑
40 – 49 4
But A = 15, ∑ = 40 and ∑ = 610 50 – 59 3
̅ = 15 + = 15 + 15.25 = 30.25 60 – 69 4

Exercises 10.13 The Histogram

1. Consider the test results of 42 form three A histogram is a graph showing the number of
students in Mathematics occurrences of items of data, and is drawn using
vertical bars, similar to that of a bar chart.Group
marks 40 – 50 – 60 – 70 – 80 – data is usually represented graphically by
49 59 69 79 89 histograms.
Freq. 1 4 19 15 3
The bars used for histograms are usually of
Taking the assumed mean as 64.5, find the actual
standard size. The width of each bar is equivalent
mean of the distribution
to the class interval representing it. Also, the area
2. The test results of 50 students were recorded as of the bars isproportional to the total frequency of
follows: the items in the class they represent

Marks 50 – 60 – 70 – 80 – 90 – Steps in Drawing a Histogram

59 69 79 89 99 I. Choose a suitable scale for both vertical and
Freq. 6 11 19 9 5 horizontal axes, when no scale is given. The scale
must be large enough to ensure that the histogram
Assuming the mean is 74.5, calculate the mean covers at least half of the graph sheet. Usually,
mark scales using multiples of 5 or 10 are the best

3.The table below shows the marks obtained by II. Draw two axes, the vertical and horizontal
students in an examination. axes. Label the vertical axes for the frequencies
Marks 0 – 9 10 – 20 – 30 – 40 – and the horizontal according to the given data.
19 29 39 49 E.g. Ages (years), Marks Height (cm), not
Freq 2 3 15 10 10 forgetting the units where applicable

Assuming the mean is 15, calculate the actual III. The bars can be drawn in two ways explained
mean mark. as follows:
a. Using the class boundaries, draw bars using the
4. Calculate the mean of the following data using
lower and upper class boundaries
the assumed mean method, given the assumed
b. Using the class marks, place the class mark at
mean as 30.5;
the center of the bars and ensure that each bar has
the same width corresponding to the class width.
Class Frequency
10 – 19 11
20 – 29 5

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IV. Use the symbol to indicate that the 6 2 1 3 4 2 5 1
graph does not start from zero on the horizontal Use the data to draw a histogram
axis. The vertical axis should start from zero
V. Give the graph a title Solution

Histogram for Ungrouped Data Score Tally Frequency

Worked Examples 1 //////// 10
1. A die was tossed 50 times and the following 2 //////// / 11
were recorded 3 //// / 6
3 2 6 4 6 2 5 1 4 5 2 5 4 1 4 //// // 7
2 6 1 3 2 6 3 1 3 5 6 3 2 4 5 //// / 6
6 //////// 10
6 4 2 1 1 6 1 2 4 2 5 6 1 6
Frequency

16
14
12
10
8
6
4
2
0
1 2 3 4 5 6

Marks (%)

Histogram for a Grouped Data i. Prepare a group frequency table for the data
Worked Examples using the intervals 5 – 9 , 10 – 14 …
1. Below is a record of the marks (out of 50) ii. From the table, determine the mean mark
obtained by 22 students in a test. iii. Draw a histogram to represent the data
14 12 8 19 13 27 22 21 32 30 35
39 36 37 42 45 40 41 44 46 44 46

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Solution

i.
Marks Class Boundaries Class Marks (x) Frequency (f) fx
5–9 4.5 – 9.5 7 1 7
10 – 14 9.5 – 14.5 12 3 36
15 – 19 14.5 – 19.5 17 1 17
20 – 24 19.5 – 24.5 22 2 44
25 – 29 24.5 – 29.5 27 1 27
30 – 34 29.5 – 34.5 32 2 64
35 – 39 34.5 – 39.5 37 4 148
40 – 44 39.5 – 44.5 42 5 210
45 – 49 44.5 – 49.5 47 3 141
∑ = 22 ∑ = 694

∑
ii. Mean ( ̅ ) = ∑
= = 31.55

iii.
frequency

4.5 9.5 14.5 19.5 24.5 29.5 34.5 39.5 44.5 49.5
Marks

2. Copy and complete the table below and use it

to draw a histogram using: Ages (years) Frequency
i. the class boundaries 5–9 3
ii. the class mark 10 – 14 2
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 15 – 19 4 Ages Class
20 – 24 5 (yrs) boundaries Frequency
25 – 29 3 5–9 4.5 – 9.5 3
30 – 34 1 10 – 14 9.5 – 14.5 2
15 – 19 14.5 – 19.5 4
20 – 24 19.5 – 24.5 5
Solution
25 – 29 24.5 – 29.5 3
i. Using class boundaries
30 – 34 29.5 – 34.5 1

5
Frequency

0
4.5 9.5 14.5 19.5 24.5 29.5 34.5 Marks (%)

ii. Using the class mark

Ages (yrs) Class Mark Frequency
5–9 7 3
10 – 14 12 2
15 – 19 17 4
20 – 24 22 5
25 – 29 27 3
30 – 34 32 1

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 5

Frequency
4

0
7 12 17 22 27 32 Marks (%)

Estimating the Mode from a Histogram perpendicular touches the horizontal axis, as
The class which has the highest frequency is the mode.
called the modal class. The actual mode is
contained in the modal class. Worked Examples
1. Determine the mode of the histogram below
The mode of can be estimated from the histogram
by following the steps below:
Frequency

I. Draw the histogram for the data.

II. Draw two diagonals from the top corners of
the adjacent bars 40
a. Carefully draw a straight line from the top right 35
corner of the tallest bar to the top right corner of 30
the bar on the left 25
b. Draw another straight line from the top left 20
corner of the tallest bar to the left corner of the 15
bar to the right. 10
5
III. Draw a perpendicular from the point of 0
35.5 45.5 55.5 65.5 75.5 85.5 95.5
intersection of the diagonals onto the horizontal
Marks (%)
axis.
Solution
IV. Read off the value of the point where the
The modal class has the tallest bar
= (55.5 – 65.5)

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 From the histogram, the modal mark = 55.9 Worked Examples
Determine the median of the histogram.
Estimating the Median from a Histogram
The Median can be estimated from the histogram
by carefully following the steps below:

Frequency
I. Draw the histogram for the distribution
II. Find the sum of the areas of all the rectangles
III. Calculate half of the sum of the areas of all 40
the rectangles and locate it vertically on the 35
histogram 30
IV. If the vertical line falls exactly on a 25
boundary, then read off the boundary value as the 20
median. However, if the vertical line does not fall 15
on a boundary, then calculate the area of the 10
additional strip and add it to the area up to the
boundary value just on the left, to obtain the 0
35.5 45.5 55.5 65.5 75.5 85.5 95.5
median.
Marks (%)

Total area = 15 + 10 + 20 + 25 +15 + 5 = 90

Frequency

5 Half of total area = = 45

Median class = (14.5 – 19.5) and (19.5 – 24.5)
4

3
Frequency

5
2
4
1
3
0
4.5 9.5 14.5 19.5 24.5 29.5 34.5 Marks (%)
2
Solution
Area of each bar = L × B,
1
B = (9.5 – 4.5) = 5
A1 = 3 × 5 = 15 A4 = 5 × 5 = 25 A1 A2 A3 A4 A5 A6
0
A2 = 2 × 5 = 10 A5 = 3 × 5 = 15 4.5 9.5 14.5 19.5 24.5 29.5 34.5 Marks (%)
A3 = 4 × 5 = 20 A6 = 1 × 5 = 5
Median = 19.5 (shown on the graph)
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2. What is the median of the histogram below?
Frequency Estimated median = 15.5 + 4.8 = 20.3

2. What is the median of the histogram below?

16

Frequency
14
12
10
16
8
14
6
12
4
10
2
0 8
0.5 5.5 10.5 15.5 20.5 25.5 30.5 35.5 40.5 6
Solution Marks (%) 4
Area of each bar = L × B, 2
B = (5.5 – 0.5) = 5 0
3 8 13 18 23 28 33 38
A1 = 2 × 5= 10 A5 = 12 × 5 = 60
A2 = 3 × 5 =15 A6 = 7 × 5 = 35 Marks (%)
A3 = 6 × 5 = 30 A7 = 4 × 5 = 20 Solution
A4 = 15 × 5 = 75 A8 = 2 × 5 = 10
Total area = 10 + 15 + 30 + 75 + 60 + 35 + 20 +
Frequency

10 = 255

Half of total area = = 127.5

16
Median class = (15.5 – 20.5) meaning after 15.5
but not up to 20.5 14
12
From the histogram, area of the shaded region or 10
rectangle = 15 × x = 15x 8
Half the area = A1 + A2 + A3 + area of shaded 6
region = 127.5 4
10 + 15 + 30 + 15x = 127.5 2
15x = 127.5 – 10 – 15 – 30 0 x
15x = 72.5 3 8 13 18 23 28 33 38

x= = 4. Marks (%)

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Area of each bar = L × B, 2. The table gives the heights measured to the
B = (8 – 3 ) = 5 nearest meter of 291 trees.
A1 = 2 × 5 = 10 A5 = 12 × 5 = 60
A2 = 3 × 5 =15 A6 = 7 × 5 = 35 Height(m) 2 3 4 5 6 7 8 9
A3 = 6 × 5 = 30 A7 = 4 × 5 = 20 No of 14 12 42 83 118 12 7 3
A4 = 15 × 5 = 75 A8 = 2 × 5 = 10 trees

Total area a. Draw a histogram to illustrate the information.

= 10 + 15 + 30 + 75 + 60 + 35 + 20 + 10 = 255 b. Calculate the mean of the distribution correct
Half of total area = = 127.5 to the nearest meter.
c. From the histogram, estimate the mode and
Median class = (18)
median of the heights.
Class boundary = = 2.5
Lower class boundary of the median class; 3. The frequency table below shows the height (to
= 13 + 2.5 = 15. 5 the nearest cm) of 26 girls in JHS 1.

From the histogram, area of the shaded region or Class interval (cm) Frequency
rectangle = 15 × x = 15x 140 – 143 1
Half the area = A1 + A2 + A3 + area of shaded 144 – 147 2
region = 127.5 148 – 151 7
10 + 15 + 30 + 15x = 127.5 152 – 155 5
15x = 127.5 – 10 – 15 – 30 156 – 159 8
160 – 163 3
15x = 72.5
x= = 4.8 a. Draw a histogram of the distribution.
Estimated median = Lower class boundary + x b. Calculate the mean of the distribution correct
= 15.5 + 4.8 = 20.3 to the nearest centimeter.
c. From the histogram, estimate the mode and
Exercises 10.15 median of the heights.
1. The marks obtained by 30 pupils in an
examination are as follows; 4.The marks obtained by some candidates in
71 65 83 78 74 63 73 87 78 80 93 an examination are;
78 88 76 80 68 77 68 70 84 90 69 27 36 40 48 50 55 695 8 57 60 61 63
75 61 66 90 95 85 77 76 61 63 65 68 65 68 67 70 55 71 69
a. Arrange these marks in a frequency distribution 73 72 75 76 78 76 80 81 83 88 89 90
table using class intervals of 61 – 65, 66 – 70, 71 i. Form a group frequency table for this data,
– 75… using the class intervals 20 – 29, 30 – 39…
b. i. Draw a histogram of the distribution ii. From the table, calculate the mean correct to 1
ii. Use your histogram to estimate the mode decimal place.
c. Calculate the mean of the distribution. iii. Draw a histogram for the data.

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5. The following are the marks scored by 50
students in a Mathematics test.

Cumm. Freq.
5

41 45 56 58 28 56 40 55 29 34
4
28 52 8 43 16 24 44 51 24 26
28 23 25 36 38 28 36 47 17 37
3
42 21 23 46 24 53 39 47 39 40
32 27 27 34 24 48 48 32 46 32
2

i. Prepare a frequency table for these data using

1
equal class intervals of 1 – 10, 11 – 20 and so on.
ii. What is the modal class of the distribution?
iii. Illustrate the data by means of a suitable 4.5 9.5 14.5 19.5 24.5 29.5 Ages (yrs)
diagram.
iv. Using an assumed mean of 35.5 or otherwise,
Solution
calculate from your table, the mean of the
I. Frequencies = Height of bars = 3, 5, 2, 4, 1
distribution.
II. Class boundaries = 4.5 – 9.5, 9.5 – 14.5, 14.5 –
19.5, 19.5 – 24.5, 24.5 – 29.5
Preparing a Frequency Distribution Table for
III. Add 0.5 to lower class boundaries and
a Given Histogram
subtract 0.5 from upper class boundaries of the
A. Class Boundary Histogram
I. On the vertical axis, identify the height of each same group to generate the class size.
bar as the frequency. ⇒5 – 9, 10 – 14, 15 – 19, 20 – 24, 25 – 29
II. For each bar, identify the first and second
numbers on the horizontal axis as the lower class Class Class
Size Boundary Frequencies
bounadary and upper class boundary respectively,
5–9 4.5 – 9.5 3
and by this technique, generate all the class
10 – 14 9.5 – 14.5 5
boundaries.
15 – 19 14.5 – 19.5 2
III. For each class boundaries, add 0.5 to the
20 – 24 19.5 – 24.5 4
lower class boundary and subtract 0.5 from the
25 – 29 24.5 – 29.5 1
upper class boundaries to generate each class
width or intervals.
B. Class Mark/Mid Point Histogram
I. On the vertical axis, identify the height of each
Worked Examples
bar as the frequency
Prepare a frequency distribution table for the
II. On the horizontal axis, find the common
histogram below;
difference (d) between the class marks, and
divide by two to obtain an answer, say A.
⇒ =A

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III. Subtract and add the answer obtained in act 17 – 2.5 = 14.5 and 17 + 2.5 = 19.5
II, that is A, from and to each class mark to obtain (14.5 – 19.5)
the lower class boundary and the upper class 22 – 2.5 = 19.5 and 22 + 2.5 = 24.5
boundary respectively of that class (19.5 – 24.5)
IV. For each class boundaries, add 0.5 to the 27 – 2.5 = 24.5 and 27 + 2.5 = 29.5
lower class boundary and subtract 0.5 from the (24.5 – 29.5)
upper class boundaries to generate each class
width or intervals. IV. Add 0.5 to the lower class boundary and
subtract 0.5 from the upper class boundaries
Worked Example (5 – 9) (10 – 14), (15 – 19), (20 – 24)
(25 – 29) to obtain class size/width

Class Class
Cumm. Freq.

5
Size Boundary Frequencies
5–9 4.5 – 9.5 3
4
10 – 14 9.5 – 14.5 5
15 – 19 14.5 – 19.5 2
3
20 – 24 19.5 – 24.5 4
25 – 29 24.5 – 29.5 1
2

The Frequency Polygon

1
It is a line graph in which frequencies are plotted
against corresponding class marks. To be precise,
7 12 17 22 27 Ages (yrs) if you mark two additional intervals on each side
Solution of a histogram and then join the mid points of
I. Frequencies = Height of bars = 3, 5, 2, 4, 1 these intervals to the mid points of horizontal
II. Difference between the first and second class sides of the rectangle with straight lines, a
frequency polygon is obtained
marks = 12 – 7 = 5, the = 2.5
III. Subtract and add 2.5 from and to each class Worked Examples
mark to obtain the class boundaries 1. Use the table below to draw a frequency
7 – 2.5 = 4.5 and 7 + 2.5 = 9.5 polygon
(4.5 – 9.5)
Age (yrs) 13 14 15 16 17
12 – 2.5 = 9.5 and 12 + 2.5 = 14.5 Frequency 1 4 2 2 1
(9.5 – 14.5)

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Solution 71 – 75 73 5

Draw a frequency polygon for the distribution

Solution
4

20
2

15
1
10
13 14 15 16 17
5
2. Given the following frequency distribution
table of marks in a language test. 53 58 63 86 73

Score Class mid – point Frequency

51 – 55 53 10
56 – 60 58 20
61 – 65 63 20
66 – 70 68 10

Exercises 10.17
The marks scored by a group of students in a test are given below;
14 62 51 37 60 42 12 53 52 3153
45 20 72 54 26 31 27 60 11 21 58 57 22 68 8 56 37 54 40 59 63 19 59 20 51
41 55 38 56 36 56 33 57
i. Group the data in tens and construct a frequency table
ii. Draw a cumulative frequency polygon for the table

2. Fifty bags of sweets were opened. The number of sweets in each bag is recorded in
the table below:

No of 18 – 23 – 28 – 33 – 38 –
sweets 22 27 32 37 42
Frequency 6 14 16 11 3

Represent this information in the form of a frequency polygon

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11 RIGID MOTION Baffour – Ba Series

Transformation Solution
Transformation is the process of changing the P 1 = P + Translation vector
shape, size and position of an object. The P1 = ( ) ( )=( ) = ( ) = (4, 9)
Methods used are;
1. Translation 2. Enlargement
3. Find the image of Q (-3, 4) under a translation
3. Reflection 4. Rotation
by the vector ( ).
5. Mapping

Translation Solution
In translation, a translation vector written in the Q 1 = Q + Translation vector.
form ( ) is added to the given point(s) or object Q1 = ( ) ( )=( ) = ( ) = (- 6, - 8)
to produce the image. For e.g, if A( ) is
4. Given A (5, 8), B (-2, 4) and C (9, 7), find the
translated by the vector ( ), then the image of A,
images A 1 B 1 C 1 of ABC respectively under a
which is A 1 = A + translation vector.
translation by the vector ( ).
That is: A 1 = ( ) + ( )= (x + a, y + b)
Solution
Properties of Translation A 1 = A + Translation vector
1. Every point moves the same distance in the A 1 =( ) ( )=( ) = ( ) = (0, 5)
same direction
2. The size of an angle remains the same.
B 1 = B + Translation vector
3. The length of the line remains the same.
4. The shape of the object remains the same. B1 = ( ) ( )=( ) = ( ) = ( - 7, 1)
5. The size of the object does not change.
C 1 = C + Translation vector
Worked Examples C1 = ( ) ( )=( ) = ( ) = (4, 4)
1. If A = (7, 5) is translated by the vector ( ), find
the image, A 1 Exercises 11.1
1. The points P( Q( and R(
Solution are translated by the vector ( ), find the images
A 1 = A+ translation vector P 1 Q 1 R 1 of PQR, such that P →P1, Q → Q 1 and
A1 = ( ) ( ) ( ) = ( ) = (10, 9) R → R1
2. Given R(7, 5), S(4, 6) and T(-2, -3), find the
2. Find the image of P(2, 6) when translated by co-ordinates of the images R1,S1,T1 of R, S, T
the vector ( ) . under a translation by the vector ( ) such that :

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R → R1, S → S1 and T → T1. P = ( ) – ( ) = (5 + 6, 9 – 7) = (11, 2)
The coordinates of the object P is (11, 2)
3. The point R(-3, 2) is translated by the vector
( ). It is then given a further translation by the
Exercises 11.2
vector ( ). Determine the final image of R 1. The respective images of ABC under
translation by the translation vector ( ) are A1(
Finding the Object Given the Image A1 and the 0, 5), B1(- 3, 8) and C1(- 5, -1). Find the
Translation Vector (Tv), coordinates of ABC.
Given the image A1 and the translation vector
(Tv), the coordinates of the object, A, is 2. K1(9, - 4) is the image of K under translation
calculated by the relation: by the translation vector ( ). Determine the
A = Image ( A1 ) – Translation vector (Tv)
coordinates of K.
A( ) = ( ) – ( ) = A (x1 – a, y1 – b )
3. If U(x, y) under translation by the translation
Worked Examples vector ( ) produced the image U1(-7, - 4). Find
1. The image of an object translated by the vector the values of x and y.
( ) is M1 (7, 5). What is the coordinates of the
object, M. 4. The image of a point under a translation by the
vector ( ) is (2, -2). Find the coordinates of the
Solution point. Ans (5, 3)
M = Image ( M1 ) – Translation vector (Tv)
M = ( ) – ( ) = (7 – 1, 5 – 4) = (6, 1) Finding the Translation Vector Given the Image
The coordinates of the object M is M(6, 1) A1 and the Object, A.
Given the image, A1 and the object A, the
2. Find the coordinates of the object P, translated translation vector (Tv) is calculated by the
by the translation vector ( ) to produce the relation:
Tv = Image ( A1 ) – Object ( A )
image P1( 4, 2).
Tv = A1( ) – A ( ) = ( )
Solution
P = Image ( P1 ) – Translation vector (Tv) Worked Examples
P = ( ) – ( ) = (4 + 3, 2 + 5) = (7, 7) 1. The image of P (4, 1) under translation by the
The coordinates of the object P is (7, 7) vector, M, is P1(7, 6). What is the coordinates of
the translation vector, M?
3. The point P(x, y) is mapped onto P1(5, 9) by
the vector ( ). Find the coordinates of P. Solution
Tv= A1( ) – A ( ) = ( )
Solution M = P1( ) – P ( ) = ( )=( )
P = Image (P1 ) – Translation vector (Tv)

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The translation vector is M ( ) 1. If B1(-5, 4) is the image of B(3, 7) under
translation, find the translation vector.
2. If V1(5, 2) is the image of V(9, 0) under
translation, find the translation vector. 2. Under a translation, a point L(-8, 2) has an
image L1(-2, 1). Find the translation vector.
Solution
3. Under a translation, (4, 7) → (1, -3). What is
Tv= A1( ) – A ( ) = ( )
the translation vector?
Tv= V1( ) – V ( ) = ( )=( )
4. A(4, 3), B(-3, -2), C(0, -7) and A1(5, 7), B1(4,
3. P1(8, -2) is the image of the point P(5, 2) by the 7), C1(1, 2) are such that A A1, B B1 and C
C1 under translation by different translation
translation vector V. Find: vectors. Identify the translation vector of each.
a. the vector, V.
b. the coordinates of the point Q which maps onto Enlargement
the points Q1(5, -2) under V. An enlargement is a transformation which
c. ⃗⃗⃗⃗⃗⃗⃗⃗⃗ increases the size of shapes by a constant k. This
d. constant (number) is called the scale factor of
enlargement.
Solution
The scale factor does not have to be a whole
a. Tv = A1( ) – A ( ) = ( ) number all the time. It could even be a number
less than one, causing the object to be smaller.
Tv = P1( ) – P( ) = ( )=( ) Under such conditions, the transformation can be
called either an enlargement or a reduction.
b. Object + TV = Image
Q + TV = Q1 Whenever an enlargement is done, one point
Q+( )=( ) =( ) ( ) =( )=( ) remains fixed. This point is called the center of
enlargement. Thus, if A( , is enlarged by the
c. P1(8, -2) and Q1(5, -2) scale factor, k, from the origin, then the image of
A, is: A1 = scale factor × A.
⃗⃗⃗⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗⃗⃗ – ⃗⃗⃗⃗⃗⃗
⇒A1 = k × ( ) = ( ) = (kx, ky)
=( ) – ( )=( )=( )

Properties of Enlargement
d. P1(8, -2) and Q1(5, -2)
Generally, in an enlargement with scale factor, k,.
= √( ( I. Each length is multiplied by the scale factor.
= √( =3 II. Each angle remains constant.
III. The shape of the figures is maintained but not
Exercises 11.3 necessarily the size.
IV. Corresponding lines are parallel.

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Worked Examples 1. Find the image of Q(3,7) under enlargement
1. Find the image of A (5, 4), under enlargement from the origin with scale factor -2.
from the origin with scale factor 2.
Solution
Solution Q1 = scale factor × Q
A1 = scale factor × A Q1 = -2 × ( ) = ( )=( ) (-6, -14)
A1 = 2 × ( ) = ( )= ( ) = (10,-8)
2. What is the image of B (-6, -1) under
2. What is the coordinates of the image of Q (8, - enlargement from the origin with scale factor -3.
12) under enlargement with scale factor, from
Solution
the origin?
B1 = Scale factor × B
B1 = -3 × ( ) = ( ) = ( ) = (18, 3)
Solution
Q1 = scale factor × Q
3. Given P (5,-3), Q (-7, 2) and R(-6, -8). Find
Q1 = × ( )=( )=( ) = (1.6, -2.4) the image P1 Q1R1 such that P P1, Q Q1 and
R R1 under enlargement with scale factor - 3
Exercises 11.4 from the origin.
1. Find the coordinates of the image of A(3, -7)
under enlargement with scale factor -2 Solutions
P1 = Scale factor × P
2. What is the coordinates of the image of P1 = -3 × ( ) = ( )=( ) = (-15, 9)
Q (-7, -11) under enlargement with scale factor,
from the origin? Q1 = Scale factor × Q
Q1 = -3 × ( ) = ( ) = ( ) = (21, -6)
3. The vertices of a quadrilateral ABCD are (-5,
4), (1, 3), (-2, 2) and (1, 2) respectively. ABCD is R1 = Scale factor × R
enlarged by a scale factor, 3 to produce R1 = -3 × ( ) = ( ) = ( ) = (18, 24)
A1B1C1D1. Find the coordinates of the new figure
formed. Exercises 11.5
1. If A(0, 5), B(-4, 7) and C(-2, -3) is enlarged by
Enlargement with Negative Scale Factor scale factor -2 from the origin, what is the co-
An enlargement with a negative scale factor turns ordinates of the image A1B1 C1 such that A → A1,
a shape upside down as well changing its size. B → B1and C → C1?
That is if A(x, y) is enlarged by – k, then A1 = -k
( ) = A1(-kx, -ky) 2. Find the co-ordinates of the image of R,S, T
under enlargement with scale factor – , such that
Worked Examples

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R → R1, S → S1 and T →T1 from the origin given 2. The triangle ABC has vertices at A(-2, 1), B(2,
(5, -3), (7, 4), (-6, -1). 3), C (0, 5). This triangle is enlarged from the
point (0,1) by a scale factor of . Write down the
3. Find the coordinates of the image of P(-9, 12) new coordinates A1, B1, C1 of ABC such that A →
under enlargement with scale factor, – from the A1, B →B1, C → C1

origin Solution
Enlargement from a Given Point/Centre
A(-2, 1), k = , center of enlargement (0, 1)
When the point A(x, y)is enlarged from a point
other than the origin, say (a, b), with scale factor, k, A1( ) = A1( )
I. Subtract the center of enlargement from the A1( ) × = A1(
⁄
* = A1( )
⁄
coordinate of the object. i.e: A1( )
A1( )+ ( ) = A1( ) = A1(-1, 1)
II. Multiply the results by the scale factor, k. That
is:A1( )k = A1( ) B1( ) = B1( )
III. Add the results to the center of enlargement to ⁄
obtain the coordinates of the image, A1. That is: B1( ) × = B1( * = B1( )
⁄

A1( )+ ( ) B1( ) + ( ) = B1( ) = B1(1, 2)

The steps are easily remembered by the C1( )= C1( )

abbreviations S.M.A, meaning Subtract, Multiply ⁄
C1( )× = C1( * = C1( )
and Add. ⁄

C1( ) + ( ) = C1( ) = C1(0, 3)

Worked Examples A1(-1, 1), B1(1, 2), C1(0, 3)
1. Find the image of M(5, 3) under enlargement
from the point (0, 2) with scale factor – 4. 3.A triangle has vertices A( B( and C
(5, 8)
Solution i. Calculate the new coordinates of these vertices
M(5, 3), k = -4, point of enlargement (0, 2) if the triangle is translated by the vector ( )
i.Subtract the point of enlargement from the ii. If instead, the triangle is enlarged using A as
coordinates of the object center of enlargement, with scale factor 2, find
M1( )= M1( ) the coordinates of the images of B and C.

ii. Multiply M1( )by the scale factor, -4 Solution

M1( )× - 4 = M1( ) = M1( ) i. A( B( and C (
Translation by the vector ( )
iii. Add the results to the center of enlargement O + Tv = I
M1( ) ( ) = M1( ) = M1(-20, -2) A( )+ ( ) = A1( ), A1 (2, 0)
B( )+ ( ) = B1( ), B1(3, 2)

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C( )+ ( ) = C1( ), C1(6, 7) Given the image A1( and the scale factor, ,
the object is calculated by the rule;
⁄
ii. Centre of enlargement A(1, 1) = `1( *=( )
⁄
Scale factor = 2
B( ) – ( ) = ( )
Worked Examples
( )× 2 = ( )
( )+ ( )= B2( ) B2(3, 7) 1. The image produced by M under enlargement
with scale factor 2 is M1(8, 12). What is the
coordinates of M?
C( ) – ( ) = ( )
( )× 2 =( ) Solution
( ) + ( )= C2( ) C2(9, 15) ⁄
= 1( *=( )
⁄

Exercises 11.6 Given M1(12, 8) and scale factor, 2

A. 1. Triangle ABC has vertices A(1, 2), B(3, 2), x = 8, y = 12 and k = 2
⁄
C(5, 0). The triangle is enlarged by a scale factor = 1( *=( ) = (4, 8)
⁄
of 2 about the point (1, 4). Write down the
coordinates of the transformed triangle. 2. The image of R when it is enlarged by a scale
factor – 3 is R1 (3, -9). Find the coordinates of the R
2. If the line segment AB with vertices A(1, 2)
and B(3, 2), is reduced by a scale factor of Solution
⁄
about the point (1, 0), find the coordinates of the = `1( *=( )
⁄
images A1, B1.
Given R1(3, -9) and scale factor, -3
x = 3, y = -9 and k = -3
3. The rectangle PQRS has corners at P (1, 1),
⁄
Q(4, 1), R(4, -5), S(1, -5). The rectangle is R = R1( *= ( ) = (-1, 3)
⁄
enlarged by the factor of 1 about the point (4, -
2). Write down the coordinates of the vertices of Exercises 11.7
the enlarged rectangle. 1. Under enlargement with scale factor 3, B →
B1. If the coordinate of B1 is (15, 6), what is the
4. Write down the coordinates of the images of coordinates of B?
the point (3, 2) under enlargement with the
following centers and scale factors; 2.The image of A when it is enlarged by a scale
factor –2 is A1 (-6, -2). Find the coordinates of R
i. (3, 2), -5 ii. (9, 8), –

3. The vertices of a triangle are PQR.. If P is

Finding the Object given the image A1 and the
enlarged by a scale factor, , Q is enlarged by a
Scale Factor, k
scale factor 4, and R is enlarged by scale factor, –

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 , such that P → P1( -6, 2), Q → Q1(0, 1) and R Q( ) → Q1( )
→R1(-9, -12), determine the coordinates of PQR
Q( ) → Q1( ) Q1(7, -9)
Reflection
Here an image is produced or formed by 4. Find the image of M(-11, 5) under reflection in
reflecting an object or point in either the x – the line y = 0
axis or y – axis. Thus, we have -:
1. Reflection in the x – axis or line y = 0 Solution
2. Reflection in the y- axis or line x = 0
M ( ) → M1( )
3. Reflection in a given line
M( ) → M1( ) M1(-11, -5)
Reflection in the x- axis (Line y = 0)
This is represented by the mapping: Solved Past Questions
( ) → ( ) 1. a. Find the image of the position vector ( )

Object Image under translation by the vector ( )

b. If A(2, 3) is reflected in the x – axis, find the

Worked Examples
image A1 of A
1. If (3, 5) is reflected in the – axis, find the
image A1
Solution
Solution a. ( ) + ( ) = ( )
A( ) → A1( )
A( ) → A1( ), A1(3, -5) b. A( ) → A1( )
A( ) → A1( ), A1(2, -3)
2. Find the image of P(-2, -4) when it is reflected
in the x- axis. Reflection in the y- axis (Line x = 0)
This is represented by the mapping,
Solution
( ) →( )
In the x- axis, ( ) → ( )
Object Image
P( ) → P1( ) P1(-2, 4)
Worked Examples
3. What is the image of Q (7, 9) under reflection 1. If A ( is reflected in the y – axis, find its
in the x- axis? image, A1

Solution Solution
Reflection in the x – axis
A( ) → A1( )

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A( ) → A1( ) A1(-3, 8) ( ) → ( )
Object Image
2. Given A ( , B( and C(-5, -4), find the
image A1B1C1 of ABC under reflection in the line Worked Example
x = 0, such that A → A1, B →B1 and C → C1. 1. Find the coordinates of the image of the point
(5, 2) under reflection in the line y = -x
Solution
For reflection in the line x = 0 or y- axis; Solution
( )→( ) For reflection in the line y = -x;

A( ) → A1( ), A1 (-3, 2) ( ) → ( )
B( ) → B1 ( ) B1 (-7, 5) ( )→( ), (-2, -5)
C( ) → C1( ), C1 (5,-4)
2. What is the coordinates of the image of
Exercises 11.8 A (-7, 5) under reflection in the line y = – x
1. The image of P(3, 0) is P1. If P is reflected
inthe y – axis, find the coordinates of P1. Solution
2. Given that U ( , V( and W ( . For reflection in the line y = – x ;
Find the images U1V1W1 of U, V, W under ( ) → ( )
reflection in the line x = 0. A( ) →A2( ), A2(-5, 7)

Reflection in a Given Line 3. Reflection in the line x = a or x – a = 0

1. Reflection in the line y = x or y – x = 0 This is represented by the mapping:
This is represented by the mapping: (
( ) → ( )
( ) → ( )
Object Image
Object Image
Worked Example
Worked Example 1. The point P(2, 1) is reflected in the line x = 3.
1. P is the point (3, 7), write down the coordinate Find the coordinates of its image.
of the point obtained by reflecting P in the line
y=x Solution
For reflection in the line x = a
Solution
(
For reflection in the line y = x or y – x = 0 ( ) → ( )
( ) → ( ) But x = 2, y = 1and line a = 3
(
( )→( ), (7, 3) Substitute in ( ) → ( )
(
⇒ P ( ) → P1( )
2. Reflection in the line y = -x or y + x = 0 P ( ) → P1( ) P1(4, 1)
This is represented by the mapping:
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 Substitute in P1( ) → P2( ( )
2. Under reflection in the line x = 7, the point
A(2, 3) is mapped onto the point A1: P1( ) → P2( ( )
i. Find the coordinates of A1 P1( ) → P2( ) P1(-3, 0)
ii. If O is the origin, find the area of ∆OAA1
2. Find the image of the point P(-12, 4) under a
Solution
reflection in the line y = -3
i. For reflection in the line x = a
(
( ) → ( ) Solution
But x = 2, y = 3 and line a = 7 For reflection in the line y = a
(
Substitute in ( ) → ( ) P1( ) → P2( ( )
A ( ) → A1( ( )
A ( ) → A1( ) A1(12, 3) But x = -12, y = 4 and a = -3
Substitute in P1( ) → P2( ( )
ii. O(0, 0), A(2, 3) and A1(12, 3)
P1( ) → P2( ( )
A = bh
P1( ) → P2( ) P1(-12, -10)
But b = /AA1/ = √( (
=√ =√ = 10
Exercises 11.9
h = y axis of /OA/
1. The coordinates of the vertices of a
h = √( =√ =3 quadrilateral are A(3, 1), B(-3, -4), C(0, -4) and
D(5, -2). ABCD is reflected in the line x = -4,
A = bh = × 10 × 3 = 15sq. units calculate the coordinates of the images of A, B, C
and D.
4. Reflection in the line y = a or y – a = 0
This is represented by the mapping: 2. A kite has vertices at the four points O(0,6),
P(-3,-2), R(0, -6) and T(3, -2)
( ) → ( ( ) i. Find the coordinates of the images of the kite
Object Image when it is reflected in the line y = 3
ii. If the image kite is reflected in the line x = 2,
Worked Example determine the coordinates of the of the vertices of
1. The point P(-3, 4) is reflected in the line the final kite obtained.
y = 2. Find the image of P.
3. A parallelogram ABCD has vertices A(1, 3),
Solution B(4, 5) C(9, 5) and D(6, 3). It is given two
For reflection in the line y = a transformations, a reflection in the x – axis,
followed by a reflection in the line x = - 2.
P1( ) → P2( ( )
Determine the coordinates of the vertices of the
But x = -3, y = 4 and a = 2 parallelogram in both positions.

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4. The vertices of a triangle are A(2,0), B(-3, 0), A( ) → A1( ), A1 (4, -5)
C(-5, 3). The triangle is reflected in the line y = -
4 and followed by a reflection in the line x = 4. B( ) → B1 ( ), B1 (6, 1)
Determine the coordinates of the of the vertices C( ) → C( ), C1(-2, 7)
of the triangle in both positions.

2. Clockwise Rotation through 1800or Anti –

Rotation clockwise Rotation through 1800 about the
Rotation is the turning of an object about or Origin.
around a fixed point or the origin. Rotation can be This is described by the mapping;
clockwise or anticlockwise through 900, 1800,
( ) →( )
2700 and 3600.
Object Image
y
Anti-Clockwise(A) Clockwise(C)
( )
Worked Example
0
90 anticlockwise ( * Find the image of P ( , Q(-5, -3) and
0
270 clockwise R( under clockwise rotation through 1800
about the origin.
x
0 0
180 clockwise 90 Clockwise Solution
0 0
180 anticlockwise 270 anticlockwise
For clockwise rotation through 1800 about the
( * ( *
origin:
( ) ( )
Clockwise and Anticlockwise Rotations about
P( ) → P1( ), P1(-7, -4)
the Origin
Q( ) → Q1( ), Q1(5, 3)
1. Clockwise Rotation through 900 or Anti –
clockwise Rotation through 2700 about the R1( )→ R1( ) R1(10, -8)
origin.
This is described by the mapping; 3. Clockwise Rotation through 2700 clockwise or
( ) ( ) Anti – clockwise Rotation Through 900 about the
Object Image origin
This is represented by the mapping:
Worked Example ( ) →( )
Find the image A1B1C1 of A(5,4), B(-1,6) and C(- Object Image
7, -2) under a clockwise rotation through 900
about the origin. Worked Examples
Given that U( ( -6, -5) and ( . Find
Solution U1, V1 and W1, the respective images of U, V,
For clockwise rotation through 900 about the and W through a clockwise rotation of 2700 about
origin;( ) ( ) the origin.

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Solution 900 clockwise. Determine the coordinates of the
For clockwise rotation of 2700 about the origin: vertices of P1, Q1 and R1 of the image triangle P1
( ) →( ) Q1 R1
ii. Triangle P1Q1R1 is also rotated through 900
clockwise to obtain triangle P2Q2R2. Write down
U( ) → U1( ) ,
the coordinates of P2Q2R2
V( ) → V1( )
W( ) → W1 ( ) Finding the Angle of Rotation Given the Object
and its Image
Solved Past Questions Given the coordinates of the object and its
The point P(-5, 1) is rotated through 1800. Find corresponding image, the angle of rotation can be
the coordinates of its image found by investigating the kind of rotation either
clockwise or anti clockwise connecting them.
Solution
Rotation through 1800 is represented by Mapping Transformation
( ) →( ) ( ) →( ) Rotation through 900
about the origin
P ( ) → P1( ) P1(5, -1)
( ) →( ) Rotation through –900
2. The point P(3, -1) is rotated about the origin about the origin
through an angle of 2700 in the clockwise ( ) →( ) Rotation through 1800
direction. Find the image of P about the origin

Solution Note that a rotation of -900denotes a clockwise

Rotation through 2700clockwise is represented rotation of 900 and a rotation of 900 means an anti
by: ( ) →( ) – clockwise rotation of 900.
(
P ( ) → P1( ) P1(1, 3) Exercises 11.11
Each of the following pair is a point and its
Exercises 11.10 image under a rotation about the origin. State
Given A(3, 4), B(5, -2) and C(-9 , -3), find: the angle of rotation in each case
1. The image A1B1C1 of ABC under clockwise 1. (6, 3) → (-3, 6) 2. (5, -2) → (-2, -5)
rotation through 900 about the origin. 3. (4, -1) → (-4, 1) 4. (-2, -3) → (3, -2)
2. The image A2 B2 C2 of ABC under a rotation
through 1800 clockwise about the origin.
3. The image A3 B3 C3 of A1B1C1 under a Rotation About a Fixed Point
clockwise rotation through 2700 about origin. Clockwise and Anti – clockwise Rotations
4. P(- 4, 3), Q(1, 2) and R (3,5) are the vertices of Through a given Angle about a Point (a, b)
a triangle. When the point A(x, y) is rotatedthrough 900about
i. The triangle is rotated about the origin through a pointsay (a, b), other than the origin, the image
A1 is found as shown below;

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I. Identify the object as A(x, y), the angle of 3. Find the image of the coordinates of the point
rotation and the center of rotation as (a, b) (-7, 2) under a clockwise rotation of 2700 with
II. Subtract the center of rotation from the given centre (-1 , - 5).
coordinate of the object. i.e. A1( )
4. Find the image of the coordinates of the point
III. Rotate the results (in II) through the given (0, 9) under an anticlockwise rotation of 1800
angle. with centre (2, -2).
IV. Add the results (in III) to the center of
rotation.
The Image of an Object undera Mapping
The processes are easily remembered by the To find the image of an object under a given
abbreviations S. R. A. where S means mapping:
“Subtract”, R means “Rotate” and A means I. Identify the x and y coordinates of the object
“Add” and the given mapping
II. Substitute the values of the x and y coordinates
Worked Examples in the rule of the mapping
Find the image of the coordinates of the point (7, III. Workout or simplify or evaluate to obtain the
8) under an anticlockwise rotation of 900 with coordinates of the image of the object
centre (1, 5).
Worked Examples
Solution 1. A point P(-6, 8) undergoes transformation
Object = (7, 8),
Angle of rotation = 900 anticlockwise under the mapping ( ) →( )
Centre of rotation = (1, 5) i. Find the coordinates of the image P1
Coordinate of the object – Centre of rotation ii. Find ⃗⃗⃗⃗⃗⃗⃗
i.e. A1( )= ( ) iii. Calculate
( ) under an anticlockwise rotation of 900
Solution
( ) →( ) i. Given P(-6, 8) ⇒x = -6, y = 8,
( ) →( ) Substitute in ( ) →( )
Add the results to the centre of rotation. P → P 1

( ) + ( )= ( ), (-2, 11) P( ) → P1( ) → P1( ) P1(-10, 6)

Exercises 11.12
ii. ⃗⃗⃗⃗⃗⃗⃗ = OP1 – OP = ( ) ( )=( )
1. Find the image of the coordinates of the point
(-3, 8) under an anticlockwise rotation of 900 with iii. = √( (
centre (2, 3). =√ =√ = 4.5 units
2. Find the image of the coordinates of the point
(-5, -7) under a clockwise rotation of 900 with 2. Find the image of (3, -7) under the
centre (-4, 8). transformation ( ) →( ).

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Solution 2. Determine the image of A(-10, 5) under the
Given (3, -7) ⇒x = 3, y = -7 substitute in mapping ( ) →( *
( ) →( )
(
( ) →( (
) Drawing the Graph of Transformation
I. Draw two perpendicular lines, Ox and Oy that
( ) →( )
divide the graph sheet into two equal
( ) →( ) (-1, - 16)
halves or unequal halves
3. Find the image of (5, -8) under the mapping II. Identify the origin or center as (0, 0)
III. Name the vertical axis as yand horizontal axis
( ) →( ).
as x
IV. Mark and number both axes according
Solution to the given scale and intervals provided.
Given (5, -8), V. Plot the given points (of the object).
⇒x = 5, y = -8 substitute in VI. Workout the image whether under translation,
( ) →( ) enlargement, reflection, rotation or mapping and
( plot and form the image (s)on the graph sheet.
( ) →( (
)
( ) →( ) (8, - 11) Worked Examples
1.Using a scale of 2cm to 2units on both axes,
4. What is the image of (-6, 4) under the draw two perpendicular lines on a
mapping ( ) →( *? graph sheet for the intervals -8 8 and -8
8
b. Plot P (3, 1), Q (3, 4), R (-2, 2) and join the
Solution
points with a rule to form triangle PQR
Given (-6, 4),⇒x = -6, y = 4 substitute in
c. Draw the image P1 Q1 R1 of triangle PQR
( ) →( * under translation by the translation vector( ),
( where P → P1, Q → Q1 and R → R1
( ) →( (
*
( ) →( ) (2, 2) Solution
P (3, 1), Q (3, 4), R (-2, 2)
Exercises 11.13 Object + Translation Vector = Image
1. Find the image of P(-4, -5) under the mapping P ( ) + ( ) = P1( ), P1(7, -2)
( ) →( ) Q ( ) + ( ) = Q1( ), Q1(7, 1)

R ( ) + ( ) = R1( ), R1(2, -1)

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 y

10 Scale : 2cm : 2 units on both axes

6
Q(3, 4)
4
R(-2, 2)
2 Q1 (7, 1)
P(3, 1)
x
-10 -8 -6 -4 -2 0 2 4 6 8 10
-2 R1 (2, -1)
P1 (7, -2)
-4

-6

-8

-10

2. Using a scale of 2cm to 2units on both axes, coordinates of the vertices of the triangles drawn
draw two perpendicular lines Ox and Oy on a
graph sheet for the intervals -8 8 and -8 Solution
8 A (3, 5), B (-1, 2) , C (4, 2).
b. Plot the points A (3, 5), B (-1, 2) and C (4, 2) Reflection in the x – axis ( ) → ( )
and join them with a rule to form triangle ABC
A( ) → A1( ) A1(3, -5)
c. Draw the image P1 Q1 R1 of triangle PQR
under reflection in the x - axis, such that A → A1, B( ) → B1( ), B1(-1, -2)
B → B1 and C → C1. Clearly indicate the C( ) → C1( ) Q1(4, -2)

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 y

10 2cm : 2 units on both axes

6
A(3, 5)
4
B(-1, 2)
2
C (4, 2)
x
-10 -8 -6 -4 -2 0 2 4 6 8 10
B1 (-1, -2) -2 C1 (4, -2)

-4

-6 A1 (3,-5)

-8

-10

3. i. Using a scale of 2cm to 2 units on both axes, B → B1 and C → C1. Clearly indicate the
draw two perpendicular lines on a coordinates of the vertices of the triangles drawn
graph sheet for the intervals -8 8 and -8 Solution
8 ii. A (3, 5), B (-1, 2), C (4, 2)
ii. Plot the points A (3, 5), B (-1, 2) and C (4, 2) iii. Reflection in the y – axis ( ) → ( )
and join them
A( ) → A1( ) A1(-3, 5)
with a rule to form triangle ABC
iii. Draw the image A1 B1 C1 of triangle ABC
under reflection in the y - axis, such that A → A1, B( ) → B1( ), B1(1, 2)
C( ) → C1( ) C1(-4, 2)

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 y

10 2cm : 2 units on both axes

A1 (-3,5) 6
A(3, 5)
4
B(-1, 2)
2
C1 (-4, 2) B1 (1, 2) C (4, 2)
x
-10 -8 -6 -4 -2 0 2 4 6 8 10
-2

-4

-6

-8

-10

4. a. Using a scale of 2cm to 1unit on both axes, of the vertices of the triangles drawn
draw two perpendicular lines on a Solution
graph sheet for the intervals -6 6 and a. A (3, 1), B (-2, 3) and C (-5, -1)
-6 6
b. Plot the points A (3, 1), B (-2, 3) and C (-5, -1) c. Anti - clockwise rotation through 900 about the
and join them with a rule to form ∆ ABC origin( ) →( )
c. Draw the image A1 B1 C1 of triangle ABC
A( ) → A1( ) A1(-1, 3)
under rotation through 900 anti – clockwise about
the origin, such that A → A1, B → B1 and C → B( ) → B1( ), B1(-3, -2)
C1. Clearly indicate the coordinates C( ) → C1( ) C1(1, -5)

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 y

5 2cm : 1 unit on both axes

4
B(-2, 3) A1 (-1, 3)
3

1 A(3, 1)

x
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1
C (-5, -1)
-2
B1 (-3, -2)
-3

-4

-5
C1 (1, -5)

Some Solved Past Questions ii. Equation of AC;

1.i. Using a scale of 2cm to 1 unit on both axes, A( and C(
draw the x and y axes for and
. Grad of AC, m = = (undefined)
ii. Plot the points A( , B( and C(
and describe the triangle ABC. iv. Reflection of ∆ ABC in the
iii. Find the equation of AC ( ) →( )
iv. Draw the triangle which is the A ( ) → A1( ) A1(-3, 1)
reflection of triangle ABC in the
B ( ) → B1( ) B1(-1, 1)
where, , and . Indicate
C( ) → C1( ) C1(-3, 5)
clearly the coordinates of , and .
v. Draw triangle which is the image of
v. Coordinates of under the mapping;
triangle ABC under the mapping( ) ( * where

, and indicating clearly the ( ) ( *

coordinates of , and . A ( ) → A2( (
) A2(6, 0)
(
B ( ) → B2( ) B2(2, 0)
Solution
(
ii.A( , B( and C( C( ) → C2( ) C2(6, 5)
Triangle ABC is a right triangle

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 y
6
C1 (-1, 5) C(3, 5)
5

1
A1 (-3, 1) B1 (-1, 1) B(1, 1) A(3, 1) A2 (6, 0)
x
-4 -3 -2 -1 0 1 2 3 4 5 6
-1 B2 (2, 0)

-2
-3

-4
-5
C2 (6, 5)
-6

2. Using a scale of to units on each axis,

draw on a sheet of paper two perpendicular Solution
axes and for the interval and i. ( , ( and (
Draw: ii. Reflection in the line
i. Triangle PQR with vertices ( , ( and ( ) → ( )
(
(
ii. The image triangle P1Q1R1 of triangle a P( ) →P1( ( ) → P1( ), P1 = (2, -4)
reflection in the liney = 2where P → P1, Q → Q1 Q( ) →Q1( ) → Q1( ), Q1 (5, -4)
(
and R → R1
iii. The image triangle P2Q2R2of triangle R( ) →R1( ( )→ R1( ), R1 (2, 0)
under a half turn about the origin
where P → P2, Q → Q2 and R → R2. iii. Half turn about the origin, O
iv. Describe precisely the transformation that will ( ) →( )
map triangle P2Q2R2 onto triangle P1Q1R1 where P( ) →P2( ) P2(-8, 2)
P2 → P1, Q2 → Q1 and R2 → R1. Q( ) → Q2( ) Q2(-8, 5)
R( ) →Q2( ) R2(-4, 2)

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 y
12
Scale = 2cm to 2units on both axes 10

8 P(2,8) Q(5,8)

6
P2 (-8,5)
4
Q(2,4)
2
P2 (-8,2) P1(-4, 2)
R1(2,0)
x
-10 -8 -6 -4 -2 2 4 6 8 10
-2
-4
P1 (2, -4) Q1(5, -4)
-6

-8

-10
-12

Exercises 11.14 2. i. Using a scale of 2cm to 2units on both axes

1. i. Draw on a sheet of graph paper two draw two perpendicular lines Ox and Oy on a
perpendicular axes for and graph sheet.
, using a scale of 2cm to 1 unit on both ii. Mark the x-axis from -10 to 10 and the y-axis
axes; from -12 to 12.
ii. Draw on the graph sheet a quadrilateral FGHJ iii. Plot P (2, 3), Q (0, 5), R (5, 3). Join the
with ( ,( ,( and ( . points with a rule to form triangle PQR. What
iii. Draw the image , where type of triangle is PQR?
and under the iv. Draw the image P1Q1R1 of triangle PQR,
mapping ( ) ( ) showing clearly the under enlargement with scale factor 2 from the
origin, such that P → P1, Q → Q1 and R→ R1.
coordinates of the vertices.
v. Draw the image P2 Q2 R2 of triangle PQR, such
iv. Describe precisely the shape of the image
that P→P2, Q → Q2 and R → R2 under reflection
in the x- axis. Show all the linesof transformation.
v. What is the ratio of the area of FGHJ to the vi. Draw P3Q3R3 of PQR the image under rotation
area of ? through 2700 anticlockwise such that P → P3, Q
vi. What two transformations will map F onto ?

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→ Q3 and R → R3 Clearly indicate all the co- 5. i. Using a scale of 2cm to 2 units on both axes,
ordinates of the triangles drawn. draw two perpendicular lines Ox and Oy on a
graph sheet.
3. i. Using a scale of 2cm to 2 units on both axes, ii. Mark the x-axis from -10 to 10 and the y-axis,
draw two perpendicular lines Ox and Oy from -12 to 12.
on a graph sheet. iii. Plot the points A (1, 3), B (6, 5) and C (4, 0) .
ii. Mark the x-axis from -10 to 10 and the y-axis Join the points to form triangle ABC.
from -12 to 12. What type of triangle is formed?
iii. Plot the points A (-9, -4), B (-7,-2), C (-5,-7) iv. Draw triangle A1B1C1, the image of triangle
and D (-3,-5) Join all the points with a rule and ABC under translation by the vector( ) such that
name the figure formed. A → A1, B → B1, and C → C1
iv. Draw the image A1B1C1D1 of figure ABCD v. Draw the image A2B2C2 of ∆ABC under
under rotation through 2700 clockwise about the enlargement with scale factor -1, from the origin,
origin, such that A→A1, B→B1, C→C1 and such that A → A2, B → B2, C →C2
D→D1. vi. With line x = 0 as mirror line, draw the image
v. Draw the image A2B2C2D2 of figure ABCD A3B3C3 of triangle ABC, such that A→ A3, B→
under enlargement from the origin with scale B3 and C→ C3 .Show all the lines transformation
factor -1,such that A → A2, B → B2, C → C2 and of triangle ABC and triangle A3B3C3
D → D2
vi. With line y = 0 as line of reflection, reflect 6. i. Using a scale of 2cm to 1 unit on x- axis and
ABCD to get A3 B3 C3 D3 such that A → A3, 2cm to 2 units on y-axis axes, draw two
B →B3, C→ C3 and D →D3 perpendicular lines Ox and Oy on a graph sheet.
ii. Mark the x - axis from -5 to 5 and the y-axis
4. a. Using a scale of 2cm to 1 unit on both axes, from -12 to12.
for – 4 ≤ x ≤ 5 and – 5 ≤ x ≤ 5, draw ∆ PQR with
iii. Plot the points, ( ,-3), ( ,-2) and C (3,-
vertices ( ( and (
5). Join A to B and B to C with a rule.
b. Draw the image ∆ of triangle PQR
iv. Draw the image A1B1C1 of ABC under
under a reflection in the line where P
enlargement with scale factor -1 from the origin,
Q and R the image ∆ of ∆ PQR
such that A → A1, B → B1 and C → C1
under a translation by the vector ( ), where
v. Draw the image A2 B2 C2 of ABC under
P ,Q and R reflection in the line x = 0,such that A →A2, B →
c. Draw the image ∆ of ∆ PQR under an B2 and C → C2.
anticlockwise rotation of 90 about the origin vi. Draw the image A3B3C3 of ABC under
where and rotation through 2700 clockwise about the origin,
d. Name two coincident image points in your such that A → A3 B →B 3 and C → C3. Clearly
diagram. indicate all the coordinates of the figure drawn.
e. Describe precisely a single transformation that vii. What single transformation maps A1B1C1 to
maps triangle unto ∆ where A2B2C2?
and

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7. i. Using a scale of 2cm to 1unit on both axes, 9. Using a scale of 2cm to 2 units on each axes,
draw two perpendicular lines Ox and Oy on a draw on a graph sheet of paper two perpendicular
graph sheet. line Ox and Oy for the intervals - 8 x 12 and
ii. Mark the x-axis from -10 to 10 and the y-axis -12 y 12
from -12 to 12. i. Draw ∆ABC with coordinates A(5, 7), B(3, 4)
iii. Plot the points P(1, 1), Q(3, 1), R(3, 3) and and C(7, 3)
S(1, 4) and join them with a ruler to form figure ii. Draw the image ∆A1 B1 C1 of ∆ABC under
PQRS. What type of quadrilateral is PQRS? translation by the vector ( ), such that A → A1,
iv. Draw image P1Q1R1S1 of PQRS, under B → B1 and C → C1
enlargement with scale factor (-1) from the iii. Draw the image ∆A2B2C2 of ∆ABC under
origin, such that P 1, Q 1, R 1 and S reflection in the line y = -2, such that A → A2, B
1 → B2 and C →C2
v. Draw image P2Q2R2S2 of PQRS under iv. Draw ∆A3B3C3 of ∆ABC the image under
clockwise rotation through 900about the origin rotation through 900 anticlockwise about the
such that P 2, Q 2, R 2 and S 2 origin, such that A → A3, B → B3 and C → C3
vi. Draw the image P3Q3R3S3 of P, Q, R, S under v. find the gradient of line B2B3
reflection in the line x = 0, such that P 3, Q

3, R 3 and S 3. Scale Factor (k)

A geometric figure can be enlarged or reduced
8. i. Using a scale of 2cm to 2 units on both axes, from a point called center of enlargement or
draw on a sheet of graph paper two perpendicular reduction respectfully.
axes Ox and Oy for the interval
and When a geometric figure is enlarged or reduced, a
ii. Plot the following points: A (2, 4), B (2, 1) and new figure is formed. The new figure formed is
C(-1, 1). Join the points to form triangle ABC. called the Image (I)of the original figure, called
iii. On the same graph sheet draw the image the Object (O).
triangle A1B1C1 under translation by vector ( ),
such that 1, 1 and 1.
The ratio of the length of the image to its
iv. Draw the image triangle A2B2C2 of triangle corresponding object length is a constant called
ABC under reflection in the y – Scale factor of enlargement or reduction, denoted
axis such that 2, 2 and 2.
by k : Thus,
v. Draw the image triangle A3B3C3of triangle k=
ABC under enlargement with scale factor 2, such
that A→A3, 3 and 3. In case of comparative areas,
vi. Draw the image triangle A4B4C4 of triangle k2 =
ABC under clockwise rotation through 2700
about the origin, such that 4, 4 and In case of comparative volumes,
4. Clearly indicate the coordinates of all the
k3 =
triangles drawn.

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Effects Of The Scale Factor i. Calculate the scale factor
I. If k is positive, the object and its image are on ii. Find the value of x
the same side of the center of enlargement.
II. If k is negative, the object and its image are on Solution
opposite sides of the centre of enlargement. i. k = K
III. If k = 1 or – 1 , it implies that the object and x
1.5m
k= =3 J
its image are of the same size.
3m 6m
II. If k ˃ 1, the image size is bigger than the
object size ii. k = , but O = 1.5m, I = x and k = 3
IV. If k ˂ 1, the object size is bigger than the
,
image size.
x = k × O = 3 × 1.5 = 4.5m OR
On the other hand, if k is a proper fraction, the x = 3 × 1. 5 = = = 4.5m
object size is reduced in a process called
Reduction. C1 Exercises 11.15
1. In the figure below, triangle OX1Y1 is an
C enlargement of OXY. Find the scale factor.
Y X1
O
A B A1 B1
O
4cm 10cm
In the diagram above,
I. O is the center of enlargement.
II. ABC is the object and A1B 1C1 is the image. X Y1
III. Scale factor, k = 2. In the diagram below,
a. Find the scale factor of the enlargement.
Worked Examples b. Calculate / OC1/ if OC = 6cm
O C C1
1. In the diagram below, rectangle AEFG is an
enlargement of rectangle ABCD. If /AB/ = 5cm 2cm
A B
and /AE/ = 15cm, what is the scale factor of
1cm
enlargement? A1 B1
1. A B E
3. In the diagrams below, fig I is an enlargement
of Fig II. Find the side EF of Fig II.
D C
F1 Fig I F Fig II
G
F 30cm
g E
Solution E G
R 4cm
k= = 3cm 1 G 1
20cm H
H1
2. In the diagram below, K is an enlargement of J.

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4. In the triangle below, OA1B1 is an enlargement is the image of the tyre of the lorry, find the scale
of triangle OAB. Find the scale factor of factor of enlargement.
enlargement.
A
3cm Challenge problem
A1
1. The area of a hockey pitch is 1408m2 and its
6cm image which is a volley pitch has an area of
88m2. If the length of the hockey pitch is 44m,
O i. Determine the scale factor of enlargement ii.
B1 B
Determine the length and breadth of the volley
pitch.
P
5. In the diagram below, PQR
is an enlargement of PAB. Find: More Worked Examples
i) the scale factor of A B
Given that triangle A1B1 C1 is an enlargement of
enlargement; 8cm triangle ABC with scale factor 3,
ii) /PB/ if /PR/ = 15cm. B
Q 24cm R 1 B

5cm 7cm
6. In the diagram below, the square PQRS is an
enlargement of square ABCD from the A C A 3cm C
center O. The area of square ABCD is 4cm2 and
Find1 the length
1 of the following;
the area of square PQRS is 9cm2. Find the scale
i) A1 B1 ii) A1 C1 iii) B1C1
factor of enlargement.
P Q Solution
A B
i. k =
O
/A1 B1/ = k /AB/
C D
S R But k = 3, and /AB/ = 5cm
/A1 B1 / = 3 × 5 cm = 15cm
7. Find the scale factor of enlargement and /OQ/
from the diagram below: ii. k =
P1
10cm k = 3 and /AC/ = 3cm
P /A1 C1 / = k /AC/ = 3 × 3cm = 9cm
4cm
o
m
5cm
iii) k = ,
Q1
But k = 3 and / B C / = 7 cm
Q
/ B1 C1 / = k / BC / = 3 × 7 cm = 21cm
8. The radius of a lorry tyre is 54 cm and that of a
motor bike is 18cm. If the tyre of the motor bike Exercises 11.16
Triangle P1Q1R1 is an enlargement of triangle

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PQR with scale factor 2. Find the following
lengths; Q A
i. P1 Q1 Q1
ii. P1 R1 L M
iii.Q1 R1 3cm

P1 R1 B 8 C
P 4cm R

Similar Figures i. Find a pair of similar triangles

If one figure is an enlargement of another, then ii. If BC = 8cm, Find LM
they have the same shape, but not necessarily the
same size. In this case they are said to be similar. Solution
ALM is a reduction of ABC, with scale factor,
If two figures are similar, then they have the same
k= (midpoint)
angles. There is also a fixed ratio between the
sides of the figures called the scale factor, k, of Hence ABC is similar to ALM.
enlargement. It follows that LM: BC is equal to
LM = 8 × = 4
In particular, if triangles ABC and DEF are
similar, ⇒< A = < D, < B = < E and < C = < F
E 2. L and M are on the sides AB and AC of the
B triangle ABC. AM = 4, MC = 5, AL = 3, LB = 9.
A
3 4
A C D F L M
Likewise, there is a fixed ratio between the sides 5
9
of ∆ABC and the sides of ∆DEF C
k= = =
B
Find the ratio LM : BC
The ratio of sides is the same for both triangles:
= : = = : = = Solution
AB = 12, AC = 9. So the ratios AL : AC and AM :
For clarity and better understanding, there is the AB are both equal to 1 : 3
need to draw the figures separately but it is not The angle A is the same in both triangles. So ABC
required all the time. and AML are similar. It follows that LM : BC =
AM : AB = 1 : 3
Worked Examples
1. L and M are the mid points of the sides AB and 3. The figure below shows that triangle ADE is
AC of the triangle ABC. an enlargement of triangle ABC. If /DB/ = 6cm,
/AD/ = 9cm, /CE/= 2cm and /DE/ = 12cm,

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calculate; i. /BC/ ii. /AE/ /AE/ = = 3 cm.
A

4. In the accompanying Z
xcm
diagram, find x, y and h W
D E
5cm hcm 6cm
Solution
A
B C X ycm Y
9 Solution
D c 12 E Redraw the triangle separately as shown:
6 m 2
Object = ABC
c c Image = ADE Z Z
B C x cm
m m 6cm
Redraw the triangle separately as shown below; 5 cm x cm
6cm
A
9 A Y W
h cm
X y cm Y
6 9
2
ZYW is equivalent to ZXY;
D E
B C 12 So,
∆ ABC and ∆ADE are similar. Thus, /AE/ ⇒
corresponds to side /AC/, /AD/ corresponds to
x (x + 5) = 6 × 6 (cross products)
/AB/ and / DE/ corresponds to /BC/ 2
x + 5x = 36
⇒ x2 + 5x – 36 = 0
(x − 4) (x + 9) = 0 (By factirisation)
Consider, x − 4 = 0 or x + 9 = 0
x = 4 or x = − 9
⇒ So x = 4 units
=
9 × /BC/ = 12 × 15 Consider triangle YZW
h2 = 62 – x2, but x = 4
/ BC / = = 20 cm
h2 = 36 – 16
h2 = 20
ii. To find /AE/,
h = √ cm
,
By Pythagoras theorem,
y2 = 52 + h2
15 × /AE/ = 9 × (/AE/ + 2)
y2 = 52 + √ 2
15/AE/ = 9/AE/ + 18
y2 = 45
15/AE/ − /9AE/ = 18
6 /AE/ = 18 y = √ cm

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5. In the diagram below, < PQR = < PSQ = 900. 6. In the diagram, VXZ is a triangle, /VZ/ = 6cm,
/PS/ = 9cm, /SR/ = 16cm and /SQ/ = xcm. Find: /WY/ = 2 cm and <VZX = XWY = 900. Find the
a. the value of x; P ratio of the area of ∆ VZX to the area of ∆ WXY
9cm
ii. < QRS correct to S V
the nearest degree; xcm 16cm
W
iii. /PQ/. 6cm
Q R
2cm

Solution Z Y X
a. Redraw the figure separately as below; Solution
R Redraw the triangle separately as shown:
Q
V Y
x
16cm
Q P 6cm 2cm
S x S 9cm
Z a cm X W a X

QRS is similar to PQS ∆ VZX is similar to ∆ WXY

⇒ = k= = =
2
x = 16 × 9 ⇒ ZX a, /WX/ = a
x2 = 144
Area of ∆ VZX = b h
x=√ = 12cm
S = × a × 6 = 3a cm2
b. From QRS,
12cm 16cm
tan θ = Area of ∆ WXY = b h = × a × 2 = a cm2
Ɵ
( ) Q R
θ=
The ratio of the area of ∆ VZX to the area of ∆
θ = 36.6
θ = 370 (correct to the nearest degeree) WXY = 3a : a = 3 :
3 × 3 : 3 × = 9 : 1 (Multiply through by 3)
c. From QRS,
/QR/2 = 122 + 162 6. A pole and a stick stand vertically on a level
/QR/ = √ =√ = 20cm ground. The stick is 25 cm long and cast a
shadow of length 20 cm. If the pole casts a
From PQR, shadow of length 5m, what is the height of the
P (9 + 16)2 = /PQ/2 + 202 pole?
9cm 252 = /PQ/2 + 202
S
/PQ/2 = 252 - 202 Solution h
xcm 16cm 25cm
/PQ/ = √ Let the height
Q R /PQ/ = √ of the pole be h 20cm
20cm 500cm
/PQ/ = 15cm

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From the similar triangles, 4. A vertical pole 6m high, cast a shadow 9m
= long at the same time that a tree cast a shadow
30m long. Find the height of the tree . A = 625 cm
⇒h = = 625 cm
5. In the figure below,
C
Exercises 11.17
1. X and Y lie on the sides AB, AC of triangle C1 80mm
ABC, so that XY is parallel A O B 120mm
B1 100mm
to BC as shown in the 32mm
figure below. X Y A
Write down a pair of A1
similar triangles. B i. Write three equal ratios of the sizes of the two
C
similar triangles.
2. L, M and N lie on the sides AB, BC, CA of ∆ ii. If /A1B1/ = 32mm, find /A1C1/ and /B1C1/.
ABC. LM, MN, NL are parallel to CA, AB, BC
respectively as shown below, D. The triangles ABC and DEF below are similar

A
N
A C D

M
L

B C E F
B
1. If AB = 4, DE = 2 and AC = 6, find DF
Write down as many similar triangles as you can.
2. If AB = 3, AC = 2 and DE = 6, find DF
B. The two triangles below are similar with the 3. If BC = 9, EF = 12 and DE = 8, find AB
marked angles equal. Use this to answer 4. If AB = , DE = and BC =1 , find EF
question 1 to 3.
Rotational Symmetry
A 18cm
C D Any figure that can be turned through an angle
27cm E (say 900, 1800, 2700 and 3600 ) about a fixed point
3cm
7cm
such that the image fits onto the original shape of
the figure is said to have rotational symmetry.
B F
The number of times the figure fits onto the
1. Write a proportion involving the sides of the original shape when turned through is called the
two triangles. order of rotational symmetry. The order of
2. Calculate /BC/ Ans.= 63cm rotational symmetry of a rectangle is verified
3. Find /DE/ Ans.= 2cm below;

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Original shape Circle Infinity
Regular 5
pentagon
Rhombus 2

Exercises 11.18
1. Name 3 shapes that have both line of
symmetry and rotational symmetry.

Since two rotations fit the original shape of the

2. Determine the order of rotational symmetry of
object, the order of rotation of a rectangle is 2.
each of the following:
a. Isosceles triangle b. Equilateral triangle
Object Order of Symmetry
c. Regular Pentagon d. Regular Hexagon.
square 4
Rectangle 2
Kite 1

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12 RATIO AND RATES Baffour – BaSeries

Meaning of Ratio Solution

Ratio is the comparison of two or more Ratio of mother‟s age to daughter‟s age
similar quantities. Thus, we compare pen to = 60 : 28 = = = 15 : 7
pen, pencil to pencil, money to money etc.
4. Express 9 days to 3 weeks as a ratio.
In general the ratio of two numbers x and y
is written as x :yread as “x is to y”.
Solution
Express the ratiosin similar units.That is:
Ratio can also be expressed as a fraction. Thus,
9 days: 3 weeks, but7 days = 1 Week
x : y = and y : x = .  3 weeks = 3 × 7 days = 21 days
The new ratio is 9 days: 21 days
All ratios must be expressed in its simplest = or 3 : 7
form with no unit. If the units are not the same,
change one unit to the unit of the other before
5. Express the ratio 20cm to 15m in the form 1: n
simplification
Solution
Worked Examples Change 15m to cm
1. There are 24 boys and 21 girls in a class, find
Thus 15m = 1500cm
the ratio of boys to girls.
New ratio is 20cm: 1500cm
= = 1 : 75
Solution
Let B represent Boys and G represent Girls.
a. Ratio of boys to girls = B : G 6. Simplify the ratio 4 : 1
= 24 : 21 = = = 8:7
Solution
2. Two sticks measure 25cm and 100cm 4:1
respectively, find the ratio of their respective
4: ( *
lengths
4×4:4× ( *
Solution
16 : 7
25cm : 100cm = = = 1 : 4
Exercises 12.1
3. A mother is 60 years and the daughter is A. Express each ratio in its simplest form;
28 years, what is the ratio of mother‟s age to the 1. 85p : Ghȼ5 2. 5cm : 20mm
daughter‟s age. 3. 500g : 2kg 4. 3 km : 600m

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B. 1. Tom and Jerry were given 124 oranges to (5x – 2y) (2x – y) = 0
share. If Tom received 24 oranges, 5x – 2y = 0
a. find the ratio of Jerry‟s oranges to that of Tom. 5x = 2y ( *
b. express Jerry‟s share as a ratio of the total
x = 2 and y = 5
oranges.
=
2. Romeo is 9 years older than Juliet. If Juliet is Therefore the ratio x : y = 2: 5
18 years old; find the ratio of the age of Juliet to
that of Romeo. 2x – y = 0
2x = y ( *
3. Express 6 days is to 3 weeks as a ratio in its
simplest form. x = 1 and y = 2
=
4. Three friends are discussing pocket money. Therefore, the ratio x : y = 1: 2
The first gets Ghȼ3.60 a week, the second gets
Ghȼ4.80 and the third gets Ghȼ6.00 a week. 2. Find the ratio y : x in 4x2 + 6xy – 4y2 = 0
Write their pocket money as a ratio in its simplest
form Solution
4x2 + 6xy – 4y2 = 0
Ratio of Two variables in an Equation 4x2 – 2xy + 8xy – 4y2 = 0
A. In a quadratic equation, (4x2 – 2xy ) + (8xy – 4y2) = 0
I. Solve it the quadratic way by finding the sum 2x(2x – y) + 4y(2x – y) = 0
of roots and products of roots to obtain the (2x + 4y) (2x – y ) = 0
factors. 2x + 4y = 0 or 2x – y = 0
II. Equate the roots obtained to zero. 2x = - 4y or 2x = y
III. Solve each of the equations to obtain the
required ratio. When 2x = - 4y ( *

B. In a linear equation, x = - 4 and y = 2

I. Group like terms at each side of the equation. = =
II. Solve to obtain the required ratio. Therefore, y : x = 1 : - 2

Worked Example
When 2x = y ( *
1. In 10x2 – 9xy + 2y2 = 0 find the ratio x : y
x= 1 or y = 2
Solution =
10x2 – 9xy + 2y2 = 0 Therefore, y : x = 2 : 1
10x2 – 5xy – 4xy + 2y2 = 0
(10x2 – 5xy) – (4xy + 2y2) = 0
3. If 9 ( ) = 1, find the ratio of x : y
5x(2x – y ) – 2y(2x – y) = 0

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Solution 2. x2 + xy – 2y2 = 0 6. x (x – y ) = y(x + 3y)
9( )=1 3. 3x2 – 7xy + 4y2 = 0
4. = 7. + =9
=1
9x – 18y = x + 2y
Proportion
9x – x = 2y+ 18y
A proportion is an equation which says that two
8x = 20y ( * or more ratios are equal. In other words, when
x = 20 and y = 8 two ratios are equated, they form a proportion.
=
A proportion is read as “ x is to y” as “a is to b”
= expressed as x : y = a : b or =
Therefore x : y = 5 : 2 The products xb and ay are called cross products.
Cross products are always equal.
4. 3x – 4 = 2(y – 2). Find , where y 0 That is, in = , x × b = a × y

Solution Proving whether Two Ratios form a proportion

3x – 4 = 2(y – 2) To prove whether two or more ratios form a
3x – 4 = 2y – 4 proportion, check their cross products
3x = 2y – 4 + 4 I. If they are equal, the ratios form a proportion
3x = 2y ( * II. If they are unequal, the ratios do not form a
proportion
x = 2 and y = 3
= Worked Examples
1. Show whether 3 : 4 and 18 : 24 form a
5. If 4(a – 2) = 3b – 8, find , b 0 proportion.

Solution
Solution
4(a – 2) = 3b – 8 =
4a – 8 = 3b – 8 3 × 24 = 4 × 18 By cross multiplication,

4a = 3b – 8 + 8 72 = 72
4a = 3b  3 : 4 and 18 : 24 are proportional
a = 3 and b = 4
= 2. Do 3 : 6 and 18 : 36 form a proportion ?

Solution
Exercises 12.2 =
Find the ratio x : yin the following: 3 × 36 = 6 × 18
( – 108 = 108
1. x2 + xy – 6y2 = 0 5. =2

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 3 : 6 and 18 : 24 are proportional Proportion with a Variable
If one side of a proportion contains a variable, the
3. What proportion of Ghȼ150 is Ghȼ5? value of the variable is workout as follows;
I. Find the cross products of the ratios.
Solution II. Divide both sides of the equation by the
This means, express Ghȼ5 as a fraction of coefficient of the variable to obtain the value of
Ghȼ150 the variable.
=
In summary, if , then the cross product is
4. 10 men and 12 women work in a factory. a  d = bc
What proportion of the work force are men and
what proportion are women? Worked Examples
1. The ratio 9 : x is equivalent to 36 : 20, what
Solution
is the value of x?
10 men + 12 men = 22 workers
Proportion of men = = Solution
Proportion of women = = 9 : x = 36 : 20
=
Exercises 12.3 9  20 = 36  x
A. Determine whether the following pair of 180 = 36x
ratios form a proportion or not: =
1. 2 : 16 and 5 : 40 2. 7 : 3 and 21 : 6
3. 12 : 4 and 9 : 3 4. 15 : 5 and 5 : 15 5=x or x=5
2. The ratio 8:12 is equivalent to y : 9. What is the
B. Answer the following: value of y ?
1. What proportion of Ghȼ25 is Ghȼ5?
2. What proportion of 10 is 2.5? Solution
3. What proportion of 5 kg is 600g? 8 : 12 = y : 9

C. 1. In a sports stadium, it is found that the ratio =

of boys to girls is 21:19. What proportions are ⇒12 × y = 8 × 9
boys and what proportions are girls? 12y = 72
y= =6
2. Of the 200 people who took their driving test
last week, 60 passed. What proportion failed the
3. If 13 : 9 = 39 : 9k, find the value of k.
driving test?
Solution
3. Nancy throws a die 600 times and finds that a
13 : 9 = 39 : 9k
“6” turns up 80 times. What proportion of the
throws were sixes? =

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13 × 9k = 9 × 39 Solution
117k = 351 3x2 : 8 = 24: 16
k= =3 =

=
4. Given that (m + 2) : 1 = 6 : 2 , find m
2 × 3x2 = 3 × 8
Solution 6x2 = 24

= x2 =

2 (m + 2) = 1 × 6 x2 = 4
2m + 4 = 6 x=√ =2
2m = 6 – 4
2m = 2, Exercises 12.4
x=1 A. Find the value of the variables;
1. y : 28 = 5 : 7
5. Solve 3: (n – 7) = 5 : 10 2. 2 : x = 12 : 30
3. 2 : 11 = a : 33
Solution
B. Find value of x in the following:
=
1. = 2. = 3. =
5 (n – 7) = 10 × 3
5n − 35 = 30 4. = 5. = 6. =
5n = 30 + 35
5n = 65
C. Find the values of the variables:
n= = 13 1. 5 : 4 = 30 : 3x 3. 32 : 14 = 2x : 7
2. 4x : 48 = 1 : 3 4. 21 : 2x = 7 : 10,
Solved past Questions
1. If a : b = 4 : 5, find a + b : 2a – b D. Find the value of the variable:
1. (3 − 2y): 4 = 5: 4 4. 2 : (5 + y ) = 4 : 2
Solution 2. 5 (2a – a) : 3 = 25 : 15
If a : b = 4 : 5 3. (3y − 3) : 6 = 6 : 4 5. 3 : 7 = 6 : (b – 2)
=
Direct Proportion
Direct proportion involves comparison of two
⇒a = 4 and b = 5
quantities such that an increase in one quantity
a+b=4+5=9 automatically causes an increase in the other
2a – b = 2(4) – 5 = 8 – 5 = 3 quantity. Likewise, a decrease in one quantity
a + b : 2a – b = 9 : 3 = 3 : 1 requires a decrease in the other. For instance, if
the cost of 3 books is Gh¢30.00, any increase in
2. If 3x2 : 8 = 24 : 16, find the value of x given the number of books will cost more than
that x > 0

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Gh¢30.00 and any decrease in the number of Solution
books will cost less than Gh¢30.00. Let y represents the weight of 7 bags of rice
By direct proportion:
To solve problems involving direct proportions, 17 : 7 = 680: y
I. Represent the required value by any preferred =
variable; examplex, y, m, n etc 17 × y = 7 × 680
II. Identify the ratios involved and form a simple 17y = 4,760
proportion.
= 280
III. Solve for the value of the variable
 The weight of 7 bags of the rice is 280kg.
Worked Examples
1. The cost of 3 crates of eggs is Gh¢21.00. How 3. If 120km trip requires 8 gallons of petrol, how
much will you pay for 6 crates of eggs? many gallons of petrol is needed for a 600km
trip?
Solution
Method 1 Solution
The two quantities involved are “eggs” and Let x represent the liters of petrol required.
“money”. Therefore, form a ratio of eggs and By direct proportion,
equate it to the ratio of money. That is: 120 km : 600km = 8 : x
3crates : 6 crates = 21 : x = ,
= 120 × x = 600 × 8
3 × x = 21 × 6 By cross multiplication,
120x = 4,800
3x = 126 x= = 40
= = 44  40gallons is needed for the 600km trip.
 The 6 crates eggs will cost Gh¢42.00
4. At Mr. Opoku‟s shop, 3 wrist watches are sold
Method 2 for Gh¢100.00. How many watches can you
By unitary method, purchase with Gh¢600.00?
First find the cost of 1 unit and multiply by the
Solution
total number required.
Let m be the number of wrist watches.
i.e 3 eggs cost Gh¢21
100 : 600 = 3 : m
 1 crate will cost = Gh¢7.00
=
1×m= 6×3
If 1 crate costsGh¢7.00;
m = 18
then 6 eggs will cost = 6 × Gh¢7 = Gh¢42.00
 18 wrist watches cost Gh¢ 600.00

2. If the weight of 17 bags of rice is 680kg. What

Exercises 12.5
is the weight of 7 bags ?
1. If 2 yards of a floor carpet cost Gh¢ 40.00, how

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many yards can be bought for Gh¢ 2,400.00? Ratio of men = 4: 8
2. Mr. Jones purchased 6 crates of eggs for Ratio of days = 6: x
Gh¢180.00, find the cost of a crate of egg? As indirect proportion,
3. Mr. Green walks 16 minutes to his shop every 4 : 8 = x : 6
day. Calculate in hours, the time he walks to his =
shop in 5 days.
8×x= 6×4
4. A bus has a capacity of 63 passengers.If 8
8x = 24
buses are hired for a trip, how many passengers
x = 3days
can make up the trip.
 8 men will use 3days to weed the same plot
5. In an organized football competition, the
duration of a match is 30minutes, how many 2. It took 12 minutes for 8 girls to sweep a class.
matches can be played in a total of 420 minutes. How many girls can sweep the same classroom at
the same work rate in 4 minutes?
Indirect Proportion
Indirect proportion involves comparison of two Solution
quantities such that an increase in one quantity Ratio of minutes = 12: 4
causes a decrease in the other quantity and vice Ratio of girls =8: x
versa. For instance, if 4 girls sweep a class in 20
minutes, provided they work at the same rate, As indirect proportion,
then a decrease in the number of girls will cause 12: 4 = x : 8
an increase in the time (more than 20minutes)
=
required to finish. Likewise, an increase in the
number of girls will cause the work to be finished 4 × x = 12 × 8
earlier, thus a decrease in the time (less than 20 4x = 96
minutes) required to finish the work. = 24 girls
 24 girls can sweep the room 4 minutes.
To solve problems involving indirect proportion,
I. Identify the two ratios. 3. If a car traveling at 80km/h takes 3 hours to
II. Form a proportion but interchange the cover a certain distance, how long will it take
positions of the terms of the second ratio. i.e if another car traveling at 60km/h to cover the same
we have a : b = c : d , then the indirect distance?
proportion is a : b = d : c
III. Solve for the value of the unknown variable Solution
Ratio of distance = 80 : 60
Worked Examples Ratio of 1 hour =3 : x
1. It takes 6 days for 4men to weed a plot of As indirect proportion,
land. How long will it take 8men working at the 80 : 60 = x : 3
same rate, to weed the same plot of land?
=
60 × x = 80 × 3
Solution

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60x = 240 i. how many buckets of water will each person
x= = 4 hours fetch?
ii. in how many minutes will the poly tank be full,
if they work at the same rate ?
4. A Lorry traveling at 120km/h took 3 hours to
cover a certain distance. How long would it take
Sharing According to a Given Ratio
another Lorry traveling at 90km/h to cover the
It involves the situation whereby a given
same distance?
quantity is shared according to a given ratio
Solution between two or more people. The steps involve
Ratio of distance = 120 : 90 are;
Ratio of time =3 : x I. Identify the sharers and their respective ratio or
As indirect proportion, part.
120: 90 = x: 3 II. Find the total ratio.
III. Identify the quantity to be shared.
=
IV. Express each ratio as a fraction of the total
90 × x = 120 × 3 ratio and multiply by the total quantity.
90x = 360 For instance, to share K oranges in the ratio
x = 4 hours. a : b, to let‟s say Jenifer and Rosemary
respectively,
Exercises 12.6
Jennifer‟s part = a, Rosemary‟s part = b
1. If 30 men can dig a pit in 21 days, how many
Total ratio = a + b
days will 14 men take to dig the same pit,
Jennifer‟s share = ×k
working at the same rate?
Rosemary‟s share = ×k
2. It took 6 students 1hour, 18 minutes to weed
their compound. If 7 students are added, how Worked Examples
long will it take them to weed the same plot of Type 1
land at the same rate? 1. Jack and Jill shared Gh¢60,000.00 between
them in the ratio 3 : 2, find:
3. Eight boys weeded a piece of farmland in 6
i. the share of each person.
hours. How long will it take 2 boys to weed the
ii. how much did Jack receive than Jill?
same piece of farmland, working at the same
rate?
Solution
i. Jack : Jill = 3 : 2
4. If 10 painters can take 8 days to paint a house,
Total ratio = 3 + 2
how long will it take 20 painters to paint the same
Total amount = Gh¢60,000.00
house working at the same rate?
Jack‟share = × 60 = Ghȼ36,000
5. Mansah fetched 84 buckets of water to fill a
poly tank in 300 minutes. If she fills the same Jill‟s share = × Ghȼ60,000 = Ghȼ24,000
poly tank with her other 5 friends,

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ii. Gh¢36,000.00 – Gh¢24,000.00 Ratio = 7 : 8 : 9.
= Gh¢12,000.00
Jack received Gh¢12,000.00 more than Jill. Total ratio = 7 + 8 + 9 = 24
Least ratio = 7
2. Pat, Tom and Ken share a profit of Least share = × Ghȼ480,000 = Ghȼ140,000
Gh¢150,000.00 in the ratio 1 : 4 : 3 respectively.
How much did each person get? Greatest ratio = 9
Greatest share = × Ghȼ480,000 = Ghȼ180,000
Solution
Pat : Tom : Ken = 1 : 4 : 3
Difference = Ghȼ180,000 – Ghȼ140,000
Total ratio = 1 + 4 + 3 = 8
= Ghȼ40,000.00
Total amount = Gh¢150,000.00
5. The sum of the ages of a man and his wife is
Pat received = × 150,000 = 18,700.50 81 years. If the ratio of their ages is 5 : 4, find the
Therefore,Pat‟s share = Gh¢18,700.50 age of the younger person.

Tom = × 150,000 = 75,000 Solution

Total ratio = 5 + 4 = 9
Therefore Tom received Ghȼ75,000.00
Total age = 81
Younger person‟s age = × 81= 36 years
Ken‟s share = × 150, 000 = 56,200.50
Therefore, Ken received = Gh¢56,250.00 6. Three children Kwabena, Esi and Yaw were
given 160 oranges to share. Kwabena got of the
3. Three people shared Ghȼ540,000.00 in the
ratio 2: 3: 4. Find the least amount received. Oranges. Esi and Yaw shared the remainder in
the ratio 3 : 2.
Solution i. How many oranges did Esi receive?
From 2 : 3: 4, the least share is 2 ii. How many more oranges did Yaw receive than
Kwabena?
Total ratio = 2 + 3 + 4 = 9
Amount shared = Ghȼ540,000 Solution
Total number of oranges = 160
Least share = × Ghȼ540,000 = Ghȼ120,000.00
Kwabena‟s share = × 160= 40 oranges
4. Three men shared Ghȼ480,000.00 in the ratio The remainder = 160 − 40 = 120 oranges
7 : 8: 9. Find the difference between the least and Esi : Yaw = 3 : 2
the greatest shares. Total ratio = 3 + 2 = 5

Solution i. Esi‟s share = × 120 = 72 oranges

Amount shared = Ghȼ480,000 Therefore, Esi received = 72 oranges

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ii. Yaw‟s share = 120 −72 = 48 Oranges = 20% of 60,000
Kwabena‟s share = 40 Oranges = × 60,000 = Ghȼ12,000.00
(48 − 40) Oranges = 8 oranges
Therefore, Yaw received 8 oranges more than Remaining amount;
Kwabena = Ghȼ(60,000 – 12,000) = Ghȼ48,000.00

7. In sharing 95 oranges with Dede, Fofo kept 45 ii. Total ratio = 3 + 2

of them and share the rest equally with Dede. Daughter‟s share = × 48,000 = Ghȼ19,200
Calculate the share of each person.
iii. Son‟s share = × 48,000 = Ghȼ28,800.00
Solution
Total oranges = 95 Difference = Ghȼ(28,800 – 12,000) = Ghȼ16,800
Fofo‟s first share = 45 oranges
Remaining = (95 − 45) = 50 oranges Type 2
50 oranges shared equally means a ratio 1:1 Worked Examples
1. The ratio of the number of boys to the number
of girls in a school of 432 pupils is 5 : 4. If the
Fofo‟s second share = × 50
number of boys increased by 12, the new ratio of
= 25 oranges boys to girls is 7 : 6. Find the increase in the
 Fofo‟s total share = 40 + 25 = 65oranges number of girls.

Dede‟s share = × 50 = 25 oranges Solution

Total number of boys and girls = 432
8. In his will, a father left an estate worth Ratio of boys to girls = 5 : 4
Ghȼ76,000.00. Out of this, Ghȼ16,000.00 was Total ratio = 9
reserved for various purposes. The rest of the Number of boys = × 432 = 240
amount was shared among his three children. The
Number of girls = × 432 = 192
eldest son received 20% of the amount. The
remaining amount was shared between the other
If the number of boys increased by 12, the
son and the daughter in the ratio 3 : 2
number of girls increased by x. Therefore, ratio of
respectively. Calculate:
boys to girls;
i. the amount that the eldest son received,
240 + 12 : 192 + x = 7 : 6
ii. the amount that the daughter received,
iii. the difference between the amounts the two ⇒ =
sons received. =
6 × 252 = 7(192 + x) Cross multiplication,
Solution 1512 = 1344+ 7x
i. Amount = Ghȼ (76,000 – 16,000) = Ghȼ 60,000 7x = 1512 – 1344
7x = 168
Eldest son‟s share, x = 24

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2. There are 5 more girls than boys in a class. If 2 Remainder = Ghȼ(300,000 – 60,000)
boys join the class, the ratio of boys to girls will = Ghȼ240,000
be 5 : 4. Find the :
i. number of girls in the class; Kojo‟s share = × Ghȼ240,000 = Ghȼ100,000
ii. total number of pupils in the class.
Esi‟s share
Solution
= Ghȼ(300,000 – 60,000 – 100,000)
i. Let the number of boys and girls be x
= Ghȼ140,000
respectively.
Ratio of Ama : Kojo : Esi
Ratio of girlsto boys = x : x
= 60,000 : 100,000 : 140,000 = 2 : 5 : 7
5 more girls than boys = x + 5 : x
The money was shared in the ratio 2 : 5 : 7

If 2 more boys join the class, the ratio of boys to

2. A man gave out Ghȼ24,000.00 to his three
girls will be 5 : 4
brothers X, Y and Z, to be shared among them. If
x+5:x+2=5:4
X takes twice as much as Y and Y is given one –
⇒ = third of what Z takes, how much did each of them
4(x + 5) = 5(x + 2) receive?
4x + 20 = 5x + 10
20 – 10 = 5x – 4x Solution
10 = x Let Z‟s share = z
x = 10 girls Let Y‟s share = z

But number of girls = x + 5 = 10 + 5 = 15 X‟s share = 2 ( ) Z

Totaal number of pupils in the class Total share = Ghȼ24,000
= number of boys + number of girls
=x+x+5 ⇒ (X + Y + Z) shares = 24,000
= 10 + 10 + 5 = 25 pupils By substitution,
z + z + 2 ( ) Z = 24,000
Some solved Past Questions
z+ + = 24,000
1. An amount of Ghȼ300,000.00 was shared
among Ama, Kojo and Esi. Ama received 3z + z + 2z = 72,000
6z = 72,000
Ghȼ60,000.00. Kojo received of the remainder,
Z= = 12,000
while the rest went to Esi. In what ratio was the
money shared? Therefore the share of Z is Ghȼ12,000.00

Solution The share of Y = (z)

Amount shared = Ghȼ300,000
= (12,000) = Ghȼ4,000.00
Ama‟s share = Ghȼ60,000

Baffour – Ba Series, Core Maths for Schools and Colleges Page 337
 ( i. How much is the share of each person?
The share of X = = = Ghȼ8,000
ii. How much did Ayeley receive than Naa?
Exercises 12.7
7. Three brothers contribute money to a savings
1. Shamo, Bako and Aspa were given
fund according to the ratio of their ages. The
Gh¢40,000.00 to share. Shamo received of the brothers are aged 16, 17 and 19.
amount and the rest was shared between Bako i. If the eldest brother has to contribute
and Aspa in the ratio 5 : 3, Ghȼ9,500.00 to the fund, how much money must
i. Calculate the share of Shamo, Bako and Aspa each of the brothers contribute?
ii. How much did Bako receive than Shamo? ii. If the younger brother has to contribute
Ghȼ5,600.00, how much money must each of the
2. If Bole and Bamboi are to share an amount of brothers contribute?
Gh¢80,000.00 in the ratio 5 : 3 respectively,
what will be the difference between the share of 8. Mr. Asamoah shared Ghȼ200,000.00 among
the two persons? his three sons Pak, Tom and Ben as follows; he
gave of the money to Pak, of the money to
3. A man gives his twochildren some spending
Tom and Ghȼ20,000.00 to Ben. He then shared
money according to the ratio of their ages. If the
the rest of the money between Tom and Ben in
children are aged 11 and 13 and he gives out a
the ratio 2:3 respectively.
total of Ghȼ40,800.00, how much does each child
i. Calculate the total share of each person.
gets.
ii. Calculate the sum of money received by Ben
and Pak.
4. A sum of money is divided between three men
iii. Which of the three sons receive the least share
X, Y and Z in the ratio 5 : 3 : 1. If Y has Ghȼ3.50
and by how much does it differ from the greatest
more than Z, calculate how much X has.
share?
5. An amount of Gh¢165,000.00 was shared
9. A brother and a sister share 2400 kola nuts in
among Okra, Harry and Regina as follows: Okra
the ratio of 5 : 3. The brother then shares his kola
received of the amount and Gh¢1,500.00 out of nut with two friends in the ratio of 3 : 2 : 1
the rest was given to Regina. Thereafter, the rest respectively. How many nuts does each friend
of the money was shared between Harry and receive?
Regina in the ratio 5 : 2.
i. Calculate the share of each person. Finding the Total Amount (Quantity) Shared
ii.How much did Okra receive than Regina? Given the Ratio and the Share of One Person
Given the ratio and the share of one person, the
6. Naa and Ayeley were given Gh¢70,000.00 to total amount (quantity) of an item shared can be
share. They agreed to give Gh¢ 6,000.00 to their calculated. Any of the following methods can be
mother and shared the rest in the ratio 3 : 5 used;
respectively.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 338
Method I II. Find the sum of the given ratio (a)
Take for instance, A and B shared a certain III. Identify the given share of one person, (y)
quantity, x, in the ratio a : b respectively and A‟s and its corresponding ratio (b)
share is m, then A‟s share can be expressed as IV. Form a proportion of the corresponding ratio
shown below; to the total ratio and the given share to
A‟s Share = ×x=m the total quantity shared. That is:
m=(
b :a = y : x OR =
Making x the subject,
(a + b ) m = a x Worked Examples
(
x= 1. Adam and Eve shared a certain amount of
money in the ratio 3:5. If Adam received
Given the share of the first person, the quantity Gh¢9,000.00, how much was shared between
shared can be calculated by the formula; them?
(
x=
Solution
Where x is the quantity shared, (a + b) is the total Method I
ratio, a is the ratio of the first person (A), and m Let the amount shared be x.
is the share of A. Adam‟s ratio (a) = 3
Eve‟s ratio (b) = 5
Similarly, if B‟s share is n, then B‟s share can be Total ratio = 3 + 5 = 8
expressed as; Adam‟s share (m) = Gh¢9,000.00
B‟s Share = ( ×x=n
(
n=( Quantity shared, x =
But a = 3, b = 5 and m = 9000
Making x the subject,
a+ b = 3 + 5 = 8
(a +b)n = bx ( (
( x= = = 24000
x=
The amount shared is Gh¢24,000.00
Given the share of the second person (B), the
Method II
quantity shared can be calculated by the formula;
Let the amount shared be x
(
x= , where x is the quantity shared, (a + Given ratio = Adam and Eve = 3 : 5
b) is the total ratio, b is the ratio of the second Total ratio = 3 + 5 = 8 and its corresponding
person (B), andn is the share of the second amount = x
person. Adam‟s ratio = 3 and its corresponding amount =
Ghȼ9000
II. (Simple Proportion) 3 : 8 = 9000 : x By simple proportion,
I. Represent the total quantity shared by any =
preferred variable ( x) 3x = 8 × 9000

Baffour – Ba Series, Core Maths for Schools and Colleges Page 339
x= = Ghȼ24,000 amount received Ghȼ50,000.00, find the amount
that was shared.
The amount shared is Ghȼ24,000.00
Solution
Method III
Method I
Adam‟s ratio = 3,
Let the amount shared be x.
Adams share = Gh¢9,000
The least ratio (a) = 3
Eve‟s share = x
The least share (n) = Ghȼ50,000
As proportion,
Total ratio (a + b + c) = 3 + 5 + 7 = 15
Adam and Eve = 3 : 5 = 900 : x
The least share = = 50,000
=
⇒ = 50,000
3 x = 5 × Ghȼ9000
3x = 15 × 50,000
= = Gh¢15,000.00
x= = Ghȼ250,000
Therefore, Eve‟s share is Gh¢15,000.00
The amount shared is Ghȼ250,000.00
But total amount shared 4. Three school children shared some oranges as
= Adam‟s share + Eve‟s share follows; Akwasi got of the total and the
= Gh¢ (9000.00 + 1,500.00) = Gh¢2,4000.00 remainder was shared between Abena and Jantua
in the ratio 3 : 2. If Jantua got 24 oranges,
2. Bole and Bamboi shared an amount of money i. find the total number of oranges shared;
in the ratio 2:5. If Bamboi received Gh¢4,500.00, ii. how many oranges did Akwasi get?
how much money did they share?
Solution
Solution i. Let the total number of oranges be x
Let the amount shared be x
Akwasi‟s share = x
Given ratio = Bole and Bamboi = 2 : 5
Total ratio = 2 + 5 = 7 and its corresponding Remainder = x – x = =
amount = x Total ratio = 3 : 2 = 5
Bamboi‟s ratio = 5 and its corresponding amount Jantua‟s share = × x = 24
= Ghȼ4,500
x = 24
5 : 7 = 4,500 : x By simple proportion,
= 4x = 24 × 15
4x = 360
5x = 7 × 4,500
x= = 90
x= = Ghȼ6,300
Total number of oranges shared was 90
The amount shared is Ghȼ6,300.00

ii.Akwasi‟s share = x, but x= 90

3. Mr. Musa shared an amount to his three sons in
the ratio 3 : 5 : 7. The one who had the least = × 90 = 30 oranges

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Solved Past Question Total ratio = 2 + 4 + 3 = 9
1. Three friends contributed to the capital of a Kwame and Abena‟s ratio = 4 + 3 = 7 and its
company in the ratio 3 : 2 : 5. If the least corresponding amount = Ghȼ420,000.00
contributor paid Ghȼ30,000.00, how much was By simple proportion,
their total contribution? 7 : 9 = 420,000 : x
=
Solution
7x = 9 × 420,000
Method I
Let the total contribution be x x= = 540, 000
Ratio = 3 : 2 : 5 The total profit shared is Ghȼ540,000.00
Total ratio = 3 + 2 + 5 = 10
Least ratio = 2 Method II
Least contribution = × x = 30,000 Let the total profit shared be x
Afia, Kwame and Abena = 2 : 4 : 3
2x = 10 × 30,000
Total ratio = 2 + 4 + 3 = 9
2x = 300,000
Kwame and Abena‟s total ratio = 4 + 3 = 7
x= = 150,000. Kwame and Abena‟s share =
The total contribution is Ghȼ150,000.00 × x = 420,000
7x = 9 × 420,000
Method II
By forming a proportion, x= = 540,000
Least ratio : Total ratio = Least contribution : The total profit shared is Ghȼ540,000.00
Total contribution
3 : 10 = 30,000 : x 2.Three partners Ali, Baba and Musa shared their
= profit in the ratio 7 : 13 : 5 respectively. At the
end of a certain year Baba had Ghȼ840,000.00
3x = 10 × 30,000
more than Ali. What was the total profit shared
x= = 150,000 by the three partners?
The total contribution is Ghȼ150,000.00
Solution
3. Afia, Kwame and Abena shared the profit they Method I
earned in a month from a joint business in the Let the profit shared bex
ratio 2 : 4 : 3 respectively. If Abena and Kwame Total ratio = 7 + 13 + 5 = 25
together got a total of Ghȼ420,000.00, what was Ali‟s share =
the total profit that the three shared?
Baba‟s share =
Solution Ali‟s share =
Method I
Let the total profit shared be x Baba had Ghȼ840,000.00 more than Ali
Afia, Kwame and Abena = 2 : 4 : 3 ⇒Baba‟s share = y + 840,000

Baffour – Ba Series, Core Maths for Schools and Colleges Page 341
 = + 840,000 3. Three candidates K, L and M were voted into
office as school prefects. K secured 45% of the
– = 840,000
votes. L had 33% of the votes and M had the rest
= 840,000 of the votes. If M secured 1,430 votes, calculate;
6x = 25 × 840,000 i. the total number of votes casted,
ii. how many more votes K received than L.
x= = 3,500,000.00
Total profit shared was Ghȼ3,500,000.00 Solution
Total percentage = 100%
Method II Percentage of M‟s votes
Let the total profit shared be x = (100 – 45 – 33)% = 22%
Ali, Baba and Musa = 7 : 13 : 5 Let the total votes be x
Total ratio = 7 + 13 + 5 = 25
M‟s votes = × x = 1,430
Let Ali‟s share = y
⇒ Baba‟s share = y+ 840,000 ⇒ = 1,430
7 : 25 = y : x 22x = 1,430 × 100
= x= = 6,500
7x = 25y
x= ………. (1) ii. K‟s vote = × 6,500 = 2,925votes
L‟s vote = × 6,500 = 2,145 votes
13 : 25 = (y + 840,000) : x
= Votes K received than L;
13x = 25( y + 840,000) = 2,925 – 2,145 = 980 votes

4. The development budget of a district council

13x = 25y + 21,000,000 includes expenditure on feeder roads, schools and
water supply. The expenditure on road, schools
x= ……………. (2)
and water supply are in the ratio 7 : 15 : 2. If the
expenditure on roads is Ghȼ28 million, find the
Equating eqn (1) and eqn (2) expenditure on;
= i. Schools,
13 × 25y = 7(25y + 21,000,000) ii. Water supply,
325y = 175y + 147,000,000 iii. What is the total budget for these three
325y – 175y = 147,000,000 projects?
150y = 147,000,000 iv. The cost of maintaining libraries is
y= = Ghȼ980,000 Ghȼ900,000 and this is net from the expenditure
on schools. What percentage, correct to three
Substitute y = 980,000 in eqn (1)
significant figures of the expenditure on schools
x= = Gh¢3,500,000.00 is spent on maintaining libraries?

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Solution
i. Road : schools : water supply = 7 : 15 : 2 Exercises 12.8
Total ratio = 7 + 15 + 2 = 24 1. Jacob and Esau were given some exercise
Let the expenditure on schools be = x books to share in a respective ratio of 4:3. If
Ratio of road = 7 Jacob had 120 books, how many books did they
Expenditure on road = 28m share?
Proportion of road and schools
Ratio = Expenditure 2. Sugri is 16 years old and Dabre is 14 years
7 : 15 = 28 : x older than Sugri. When they shared a certain
= yards of clothing in the ratio of their ages, Dabre
had 15 yards. How many yards of cloth did they
7x = 28 × 15
share?
x= = 60 million
The expenditure on schools is Gh¢60 million 3. Sampson and Delilah shared some money in
the ratio 3:5 respectively. Calculate the total
ii. Let the expenditure on water supply = y amount shared if Sampson had Gh¢150,000.00.
Proportion of road and water supply
4. A number of pens were shared among
Ratio = Expenditure Shadrack, Mershack and Abednego in the ratio 3
7 : 2 = 28 : y : 1 : 4 respectively. If Merchack had 14 pens,
= find the total number of pens shared.
7y = 2 × 28
5. The ratio of the population of Secondi and
y= = 8 million
Takoradi is 3 : 4 respectively. If the population of
Takoradi is 17,540 people, what is the population
iii. Let the total expenditure be m
of Secondi?
Proportion of Road and Total expenditure
7 : 24 = 28million : m
Challenge Problem
= 1. Baffour Ba shared some money among his
7m = 24 × 28 three family members namely; Ananse, Aso and
m= = 96 million Ntekuma. Ananse received of the money and
The total expenditure spent on the three projects the remaining was shared between Aso and
is Ghȼ96 million Ntekuma, where Ntekuma receive Ghȼ5,200.00
constituting of the remaining amount.
iv.Expenditure on schools = 60 million
Cost of maintaining the library = Ghȼ900, 000 i. Find the total amount of money shared by the
Percentage of expenditure on schools spent on three family members.
maintaining libraries ii. Find the total share of each person.
iii. How much did Aso received than Ananse?
= × 100% = 1.5%

Baffour – Ba Series, Core Maths for Schools and Colleges Page 343
Rate Solution
A rate is the ratio of two different quantities. This Rate = = = 400 liters per day
implies that rate is the comparison of two
different quantities.
4. Jojo bought 3kg of meat for Gh¢810.00. Find
the cost per kilogram.
A rate has a unit expressed as the number of
units of the first quantity to that of the second
Solution
quantity. For example, if a household consumes
Cost per kilogram =
20units of electricity in 5days, then the rate as
explains above requires a comparison of the two But cost = Gh¢810.00 and kilogram = 3kg Cost
quantities , the first being “ 20 units of electricity per kilogram = = Gh¢270 per kg
“ and the second “ 5days” ; so as to arrive at “ a
per unit” of the second quantity.
5. A certain man pays Gh¢6,000.00 for staying in
⇒ Rate = = a hotel for 8 nights. What was the rate per night?
Rate = 4 units per day written as:“4 units/day”
Solution
Worked Examples Rate = = Gh¢750
1. A car travels 120km in 2 hours. Find: The rate per night is Gh¢750.00
i. the rate at which the car travels in km/h.
ii. the rate at which the car travels in m/s 6. Musa works at a restaurant for 4 hours.
He is paid Gh¢250.00 per hour. How much
Solution does he earn?
i. Rate = = = 60km/h
Solution
Gh¢250.00 is for 1 hour.
ii. 120km = 120 × 1,000m = 120,000m
Therefore, 4 hours = 4 × 250 = Gh¢1,000.00
2hr = 2 × 3600s = 72,000s
He earns Gh¢1,000.00
Rate = = 16.6m/s
7. If 154 pages of a story book was read by
2. A man earns Gh¢400.00 for working 8 Rachel in 2 weeks. How many pages did she read
hours. How much does he earn per hour? per day?

Solution Solution
Rate = = = Gh¢50 per hour Rate =
Therefore, he earns Gh¢50.00 per hour.
But 1st quantity = 154 pages
3. A household consumes 2,800 liters of water 2nd quantity = 2 weeks = 14 days
per week. Find the daily rate of consumption. Rate = = 11pages/day.

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Exercises 12.9 Average Speed (S)
1. A factory worker earned Gh¢1,200.00 for 4 The average speed(s) is the rate of the total
hours work. How much does he earn per hour? distance (d) to the total time (t). That is:
(
Average speed(s) = (
2. A girl paid Gh¢1,300.00 for fetching 13 tanks
of water. Find the rate per bucket. Simply, s=

3. A shopkeeper is marked Gh¢3.20 per The average speed (s) can be in

hour. How much does he receive for working for: m/s,km/h,km/minetc
i. 10 hours ii. 4 hours a day for 5 days?
Worked Examples
4. A man receives an annual salary of Gh¢ 1. A car travels 72 kilometers in an hour.
3,204.00. How much does he receive at the end of Find its speed in meters per seconds
every Month?
Solution
5. Mr. Green filled his tank with 42 gallons of S = ? t = 1hr, d = 72km/h
petrol at a cost of Gh¢2,940.00. Find the rate per Since the speed is required in meters per
gallon. seconds, express the distance in meters and
the time in seconds.
6. At Mammon Shoe Factory, each worker is paid 1 hour = 60 min, but 1 min = 60 seconds
Gh¢24.80 for 12 hours. Find the pay per hour of ⇒60 minutes = = 3,600s
each worker.
t = 3,600s
7. A school boy wrote 374 words to cover
If 1km = 1,000m
a page of a sheet of paper. If the sheet contains 17
lines, how many words filled a line? ⇒, 72km = = 72,000m
S= = = 20m/s
8. At a certain hospital, it was reordered that 138
H.I.V. patients die every 6 months. What is the
3. A van travels 154km in 1 hour. Find its speed
monthly rate of death?
in km/h.
Challenge Problems
1. A typist charges Gh¢20.00 for the first 8 Solution
pages and 50p for each additional page typed. S = ?, d = 154km, t = 1 h = h
Find the cost of typing 93 pages. S= = = = 88km/h

2. At his communication centre, Mr. Green

charges 150p for the first 15 minutes call and 4. A cyclist covers 900m in 5minutes. What is his
130p for each additional minute. How much will average speed in m/s?
you pay for a 50 minutes call?

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Solution 5. A car travels 72km in 1 hours and the travels
S = ?, d = 900m and t = 5mins.
at a constant speed of 28 km/h for a further 1
But 1 min = 60sec
hours. Calculate:
 5mins = = 300s.
i. the average speed for the first 1 hours;
S= = = 3m/s
ii. the average speed for the first 2 hours;
5. An athlete runs 100 meters in 10 seconds. iii. the average speed for the whole journey.
Find his speed in kilometers per hour.
Finding the Time Given the Speed and Distance
Solution Given the average speed, s, and the total time, t,
the distance, d, is calculated by the formula:
S= (
Total time (t) =
But d = 100m = km = 0.1km (

Simply, t =
t = 10s = ,
S= = = 36km/h The total time can be measured in seconds,
⁄
minutes, hour etc.
6. A car travels 60km in 48minutes. Find the
average speed of the car in km/h. Note:
The phrase“how long”means, find the time.
Solution
d = 60km, Worked Examples
t = 48min = h= 0.8h 1. A train travels at a speed of 80km per hour.
How long will it take to travel a distance of
S= = = 75km/h 320km?

Exercises 12.10 Solution

1. A cyclist covers 700km in 2hours. What is his S = 80km/h, d = 320km and t =?
average speed in: i. km/h ii. m/s t= = = 4 hours.

2. A 5,000 meter runner covered such a distance

2. What is the time in minutes taken to
in 25 minutes. Calculate his rate of speed in:
cover 81km at an average speed of 27km/h?
i. m/s ii. km/min.
Solution
3. A runner covered ⁄ of a 4,000km marathon
race in 2 hours and covered the rest in 5 hours. S = 27km /h, d = 81km and t =?
What is the average speed of the runner in km/h? But, t = = = 3h
But 1 hour = 60mins
4. A van travels 344km in 2 hours. Find its speed in:
 3 hour = = 180mins
i. km/h. ii. m/s

Baffour – Ba Series, Core Maths for Schools and Colleges Page 346
Exercises 12 .11 2. How far will an aero plane with speed 415km/h
1. Jude traveled a distance of 12km on a go in 3hrs?
bicycle at an average speed of 8km/h.
How long did it take him to make the trip? Solution
S = 415km/h, t = 3hrs and d = ?
2. A boy cycles 75km to school at a speed of d = s × t = 415km/h × 3h = 1, 245 km
25km/h. How long did it take him to reach the
school? 3. The speed of an athlete is 12m/s. What
distance in kilometers can he cover in a time of
3. Mr. Jimmy traveled a distance of 50km on a 9.5 seconds?
bicycle at an average speed of 10km/h.
Find the time he used to make the trip in minutes? Solution
S = 12m/s, t = 9.5s and d = ?
4. A train traveling at 105km/h goes through d = s × t = 12m/s × 9.5 s = 114m
a tunnel 630km long. Calculate the time in d= = 0.114km
seconds, a passenger in the train spends inside the
tunnel. 4. Kwame rode a bicycle for a distance of x km
and walked for another ⁄ an hour at a rate of
Unknown Distance Given the Speed and Time 6km/hour. If Kwame covered a total distance of
Given the average speed, s, and the total time, t, 10km, find the distance x he covered by bicycle.
the distance, d, is calculated by the formula:
Total distance = Speed × Time taken Solution
Simply: d = s × t Total distance = 10 km
Bicycle = xkm
The total distance, d, can be measured in cm, m,
Walking = h× 6 km/h = 3 km.
km etc.
Total distance = bicycle + walking
Note: The phrase “how far” means find the 10km = xkm + 3 km
x = (10 – 3) km = 7 km
distance
The distance covered by the bicycle is 7 km
Worked Examples
1. A car is traveling at 40 km/h. How far does it Exercises 12.12
1. Calculate the distance covered by a car in 9
travel in 2 hours?
hours at a speedof 80km/h.

Solution 2. A Boeing 747 aircraft has a speed of 225km/h.

S = 40km/h, t = 2 hrs and d = ? What distance can it cover in 12hours?
But d = s × t
= 40 × 2 3. What distance will an athlete with speed 10
= 40 × = 100km m/s cover in 8 minutes?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 347
4. A Yutong bus travels from Kumasi to Tumu at d2 = 60km, S2= 50 km/h
a speed of 80km/h. If the bus takes off from t2 = = 1.2h
Kumasi at 7 : 00 a.m and reaches its destination
at 11 : 30 a.m, how far is Tumu from Kumasi?
Total time, t = t1 + t1
t = 0.8h + 1.2h = 2h
5. A train traveled from Tarkwa to Obuasi at a
speed of 120km/h in 4 hours. It then travelled
from Obuasi to Ejisu at a speed of 90km/h in ii. Average speed =
5hours. Calculate the total distance covered by Total distance = (56 + 60) km = 116km
the train from Tarkwa to Ejisu.
Total time = 2hr
Solved Past Question Average speed = = 58km/h
1. An aero plane leaves Barcelona at 10 : 10 p.m
and reaches Accra 4,415 km away at 5 : 50 a. m Scale Drawing (Scale of a Map)
the next morning. Calculate, correct to the nearest The scale of a map is the ratio of the distance on a
whole number, the average speed of the aero map to theactual corresponding distance on the
plane in km/h. ground. Simply put,
Scale =
Solution
S= , For similar areas and volumes, if the ratio of
But d = 4,415km lengths is a : b, the ratio of the areas is a2 : b2,
t = 10 : 10 p.m – 5 : 50 a. m and the ratio of the volumes is a3 : b3
t = 7h 40 min
On a map, the ratio map length : ground length is
40 min to hr = h = 0.67h known as representative fraction
 t = 7h + 40 min = (7 + 0.67) h = 7.67h
⇒S = = 576km/h The scale is usually expressed as a ratio in the
form 1 : n . For instance, if the scale of a map is
given as 1 : 10,000, it means 1 unit on the map
2. A bus travelled a distance of 56km at an
represents 10,000 of the same units on the
average speed of 70 km per hour. It travels a
ground.
further 60km at an average speed of 50 km per
hour. Calculate for the whole journey: Finding the Actual Size/Distance Given the
i. the total time taken, ii. the average speed. Scale andSize/ Distance on the Map
To find the actual size or distance (x) of an object
Solution given its size (y) on the map and the scale:
i. t = I. Represent the actual size/distance by any
d1 = 56km, s1 = 70km/h preferred variable, say x
t1 = = 0.8h II. Identify the object‟s size on the map, say y

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III. Identify the given scale, say a : b is 0.22m2, what is the area of the aircraft
IV. Equate the given scale to the ratio of thesize iii. The air crafts carries 30 m2of fuel. What is the
on the map to the actual size. That is : capacity of the model‟s fuel tank?
a : b = y : x OR =
Solution
V. Solve for the value of x, to obtain the actual
i. Scale of lengths = 1 : 20 = x : 25
size of the object
=
Consider the scale drawing of the football field 20x = 25
below; x = 1.25m
11cm
8cm ii. Scale of areas = 1 : 202 = 1: 400
1 : 400 = 0.22 : y
Given a scale of 1 : 1,000, it means that =
1cm on the map represents 1,000cm on the
y = 400 × 0.22 = 88m2
ground. Therefore the actual size of the field
length is:
iii. Scale of volumes = 1 : 203
1 : 1,000 = 11cm :xcm 1 : 8,000 = n : 30
, =
x = 11× 1000 = 11,000cm n= = 3.75 × m

The actual size of the field width is:

1: 1,000 = 8cm : y cm 2. If the scale of a map is given as 1 : 900, what
, is the distance represented by 4cm on the map?
y = 8 × 1,000 = 8,000cm
Therefore, the actual size of the field on the Solution
ground is shown below: Method 1
11,000cm Let the actual distance be x
If 1 : 900
8,000cm ⇒ 4 cm = = 3,600cm

This implies that the actual area of the field

Method 2
= 11, 000cm × 8, 000cm Let the actual distance be x
= 88, 000, 000 cm2 1 : 900 = 4cm : xcm

Worked Examples
The model of an aircraft is made on a scale of 1 : 20 x = 900 × 4cm = 3,600cm
i. If the aircraft is 25m long, how long is the
model? 3. On a map, 1cm represents 120km. The
ii. If the area of the wing surface is on the model distance between two towns on the map

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is 5cm. Find the actual distance between 6. On a map, two towns P and Q are 15.5cm apart.
the two towns. the scale of the map is 1 cm : 4 km. Calculate the
actual distance between P and Q.
Solution
Method 1 Solution
If 1cm represents 120km, Method 1
Then 5cm = = 600km 1cm : 4km

 The distance between the two towns is 600km. 15.5cm = = 62km

 The distance between P and Q is 62km.
Method 2
Scale 1 : 120 = 5 : x 7. On a map, 1cm represents 4.5km. What is
the actual distance between two towns which are
4cm apart on the map?
x = 120 × 5 = 600 km
Solution
4. The scale of a map is 1cm to 100km. The
If 1cm represents 4.5km
distance between city P and city Q on the map is
23cm. What is the actual distance between city P Then 4cm = = 18 km
and city Q.  The distance between the two towns is 18km

Solution Exercises 12.13

Method 1 1. The scale of a map is 1 cm to 1,000 km. What
If 1cm represents 100km, is the distance between two towns which are:
then 23cm = = 2,300 km i. 8cm apart on the map

 The actual distance between city P and Q ii. 3 cm apart on the map
is 2,300km. iii. 103 mm apart on the map
iv. 2.8 cm apart on the map
5. The map of a large town is drawn to the scale
1 : 100,000. What is the distance in kilometer 2. a. The scale of a map is 1 cm to 5km. Write the
represented by a line segment 4cm long on the scale as a ratio in the form 1 : n
map? b. What is the distance on the map between two
towns which are:
Solution i. 15km apart ii. 28 km apart,
If 1cm represents 100,000cm iii. 47.5 km apart.
Then 4cm = = 400,000cm
3. The scale of a map is 1 : 100,000. What is the
But 100,000cm = 1km distance in kilometers between two towns that
 400,000cm = × 1km = 4 km are 12cm apart on the map?

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4. The area of a football field is represented on a III. Form a ratio of the size on the map to the size
map as 8cm × 5cm. Find the actual size and the area on the ground. That is :
of the field if the map is drawn to the scale 1 : 900. (
Scale = =a:b
(

5. The map of a region is drawn on a scale of 5cm

to 1 km. IV. Simplify the ratio in the form 1 : n to obtain
the scale in the same unit
i. Express this scale as representative fraction of
the map.
Worked Examples
ii. Find the area of the region in km2 which
1. The length of a pole 24,000cm is represented
corresponds to 125 cm2
on a map as 8cm. What is the representative scale
of the map?
6. The scale of a map is 1 : 100,000.
i. What distance in km, is represented by 2.2 cm
Solution
on the map?
ii. If the length and breadth of a rectangular Scale =
region on the map are measured correct to the = = = 1 : 3,000
nearest cm as 2.2 cm and 0.7 cm, what are the
The scale of the map is 1 : 3,000
limits between which the exact length and
breadth must lie? 2. The distance between two towns on a map is
2cm. If the actual distance between the towns is
7. A reservoir is to be constructed to hold 3.2 ×
8km, find the scale of the map.
107 m3 of water when full, an accurate model of it
is built to a scale of 1 : 200. Solution
i. When the model is full of water, the greatest
Scale =
depth is 18cm. What will be the greatest depth in
the reservoir when full? = = = 1cm : 4km
ii. If the surface area of the water in the model is
40 m2, calculate the corresponding surfacearea of But 4 km = 4 × 1,000 m = 4,000m
the water in the reservoir.
= cm = = 400, 000 cm
iii. Write down the volume of water in the ⁄

reservoir when full. The scale of the map is 1 : 400,000

Finding the Scale Given the Size of an Object Exercises 12.14

on the Map and the Actual Size of the Object on 1. On a scale drawing, the length of a ship is
the Ground 0.42cm. If the actual length of the ship is 84m,
To find the scale to which an object is what is the scale?
represented on a map;
I. Identify the size of the object on the map, (a) 2. The length of a field, 1.2km long is represented
II. Identify the actual size of the object on the on map by line 40mm long. What is the scale of
ground (b) the mapping?

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3. The actual length of a pole is 15,000 cm. If it is ⇒ Gh¢600 = = $400.00
represented on the map as 3 mm, to which scale
ii. If Gh¢1.50 = $1
is the map drawn?
⇒ Gh¢30 = = $20.00
4. On a photograph, a man whose height is 170 iii. If Gh¢1.50 = $1
cm appears as 2cm. Find the scale of the
photograph. ⇒ Gh¢9,000 = = $ 6,000.00

iv. If Gh¢1.50 = $1
5. Two towns are 100km apart. On a scale
drawing, these towns are 2.5cm away from each ⇒ Gh¢120 = = $80.00
other. Determine the scale of the drawing.
2. If $1.00 = Gh¢1.34, what is the cedi value of
Foreign Exchange Conversion
an article which cost $ 1.65
In order to transact business with other people in
other countries, there is the need to buy (by
conversion), the currency of that country. This is Solution
done at the forex Bureau, Banks or “Black Let xbe the cedis value.
market” that set to exchange rates namely; „sell By method of direct proportion,
at‟ and „buy at‟ rates for every currency. Dollar = Cedi
1 : 1.65 = 1.34 : x
To change or convert one currency to another, =
I. Identify the exchange rate x = 1.65 × 1.34 = Gh¢2.21
II. Identify the given currency
III. Represent the required currency by any 3. If £ 1.00 = Gh¢ 2.20, how much cedis will you
preferred variable. get for converting £ 40.00?
IV. Form a ratio of the exchange rate and equate
to the ratio of the given currency to the required Solution
currency(form a proportion). Let y be the cedi value.
V. Work out for the value of the variable to By method of direct proportion,
obtain the required currency. £ = Gh ¢
1:40 = 2.20: y
Worked Examples =
1. If $1 = Gh¢2.50, calculate the dollar equivalent y = 40 × 2.20 = Gh¢ 88.00
i. Gh¢600.00 ii.Gh¢30.00
iii. Gh¢9,000.00 iv.Gh¢120.00 4. You require €250 (euro) to purchase a
lap top computer. If the current conversion
Solution rate is €1.00 = Gh¢1.74. How much
Using simple proportion, money in cedis will you need to buy the
i. If Gh¢1.50 = $ 1 laptop?

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Solution I. Name thex – axis as time and the y – axis as
Letx be the money require in cedis distance
By method of direct proportion, II. Number the axes according to the given scale.
€ = Gh¢ If the scale is not given, use a convenient scale.
1 : 250 = 1.74 : x III. Identify the time movement and its
= corresponding distance movement and plot on the
graph. i.e. (time, distance)
x = 1.74 × 250 = Gh¢435.00
d
Exercises 12.15 (ahr, b km)
A. Change the following amount in cedis to b

pounds given that £1.00 = Gh¢6.70

1.Gh¢51,000.00 3. Gh¢17,000.00 a t
2. Gh¢102,000.00 4.Gh¢680,000.00 IV.Join the points of movements by a slanted line
such as ∕(usually beginning from the origin)
B. Change the following amount in cedis to d

dollars, if $1.00 = Gh¢4.75 (ahr, b km)

b
1. Gh¢35,700.00 3.Gh¢15,980.00
2. Gh¢119,000.00 4.Gh¢79,900.00
a t
Travel Graph V. Join rest periods or periods of no movement
It is a graph that shows the relationship between by a horizontal line such as ─.
distance and time, for a journey. VI. A return to the starting point(finally at rest) is
For most travel graphs, distance is usually represented by a slanted line such as \ to touch the
represented on the vertical axis (y - axis) and time axis.
time, on the horizontal axis (x – axis). Rest period (c – a)
d
dis tan ce Return to rest
d2 b or starting point

d1
a c k t
time Worked Examples
t1 t2
1. Mr. Brown travelled with his car at 6 : 30 a.m
From the distance – time graph above, and for the first 2 hours covered a distance of 20
Average speed (S) = = km to reach town A. He stopped for 30 minutes at
town A and continued the journey travelling a
distance of 15 km in an hour to town B, where he
Drawing a Travel Graph
rested for an hour. He then travelled from town B
Observe the following when drawing a travel
to town C, 10km apart in a time of 1 ⁄ hours.
graph:

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a. Using a scale of 2 cm to 1 hour on the time axis v. the time he reached town C
and 2 cm to 5 km on the distance axis, draw the
graph of Mr. Brown‟s journey c. Find the speed of Mr. Brown in moving:
i. from his house to town A
b. From the graph, find: ii. from town A to town B
i. the time he reached town A iii. from town B to town C
ii. the time he left town A iv. His average speed in travelling from his house
iii. the time he reached town B to town
iv. the time he left town B

Solution
a. A = (8: 30, 20km), Rest = (8 : 30 to 9: 00),
B(10:00, 35km), Rest = (10:00 to 11: 30), C(1:30, 45km)
Distance (km)

x = 2 cm to 1h
y = 2cm to 5km

50
C
45

40
B B
35

30

25
A A
20

15

10

6:30 7:30 8:30 9:30 10:30 11:30 12:30 1:30 2:30 Time (h)

b. From the graph: s= = = 10 km/h

i. he reached town A at 8 : 30 a.m
ii. he left town A at 9 : 00 a.m
ii. Speed from town A to town B;
iii. he reached town B at 10 : 00 a.m
iv. he left town B at 11 : 30 a.m s= = = 15 km/h
v. he reached town C at 1 : 30 pm
iii. Speed from town B to town C;
c. i. Speed from his house to town A;

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s= = = 5 km/h minutes back to his house,
a. draw the travel graph of the boy using a scale
v. His average speed in traveling from his house of 2cm to 20 minutes on the time axis
to town C. and 2cm to 2 km on the distance axis
( b. i. Find the speed of the boy before he reached
= = km/h = 10km/h
town P
ii. At what time did he leave town P?
2. A boy left home at 8 : 00 a. m. and travelled on iii. Find the speed of the boy in moving from
foot to visit his grandma at town Q, 14km away town P to town Q
from his house. In a time of 30 minutes, he iv. At what time did he leave his grandma‟s
covered a distance of 6 km to town P. He rested place?
and after 20 minutes continued his journey to his v. Find the exact time the boy reached home
grandma‟s place in 40 minutes. If he spent 40 vi. Find the average speed of the boy in walking
minutes at his grandma‟s place before walking 50 from his house to town.

Solution
Distance ( km
)

18

16
Q
14

12

10

8
P
6

8:00 8:20 8:40 9:00 9:20 9:40 10:00 10:20 10:40 11:00 Time (min)

b. i. The speed of the boy before he reached town P iii. The speed of the boy in moving from town P
s= = = 0.2 km/min to town Q
ii. He left town P at 8 : 50 a.m s= = = 0.2 km/min

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iv. He left his grandma‟s place at 10: 10 a.m km/h, at which Kojo cycled, correct to 2 s.f
v. The boy reached home at 11 : 00 a.m
vi. Find the average speed of the boy in walking Solution
from his house to town Q Total distance = 14 km
= 0.2 km/min and = 0.2 km/min
( Plot 30 min from 7 : 00 a.m. against 2.5 km
Average speed = km/min = 0.2km/min
from 0km = (30 min, 2.5km)
Some Solved Past Questions Remaining distance = (14 – 2.5)km = 11.5km
1. a. Kwame left his home in Takoradi at 7 : 00 Let x represent the time in minutes used by the
a.m to walk to Apramdu, 14 km away. After car to cover 11.5km
But the car travelled at an average speed of 40km /h
walking 2.5 km during the first 30 minutes, he
⇒the car traveled40km every 1 hour, 11.5km will
was given a ride in a car to Apramdu. The car
be covered in x hours. That is:
travelled at an average speed of 40km/h. Draw
40 km = 1h
Kwame‟s travel graph using a scale of 2cm to 10
11.5km = x h
minutes on the time axis and 2cm to 2km on the
distance axis =
b. Kojo also left Kwame‟s house at 7 a.m and
x= = 0.2875 h
cycled to Apramdu, arriving there at the same
time as Kwame. Using the same axes and scale as 0.2875 h = 0.2875 × 60 min = 17.25min
in (a), draw Kojo‟s travel graph Plot 17.25 minutes from 7 : 30a.m against
c. Use your graph to determine the speed in 11.5km from 2.5km = (7 : 47 a.m, 14km)
Distance ( km
)

18

16
Apramdu
14

12

10

8 Kojo
Kwame
6

2
Kwame

7:00 7:10 7:20 7:30 7:40 7:50 8:00 8:10 8:20 8:30 Time (min)

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c. Kojo‟s speed = continued his journey at 12.5km/h to reach town
B. Yaw started from town A at 9:00 a.m and
overtook kojo 30 minutes later. Yaw continued
Distance covered = (14 – 0)km = 14 km
with the
same speed till he got to town B.
Time taken = 7 : 47 – 7: 00 = 47 min
a. using a scale of 2cm to 30 minutes and 2cm to
Time taken = h = 0.78h 5 km, draw the distance - time graph for these
Kojo‟s speed = = 18km/h friends.

3. Two friends Kojo and yaw travelled the same b. Use your graphs to find:
route from town A to town B, a distance of 53km. i. When Kojo reached town B
Kojo‟s started at 6:30a.m. and for the first one ii. When Yaw reached town B
and half hours, he moved at a constant speed and iii. Yaw‟s speed
covered 28 km. He stopped for 30 minutes and

Solution
Distance ( km )

55
B B

50

45

40

35 Yaw
Kojo
30 A
A
25

20

15

10

6:30 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 Time (min)

= = 25 km/h
b. i. Kojo reached town B at 10 : 30
ii. Yaw reached town B 10 : 00 4. Mr. Brakatu and Mr. Jojo walk from house A to
iii. Yaw‟s speed = B, 11.2km away. Mr. Jojo starts at 10.00 a.m. and

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walks at a constant speed of 4.8km/h, but makes a = h= × 60 min = 80 min
call lasting 30 minutes at a point on the journey
Plot 80 min from 10 : 00 a. m. and 6.4km from 0km
6.4km away from house A. Mr. Brakatu starts at
= (11:20 a.m, 6.4km)
11.00 a.m. and walks at a constant speed of
Remaining distance = (11.2 – 6.4) km = 4.8km
6.4km/h.
a. Taking 2cm to represent 20 minutes on the time
Call time = 30 minutes from 11: 20 a. m. and ends
axis and 2cm to represent 1km on the distance axis,
at 11:50 a. m.
draw the travel graphs of Mr. Brakatu and Mr. Jojo
Remaining distance, 4.8 km was covered in the next
using the same axis
1 hour. Plot 1 hour from 11:50 a.m. and 4.8km from
b. Use your graphs to find;
6.4km = (12:50p.m, 11.2km)
i. where Mr. Brakatu overtakes Jojo,
ii. the times Mr. Brakatu and Mr. Jojo arrived at Mr. Brakatu‟s starting time = 11: 00 p.m.
their destination. Walks at a constant speed of 6.4km/h
⇒ 6.4km covered in 1 hour
Solution
Total distance = 11.2km 11.2km will be covered in h= h × 60 min
Mr. Jojo‟s starting time = 10: 00 a.m = 105 minutes = 1hr 45min
He travelled a constant speed 4.8km/h
⇒ 4.8km is covered every 1 hour Plot 1 hr 45min, from 11:00 p.m and 11.2km from
6.4km will be covered in the time 0km = (12:45 p.m, 11.2km)
Distance

11

10

6
Mr. Brakatu
5 Mr. Jojo

10 : 00 10 : 20 10 : 40 11 : 00 11 : 20 11 : 40 12 : 00 12 : 20 12: 40 01 : 00
Time (Min)

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b. i. At the time 12:35 p. m. and a distance of graph of his journey and find how far he is from
10km Oforikrom at 9: 30 a. m. (Take as horizontal scale
ii. Mr. Brakatu arrived at 12 : 45 p.m. and Mr. 2cm to represent 20 minutes, and as vertical scale
Jojo arrived at 12 : 50 p.m. 2cm to 1km
b. If the distance and walks at 7km/h towards
Exercises 12.16 Oforikrom.
1. A cyclist set out at 9 : 00 a. m. for a destination
40 km away. He cycled at a speed of 15km/h 4. A cyclist starts a 30 kmjourney at 9: 00 a.m.
unto10:30, when he rested for half an hour. He He maintains an average speed of 20km/h for the
then completed his journey at a speed of 20km/h. first three – quarters of an hour and then stops.
Using a suitable scale, draw a distance – time Subsequently he continues his journey at an
graph to represent the journey and use your graph average speed of 30 km/h, arriving at his
toestimate the time at which the cyclist reached destination at 11: 00. Draw the distance - time
his destination. graph and state, in minutes, the duration of his
stop.
2. A cyclist P starts from his house at 7 : 30 a.m
and rides to his office at an average speed of 5. A boy rows upstream for 1 km at 2 km/h,
12km/h. A motorist Q starts from P‟s house at 8 : waits half an hour and then rows back with the
00 a. m. and travels in the same direction as P at stream, taking 25 minutes. Another boy walks up
an average speed of 50km/h. After travelling the river bank at 3 km/h, starting from the same
5km, the motorist meets a friend and stops for 9 place as the first boy but 1 hour later.
minutes. He then continues his journey at the i.Draw a graph, with time on the horizontal axis,
same average speed as before. showing the distances of the two boys from their
i. Using a scale of 2 cm to 15 minutes on the start
horizontal (time) axis and 2 cm to 5km on the ii. Find where and when the two boys meet.
vertical (distance) axis, draw thegraphs of the iii. Find the speed of the boat downstream, in
journeys of P and Q km/h
ii. Use your graph to find; iv. What is the speed of the stream?
a. When the motorist Q overtakes the cyclist P
b. How far P is from his house when Q overtakes 6. A train leaves Onu station at 6 : 00 a. m. for
him. Krom station. It travels at an average speed of
c. The distance between P and Q at 8:15 a.m. 48km/h throughout stopping for 6 minutes when
it is 30 km from Onu. Another train leaves Onu at
3. At 8 : 00 a.m. a man set out on his bicycle from 6 : 25 a.m. by the same route and travels non-stop
Oforikrom on the main road to Femesua. He rides at an average speed of 80 km/h. If both trains
at a constant speed of 12km/h for 45 minutes and arrive at Kron at the same time,
then stops and talk to a friend. After he has been a. draw the graphs of the journeys of the two
talking for 24 minutes, he suddenly realizes that trains, using a scale of 2 cm to 10 minutes on the
he has forgotten something and returns horizontal axis and 2 cm to 5 km on the vertical
immediately to Oforikrom at 18km/h. Draw the axis.

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b. Use your graphs to find: S = A
i. the distance between the two trains at 6 : 40 10
But d = d2 − d1 = 10 – 0=10km
a.m.
t = t2 − t1 = 5 – 0 = 5min d
ii. the time the two trains arrive at Kron station
iii. how far Kron station is from Onu station. ⇒S = = 2km/min
O t 5
7.Manjang left Banjul on his motor bicycle at ii. To calculate the speed of the body in moving
9.00 a. m to travel to Lamin, 24 km away. He was from B to C
expected to arrive at Lamin at 10: 12 a. m but had S = C 20
to stop 8km from Banjul for 20 minutes. Draw a d = d2 − d1 = 20 – 10 = 10 km
graph for the journey t = t2 − t1 = 20 – 10 =10 min
i. Use your graph to find:
⇒S = = 1km/min B
10
10
a. his average speed for the last stage of the 20
journey,
iii. To calculate the speed (s) of the body in
b. his average speed for the last stage of the
journey if he was to arrive at Lamin at 10:12 a.m
moving from D to E.
S= , but d = d2 − d1
D 20
Interpretation of Travel Graph d = 20 − 0 = 20km
Consider the travel graph graph below; t = t2 − t1 = 35 − 25 = 10min
Distance (km) ⇒S = = 2km/min
0 35 E
25 25
C D 2. Consider the graph below which shows the
20
journey made by two friends Asamoah and
15
A B Koduah
10
a. At what time do they meet each other?
5 b. What is Asamoah‟s average speed?
E c. What is Koduah‟s average speed?
0
5 10 15 25 35
20 30 Time (s) Distance ( km)

I. A body starts from rest 0 and moved a 30 Town B

distance of 10km in 5minutes to reach A. 25

II. At A, the body rested for 5 minutes 20 Koduah
(A− B) before moving from B.
15
III. From B, the body moved 10km to C where it 13 Asamoah
rested for 10 minutes (C−D) before moving from 10
D at a distance of 20km in 10minutes to rest at E. 5

Calculations 0
1 2 3 4 5 6 7
i. To calculate the speed(s) of body in moving Time
Town A
(h)
from O to A, it follows that; ( pm )

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Solution
a. From the graph, they meet each other at The above travel graph describes the journey of a
4 : 00pm cyclist from Town X to Y.
b. Asamoah‟s average speed, S = a. What is the average speed for the return
journey from Town Y to Town X?
But d = d2 d1
b. State the period within which the cyclist
d = (13 – 0) km = 13km
traveled to Town Y after his first rest.
t = t2 t1 = (5 0) = 5h
c. How many minutes did the cyclist spend in
⇒S = = 2. 6km/h town Y?

c. Koduah‟s average speed, S = Solution

But d = d2 d1 = (30 0)km = 30km i. S =
t = t2– t1 = (6 0) h = 6h Distance from Town Y to Town X,
⇒S = = = 5km/h d = d2 d1 = (50 0) km = 50km.

Time traveled from Town Y to Town X,

Worked Examples
t = t2 t1 = (1 – 0) hr = 1 h
Distance in km
Town Y
Now S = = = 33.33km/h
50
40 ii. Period of first rest = 1.00pm to 1:30pm
30 iii. Time spend in town Y,
20 t = t2– t1= 2:00pm 1: 30pm = 30mins
10
Exercises 12.17
Town 0 1. The travel graph below shows the journey of
1 2 3 4 Mr. Brown who travelled in his car to his
Time in hours
X
mother‟s place at town B, 25km from his house.

30
B
25
20
15
10 A

5
0
73
7:30 77
7:50 8:10
81 85
8:30 89
8:50 93
9:10 9:30
97 101
9:50 105
10:10

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Determine from the graph, Solution
i. the time he spent in town A Population density =
ii. the time he left town A
iii. the time he spent in town B
520 people/km2 =
iv. the time he reached town A
v. his speed in travelling from town A to B ⇒Population = 520 people/km2 × 3.3km2
Population = 1,716 people
Population Density
It is the degree to which an area is filled with Exercises 12.18
people. Mathematically, it is the ratio of the 1.Oseikrom is a town that coversan area of 250
number of people in a particular area to the size hectares. If it is populated with 5,000 people,
of the area. calculate the density of the population correct to
( the nearest whole person.
Population density =

2. A village has a population of 1,712people and

Worked Examples an area of 214 hectares. Calculate the population
1. A village has an area of 80 hectares and a density of the town in people/ hector correct to
population of 2,500 people. Calculate the the nearest whole person.
population density of the town in people/ hector
correct to the nearest whole person 3. The population density of a village is 172
people/km2. If the area of the village is about
Solution 33km by 20 km, find its population to the nearest
Population density = = 100 people.
Population density = 31people /hectare
4. The population density of a village is 316
2. The population density of a village is 520
people/km2. If the area of the village is about 345
people/km2. If the area of the village is about 3.3
km2, find its population to the nearest 100 people.
km2, find its population to the nearest 100 people.

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13 PERCENTAGES 1 Baffour – Ba series

Definition Solution
A percentage is a fraction with a denominator of 1. 28% = 28 = =
100. For example, = 3%. Therefore, to
2. 35% = 35 = =
express anything as a percentage is to divide it
into 100 equal parts. The 100 equal parts
Exercises 13.2
becomes the denominator and the portion taken
Convert the following percentages to
out of the 100 is the numerator. If say “a” is the
fractions in its lowest term;
part taken out of the 100, then we have “a”
1. 72% 2.80% 3.0.5% 4.112 %
percent. That is: = a%, read as: “a percent”.
For eg, = 35%, = 87% etc Fractions and Decimal as Percentages
To convert a fraction to a percentage, multiply
Worked Examples the fraction by 100. That is:
Write the following as percentages;
as a percentage = =
1. 2.
To write a decimal as a percentage,express the
Solution decimal as a number with a denominator of 100
1. = 15% 2. = 40% and write the numerator as the percentage. For
e.g. 0.21 as a percentage; 0.21 = = 21%
Exercises 13.1
Write the following as percent; Worked Examples
1. 2. 3. 4. A. Change the following to percentages;
1. 2. 3.
Percentages as Common Fractions
To convert percentages to common fractions and
vice – versa, take note of the fact that, the B. Write the following as percentages;

percentage sign, % = . This means that: 1) 0. 475 2) 0.3 3) 0.15

a% = a × , expressed in the simplest

Solutions
form, to obtain the fraction.
A. Multiply each fraction by 100

Worked Examples 1. × 100 = = 40%

Express the following percentages as a fraction in
its lowest term; 2. × 100 = = 14%
1. 28% 2.35%
3. × 100 = = 75%

Baffour – Ba Series, Core Maths for Schools and Colleges Page 363
B. Express each decimal as a fraction with a
denominator of 100 and pick the numerator as
the percentage;
1. 0.475 = = 47.5% Comparing the products,
2. ⇒125
Descending order: , 54%, 0.52
3.

Method 2
Exercises 13.3
Express each number with a common
A. Express the fractions as percentages;
denominator of 100
1. 2. 3. 2 4. 5.1
54% =
B. Write the decimals as percentages; = =
1. 0.625 2. 0.09 3. 0.72
0.52 =

Ordering a Combination of Fractions, = =

Decimals and Percentages
To order a combination of common fractions, Comparing numerators,
decimal fractions and percentages, it is advisable > > >
to:
Method 1 In descending order ; 54%, 0.52%
Multiplying each number by 100% and order
them accordingly. 2. Arrange the following in descendingorder:
53%, , 0.52
Method 2
Express all the numbers as a fraction with Solution
common denominator, usually 100 or a power of Multiplyingeach number by 100%
10 if possible and order them according to the
53% = = 53%
value of their numerators.
= % = 60%
Worked Examples 0. 52 100% = 52%
1. Arrange in descending order; 54%, , 0.52, Comparing the products,
60 > 53 > 52
Solution In descending order; 53%, 0.52
Method 1
Multiply each number by 100% 3. Arrange in ascending order; 0.72, 87%,
54 % = × 100% = 54%
Solution
Multiplying each number by 100%.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 364
1. 0.72 = 72% 1. Find the percentage error in approximating 2.7
2. 87 % = × 100% = 87% as 3.

3.
Solution
Comparing the products, Absolute error = 3 – 2.7= 0.3
87 72 70 Original value = 2.7
In ascending order we have , 0.72, 87% Percentage error = 100%

= × 100% = 11.1 %
Exercises 13.4
A. Arrange in ascending order;
1. 0.6, , 0.15, 4. 51%, 0.7, 2. Find the percentage error in approximating
as 6.
2. 40%, 0.37, , 5. , 0.22, , 33%

Solution
B. Arrange in descending order;
Absolute error = 6 – 4 . But 4 = 4.5
1. 0.9, , 85% 3. , 0.31, 65%, 2.
Absolute error = 6 – 4.5 = 1.5
78%, , 0.77, 4. , 0.55, 25%, Original value = 4.5

Relative Error and Percentage Error Percentage error = 100%

Relative error (often used in measurement) is a
= × 100% = 33.3 %
comparison or ratio of the absolute error to the
actual / original value. Exercises 13.44
Find the percentage error :
Mathematically, 1) 9.8 as 10 2) 1.2 as 1.8
3) 15.2 as16.0 4) 5.4kg as 5.8kg
Relative error = ;
5). 3 as 4.2 6) 5 as 5
where absolute error is the difference between the
two values.
Percentage of a Given Quantity
When the relative error is multiplied by 100%, it To find the percentage of a given quantity (y),
is called percentage error.
express the percentage (x%) as a fraction and
Mathematically: multiply by the given quantity.
⇒x% of y = ×y
Percentage error = × 100%

Worked Examples
Percentage error = Relative error × 100%
1. Find 35% of Ghȼ400.00

Worked Examples
Solution

Baffour – Ba Series, Core Maths for Schools and Colleges Page 365
x% of y = ×y
B. Find the following
35% of Gh¢400 = × 400 = 140
1. 15 % of Gh¢40,000.00
35% of Gh¢400.00 = Gh¢140.00
2. What is 25% of 80 books? 2. 14 % of Gh¢33,400.00
3. % of Gh¢16,000.00
Solution
25% of 80 books 4. 2 % of Ghȼ10,000.00
× 80 = 20
Expressing one Quantity as a Percentage of
25% of 80 books = 20 books
Another
To express a quantity, as a percentage of a similar
3. Find 2 % of Gh¢20.00. quantity, express the quantity as a fraction of the
total quantity and multiply by 100 percent. That is:
Solution x as a percentage of y = × 100%
Change 2 % = %
Worked Examples
% of Gh¢20.00 = × 20 = × 20 1. Express 25 as a percentage of 75.
= Gh¢0.50 = 50p
Solution

4. Given that 7 % of k is 33, find k. × 100% = = 33.3%

2. What percentage of 5 is 0.25?

Solution
7 % of k is 33 Method 1
⇒7 % × k = 33 If 5 = 100%
⁄ ⇒ 0.25 = = 5%
× k = 33
Method 2
× k = 33
× 100% = 5%
= 33
15k = 200 × 33 3. What percentage of Gh¢250.00 is Gh¢50.00?
k= = 440
Solution
Exercises 13.5 × 100% = 20%
A. Find the following;
1. 26% of 5,000 pupils 4. Express 190 girls as a percentage of 250
2. 30% of 4,020 sheep pupils.
3. 90% of Gh¢20,000.00
4. Find 12 % of Gh¢80,000.00 Solution

Baffour – Ba Series, Core Maths for Schools and Colleges Page 366
 × 100% = 76% If y is the given quantity increased by x%,
1. Then the increment, I =
5. The population of Ghana was 5,000,000 in 2. New value = y + ( )
1957. The population in 1998 was estimated to be
17,000,000. Find the percentage increase in
Worked Examples
population from 1957 to 1998.
1. Increase Gh¢500.00 by 30%.
Solution
Solution
Population in 1957 = 5,000,000
Method 1
Population in 1998 = 17,000, 000 (
New value = ,
Increase in population But x = 30 and y = 500
= 17,000,000 – 5,000,000 = 12,000,000 (
New value=
= × 100 % = 240 %
New value = = Gh¢650.00

Exercises 13.6
1. What percentage of Gh¢1,250.00 is Gh¢50.00? Method 2
New value = ( )
2. A tank contains 400 liters of water. If 100 But x = 30 and y =Ghȼ500
liters is used, what percentage is left?
New value =( ) +Ghȼ500
3. Mr. Green spent Gh¢3,000.00 out of his New value = Ghȼ(150 + 500) = Gh¢650.00
Gh¢40,000.00. What percentage of his money
was spend? 2. If an item which cost Ghȼ250.00 is increased
by 60%, what is the new price of the item?
4. If 300 candidates sit for an examination and
180 pass, what percentage of the candidates fail? Solution
(
New value =
5. A candidate who gets 20% marks failed by 10
But x = 60 and y = Ghȼ250
marks but another candidate who gets 42% marks (
gets 12 % more than the passing mark. Find the
maximum marks. (
New value = = Ghȼ400.00

Increasing a Quantity by a Given Percentage

3. The population of Asuofua J.H.S was known to
Method 1
be 540 in 2010. Two years later, it increased by
If y is increased by x%, then: 20%. Find the student population in 2012.
(
New value =
Solution
(
Method 2

Baffour – Ba Series, Core Maths for Schools and Colleges Page 367
But x = 20 and y = 540 (
New value = ×y
(
= 648 people But x = 20 and y = 240
( – )
New value = × 240
Decreasing a Quantity by a Given Percentage
Method 1 New value = × 240 = 192 patients
If y is decreased by x%, then;
( 3. The cost of a rice cooker is Ghȼ1,440.00. If
New value = ×y
the cost decreases by 24%, find the new price of
the rice cooker.
Method 2
If y is decreased by x%, Solution
(
1. then the decrement, D = ×y New value = ,
But x = 24 and y = Ghȼ1440
2. New value = y – ( )
( – )
New value = × 1440
Worked Examples New value = × 1440 = Ghȼ,1094.40
1. Decrease Gh¢400.00 by 10%
Exercise 13.7
1. A 10% service charge is added to a restaurant
Solution
bill of Ghȼ200.20. What is the total amount paid?
Method 1
(
New value = , 2. An article cost Ghȼ720.80. If the cost is
But x = 10 and y = Ghȼ400 increased by 25%, find the new cost.
( – )
New value =
3. In a school of 800 boys, each boy plays either
New value = = Gh¢360.00 football or cricket or both. Given that 62% of the
boys play football and 58% play cricket, calculate
Method 2 the number of boys who play both football and
New value = y – ( ) cricket.
But x = 10 and y = Ghȼ 400
4. i. The area of a square is 400 cm2. Find the
New value = Ghȼ400 – ( )
length of a side.
New value = Ghȼ(400 – 40) = Ghȼ360.00 ii. If the length is now decreased by 20%, find the
length of the new square
2. The number of patients treated at a hospital on
a certain day was recorded as 240. If the number 5. A student was given Ghȼ2,300.00 by his
is decreased by 20% the next day, determine the father. If his mother gave him 28% of what his
number of patients that were treated the next day. father gave him in addition, find the total money
of the student.
Solution
Baffour – Ba Series, Core Maths for Schools and Colleges Page 368
6. A vendor sells 60% of apples he has and ii. New price = Original price – Discount
throws away 15% of the remainder. The next day, New price = Gh¢ (5,300 – 530) = Gh¢4,770.00
he sells 50% of the remainder and throws away
the rest. What percentage of his apples does he Method 2
throws away? New price = original price × discount rate
Discount rate = 100% − 10% = 90%
7. In an election contested by two parties A and B,
party A secured 12% of the total votes more than Discount rate =
party B. If party B got 132,000 votes and there are  New price × 5300 = Gh¢4,770.00
no invalid votes, by how many votes did it lose Discount = Gh¢(5300 – 4770) = Gh ¢530.00
the election?
2. Mr. Red paid Gh¢270.00 for T.V. set after he
Discount had been given a discount of 10%. Find the
It is the reduction in the original price of an item. marked price of the T.V set.
The discount or reduction is calculated
as a percentage of the original price. Solution
To find the discount, you require; Discount = 100% − 10% = 90%
a. Original price / Marked price ⇒Mr. Red paid 90% = Gh¢270.00.
b. Discount rate ⇒the original price which is 100% will be x.
c. New price
By proportion,
Discount = Rate × Original price i.e. 90 : 100 = 270 : x
New price = Original price – Discount
90 × x = 270 × 100
Worked Example 90 x = 27,000
1. Mr. Brown wants to buy a calculator whose x= = Gh¢300.00
price is quoted as Gh¢5300.00 with 10%
discount. Find:
3. Find the discount of 15% on Ghȼ1,600.00
i.the discount,
worth of goods. Find how much you will pay
ii. how much he will pay for the item.
for the goods.

Solution Solution
Method 1 Discount = Rate × Original price
Original price = Gh¢5,300.00 Rate = 15% =
Original price = Gh¢1,600.00
Rate = 10% =
Discount = × 1600 = Gh¢240.00
i. Discount = Rate × Original price
Discount = × 5300 = Gh¢530.00 New price = Original price –Discount

Baffour – Ba Series, Core Maths for Schools and Colleges Page 369
= Gh¢(1600 – 240 ) = Gh¢1,360.00 1. 7 % discount on on Ghȼ22, 000.00
2. 12 % discount on Ghȼ210, 000.00
4. What is the discount of 25% on a computer
that cost Gh¢5,600.00? Find the cost of the 3. 15 % discount on Ghȼ120, 000.00
computer.
4. 10 % discount on Ghȼ36, 000.00

Solution
C.1. A shop is offering a discount of 15% on all
Discount = Rate × Original price,
goods. Find the sale price of goods normally
where Rate = 25% = priced at: i. Ghȼ1650 ii. 9600 p
Original price = Gh¢5,600.00
Discount = × 5600 = Gh¢1,400.00 2. A water kettle and a rice cooker are sold at
Ghȼ410.00 and Ghȼ528.00 respectively. If a
discount of 20% is given on the cost of the kettle
The cost of the computer , and a discount of 25 % is given on the cost of the
= Original price – Discount rice cooker;
= Gh¢(5,600 – 1,400) = Gh¢4,200.00 i. how much will you pay for 5 kettles?
ii. how much will you pay for 7 rice cookers?
5. Calculate a discount of 12 % on an item that iii. find the total cost of buying 5 kettle and 7
cost Ghȼ24,000.00. Find the new price of the cookers at the given discount.
item.
3. If a shop reduces all its prices by 5 %,
Solution i. by how much is a generator that cost
Rate = 12 % = % Ghȼ72,000.00 reduced?
ii. how much will you pay for the generator?
Original price = Ghȼ24,000
⁄ Finding the Discount Rate Given the Marked
Discount = × Ghȼ24,000 = Gh¢3,000.00
Price and the Discount Value
Given the marked price and the discount value,
The cost of the item;
the discount rate is calculated by the formula:
= Original price – Discount
= Gh¢(24,000 – 3,000) = Gh¢21,000.00 Rate = × 100%

Exercises 13.8 Worked Examples

A. Find the following discounts: 1. A discount of Ghȼ480.00 was given on an
1. 14% on Ghȼ50,000.00 article marked Ghȼ24,000.00. What was the
2. 17% on Ghȼ10,000.00 percentage discount?
3. 12% on Ghȼ150,000.00
4. 19% on Ghȼ110,000.00 Solution
Rate = × 100%
B. Find the following:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 370
But discount value = Ghȼ480.00 and Marked Marked price = = = Gh¢20.00
(
price = Ghȼ24,000.00
Rate = 100 = 2% 2. Kojo paid Gh¢480.00 for T.V. set after he has
been given a discount of 20%. Find the
2. The original price of a calculator is Ghȼ600.00. marked price.
If it is reduced to Ghȼ420.00, find the rate of
discount. Solution
Marked price = ( –
Solution
But new price = Gh¢480.00
Rate = × 100% Discount rate = 20%
But discount value = Ghȼ(600 – 420)
= Ghȼ180.00 Marked price = = = Gh¢600.00
(
Original price = Ghȼ600.00
Rate = 100 = 30% Exercises 13.10
1. A bookshop reduced the price of its books by
Exercises 13.9 15%. Janet bought a book for Ghȼ2,700.00
1. If goods normally sold at Ghȼ112.00 are during the reduction sale. What was the original
offered at a sale at Ghȼ98.00 express the discount price of the book?
as percentage of the normal price.
2. A company reduced the prices of its goods by
Finding the Original Price / Marked Price 13% during promotional sales. A product was
Given the Discount Rate and the New Price. sold for Ghȼ43,500.00 during the promotional
Given the discount rate and the new price, the sales. What was the price before the promotional
original or marked price of an item is calculated sales?
by the formula:
Commission
Marked price = ( – It is the share or portion of a total amount of sales
(money) usually given to the sales person. It is
Worked Examples expressed as a percentage of the total sales.
1. A store gives 10% discount for an article
bought and paid for immediately. If a girl pays Mathematically:
Ghȼ18.00 cash for a dress, what is the marked Commission = Rate × Total Amount.
price of the dress? ⇒Total sales =

Solution Worked Examples

Marked price = 1. A shopkeeper receives a commission of 20%
( –
But new price = Gh¢18.00 and rate = 10% on her daily sales. If she makes a daily sales of
Gh¢1,050.00, calculate her commission.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 371
Solution Total sales of 1 bottle = Gh¢2.00
Rate = 20% = , total sales = Gh¢1050.00
Commission on one bottle
Commission = Rate × Total sales
= × Ghȼ2 = Gh¢0.04 = 4p
Commission = × Ghȼ,1050 = Gh¢210.00
Commission on 24 bottles = 24 × 4p = 96p
2. A house agent‟s commission on the sales of a
house is 7%. If he sell the house for
5. A crate of egg cost Gh¢2.50. If Naa sells 10
Gh¢12,500.00, find his commission.
crates at a rate of 30%, find her commission.

Solution
Solution
Commission = Rate × Total sales
Commission = Rate × Total amount
Rate = 7% = , amount = Gh¢12,500.00
Rate = 30% =
 Commission = Ghȼ12,500 = Gh¢875.00 Total amount =10 crates × Gh¢2.50 = Gh¢25.00

3. A sales girl is paid her basic monthly salary of Commission = × Ghȼ25 = Ghȼ7.50
Ghȼ15,000.00. In addition, she earns a
commission of 2% on all her sales. In a particular
Exercises 13.11
month, her sales amounted to Ghȼ1.5 million.
A. Calculate the following commissions;
Calculate her gross income for that month.
1. 26% on Gh¢5,440.00
2. 13% on Gh¢705.00
Solution
3. 17 ¼% on Gh¢108.00
Monthly salary = Ghȼ15,000
Total sales = Ghȼ1.5 million 4. 14 % on Gh¢4,800.00
Commission = 2% of Ghȼ1.5 million 5.1 % on Gh¢1,140.00
Commission = × 1,500,000 = Ghȼ30,000.00
B. 1. A book seller was given 20% commission
Gross income for that month; on his total sales of Gh¢62,500.00. Calculate his
= Monthly salary + Commission commission.
= Ghȼ(15,000 + 30,000) = Ghȼ45,000.00
2. A car dealer receives 15% commission on the
4. A bottle of soft drink cost Gh¢2.00. The sales of each car. If he sells a car worth
commission paid on one bottle is 2% of the cost Gh¢7,226.00, how much commission will he
price. Find the commission on 24 bottles of the receive?
soft drink.
3. A house agent is paid a commission of 9% on
Solution the sale of a semi – detached house. Calculate his
Commission on 1 bottle = Rate × total sales commission received on five semi – detached
But rate = 2% = building each at a cost of Gh¢2,200.00.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 372
4. Mr. Kokotii is a car dealer and a house agent as Solution
well. If he sells a car on behalf of his client, he Commission = Gh¢3,000.00
takes a commission of 16% on the total cost and Rate = 10% =
if he sells a house on behalf of his client, he takes
commission of 21% on the total cost. Calculate
his total commission if he sells three cars each at Total sales =
Gh¢7,650.00 and two houses each at a cost of Total sales = = = Gh¢30,000.00
Gh¢2,550.00

3. Find total amount if the commission on 5% of

5. Jonadab wanted Thammar as his girlfriend,
some goods sold is Gh¢110.00.
but he could not approach her due to some
fearfulness. He lured the services of his bosom
Solution
friend, Ahitofel who agreed to take a commission
of 8%on Jonadab‟s weekly wages of Rate = 5% =
Gh¢20,820.00. Find the amount received by Commission = Gh¢110.00
Ahitofel. Total sales =

Finding Total Sales Given the Rate and Total sales = = = Gh¢2,200.00
Commission
Total Sales = 4. At what amount will 14% give a commission
of Gh¢160.00.
Worked Examples
Solution
1. Aku received a commission of Gh¢9,000. 00
on some books sold. Find her total sales if she Rate =14% = , Commission = Gh¢160.00.
was given 12% commission. Total amount =
Total amount = = = Gh¢1,142.86
Solution
Total sales =
Exercises 13.12
Rate = 12% = A.1. Kofi received Gh¢8,000.00 commission on
Commission = Gh¢9,000.00 the sales he made in a particular day. The
Total sales = = 75,000.00 commission is 20% of the sales he made.
Calculate his total sales for the day.
 Total sales = Gh¢75,000.00
2. A trader received a commission of 12 % on the
2. A shopkeeper received Gh¢3,000.00
commission for selling some books If she had sales she made in a certain month. If the
10% commission find the total sales she made. commission she received was Gh¢30,000.00,
what is the total sales she made?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 373
3. A trader received a commission of 10% on the Exercises 13.13
sales made in a month. His commission was 1. A shopkeeper received Gh¢50.00 commission
Gh¢35,000.00. Find the total sales made for the from a publisher by selling 500 copies of a book
month. each at GH¢5.00. At what rate was the
commission calculated?
4. A salesman received a commission of 5% on
sales in a month. If his commission is 2. A shopkeeper received Gh¢3,000.00
Gh¢1,150.00, calculate his total sales in a month. commission from a publisher by selling 500
copies of a book at Gh¢20.00 each. What was his
Finding the Rate, Given the Total Sales and rate of commission?
Commission
Rate = 3. A book seller received Gh¢2,500.00 on a
Gh¢20,000.00 sale of books on behalf of a
publisher. What was the rate of commission?
Worked Examples
1. A shopkeeper receives Gh¢500.00 on 4. A house agent makes a commission of
Gh¢2,000.00 total sales of books from a Gh¢103,500 when he sells a house for
publisher. What is the rate of commission? Gh¢690,000.00. Calculate the percentage of his
commission.
Solution
Commission = Gh¢500.00, ChallengeProblems
Total sales = Gh¢2,000.00 1. Every week, a sales man is paid a basicsalary
of Ghȼ22.00. In addition, he receives a
Rate = = 25% commission at arate of on the first
Ghȼ1,000.00 of his weekly sales and at the rate of
2. At what rate will Gh¢720,000.00 sales yield a 2% on weekly sales over Ghȼ1,000.00. Calculate:
commission of Gh¢36,000.00? i. his earning in a week in which his sales totaled
Ghȼ1,600.00
Solution ii. his sales in a week when his total earnings
Commission = Gh¢36,000.00 amounted to Ghȼ35.00
Total amount = Gh¢720,000.00
Rate = = = 5% 2. A salesman is paid a commission at a rate of
1 % on all sales over Ghȼ10,000.00, plus a
3. A book seller received Gh¢500.00 commission weekly wage of Ghȼ22.50. Find his average
on a Gh¢4,000.00 sale of books on behalf of a weekly earnings if his sales for the year amounted
publisher. Calculate the rate of commission? to Ghȼ88,000.00.

Solution Profit and Loss

Profit (P) -: It is how much or how far the price
Rate = = = 12.5%
of an item exceeds the cost.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 374
Loss (L) -: It is how much or how far the price of C. P. of 1 orange = = Ghȼ12.00
an item recedes the cost.
Oranges sold = 458 – 62 = 396 oranges
The price at which goods are bought is called cost
C.P of 396 oranges at Ghȼ12.00;
price(C.P) and the price at which the goods are
= 396 × Ghȼ12 = Ghȼ4,752.00
sold is called selling price (S.P).
S.P of 396 oranges each at Ghȼ16.00;
Profit occurs when the S.P. is greater than the C.P
= 396 × Ghȼ16 = Ghȼ6,336.00
and loss occurs when the S.P is less than the C.P.
P = S. P – C. P
Mathematically:
P = Ghȼ(6,336 – 4,752) = Ghȼ1,584.00
a. Profit = Selling price – Cost price
P = S.P − C.P
Exercises 13.14
b. Loss = Cost price – Selling price
1. Mr. Oteng bought 210 oranges for Gh¢10.50
L = C.P − S.P
and sold them at 3 for 20p. How much profit did
he make?
Worked Examples
1. Mansah bought some oranges at Gh¢240.00
2. Mr. Brown bought some goods for
and sold them for Gh¢300.00 Find her profit.
Gh¢4,261.00 and sold it for Gh¢5,972.00. Find
his profit.
Solution
C.P = Gh¢240.00 and S.P = Gh¢300.00
3. The cost price of a refrigerator is
P = S. P − C.P
Gh¢60,303.00. If it is sold for Gh¢52,664.00,
P = Gh¢ (300 −240) = Gh¢60.00
calculate the loss incurred.

2. A radio which cost Gh¢118.00 was sold for

4. A bag of cement cost Gh¢23.00. If Mr. White
Gh¢96.00. Find the profit or loss.
buys 115 bags and sells them at Gh¢25.00 each,
calculate his profit.
Solution
C.P = Gh¢118.00 and S.P = Gh¢96.00
Profit and Loss Percentage
L = C.P – S.P
The profit or loss percent can be expressed as the
L = Gh¢ (118 – 96) = Gh¢22.00
percentage of the cost price (C.P). Thus, the
profit percent is calculated by,
3. A woman trader bought 458 oranges for
Ghȼ5,496.00. She kept 62 oranges for her family P% = × 100% =  100%
and sold the rest at Ghȼ16.00 each. Calculate to
one decimal place, her profit. The loss percent is calculated by,
L% =  100 % =  100 %
Solution
C. P. of 458 oranges = Ghȼ5,496

Baffour – Ba Series, Core Maths for Schools and Colleges Page 375
Worked Examples 20% of the eggs broken = × 30 = 6 eggs
1. A shoe which cost Gh¢16.00 was sold for Gh¢
The rest = 30 – 6 = 24 eggs
10.00. Find the loss percentage.
12 eggs = 1 dozen
Solution
24 eggs = = 2 dozens
C. P = Gh¢16.00 and S. P = Gh¢10.00
L = C.P − S.P. But 1 dozen = Ghȼ30.00
L = Gh¢(16 − 10 ) = Gh¢6.00 2 dozens = 2 × Ghȼ30 = Ghȼ60.00

C. P = Ghȼ45 and S. P. = Ghȼ60

L% = × 100% = × 100% = 37.5% P = S.P – C. P
P = Ghȼ(60 – 45) = Ghȼ15.00
2. An item which cost Gh¢200.00 was sold for
Gh¢250.00. Calculate the profit percent. P% = × 100% = × 100% = 33%

Solution
Solved Past Question
C.P = Gh¢200.00 and SP = Gh¢250.00
A woman bought 130kg of tomatoes for
P = S.P – C.P
Ghȼ52,000.00. She sold half of them at a profit of
P = Gh¢ (250 – 200) = Gh¢50.00
30%. The rest of the tomatoes began to go bad.
She then reduced the selling price per kg by 12%.
P% =  100% = × 100% = 25% Calculate:
i. the new selling price per kg
3. The cost of a table is Gh¢30.00. If it is sold for ii. the percentage profit on the whole transaction
Gh¢36.00, Find the percentage profit or loss. if she threw away 5kg of bad tomatoes.

Solution Solution
C.P = Gh¢ 30.00 and S.P = Gh¢36.00. i. C.P. of 130 kg of tomatoes = Ghȼ52,000.00
P = S.P − C.P
P = Gh¢(36 – 30) = Gh¢6.00 Cost of Half (65kg) = Ghȼ26,000
Profit of 30% on Ghȼ26,000
P% =  100% = × 100 = 20% P= × Ghȼ26,000 = Ghȼ7,800

4. A man buys eggs at Ghȼ45.00 per crate of S.P = Ghȼ(26,000 + 7,800) = Ghȼ33,800
30 eggs. He finds that 20% of the eggs are broken ⇒S. P of 1 kg = = Ghȼ520.00
but sells the rest at Ghȼ30.00 a dozen. Find his
percentage profit to nearest whole number. The new selling price per kg;
= 12% reduction in the S. P of 1 kg
Solution
= Ghȼ520 – ( )
Total number of eggs = 30
Cost price = Ghȼ45.00 = Ghȼ520 – Ghȼ62.4 = Ghȼ457.60

Baffour – Ba Series, Core Maths for Schools and Colleges Page 376
The new S.P per kg is Ghȼ457.60 per kg = S.P of 500 + S.P of 100
= Ghȼ2,975 + Ghȼ300 = Ghȼ3,275.00
ii. S.P of 65kg = Ghȼ33,800
Remaining = (65kg – 5kg of bad ones) = 60 kg P = S.P – C. P
P = Ghȼ3,275 – Ghȼ2,700 = Ghȼ575.00
60 kg sold at Ghȼ457.60 each
= 60 × Ghȼ457.60 = Ghȼ27,456.00 P% = × 100 % = 21.3%

Total selling price (S.P);

Exercises 13.15
= S.P of 65kg + S.P of 60 kg
1. Dansowaa bought a sewing machine for
= Ghȼ(33,800 + 27,456) = Ghȼ61,256.00
Gh¢50,000.00 and sold it for Gh¢92, 000.00.
Find the percentage of her profit.
P = S.P – C.P.
P = Ghȼ61,256 – Ghȼ52,000 = Ghȼ9,256.00
2. Mr. Green bought 300 crates of eggs each at a
cost of Gh¢6.20. After selling all the eggs, he had
P% = × 100 % = × 100 % = 17.8%
Gh¢1,200.00. Find his profit or loss percent.

2. A retailer bought 600 copies of a book at 3. Mr. White buys an old car for Ghȼ4,700.00
Ghȼ4.50 each. He sold 500 copies at Ghȼ6.00 and spends Ghȼ800.00 on its repairs. If he sells
each but gave a discount of 5p in the cedi. He the car for Ghȼ5,800.00, what is his gain percent?
sold the remainder of the books half the selling
price but without discount. Calculate the retailer‟s 4. Mr. Brown is a shopkeeper. He purchased 15
percentage gain, correct to 3 significant figures. boxes of key soap each at a cost of Gh¢12,000.00
and 17 boxes of voltic mineral water each at a
Solution cost of Gh¢8,000.00. After selling all the goods,
C.P. of 600 copies of a book at Ghȼ4.50 each he had Gh¢237,000.00. Calculate his profit
= 600 × Ghȼ4.50 = Ghȼ2,700.00 percent or loss percent.

S.P. of 500 copies at Ghȼ6.00 each but gave a 5. The total cost of 544 oranges is Ghȼ11,968.00.
discount of 5p in the cedi If 102 oranges are kept and the rest sold at
⇒ A copy = Ghȼ6.00 – 5p = Ghȼ5.95 Ghȼ20.00 each, find the percentage of the profit
or loss.
S.P of 500 copies at Ghȼ5.95 each
= 500 × Ghȼ5.95 = Ghȼ2,975.00 6. A woman bought 412 oranges for
Ghȼ7,416.00. She kept 80 of them and sold the
Remaining copies = 600 – 500 = 100 rest at Ghȼ23.00 each. Find her profit percent.
100 copies at half of Ghȼ6.00 (No discount)
100 × Ghȼ3 = Ghȼ300.00 7. A television set cost a retailer Ghȼ500.00. He
offers to sell it either for a cash price of
Total selling price (S.P) Ghȼ650.00 or a down payment of 25% of the
Baffour – Ba Series, Core Maths for Schools and Colleges Page 377
cash price followed by 24 monthly installments I. Identify the given percentage profit (P%) and
of Ghȼ20.50. Calculate his percentage profit the selling price (S.P)
calculated on his cost price and his selling price II. Substitute the values of P and S.P. in the
for each method of selling Ans: 30%, 23.1% formula: C.P. = × S.P.
III. Simplify and obtain the answer in the unit of
Challenge Problem
the given currency.
1. A merchant spent Ghȼ2,250.00 in buying
1,000 articles. He fixed the selling price to allow
himself a profit of 20% on his cost, and sold four Method 2
– fifths of the articles at this price. He then This is equivalent to increasing the cost price by
reduced his selling price by one – third and sold the given percentage. That is :
the remainder of his stock at this new price. C.P .+ (P% of C.P.) = S.P.
Calculate his profit as a percentage of his outlay. I. Identify the given percentage profit (P%) and
Ans: 12% the selling price(S.P)
II. Substitute the values of P and S.P in the
2. In a certain store, the profit is 20% of the cost. formula: C.P. + ( ) = S.P.
If the cost increased by 25% but the selling price
III. Solve for C.P and obtain the answer in the
remains constant, approximately what percentage
unit of the given currency.
of the selling price is the profit?Ans : 16.67%
Worked Examples
3. Potatoes are bought at Ghȼ20.00 per 50kg and 1. A set of furniture was sold for Gh¢300.00 at a
are sold at Ghȼ3.50 per kg. Find, correct to the profit of 20%. Findthe cost price.
nearest whole number, the profit as a percentage
of: Solution
a. the cost price Ans : (40%) Method 1
b. the selling priceAns : 28.6% S.P. = Gh¢300.00,P % = 20%, C.P. =?
C.P. = × S.P.
4. A shop marks an article so as to make a profit
of 30% on the cost price. In a sale, a discount of C.P = × Ghȼ300 = Gh¢250.00
10% was allowed off the marked price. If the
article was sold in the sale, state the actual Method 2
percentage profit made by the shop. S.P. = Gh¢300.00,P % = 20%, C.P. = x
C.P. + ( ) = S.P
Finding the Cost Price or Selling Price Given
the Percentage Profit x+( ) = Ghȼ300
A. Finding the Cost Price x + = 300
Given the profit percentage and the selling price,
5x + x = 5 × 300
the cost price is calculated as follows:
6x = 5 × 300
x= = Ghȼ250
Method 1

Baffour – Ba Series, Core Maths for Schools and Colleges Page 378
The cost price is Ghȼ250.00 Worked Examples
1. A bag of rice which cost Gh¢50.00 was sold at
2. A kettle was sold for Gh¢575.00 at a profit of a profit of 20%. Find the selling price.
15%. Find the cost price of the kettle.
Solution
Solution Method 1
S.P. = Gh¢575.00, Profit % =15% C.P. =? C.P. = Gh¢50.00, P% = 20%, S.P. =?
C.P. = × S.P. S.P. = × C.P.
C.P. = × Ghȼ575 = Ghȼ500.00 S.P. = × Ghȼ50 = Gh¢ 60.00

3. A wall clock is sold for Gh¢210.00 at a profit

Method 2
of 5%. What is the cost price of the wall clock?
C.P. = Gh¢ 50.00 and, P% = 20%, S.P. =?
Solution S.P = C.P + ( )
S.P. = Gh¢210.00 and P% = 5% C.P.= ? S.P = 50 + ( ) = Ghȼ60
C.P. = × S.P
C.P. = × Ghȼ210 = Gh¢200.00 2. A bicycle was boughtat Gh¢300.00. If it was
sold at a profit of 24%, find the selling price of
B. Finding the Selling Price the bicycle.
Given the profit percentage and the cost price, the
Solution
selling price is calculated as follows:
C.P. = Gh¢300.00 and P % = 24%, S.P.= ?
I. Identify the given percentage profit (P%) and
the cost price (C.P) S.P. = × C.P.
II. Substitute the values of P and C.P. in the S.P. = × Ghȼ300 = Gh¢372.00
formula: S.P = × C.P.
III. Simplify and obtain the answer in the unit of 3. A lorry tyre which cost Gh¢525.00 was sold at
the given currency. a profit of 12%. What was the selling price of the
tyre?
Method 2
The S.P is equal to increasing the C.P by the Solution
given percentage (P%). That is: C.P. = Gh¢525.00 and P% = 12% S.P. = ?
S.P. = C.P. + (P% of C.P.) S.P. = × C.P.
I. Identify the given percentage profit (P%) and
S.P. = × Ghȼ525 = Gh¢588.00
the cost price (C.P)
II. Substitute the values of P and C.P in the
Solved Past Question
formula: C.P +( ) = S.P. A trader buys some goods whose mark price is
III. Simplify and obtain the answer in the unit of Ghȼ100,000.00 at a discount of 2.5%. After
the given currency. selling the goods, the trader is required to pay tax

Baffour – Ba Series, Core Maths for Schools and Colleges Page 379
at a rate of 20% on the profit she makes in excess Method 2
of Ghȼ10,000.00. How much should she sell the This is equivalent to decreasing the cost price by
goods so that after the tax she would make a net the given percentage. That is :
profit of Ghȼ20,000.00? C.P – (P% of C.P) = S.P
I. Identify the given percentage loss (L%) and
Solution the selling price(S.P)
Marked price = Ghȼ100,000.00 II. Substitute the values of L and S.P in the
Discount = × Ghȼ100,000 = Ghȼ2,500.00 formula: C.P – ( ) = S.P
C.P = Ghȼ100,000 – Ghȼ2,500 III. Solve for C.P and obtain the answer in the
C.P = Ghȼ97,500.00 unit of the given currency.

Let the profit before tax be GhȼP Worked Examples

20% of profit paid as tax leaving a net profit of 1. Mr. Owusu sold his used T.V. set for
Ghȼ20,000 Gh¢255.00. If his loss was 15% , how much did
P–( ) = Ghȼ20,000.00 he buy the T.V. set?
P – = 20,000
Solution
5P – P = 5 × 20,000
Method 1
4P = 100,000
S.P. = Gh¢255.00 and C.P = ? L % = 15%
P= = Ghȼ25,000
C.P = × S.P

P = S.P – C.P C.P = × Ghȼ255

S.P = P + C.P C.P = × Ghȼ225
S.P = Ghȼ25,000 + Ghȼ97,500 = Ghȼ122,500.00
C.P = Gh¢300.00

Finding the Cost Price or Selling Price Given

the Percentage Loss Method 2
A. Finding the Cost Price S.P. = Gh¢255.00 and C.P = x,L % = 15%
Given the loss percentage and the selling price, C.P – ( ) = S.P
the cost price is calculated as follows:
Method 1 x–( ) = Gh¢255
I. Identify the given loss percentage (L%) and x– = 255
the selling price (S.P)
100x – 15x = 100 × 255
II. Substitute the values of L and S.P in the
85x = 100 × 255
formula: C.P = × S.P x= = Ghȼ300.00
III. Simplify and obtain the answer in the unit of
the given currency. 2. A mobile phone is sold for Ghȼ720.00 at a loss
of 20%. What is its cost price?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 380
Solution Method 2
S.P. = Gh¢ 720.00 and C.P = ? L % = 20% C.P = Ghȼ350, L % = 40 and S.P = ?
C.P = × S.P S.P = C.P – ( )

C.P = × Ghȼ720 = Ghȼ900.00 S.P = Ghȼ350 – ( ) = Ghȼ210.00

B. Finding the Selling Price 2. A man claims to have sold his car at a loss of
Given the loss percentage (L%) and the cost 24%. How much did he sell the car if he actually
price, the selling price is calculated as follows: bought it for Ghȼ7,500.00?
I. Identify the given percentage loss (L%) and
the cost price (C.P) Solution
II. Substitute the values of L and C.P in the C.P = Ghȼ7,500, L % = 24% and S.P = ?
–
– S.P = × C.P
formula: S.P = × C.P
–
III. Simplify and obtain the answer in the unit of S.P = × Ghȼ7,500
the given currency.
S.P = × Ghȼ7500 = Ghȼ5,700.00
Method 2
The S.P is equal to decreasing the C.P by the Solved Past Questions
given percentage (P%). That is:
1. A trader sold 1,750 articles for Ghȼ525, 000.00
S.P = C.P + (P% of C.P)
and made a profit of 20%.
I. Identify the given percentage loss (P%) and
i. Calculate the cost price of each article.
the cost price (C.P.)
ii. If he wanted 45% profit on the cost price, how
II. Substitute the values of L and C.P in the
much should he have sold each of the articles?
formula: S.P. = C.P. –( )
III. Simplify and obtain the answer in the unit of Solution
the given currency. i. C.P = × S.P
P = 20% and S.P = Ghȼ525,000
Worked Examples
C.P = × 525,000 = Ghȼ437,500.00
1. At what price must an item bought for
Ghȼ350.00 be sold so as to incur a loss of 40%.?
Cost price of one item;
Solution = = Gh¢250.00
Method 1
C.P = Ghȼ350, L % = 40 and S.P = ? ii. Increase the cost price by 45% ;
– = 437,500 + ( )
S.P = × C.P
– = 437,500 + 196,875 = Ghȼ634,375.00
S.P = × Ghȼ350
S.P = × Ghȼ350 = Ghȼ210.00 Each article should be sold for ;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 381
= = Ghȼ362.50 i. The manufacturer increased the cost,
Ghȼ6,000.00 by 10% and sold to the wholesaler.
2. A manufacturer makes a wireless set at a cost Therefore, the selling price or cost to the
of Ghȼ6,000.00 and sold it to a wholesaler at a wholesaler;
profit of 10%. The wholesaler sold it to a retailer = 6,000 + ( )
at a profit of 25%. Find: = 6,000 + 600 = Ghȼ6,600.00
i. the cost to the wholesaler;
ii. the selling price of the wholesaler; ii. The wholesaler increased the cost, Ghȼ6,600
iii. if a customer who paid cash had the price by 25% and sold it to the retailer.
reduced to Ghȼ6,200.00, find the percentage The selling price of the wholesaler is;
discount allowed on the marked price to the = 6,600 + ( )
customer, to two significant figures.
= 6,600 + 1,650 = Ghȼ8,250.00
Solution
iii. Discount = Marked price – New price
Method 1
Marked price for the customer is Ghȼ8,250.00
i. The cost to the wholesaler is the same as the
and the new price for the customer is
selling price of the manufacturer;
Ghȼ6,200.00
S.P = × C.P
Discount = Ghȼ(8,250 – 6,200) = Ghȼ2050.00
P = 10% and C. P = 6,000
S.P = × 6,000 = Ghȼ6,600.00 Rate = × 100%

Rate = × 100% = 24.8%

ii. The wholesaler bought the wireless set at
Ghȼ6,600 and sold to the retailer at 25%.
Therefore, the selling price is: 4. A publisher prints 30,000 copies of an edition
of a book. Each copy of the book costs the
S.P = × C.P
publisher Ghȼ0.45 and it is sold to the public for
S.P = × 6,600.00 = Ghȼ8,250.00 Ghȼ0.76. The publisher agrees to pay the author
10% of the selling price of the first 6,000 copies
iii. Discount = Marked price – New price sold, and 12% of the selling price for all copies
Marked price for the customer is Ghȼ8,250.00 sold in excess of 6,000.00. Altogether, 25,000
and the new price for the customer is copies of the books are sold. Calculate correct to
Ghȼ6,200.00 the nearest cedis;
Discount = Ghȼ(8,250 – 6,200) = Ghȼ2,050.00 i. the total amount received by the author;
ii. the net profit the publisher makes after paying
Rate = × 100%
the author (Assume that the unsold copies are
Rate = × 100% = 24.8% donated to various libraries);
iii. the authors total receipts as a percentage of the
Method 2 publishers net profit, correct to one decimal
place.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 382
Solution iii. Authors total receipts as a percentage of the
i. S. P of a book = ȼ0.76 publisher‟s net profit;
S.P of first 6,000 books; = × 100%
= Ghȼ0.76 × 6,000 = Ghȼ4,560.00
= × 100% = 66.1%
Authors share of the first 6,000 copies;
= 10% of the selling price 5. A trader buys 3,150 articles at a cost of
= × Ghȼ4,560 = Ghȼ456.00 Ghȼ6.00 per article. He fixes the selling price so
that, if only 3,000 articles are sold, he will make a
profit of 40% on his total cost.
Remaining copies of books;
i. Calculate the selling price of one article
= 25,000 – 6,000 = 19,000
ii. If in fact, he sells 3,150 articles at this price,
find the actual profit as a percentage of the total
S. P of a copy of the remaining copies;
cost price
= Ghȼ0.76 per copy
iii. If he had wanted 100% profit, how much
should he have sold the 3,150 articles?
S.P of the remaining 19,000 copies;
= Ghȼ0.76 × 19,000 = Ghȼ14,440.00
Solution
i. C.P. of the 3,150 articles at Ghȼ6.00 each;
Authors share of the remaining 19,000 copies; =
= Gh¢18,900
12% of the selling price;
= × 14,440 = Ghȼ1,732.80 C.P. of the 3,000 articles at Ghȼ6.00 each;
= 3,000 × Ghȼ6.00 = Ghȼ18,000.00
Total amount received by the author;
= Ghȼ(456.00 + 1,732.80) = Ghȼ2,188.80 C.P. = Ghȼ18,900.00, P% = 40%, S.P. = ?
S.P. = × C.P.
ii. Publishers share of the first 6,000 copies;
= Ghȼ(4,560 – 456) = Ghȼ4,104.00 ⇒S.P. = × Ghȼ18,900 = Ghȼ26,460.00

Publishers share of the remaining 19,000 copies; Cost of 1 article = = Ghȼ8.82

= Ghȼ(14,440.00 – 1732.80) = Ghȼ12,707.20

Publishers total share of the 25,000 copies; ii. Selling price of 3,150 articles at Ghȼ8.82;
= Ghȼ(4,104.00 + 12,707.20) = Ghȼ16,811.20 = 3,150 × Ghȼ8.82 = Ghȼ27,783.00

But cost of printing 30,000 copies at Ghȼ0.45 P = S. P – C. P

each = Ghȼ0.45 × 30,000 = Ghȼ13,500.00 P = Ghȼ(27,783 – 18,900) = Ghȼ8,883.00

Net profit = Total share – Printing cost The actual profit as a percentage of the total cost
= Ghȼ(16,811.20 – 13,500.00) price = × 100 % = 47%
= Ghȼ3,311.20

Baffour – Ba Series, Core Maths for Schools and Colleges Page 383
iii. C.P. = Ghȼ18,900, P% = 100%, S.P = ? 1. A bag of rice which cost Gh¢4,000.00 was sold
S.P. = × C.P at a profit of 20%. Find the selling price.

S.P. = × Ghȼ18,900 = Ghȼ37,800

2. A trader bought 200 oranges at 8 for Gh¢10.00.
At what price must she sell all of them to make a
6. A trader bought a radio for Ghȼ68.00 and sold profit of 8%?
it at a profit of 7 %.The following week, the cost
price of the same type of radio increased by 8%. 3. Find the selling price of goods bought at
By what percentage, correct to two significant Ghȼ2.80 and sold at :
figures, must the trader increased his selling price a. a profit of 15% of the cost price,
in order to make the same profit. b. a loss, of 15% of the cost price.

Solution 4. Maamele bought a car for Gh¢1,500.00. He

C.P = Ghȼ68.00 later sold it for at a profit of 20%. What was the
selling price?
Profit =7 % of Ghȼ68.00
Profit = × Ghȼ68 = Ghȼ5.10 5. A merchant buys two articles for Ghȼ600.00.
He sells one of them at a profit of 22% and the
S.P. = P + C.P other at a loss of 8% and makes no profit or loss
S.P. = Ghȼ5.10 + Ghȼ68.00 .= Ghȼ73.10 in the end. What is the selling price of the article
that he sold at a loss?
Increased in C.P(Ghȼ68.00) by 8% ;
B. 1. A radio set was sold for Ghȼ 250.00 at a
= 68 + ( ) = Ghȼ73.44
profit of 25%, Find the cost price.
S.P = P + C. P
S.P = Ghȼ5.10 + Ghȼ73.44 = Ghȼ78.54 2. A car was sold for Ghȼ 44,080.00 at a loss of
The new selling price = Ghȼ78.54 12%. Find the cost price.

Let x represent the percentage increase in the old 3. A man sold some articles for Ghȼ 18,000.00
S.P (Ghȼ73.10) to the new S.P (Ghȼ78.54) and made a profit of 20%. Find the cost price of
73.10 + ( ) = 78.54 the articles.
⇒ 73.10 + = 78.54
4. A refridgerator was sold for Ghȼ3,600.00 at a
⇒ 7310 + 73.10x = 7854 loss of 10 %. Find the cost price.
73.10x = 7854 – 7310
73.10x = 544
Simple Interest
x= = 7.4% (2 s.f.) Definition of Terms
She has to increase the selling price by 7.4% Simple Interest (I): When money is borrowed or
deposited at the bank, for a certain period, it
Exercises 13.16 attracts an extra amount at a given rate. This extra

Baffour – Ba Series, Core Maths for Schools and Colleges Page 384
money is called simple interest or interest (I) for But, I = = = Gh¢75. 00
short.
2. A man deposited Gh¢1,520.00 at a bank for
The actual money borrowed or deposited at the
6years at a rate of 10% p.a. Calculate the
bank is called Principal (P).
interestearned at the end of the period.
The period for which the money is borrowed or
Solution
deposited at the bank is called time (T). The time
P = Gh¢1,520.00, T = 6yrs, R =10%, I =?
can be in years, months, weeks or days.
I= = = Gh¢ 912.00
The percentage at which the interest on the
amount (money) borrowed or deposited is 3. Kusi borrowsGh¢24,000.00 from a bank. He
calculated is called the rate of interest. pays after 2 years at a rate of 12% simple interest.
Find the interest and the amount to be paid at the
Mathematically: end of 2years.
Interest =
Solution
Simply put, I = ……….. (1)
P = Gh¢24,000, T = 2yrs, R = 12%, I =?
I= = = Gh¢5,760.00
From eqn(1),
I= , P,T and R can be made the subject to
A = P + I,
get the following equations respectively; But P = Gh¢24,000 and I = Ghȼ5,760
P= , T= , R= A = Ghȼ24,000 + Ghȼ5,760 = Gh¢29,760.00

Amount (A) 4. Kate borrowed Gh¢8,260.00 from a bank at a

Amount is the sum of principal and interest. rate of 12% per anum simple interet.Calculate;
Amount = Principal + Interest i.the interest at the end of 7 years.
A=P + I
ii. the amount repaid at the end of 7 years.
For repayment on instalment: iii. the amount paid per month, if she agreed to
repay the loan amount at equal monthly
1. Monthly instalment =
installment over the 7 year period.
2. Yearly instalment =
Solution
Worked Examples P = Gh¢8,260, T = 7 = yrs, R = 12, I = ?
1. Find the simple interest on Gh¢500.00 for 3
years at a rate of 5% per anum i. I = = = Gh¢7,434.00

Solution ii. A=P + I

P = Gh¢500.00, T = 3 years, R = 5 % I = ? But P = Gh¢8,260 and I = Ghȼ7,434
Baffour – Ba Series, Core Maths for Schools and Colleges Page 385
A = Ghȼ8,260 + Ghȼ7,434 = Gh¢15,694.00 1. Find the simple interest on Gh¢500,000.00 for
5 years at a rate of 2% years at 5 % per anum.
iii. Monthly instalment =
Amount = Gh¢15,694.00 2. Find the simple interest on Gh¢8,000.00 for 2
Number of months = 7 years = 90 months years at 5 %.
Monthly Ins = = Gh¢174.38
3. Find the interest on Gh¢480,000.00 for 5
5. Find the simple interest on Gh¢2,080.00 at 3 years at 12 % per anum.
% per anum for 6months.
4. a. Find the total amount to be paid at the end of
Solution 4 years if Mrs. Diana borrows Gh¢31,520.00 at
P = Gh¢2,080.00, R = 3 % = % 22 % per anum with simple interest.
T = 6months = years and I = ? b. If she repays the loan at equal monthly
I= = = Gh¢520.00 instalment over the period, calculate the amount
paid at the end of each month.

6. A trader was charged 2 pesewas per month for To Calculate for the Principal
every Ghȼ1.00 he borrowed from a bank. To calculate for the rate,
i. At what percentage rate per anum was the
I. Use the formula, P =
interest charged?
ii. How much would the trader pay as interest on II. Substitute the values of I, T and R, in the
a loan of Ghȼ5,000.00 for 6 months? formula and simplify to obtain the value of P.

Solution Worked Examples

i. Per anum means by the year (12months) 1. The interest on a certain amount deposited at
⇒ P = 1.00 = 100p, I = 2p, T = 1 month the bank for 4years at a rate of 5% p.a. is
Gh¢1,600.00. How much money was deposited?
R= = = 2% Solution
T = 4yrs, R = 5%, I = GHȼ1,600.00, P = ?
2% per month for 12 months,
P= = = Gh¢8,000.00
= 2% × 12 = 24% per anum

2. What amount will yield an interest of

ii. P = Ghȼ5,000, I = ?, R = 24%,
Gh¢1,560.00 for 6 years at a rate of 20% p.a?
T = 6 months = year
⁄ Solution
I= = = Ghȼ600.00
I = Ghȼ1,560, T = 6yrs, R = 20%, P =?

Exercises 13.17 P= = = Gh¢1,300.00

Baffour – Ba Series, Core Maths for Schools and Colleges Page 386
Exercises 13.18 3. The simple interest on Ghȼ60,000.00 for 3
1. What principal will yield an amount of
years is Ghȼ5,625.00 Find the rate percent per
Gh¢24,400.00 in 4 years at 20% per anum simple
anum.
interest.
2. Find the principal that will earn a simple
Solution
interest of Gh¢25,000.00 in 6 years at 17 %
P = Ghȼ60,000.00, T = 3 years =
per anum.
I = Ghȼ5,625.00, R = ?
3. A man deposited an amount of money in his
savings account for 5 years. The rate of interest R= = = = 2.5%
⁄
was 14% p.a. If the interest was Gh¢35.00, find
the amount deposited. Exercises 13.19
4. Find the sum of money on which the simple 1. At what rate of interest will Gh¢8,000.00 yield
interest for 8 years at 4 % per anum is a simple interest of Gh¢3,840.00 in 6 years.
Gh¢360.00.
2. A what rate of interest will Gh¢2,650.00 yield
To Calculate for Rate a simple interest of Gh¢4,770.00 in 3 years?
To calculate for rate, use the formula:
3. An amount of Gh¢2,500.00 invested for 5
R=
years yielded a simple interest of
Gh¢3,750.00.Find the rate of interest per anum.
Note:
The rate is always expressed as a percentage. To Calculate for Time
The time is calculated by the formula,
Worked Examples
T=
1. At what rate will Gh¢5,000.00 attract an
interest of Ghȼ300.00 for 2years?
Worked Examples
1. What time will Gh¢5,000.00 yield an interest
Solution
P = Gh¢5,000, I = Gh¢300, T = 2 yrs and R = ? of Gh¢1,000.00 at a rate of 5% p.a?

R= = = 3%
Solution
P = Gh¢5,000, I = Gh¢1,000,R = 5%, T = ?
2. The interest on Gh¢1,500.00 deposited at a
T= = = 4 years
bank for 3years was Ghȼ90.00 Find the interest
rate per anum.
2. Calculate the time in which Gh¢6,000.00 will
Solution attract an interest of Ghȼ1,200.00 at 20% per
T = 3years, I = Gh¢90, P = Gh¢1,500, R =? anum?
R= = = 2%
Solution
P = Gh¢6,000, I = Ghȼ1,200, R = 20%, T=?
Baffour – Ba Series, Core Maths for Schools and Colleges Page 387
T= = = 1 year When Asamoah receives Ghȼ48,000.00
⇒ 10% of x + 3% interest on Ghȼ900
= Ghȼ48,000
Exercises 13.20
1. How many years will Gh¢5,000.00 yield a = × x + I = Ghȼ48,000
simple interest of Gh¢1,000.00 at a rate of 5% per ⇒ + I = Ghȼ48,000
anum?
But 3% interest on Ghȼ900 for 1 year
2. In how many years will Gh¢50,000.00 yield a
P = 900, R = 3%, T = 1 year, I = ?
simple interest of Gh¢100,000.00 at a rate of 5%
I= = = Ghȼ27.00
per anum?

3. A man borrows Ghȼ4,000.00 and at the end of Put I = Ghȼ27 in + I = Ghȼ48,000

each of the three following years he pays the
+ Ghȼ27 = Ghȼ48,000
lender Ghȼ400.00; part of this is in payment of
interest at 5% per anum on the amount of his debt x + Ghȼ270 = Ghȼ480,000
during the year, and the rest reduces the debt. x = Ghȼ480,000 – Ghȼ270 = Ghȼ479,730.00
How much does he still owe at the end of the
third year? Oppong‟s payment = 15% of x
But x = Ghȼ479,730
Solved Past Question
Atiamo borrows Ghȼ900.00 from Oppong and Oppong‟s payment = × Ghȼ479,730
the same amount from Asamoah to start a
= Ghȼ71,959.50
business. He agrees to pay Oppong 15% of the
profit of the business each year. He also agrees to
Ratio and Percentages Combined
pay Asamoah 10% of the annual profits in
Worked Examples
addition to 3% interest on his loan each year.
1. A school presents 200 candidates for
Find how much Oppong is paid in a year which
examination. The ratio of boys to girls was 5 : 3
Asamoah receives Ghȼ48,000.00.
respectively.
i. How many boys were presented for the
Solution
examination?
a. Let the profit for the year be x,
ii. If 24% of the girls and 28% of the boys passed
Amount borrowed from Oppong = Ghȼ900.00
with distinction, what percentage of the
Amount borrowed from Asamoah = Ghȼ900.00
candidates obtained distinction?

Agreement with Oppong; Solution

= 15% of the profit of the business each year i. Total number of candidates = 200
Ratio of boys to girls = 5 : 3
Agreement with Asamoah;
Total ratio = 5 + 3 = 8
= 10% of the annual profits in addition to 3%
interest on his loan each year

Baffour – Ba Series, Core Maths for Schools and Colleges Page 388
Number of boys = × 200 = 125 boys C.P = × Ghȼ1,530 = Ghȼ1,224.00

ii. Number of boys = 125 The factories cost price;

Number of girls = 200 – 125= 75girls S.P = Ghȼ1,224.00, P % = 27.5%, C.P. = ?
C.P = × S.P
Girls who passed with distinction; C.P = × Ghȼ1,224 = Ghȼ960.00
24% of the girls = × 75 = 18 girls
Cost of materials : cost of labor = 3 : 5.
Boys who passed with distinction; Total ratio = 3 + 5 = 8
28% of the boys = × 125 = 35 boys i. Cost of materials = Ghȼ960 = Ghȼ360.00

Students who passed with distinction; ii. Cost of labor = Ghȼ960 = Ghȼ600.00
= 18 + 35 = 53

b. 20% of labor cost ( Ghȼ600.00)

Percentage of candidates who obtained
= Ghȼ600 = Ghȼ120.00
distinction = × 100% = 26.5%

The cost of labor;

2. A soap factory finds that the cost of materials
= Ghȼ600 + Ghȼ120 = Ghȼ720.00
and labor to produce a certain brand of soap are
in the ratio 3 : 5. The factory sells to the
New cost of materials and labor;
wholesaler at a profit of 27 % and the = Ghȼ(360 + 720) = Ghȼ1,080.00
wholesaler sells to a retailer at a profit of 25%
a. If the retailer pays Ghȼ1,530.00 for each box of New Profit for the factory;
soap, calculate how much it cost the factory in: S.P. = Ghȼ1,224, C..P = Ghȼ1,080
i. materials; P = S.P – C.P
ii. labor to produce a box of soap; P = Ghȼ1,224 – Ghȼ1,080 = Ghȼ144.00
b. the labor cost went up 20% (with no increase
in the cost of materials), but the factory decided P% = × 100 = 13.3%
not to increases the price charged to the
The new percentage profit is13.3%
wholesaler. Calculate, correct to the nearest
whole number, the new percentage profit the 3. A manufacturer finds that the cost materials
factory made. and labor to make a certain articles are in the ratio
3 : 5 respectively. The manufacturer sells to a
Solution
a. The wholesaler‟s cost price; retailer at a profit of 27 % and theretailer sells
S.P = Ghȼ1,530, P% = 25% and C.P. = ? to a customer at a profit of 25%. The customer
C.P = × S.P pays Ghȼ6,375.00,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 389
a. Calculate how much the article cost the S.P = Ghȼ5,100, C.P = Ghȼ4,500
manufacturer for materials and for labor. P = S.P – C.P
b. If there is 20% rise in labor cost but no P = Ghȼ5,100 – Ghȼ4,500 = Ghȼ600.00
increases in the cost of materials and the
manufacturer decides not to increase the price P% = × 100 = 13.3%
charged to the retailer, calculate the percentage The factory makes a new percentage profit of
profit which the manufacturer then makes. 13.3%

Solution 4. The estimated cost of a house was Ghȼ

a. The retailer‟s cost price; 6,400.00. It was made up of the cost of labor,
S.P. = Ghȼ6,375, P% = 25% and C.P. = ? materials and the contractor‟s charge in the ratio
C.P. = × S.P. 12 : 15 : 5 respectively. During construction, as a
result of inflation, the cost of labor increased by r
C.P. = × Ghȼ6,375 .= Ghȼ5,100.00
% and the cost of materials by 2r% while the
contractors charge remains the same. If after the
The manufacturer‟s cost price;
increase, the cost of labor was two – thirds of the
S.P. = Ghȼ5,100.00, P % = 27.5%, C.P. = ?
cost of materials, find:
C.P. = × S.P. a. the value of r,
C.P. = × Ghȼ5,100 = Ghȼ4,000.00 b. the new cost of the house.

Solution
Cost of materials : cost of labor = 3 : 5. a. Estimated cost = Ghȼ6,400.00
Total ratio = 3 + 5 = 8
Labor : materials : contractor‟ charge
i. Cost of materials = Ghȼ4,000 12 : 15 : 5
= Ghȼ1,500.00 Total ratio = 12 + 15 + 5 = 32
Labor charge = × Ghȼ6,400
ii. Cost of labor = × Ghȼ4,000 = Ghȼ2,400.00
= Ghȼ2,500.00 Materials charge = × Ghȼ6,400
= Ghȼ3,000.00
b. 20% of labor cost ( Ghȼ2,500.00)
Contractors charge = × Ghȼ6,400
= × Ghȼ2,500 = Ghȼ500.00
= Ghȼ1,000.00

The cost of labor; Increase in labor by r%

= Ghȼ2,500 + Ghȼ500 = Ghȼ3,000.00
= 2,400 + ( )
New cost price of materials and labor; = 2,400 + 24r
= Ghȼ(1,500 + 3,000) = Ghȼ4,500.00
Increase in materials by 2r% ;
New Profit of the manufacturer; = 3,000 + ( )

Baffour – Ba Series, Core Maths for Schools and Colleges Page 390
= 3,000 + 60r b. find the new price of the article if the original
profit is maintain.
After the increase,
⇒ Cost of labor = (cost of materials) Solution
a. Price of the article = 6,000.00
2,400 + 24r = (
Cost of material = 20%,
3(2,400 + 24r) = 2(3,000 + 60r) Cost of manufacture = 50%
7,200 + 72r = 6,000 + 120r Profit = (100 – 20 – 50)% = 30%
7,200 – 6,000 = 120r – 72r
1,200 = 48r Cost of materials = × Ghȼ6,000
r= = 25%
= Ghȼ1,200.00

b. The new cost of the house; Cost of manufacture = × Ghȼ6,000

Labor charge = 2,400 + ( ) = Ghȼ3,000.00
= 2,400 + 24r
= 2,400 + 24(25) Profit = × Ghȼ6,000 = Ghȼ1,800.00
= Ghȼ3,000.00
i. 10% fall in the cost of materials;
Materials charge = 3,000 + ( ) = Ghȼ1,200 –( )
= 3,000 + 60r
= Ghȼ1,080.00
= 3,000 + 60(25)
= Ghȼ4,500.00
10% increase in the cost of manufacture;
Contractors charge = Ghȼ1,000.00 = Ghȼ3,000 +( )
= Ghȼ3,300.00
The new cost of the house ;
= Ghȼ3,000 + Ghȼ4,500 + Ghȼ1,000
Material + Manufacture + Profit = Price
= Ghȼ8,500.00
Ghȼ1,080 + Ghȼ3,300 + Profit = Ghȼ6,000
Profit = Ghȼ6,000 – Ghȼ1,080 – Ghȼ 3,300
5. The price of an article is Ghȼ6,000.00. It is
Profit = Ghȼ1,620.00
made up as follows:
Cost of material = 20%,
Exercises 13.21
Cost of manufacture = 50% and the rest is
1. A manufacture sells an article to a retailer at a
profit. If the cost of materials falls by 10% and
profit of 10%. The retailer in turn sells the article
the cost of manufacture increases by 10% but the
to a customer at a profit of 25%. If the customer
price remains the same.
pays Ghȼ4,400.00, find the cost of manufacturing
a. Find:
the article.
i. the new profit,
ii. the percentage change in profit.
2. An article cost Ghȼx to produce and this is

Baffour – Ba Series, Core Maths for Schools and Colleges Page 391
divided between labor, materials and overheads 1. For example, average time for 6months equal
in the ratio 5 : 4 : 1 instalments = = months = 3.5 months
a. Express in terms of x the cost of materials for
If 12 months = 1 year
making 1,000 articles.
Labor cost increased by 8% and materials by ⇒3.5 months = ( )years
20%, but overheads are reduced by 20% through 6. Interest rate is calculated on the balance, using
increased production. formula; R =
b. Write down an expression for the new cost
per article. Worked Examples
c. Hence find the overall percentages increase in 1. The cash price of a mobile phone is
the cost of producing the articles. Ghȼ1,800.00. If Mr. Brown pays 44% of the cash
price as deposit and agrees to pay the rest in a
Hire Purchase
monthly installment of Ghȼ120.00 for 10 months,
It is the system by which the cost of goods or
find;
items is paid by installments, after initial deposit
i. the total amount he paid for the phone,
has been made. The deposit is usually a
ii. the interest paid,
percentage of the total cost of the goods or items.
iii. the approximate interest rate.
Note the following:
Solution
1. Deposits
Cash price = Ghȼ1,800.00
= Percentage deposited × Cash price
= × cash price Cash deposit = × Ghȼ1800 = Ghȼ792.00

2. Balance = Cash price – Deposit Balance (P) = Cash price – cash deposited
= Ghȼ1,800 – Ghȼ792
3. Total amount paid for goods = Ghȼ1,008.00
= Deposit + Total monthly installments
4. Interest Ghȼ120 monthly installments for 10 months;
= Total monthly installments – Balance = Ghȼ120 × 10 = Ghȼ1,200.00

5. When finding the rate of interest, average time Total amount paid;
is taken from the period where deposit is made = Deposit + Total monthly installment
and should be in years. This is because the first = Ghȼ792 + Ghȼ1,200 = Ghȼ1,992.00
half of the average time is used to settle the
principal/balance and the second half is used to ii. Balance = Ghȼ1,008.00
settle the interest. The addition of one resultsfrom Total monthly installment = Ghȼ1,200.00
the fact that from the beginning of the first
Interest = Total monthly installments – Balance
month, to the end of the last month, n is equal
Interest = Ghȼ(1,200.00 – 1,008.00)
ton + 1. It is also explained as the month of
I = Ghȼ192.00
deposit (1) + the months of instalment (n) = n +

Baffour – Ba Series, Core Maths for Schools and Colleges Page 392
Alternatively, = Ghȼ15,000 + Ghȼ48,990 = Ghȼ63,990.00
Interest = Total amount paid – cash price
Interest = Ghȼ1,992.00 – Ghȼ1,800.00 ii. Interest = Total payment – cash price
I = Ghȼ192.00 I = Ghȼ63,990 – Ghȼ60,000.00 = Ghȼ3,990.00

iii. R = ?, I = Ghȼ192, P = Ghȼ1,008 iii. R = ?, I = Ghȼ3,990, P = Ghȼ45,000

Average time, T = = months Average time, T = = months = ( ) years
If 12 months = 1 year R= = = 30.4%
⁄
months = ( ) = ( ) years
The approximate rate of interest is 30.4%

I= , 3. A man bought a boy‟s bicycle costing

R= = = = 41.5% Ghȼ24,000 on hire purchase. He ended up paying
⁄
6 % more than the cash price. If he made an
The approximate rate of interest is 41.5%
initial deposit of 25% of the cash price and then
2. The cash price of a gas cooker was paid the rest in six equal monthly installments,
Ghȼ60,000.00. A man paid 25% of the cash price find;
as deposit. He then paid Ghȼ8,165.00 a month for i. the initial deposit,
6 months. ii. the approximate rate of interest,
i. How much did he pay altogether for the iii. the amount of each installment.
cooker?
ii. Find the interest charged. Solution
iii. Find the approximate rate of interest. i. Cash price = Ghȼ24,000

Solution Deposit = 25% of the cash price

Cash price = Ghȼ60,000.00 = × Ghȼ24,000 = Ghȼ6,000.00
Cash deposit = 25% of Ghȼ60,000
Balance (P) = Ghȼ24,000 – Ghȼ6,000
= × Ghȼ60,000 = Ghȼ15,000.00
= Ghȼ18,000.00

Balance (P) = Cash price – cash deposit Total payment;

= Ghȼ(60,000 – 15,000) = 6 % more than the cash price
= Ghȼ45,000.00 = Ghȼ24,000 + (
⁄
*= Ghȼ25,500.00

Monthly installments of Ghȼ8,165.00 for 6

Interest = Total payment – Cash price
months = 6 × Ghȼ8,165 = Ghȼ48,990.00
= Ghȼ25,500 – Ghȼ24,000 = Ghȼ1,500.00
Total payment;
Balance (P) = Cash price – Cash deposit
= Deposit + Total monthly installment
= Ghȼ24,000 – Ghȼ6,000 = Ghȼ18,000
Baffour – Ba Series, Core Maths for Schools and Colleges Page 393
P = Ghȼ18,000, I = Ghȼ1,500, R = ? installment. If you wish to purchase this computer
T =( ) = 3.5 months = ( ) years sold at a cash price of Ghȼ1,480.00, calculate:
i. the amount you will pay at the end of every
R= = = 28.6% month.
⁄
ii.the total amount you will pay for the computer
iii. Amount of each installment; iii. the percentage profit the owner will earn if he
= = = Ghȼ3,250 bought the computer for Ghȼ1,500.00

Solution
Calculating the Monthly Installments at a Given
Cash price of the computer = Ghȼ1,480.00
Interest Rate per Anum
If after the initial deposit, the balance is paid at a Deposit = × Ghȼ1480 = Ghȼ370.00
given interest rate p. a. at a given monthly
installments; Balance (P) = Cash price – Deposit
I. Identify the balance as the principal (P),identify = Ghȼ1,480 – Ghȼ370 = Ghȼ1,110.00
the given rate (R%) andidentify the time (T) as
20% interest (I) on balance at 12months (1year),
the period of instalments in years.
I=?
II. Substitute the values of P, T (not the average)
P = Ghȼ1,110, T = 12months = 1yr, R= 20%
and R in I = to determine the interest on the
I= = = Ghȼ222.00
balance.
III. Find the sum of the principal and interest to
Amount to be paid in 12 months (total monthly
obtain the amount paid during the period of
installments)
installments. That is : A = P + I
A=P+I
A= Ghȼ1,110 + Ghȼ222 = Ghȼ1,332.00
IV. Divide the amount by the period of
installment to obtain the monthly installments. Amount to be paid every month (Monthly
( installments)
⇒Monthly installment =
= = = Ghȼ111.00
V. Obtain the total payment by finding the sum of
the initial depositand the total monthly ii. Total amount to be paid (Total payment)
installments. That is:
= Deposit + Total monthly installment
Total Payment = Deposit + Principal + Interest
= Ghȼ370 + Ghȼ1,332 = Ghȼ1,702.00
VI. Determine the percentage cost of increase
by the formula: P% = × 100% iii. C.P = Ghȼ1,500
S.P under hire purchase = Ghȼ1,702.00
Worked Examples Profit (P) = S.P – C. P
1. The conditions of hire purchase on the sale of a P = Ghȼ1,702 – Ghȼ1,500 = Ghȼ202.00
computer by Mr. Brown are as follows: initial
payment of 25% of the cash price and the balance P% = × 100% = × 100% = 13.5%
paid at 20% per anum in 12 months equal

Baffour – Ba Series, Core Maths for Schools and Colleges Page 394
2. Jones bought a car for Ghȼ6, 800.00. He later P = Ghȼ9,856 – Ghȼ6,800 = Ghȼ3,056.00
put it for sale at Ghȼ8,800.00. He agreed to sell it
to Ruby under the following hire purchase terms: P% = × 100% = × 100% = 44.9%
an initial payment of 20% of the price and the
balance paid at 15% per simple interest per anum 3. Mr. Amos bought a car for Ghȼ2.5 million
in twelve monthly equal installments. Calculate; cedis. He paid 40% of the cost and paid the rest in
a. the amount paid at the end of every month. equal monthly instalments. He took 8 years to
b. the total amount Ruby paid for the car. make full payment for the car. Interest was
c. the percentage profit Jones made on the cost charged at 18% simple interest. Calculate;
price of the car. i. the monthly installments,
ii. the total amount he paid for the car,
Solution iii. the percentage increase in the cost of the car.
a. Cash price = Ghȼ8,800.00
Deposit = 20% of Ghȼ8,800 Solution
= × 8,800 = Ghȼ1,760.00 Cash price of the car = Ghȼ2,500,000.00
Deposit = 40% of Ghȼ2,500,000.00
Balance = Cash Price – Deposit = × Ghȼ2,500,000 = Ghȼ1,000,000.00
= Ghȼ8,800, – Ghȼ1,760 = Ghȼ7,040.00
Balance (P) = Cash price – Deposit
15% interest on Balance at 12 months equal = Ghȼ2,500.000 – Ghȼ1,000,000
installments ; = Ghȼ1,500,000.00
P = Ghȼ7,040, T = 1, R = 15%, I = ?
I= = = Ghȼ1,056.00 18% interest (I) on balance for 8 years;
P = Ghȼ1,500,000,T = 8years,R = 18%
Amount paid in 12 months, I= = = Ghȼ2,160,000.00
(Total monthly installments), A = P + I
A = Ghȼ7,040.00 + Ghȼ1,056.00 = Ghȼ8,096.00 Total monthly installments;
A=P+I
Amount paid at the end of every month A= Ghȼ1,500,000 + Ghȼ2,160,000
(Monthly Installments) A = Ghȼ3,660,000.00
= = = Ghȼ674.67
8 years = 96 months
Monthly installments for 8 years;
b. Total amount paid(Total payment)
= Deposit + Total monthly installment = Ghȼ38,125.00
= Ghȼ(1,760.00 + 8,096.000) = Ghȼ9,856.00
ii. Total amount paid(Total payment);
c. C.P = Ghȼ6,800.00 = Deposit + Total monthly installment
S.P under hire purchase = Ghȼ9,856.00 = Ghȼ1,000,000 + Ghȼ3,660,000
Profit (P) = S.P – C. P = Ghȼ4,660,000.00
Baffour – Ba Series, Core Maths for Schools and Colleges Page 395
iii. C.P. = Ghȼ2,500,000.00 = Deposit + Total monthly installment
S.P under hire purchase = Ghȼ4,660,000 = Ghȼ8,190 + Ghȼ41,041.00 = Ghȼ49,231.00
Profit (P) = S.P – C. P
P = Ghȼ4,660,000 – Ghȼ2,500,000 iii. C.P = Ghȼ45,500.00
P = Ghȼ2,160,000.00 S.P under hire purchase = Ghȼ49,231.00
Profit (P) = S.P – C. P
P% = × 100% = × 100% = 86.4% P = Ghȼ45,500.00 – Ghȼ49,231.00 = Ghȼ3,731

4. Mr. White bought a plot of land whose cash P% = × 100% = × 100% = 8.2%
price was Ghȼ45,500.00. He made an initial
deposit of 18% of the cash price and paid the rest Exercises 13.22
at 30% rate of interest per annum in 4 months 1. The cash price of a DSTV decoder is
equal installment. Calculate: Ghȼ20,000.00. If a lady pays 25% of the cash
i. the monthly installments, price as deposit and agrees to pay Ghȼ3,000.00 a
ii. the total amount he paid for theland, month a for 6 months, find;
iii. the percentage increase in the cost of the land. i. the total amount she paid for the decoder,
ii. the interest paid.
Solution iii. the approximate interest rate.
Cash price of the land = Ghȼ45,500.00
Deposit = 18% of Ghȼ45,500.00 2. The proprietor of a certain boarding school
= × Ghȼ45,500 = Ghȼ8190.00 purchased some student mattresses for which the
total cash price was Ghȼ90,000.00. He made an
Balance (P) = Cash price – Deposit initial deposit of 33% of the cash price and
= Ghȼ45,500 – Ghȼ8,190 = Ghȼ37,310.00 agreed to pay the rest in monthly installments of
Ghȼ6,200.00 for 10 months. Find the
30% rate of interest (I) charged on balance for 4 approximate rate of interest.
months
P = Ghȼ37,310, T = 4 months = ( ) yrs R = 30% 3. The cash price of a VW saloon car was
Ghȼ14,000.00. If Mr. Brown paid 50% of the
I= = = Ghȼ3,731.00 cash price as deposit and then agreed to pay
Ghȼ1,700.00 every month for 6 months.
Total monthly instalments; i. How much did he pay altogether?
A=P+I ii. Find the interest charged.
A = Ghȼ37,310 + Ghȼ3,731= Ghȼ41,041.00 iii. Find the approximate rate of interest.

Monthly installments for 4 months; 4. How much interest is paid on Ghȼ4,500.00

= Ghȼ10,260.25 borrowed from a financial institution and repaid
in 30 months installment of Ghȼ220.00 per
ii. Total amount paid (Total payment); month?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 396
5. Mr. Okraku bought a car at Ghȼ8,800.00 and 9. The conditions of hire purchase on the sale of a
sold it later to Mr. Kwantabisa, for a cash price of house are as follows: initial payment of 45% of
Ghȼ10,200.00. If Mr. Kwantabisa paid 33 % of the cash price and the remaining paid at 25% per
the price and paid the balance at 20% simple anum in 6 months equal installment. If the cash
interest per anum in 6 monthly equal installments, price of the land is Ghȼ78,480.00, calculate:
calculate; i. the charge per month,
i. the amount paid every month, ii. the selling price of the house,
ii. the total amount paid by Mr. Kwantabisa, iii. the percentage profit charged on the cost
iii. the percentage profit Mr. Okraku made on the of the house.
cost price of the car.
10. A man buys a house at Ghȼ20,000.00. He
6. Mr. Dadzie agreed to buy a car that cost pays 30% of the cost out of his own resources and
Ghȼ25,000.00. He paid 40% of the cost price of takes a loan for the remainder at 2 %
the car and agreed to pay the rest at an interest of simple interest per annum. Calculate;
18% per anum for 12 months. Calculate; i. the total amount the man pays for the house if
i. the monthly installment, he takes 8 years to settle the loan ans:
ii. the total amount he paid for the car, Ghȼ22,800.00
iii. the percentage increase in the cost of the car. ii. the percentage increase in the cost of the house
to the man as a result of the loan.
7. A woman buys furniture for which the cash iii. his gain percent if, after settling the loan he
price is Ghȼ180,000.00. She pays 33 % cash renovates the house at a cost of Ghȼ2,200.00 and
deposit and agrees to pay Ghȼ13,500.00 a month then sells it for Ghȼ33,000.00
for 10 months. Find the approximate rate of
interest. 11. Mr. Ababio bought a refrigerator costing
Ghȼ160,000.00 from a consumer credit union on
8. A company was granted a loan of hire purchase. He made an initial deposit of
Ghȼ750,000.00. The loan was to be repaid in four Ghȼ100,000.00 and agreed to pay the rest in 15
quarterly monthly installments of Ghȼ220,000.00 monthly installments of Ghȼ5,000.00 each. What
starting three months after the loan has been is the approximate rate of interest charged?
granted. Find the approximate rate of interest.

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14 MODULO ARITHMETIC Baffour – Ba Series

Defining Modulo Assuming the time is 8 O‟ clock now. In the next

Modular arithmetic, also known as remainder 7hours,which means moving in the clockwise
arithmetic is a type of arithmetic in which an direction 7 hours from 8 (i.e. 8hr + 7hr), the time
integer is represented by its remainder when will be 3 O‟ clock instead of 15 O‟ clock.This
divided by another integer. In other words, the isexplained as: 8hr + 7hr = 15hr – 12hr = 3hrs.
modulo of any integer is found by dividing the Since the numbers maximize at 12hours and
integer by the given modulo and the remainder begin the cycle, any sum greater than 12 produces
identified as the answer. an answer which is the difference between the
sum and 12, conditioned to be less than 12,
Modulo is written as “mod” for short. For otherwise continue tosubtract 12, till a number
instance, 23 (mod 7) means the remainder when less than 12 is obtained
23 is divided by 7. That is:
1. 23 7 = 3 × 7 + 2 = 2 Similarly, if it is 8 O‟ clock now, 9 hours ago, the
2. 23 7 = 3 R 2 = 2 OR time was 11 O‟clock, explained asmoving in the
3. 23 – 7 – 7 – 7 = 2. anti – clockwise direction 9 hours from 8 (8hrs –
9 hrs.), the answer conditioned to be a positive
Similarly, 35 (mod 8) could fit any of the number less than 12, otherwise, add 12
following: successfully until the first positive number is
1. 35 8 = 4 × 8 + 3 = 3 obtained. i.e. 8hrs – 9 hrs= -1 + 12 = 11 hrs
2. 35 ÷ 8 = 4 = 3
Thus, on theanalogue clock,the hands of the clock
3. 35 – 8 – 8 – 8 – 8 = 3
pass over the numbers, wrap up at 12, and
thereafter, begin the cycle again. The clock
Note that a particular interest is attached to the
therefore operates under modulo 12. The
value of the remainder and not the quotient. Thus,
difference is that, 12 is used in place of 0.
23(mod 7) = 2 and 35 (mod 8) = 3
The fact that one can move back on the face of
the clock to determine a previous time and also
Linear and Cyclic Variables
move forward to pre – determine the time after
1. The Analogy of the Clock
a given hours, the numbers are said to be linear
The numbers displayed on the analog clock
variables. The fact that the numbers cycle around
representing the hours are:
themselves andespecially around a maximum
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 point (12), makes them cyclic variables

11 12 1 Worked Examples
These aredisplayed on 10 2
1. If it is 9 O‟ clock now, what will be the time
the face of the clock below: 9
8
∙ 3
after the following hours?
4 i. 5hours ii.11hours iii. 23 hours
7 5
6

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Solution If P = {0, 1, 2, 3, 4, 5, 6}, then:
i. 9hrs + 5hrs = 14hrs – 12hrs = 2 O‟clock I. x number of days after P is determined as
ii. 9hrs + 11hrs = 20hrs – 12hrs = 8 O‟clock P + x. For e.g. 5 days after Wednesday is
iii. 9hrs + 23hrs = 32hrs – 12hrs – 12hrs determined as 3(Wed) + 5 = 8 – 7 = 1 represented
= 8 O‟clock by Monday
II. x number of days before P is determined as P –
2. If it is 4 O‟ clock now, what was the time the x. For e.g. 4 days before Wednesdayis determined
following hours before 4 O‟clock? as 3(Wed) – 4 = -1 + 7 = 6 representing Saturday
i. 13 hours ii. 27 hours iii. 35 hours III. Because P can move in both directions, it is
called a linear variableand because P moves in a
Solution definite cycle, it is called a cyclic variable
i. 4hrs – 13 hrs = - 9hrs + 12 = 3 O‟ clock
ii. 4hrs – 27hrs = -23hrs + 12 + 12 = 1 O‟ clock Worked Examples
iii. 4hrs – 35hrs = -31 + 12 + 12 + 12 = 5 O‟clock 1. What day of the week is it 6 days after
Saturday?
Exercises 14.1
A. Assuming the time is 6 O‟ clock now, what Solution
will be the time in the following hours 6 days after Saturday;
1.7 hours 2.9 hours 3.18 hours = 6 (sat) + 6 = 12 – 7 = 5(Fri)
4.26 hour 5.37 hours 6. 44 hours 6 days after Saturday is Friday

B. If it is 10 O‟ clock now, what was the time the 2. What day is 28 days after Wednesday?
following hours before 10O‟clock?
1. 14 hours 2.25 hours 3. 52 hours Solution
28 days after Wednesday;
2. Analyzing the Days of the Week = 3 (Wed) + 28 = 31 – 7 – 7 – 7 – 7 = 3(Wed)
Let the days of the week namely: Sunday, 28 days after Wednesday is Wednesday
Monday, Tuesday, Wednesday, Thursday, Friday
and Saturday be represented by the numbers 0, 1, 3. What day was 17 days before Friday?
2, 3, 4, 5, 6 respectively.
Below is a circular representation of the days and Solution
their numbers. 17 days before Friday;
P+x
P–x Sun = 5 (Fri) – 17 = -12 + 7 + 7 = 2 (Tue)
Mon 17 days before Friday was Tues
Sat 0
6 1
Exercises 14.2
2 Tue
Fri 5 A. Find the modulus in each of these;
4 3 1.The system of the days of the week
Wed 2. The system of the month of the year
Thur

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B. If today is Tuesday, what day will it be in Worked Examples
the next: 1. What month was it 88 months, before October?
1. 16 days 2.39 days 3. 73days
Solution
C. If today is Friday, what was the day, the Oct ( 10) + 88 = 98 ( mod 12)
following days before Friday; = 98- 12 – 12 -12 – 12 – 12 – 12 – 12 – 12
1. 17 days 2. 33 days 3. 71 days = 2 (Feb)

3. Analizing the Months of the Year Alternatively;

Below is the table of the list of months of the year Oct ( 10) + 88 = 98 ( mod 12)
and the number of days; 98 12 = 8
R = 2 ( Feb)
Month Days Month Days
Jan 31 July 31 2. What month is it 94 months after February?
Feb 28/29 Aug 31
(leap yr)
Mar 31 Sept 30 Solution
Apr 30 Oct 31 Feb (2) + 94 = 96 (mod 12) = 0 (December)
May 31 Nov 30
June 30 Dec 31 3. In a particular year, August 11 was a Tuesday.
What day was 3rd October the same year?
Assign numbers to the months as shown below:
Solution
Jan (1) Feb (2) Mar (3) Apr (4) Tue (2) + 20 days of Aug + 30 days of Sept + 3
May (5) Jun (6) Jul (7) Aug (8) days of Oct
Sept (9) Oct (10) Nov (11) Dec (0) = 55 mod 7 = 6 (Representing saturday)

Similar to the number of days, we move forward 4. In a particular leap year, 19th March, was a
(clockwise direction) to determine a month Friday. What day was it December 21st the
ahead (after a given month) and backward previous year.
(anticlockwise direction) to determine past
months (before a given month). For example, Solution
1. 9 months after September is determined as: Fri (5) – 19 days of Mar – 29 days of Feb – 31
Sept (9) + 9 days of Jan – 10 days of Dec
= 18 (mod 12) = 18 – 12 = 6 (representing June) = - 84 (mod 12)
= - 84 + 12 + 12 + 12 + 12 + 12 + 12 + 12
2. 25 months before May is determine as : = 0 (Representing Sunday)
May (5) – 25
= -20 (mod 12) = - 20 + 12 + 12 = 4 5. The days of a non leap year are numbered in
(representing April) the scale of seven: i.e. 5th Jan is day 5, 8th Jan is
day 11, 28th Jan is day 40.

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i. What number is given to: Day 52
a. 20th Feb b. 31st Dec? = 1(Tue) + (5 × 7) + 2 = 38(mod 7) = 3 (Wed)
ii. If 1st January is a Tuesday, what day of the
week is: b. Day 142
a. day 52 b. day 142? = 1 (Tue) + (1 × 7 × 7) + (4 × 7) + 2
= 1 (Tue) + 49 + 28 + 2 = 80 (mod 7) = 3 (wed)
Solution
i. a. 5th Jan is day 5, ⇒5(base 7) = 0 R 5 = 5 Exercises 14.2B
1. The month is now May. What month will it be
8th Jan is day 11, ⇒8 (base 7) = 1 R1 = 11 in; a. 33 months time? b. 119 months time?

28th Jan is day 40, ⇒28 (base 7) 4 R 0 = 40 2. What was the month, the following months
before the given months?
⇒ 20th Feb a. 41 months before November?
= 28 days (Jan) + 3 days(Jan) + 20 days (Feb) b. 103 months before June?
= 51 days
= 51 (base 7 ) = 7 R 2 = 72 3. In a certain leap year, April 3 was a Monday.
But the scale is 7 ⇒ 7 = 10 What day was the following days;
Therefore, 72 in scale 7 = 102. a. 15th March of the same year;
The number given to 20th Feb is 102 b. January 24th , the same year;
c. November 1, the previous year;
b.31stDec d. June 14, the same year.
= 28 days (Jan) + 3 days (Jan) + 28 days (Feb) +
31 (Mar) + 30 (Apr) + 31 (may) + 30 (Jun) + 31 4. Jenifer was born on Friday 15th February, 2012
(Jul) + 31(Aug) + 30(Sept) + 31(Oct) + 30 (Nov) which happened to be a leap year.
+ 31 (Dec) a. If Martha was born on 29th July 2013, on which
= 365days day of the week was she born?
= 365 (base 7) b. If Joe was born 73 days after Martha, on which
= 52 (base 7) R 1 day was he born?
=7R3R1 c. If Linda was born on 1st October, 2011, find
= 731 the day on which she was born.
d. If someone is born on 10th May 2020, find the
But the scale is 7 ⇒ 7 = 10 exact day in which the birthday falls
So 731 = 1031
The number given to 31st Dec is 1031 5. The days of a non leap year are numbered in
the scale of seven: i.e. 5th Jan is day 5, 8th Jan is
ii. a. If 1st Jan is a Tues, day 11, 28th Jan is day 40.
⇒ 1(mod 7) = 0 R 1 = 1 i. What number is given to:
Tuesday = 1 a. 3rd March b. 16th Oct?

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ii. If 1st January is a Friday, what day of the week The sequence of days can also be generated as
is: a. day 60 b. day 100? follows:
1 + 4(mod 7) = 5 → 5 + 4(mod 7) = 2 →
Other Applications 2 + 4(mod 7) = 6 → 6 + 4(mod 7) = 3 →
Modulo is applicable to situations where an 3 + 4(mod 7) = 0 → 0 + 4(mod 7) = 4 →
activity occurs at regular intervals. For example, 4 + 4(mod 7) = 1
organization of market days and the
days at which a group of people take turns to Sequences of days on duty from Monday
engage in a daily or timely activities. 1, 5, 2, 6, 3, 0, 4, 1
I. Identify the particular activity, whether market
days or duty days. Diagram of the sequence
II. Represent the days of the week as follows: Sun Su
= 0, Mon= 1, Tue = 2, Wed = 3, Thur = 4, Fri = Sa x0 Mo
5, Sat = 6 x 6 y 1 n

III. Make a “sequence of days” on which the y  2 Tu

Fr 5
activity repeats or happens.
4 3
IV. Represent the days of the week on the face of We
Th
the clock and match according to the sequence. Do x
From the sequence, his fourth duty will be on
V. Use the “sequence of days” to answer all m y
Saturday (6).
questions. ain
ii. CountingCofour times after Monday (1) his next
Worked Examples
four duties -will fall on Wednesday(3)
1. Four security officers (Nii, Noi, Ayittey and do
Odame) take turns towatch a factory. Nii is on m sequence,
iii. From the
duty on a Monday this week. Difference ain
between the days = 4 days
i. When will he be on duty the fourth time? Monday (1) to Monday (1) = 7 duties
ii. When will his next four duties fall? Number of days he will be on duty on Monday
iii. How many weeks will it be before he is next = 4 days × 7 = 28 days
on duty on a Monday?
iv. How many days will he be on duty on Number of Weeks;
Saturday after Monday? If 7 days = 1 week,
28 days = 4 weeks
Solution
i. Let Sun = 0, Mon = 1, Tue = 2, Wed = 3, iv. Difference between the days = 4 days
Thu = 4, Fri = 5, Sat = 6 Monday (1) to Saturday (6) = 3 duties
Number of days he will be on duty on Saturday
Sequences of days on duty from Monday after Monday = 4days × 3 = 12 days
1, 5, 2, 6, 3, 0, 4, 1
2. At a certain village, market days are organized
Alternatively, every ten days. If a market day falls on Tuesday;

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i. on which day will the fourth market day fall? 2. A town has a market day everynine days. If a
ii. how many days will it take a market day to fall market day falls on Tuesday this week, find how
on Tuesday again? many days it will be before a market day falls on;
i. Thursday ii. Wednesday iii. Monday
Solution
i. Let Sun = 0, Mon = 1, Tue = 2, Wed = 3, 3. There are five traffic officers at a station. One
Thu = 4, Fri = 5, Sat = 6 traffic officer takes turn to be on duty every day
Sequence of his duty day from Tuesday of the week. Mawusi, who is one of the traffic
2, 5, 1, 4, 0, 3, 6, 2 officers, is on duty on Friday this week;
i. what will be his next day on duty?
Diagram of the sequence ii. how many weeks will it be before Mawusi, is
Sun
on duty on a Friday again?
Mo
Sat 0
n
6 1 4. In a village, market days are held every four
days. This week, a market day falls on a Monday.
Fri 5
 2 Tue
i. List the sequence of market days.
4 3 ii. When will the next market day fall after
Thu We tuesday?
r d iii.When will the next fivemarket days fall after
i. From the sequence, the fourth market day will monday?
fall on Thursday (4) iv. When will the tenth market day fall after
monday?
ii. From the sequence,
v. How many days will it take a market day to fall
Difference between the days = 10 days
on Monday again?
Tuesday (2) to Tuesday (2) = 7 market days
Number of days a market day will be on Tuesday
5. In a particular year, August 16 was a
= 10 days × 7 = 70 days
Wednesday. On which day of the week was
October 11, the same year?
3. At a transport station, a bus leaves every 2
hours. If the first bus leaves at 5: 00 am, when
6. In Boateng‟s nursery, some seeds were sown
will the 10th bus leave?
on Saturday. The seedlings were transplanted 23
days later. If the plants were pruned after 86 days,
Solution on which day of the week were they pruned?
5 + (2 × 9) mod 12
= 5 + 18 (mod 12) = 23 (mod 12) = 11: 00 pm
The Modulo Equation
Take =6 for instance. This is expressed in
Exercises 14.3
1. A train regularly leaves a station at every 3 modulo arithmetic as 20 (mod 3) = 2 (mod 3).
hours. If the first train leaves at4 : 00 pm, at what But the mod is supposed to be at only one side of
time will the eighth train leave? the equation. Therefore, the statement: 20 (mod

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3) = 2 (mod 3) can be simplified as shown below, The Modulo of a Negative Number
any of which is acceptable: For all negative numbers, the modulo is
I. 20 (mod 3) = 2: keeping the mod at the left side determined by addingthe modulo number
of the equation successfully until a positive number is obtained
II. 20 = 2 (mod 3): keeping the mod at the right as answer.
side of the equation. These are called equivalent
modulo Worked Examples
Simplify each of the following in the modulo
Integers for a Given Modulo given.
Take for instance, the division of any integer by 1.-8(mod 3) = 3. - 61 (mod 9) =
4. It is seen that the possible remainders, when 2. -38 (mod12) = 4. -117 (mod 36) =
any integer is divided by 4 are 0, 1, 2, 3. Thus,
the set of integers (remainders), representing Solution
division of integers by 4 is reduced to {0, 1, 2, 3}. 1. – 8(mod 3) = -8 + 3 + 3+ 3 = 1
The divisor, (4) in this case, is called the modulus -8(mod 3) = 1
and the arithmetic is said to be in modulo 4.
Modulo 4 can therefore take one of the values {0, 2. -38 (mod12)
1, 2, 3}. Likewise, modulo 6 can take one of the = -38 + 12 + 12 + 12 + 12 = 10.
values {0, 1, 2, 3, 4, 5} -38 (mod12) = 10

In general, modulo n can take the values 0 to (n – 1) 3. -61 (mod 9)

= -61 + 9+ 9 + 9+ 9+ 9 + 9 + 9+ 9 = 2
Exercises 14.4 -61 (mod 9) = 2
Write out the set of integers for each of the
following modular system: 4. -117 (mod 36)
1. mod 7 2. mod 8 3. mod 9 4. mod 11 = -117 + 36 + 36 + 36 + 36 = 27
-117 (mod 36) = 27
Hints for Finding the Modulo
I. If the dividendis less than the divisor (modulo Exercises 14.5
number), the dividend remains the answer. For A. Simplify:
e.g. 5(mod 7) = 5, 3(mod 8) = 3 etc 1. -68 (mod 7) =. 5. -57(mod 13) =
II. Modulo 5 of all numbers ending in 0, 1, 2, 3, 2. -17 (mod 4) = 6. -88(mod 15) =
4 is the last or ending digit. e.g. 12(mod 5) = 2, 3. -33 (mod 11) = 7. -221(mod36) =
170(mod 5) = 0, 1094(mod 5) = 4 etc 4. -105 (mod 44) = 8. -175(mod 30) =

III. Modulo 5 of all numbers ending in5, 6, 7, 8, 9 Equivalent or Congruent Moduli

is found by subtracting 5 from the last or ending 1. In modulo arithmetic, two or more integers are
digit. e.g. 19(mod 5) = 9 – 5 = 4. Therefore, said to be equivalent if they give the same
19(mod 5) = 4 remainder under the same mod. For e.g,

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23(mod 7) = 2 and 16(mod 7) = 2 are said to be A. Which of the following pairs of numbers
equivalent written as: are equivalent in the modulo specified
23(mod 7) 16(mod 7) = 2 1. 17 and 31 (mod 7) 2. 19 and 25 (mod 6)
3. 67 and 87 (mod 9) 4. 42 and 63 (mod 8)
2. Similarly, in 8(mod 5) = 3, the equivalent is 5. 44 and 80 (mod 9) 6. 57 and 96 (mod 13)
written as8 3(mod 5)
Modulo Operations (Arithmetic)
Worked Examples (Addition, Subtraction, Multiplication and
1. Simplify the following: Division)
i. 14 (mod 4) ii. 20 (mod 8) The act of performing the operations: Addition,
Subtraction, Multiplication and Division
Solution involving modulo are called Simplification of
i. 14 (mod 4) = 14 4 = 3 = 2 modulo arithmetic.
14 2 (mod 4)
To simplify modulo arithmetic:
ii. 20 (mod 8) = 20 8=2 =4 I. Perform the operation (+, - , ×, )
20 4 (mod 8) II. Convert the answer to the given modulo

2. Determine whether the following pairs of Worked Examples

numbers are equivalent in the moduli given Simplify the following:
i. 29 and 77(mod 6) 1. 14 + 4 (mod 4) = 5. 13 × 33(mod 5) =
ii. 19 and 30 (mod 7) 2. 13 + 33(mod 8) = 6. 300 12 (mod 11) =
3. 26 – 9 (mod 6) = 7. 78 2(mod 4) =
Solution 4. 12 – 34 (mod 8) =
i. 29(mod 6) = 29 6=4 =5
Solution
77(mod 6) = 77 6 = 12 = 5 1. 14 + 4 (mod 4) = 18(mod 4) = 2 or
29(mod 6) = 77(mod 6) = 5 = 2 (mod 4)
2. 13 + 33(mod 8) = 46 (mod 8) = 6 or
Therefore 29 and 77(mod 6) are equivalent = 6 (mod 8)
3. 26 – 9 (mod 6) = 17 (mod 6) = 5 or
ii. 19(mod 7) = 19 7=2 =5 = 5(mod 6)
4. 12 – 34 (mod 8) = -22(mod 8)
30(mod 7) = 30 7=4 =6 = -22 + 8 + 8 + 8
19(mod 7) 30 (mod 7) = 2 or = 2 (mod 8)
Therefore 19(mod 7) 30 (mod 7) 5. 23 × 2 (mod 9) = 46(mod 9) = 1 or
are not equivalent = 1 (mod 9)
6. 13 × 33(mod 5) = 429 (mod 5) = 4 or
Exercises 14.6 = 4 (mod 5)
7. 300 12 (mod 11) = 25(mod 11) = 3 or

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 = 3(mod 11) 1. If 2 = 12 (mod x), find the value of x
8. 78 2(mod 4) = 39(mod 4) = 3 or
= 3 (mod 4) Solution
2 = 12 (mod x),
Exercises 14.7 I. Difference between the numbers
A. Simplify the following: 12 – 2 = 10
1. 57(mod 7) 2. 94(mod 8) 3. 66(mod 14) II. Factors of 10 = {1, 2, 5, 10}
4. 32(mod 12) 5. 98(mod 15) 6. 72(mod 17) III. By substitution,
0 = 12(mod 1), x 1
B. Simplify in the modulo given; 0 = 12(mod 2), x 2
1. -8 + 11 (mod 13) = 2 12(mod 5), x=5
2. -50 + 99 (mod 5) = 2 12(mod 10), x = 10
3. 120 + 77 (mod 23) = By comparison,
4. 331 + 111 (mod 35) = 12(mod x) = 12 (mod 5) = 12 (mod 10).
5. 168 + 133 (mod 16) =  x = 5 or x = 10
6. 14 + 63 + 28(mod 8) = 2. Find the value of x in 24(mod x) = 3

C. Find the following products:

Solution
1. 13 × 11 (mod 7) = 4. 89 × 37 (mod 55) =
Difference between the numbers
2. 64 × 17 (mod 28) = 5. 27 × 19 (mod 26) =
24 – 3 = 21
3. 37 × 32 (mod 42) = 6. 29 × 39 (mod 22) =
Factors of 21 = {1, 3, 7, 21}
By substituting the factors,
The Modulo Number in a Given Equation
24(mod 1) = 0, x
When the modulo number is represented by a
24(mod 3) = 0, x
variable as in 2 = 12 (mod x), the value of the
24(mod 7) = 3, x=7
variable is found by going through the following
24(mod 21) = 3, x = 21
process:
24(mod x) = 24(mod 7) = 24 (mod 21) = 3.
I. Find the difference between the numbers.
 x = 7 or x = 21
II. Find all the factors of the difference identified.
III. Substitute each factor of the difference in the
3. Determine the modulo in which 6 + 8 = 4 was
modulostatement provided and identify the one
performed.
that makes the statement true as the value of the
variable.
Solution
Let the mod in which the operation was
Note:
performed be x
The answer obtained from the difference satisfies
6 + 8 (mod x) = 4
the statement and therefore is a truth set of the
14(mod x) = 4
variable under no given conditions
x is a factor of 14 – 4 = 10
Worked examples Factors of 10 = {1, 2, 5, 10}

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By substitution, Again, in the statement 5x = 2 (mod 4) for
14 (mod 1) = 0, x instance, considering the R.H.S. of the equation,
14(mod 2) = 0, x the modulo number, 4 is greater than 2, and this
14(mod 5) = 4, x=5 gives the impression that the divisor is 4 as usual,
14 (mod 10) = 4, x = 10 the remainder is 2, and the dividend is 5x at the L.
⇒x= 5 orx =10 H. S. The statement 5x = 2 (mod 4) can therefore
Therefore, the operation was performed in be written as 5x (mod 4) = 2.
modulo 5 or modulo 10.
Worked Examples
4. Perform 50 2 = 7(mod n), where 1 n 18 1. If 7x = 3 (mod 6), find x

Solution Solution
Under the condition 1 n 18, n = 18 Method 1
50 2 = 7(mod n) In mod 6, x can take one of the values of 0, 1, 2,
25 = 7(mod n) 25(mod n) = 7 3, 4, 5. By substitution, investigate to get the
25 – 7 = 18 value of x that satisfies7x =3mod 6
Factors of 18 = 1, 2, 3, 6, 9, 18 When x = 0, 7(0) = 0 (mod 6)
When n = 1, 25(mod 1) = 0, n 1 When x = 1, 7(1) = 7(mod 6) = 1
When n = 2, 25(mod 2) = 3, n 2 When x = 2 7(2) = 14(mod 6) = 2
When n = 3, 25(mod 3) = 4, n 3 When x = 3 7(3) = 21(mod 6) = 3
When n = 6, 25(mod 6) = 1, n 6 When x = 4 7(4) = 28(mod 6) = 4
When n = 9, 25(mod 9) = 7, n 9 When x = 5 7(5) = 35(mod 6) = 5
Under the given condition, x :x = 9  x = 3, satisfies 7x = 3(mod 6)

Exercises 14.8 Method 2

Determine the modulo: The statement 7x = 3 (mod 6) can also be written
1. 13 = 3 (mod x) 2. 25 = 4(mod x) as 7x (mod 6) = 3
3. 54 = 10(mod x) 4. 42 =12(mod x) Mod 6 = {0, 1, 2, 3, 4, 5}
5. 38 = 5(mod x) 6. 60 =15(mod x) when x = 0
7(0) mod 6 = 0 (mod 6) = 0
Finding the Number under Modular when x = 1,
Operation 7(1) mod 6 = 7(mod 6) = 1
The number whose modulo is being operated may when x = 2,
be represented by a variable as in 5x = 2(mod 4) 7(2) mod 6 = 14(mod 6) = 2
Take note of the fact that modular n uses the When x = 3,
numbers 0 to (n – 1). For instance modulo 6 uses 7(3) mod 6 = 21(mod 6) = 3
the numbers 0, 1, 2,3,4,5. when x = 4,
By substitution, the values of 0 to (n – 1) that 7(4) mod 6 = 28(mod 6) = 4
makes the statement true is the value of the when x = 5,
variable. 7(5) mod 6 = 35(mod 6) = 5

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 x = 3, satisfies 7x (mod 6) = 3 5. Find the truth set of 34 = x (mod 6)

2. Solvex + 9 = 3 (mod 5) Solution

The statement 34 = x (mod 6) can be written
Solution as 34(mod 6) = x.
x + 9 = 3 (mod 5) can be written as But 34(mod 6) = 4
x + 9 (mod 5) = 3 x = 3
mod 5 = {0, 1, 2, 3, 4},
Substitute the values {0, 1, 2, 3, 4}, in Exercises 14.9
x + 9 (mod 5) = 3 A. Find the solution set in each case;
1. 2x = 2(mod 4) 5. 3x = 1(mod 4)
When x = 0, 2. 2x + 1 = 1(mod 3) 6. 3x + 4 = 5(mod 7)
2
0 + 9 (mod 5) = 9 (mod 5) = 4 3. x = 4 (mod 6) 7. x3 = 3(mod 5)
When x = 1, 4. x2 + 1 = 0(mod 5) 8. x2– 5 = 4 (mod 8 )
1 + 9 (mod 5) = 10 (mod 5) = 0
When x = 2,
B. Find the values of the variables
2 + 9 (mod 5) = 11 (mod 5) = 1
1. 7 × 3 = y (mod 8) 2. 6 × 5 = y (mod 7)
When x = 3,
3. 3 × 3 = y (mod 5) 4. 5 × 5 = y (mod 6)
3 + 9 (mod 5) = 12 (mod 5) = 2
5. 4 × 3 = y (mod 5)

When x = 4,
C. Find the solution set of the following;
4 + 9 (mod 5) = 13 (mod 5) = 3
1. 6 + x = 2(mod 8) 2. 8 + x = 3(mod 12)
 x = 4, satisfies x + 9 (mod 5) = 3
3. 4 + x = 3(mod 12) 4. 4 + x = 2(mod 6)
5. x + 5 = 3(mod 7)
3. Find the value of x in 52 + x2 = 2 (mod 4)
Modulo ArithmeticTables
Solution
Modular arithmetic tables are usually constructed
52 + x2 = 2 (mod 4) can also be written as
under addition and multiplication ⊗ for a
52 + x2 (mod 4) = 2
given modulo.
mod 4 = { 0, 1, 2, 3}
When x = 0,
Given modulon, the table is constructed using the
52 + (0)2 (mod 4) = 25 (mod 4) = 1
numbers 0 to (n – 1). Thus, for modulo 5, the set
When x = 1,
of values are {0,1, 2, 3, 4} and for modulo 6, the
52 + (1)2 (mod 4) = 26 (mod 4) = 2
range of values are 0 , 1, 2, 3, 4, 5. For large
When x = 2,
modulo, a set of values may be given for
52 + (2)2 (mod 4) = 29 (mod 4) = 1
operation.
When x = 3,
52 + (3)2 (mod 4) = 34 (mod 4) = 2 On the table, the range of values for the mod or
Therefore x = 1 and x = 3 satisfies the statement the 0 to (n – 1) values occupy the first column
52 + x2mod 4 = 2 and the first row, whilst the operator, whether +

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or × is circled to look like or ⊗ respectively,
and placed at the left top corner of the table as i. From the table:
shown below: (2 5) = 1 and (5 2) = 1
(2 5) (5 =1
0
0 ii. 4= 2
iii. 4 =0

⊗ 0 From the table,

0 When k = 0, 4 =4
When k = 1, 4 =5
Thereafter; When k = 2, 4 =0
I. Perform the operation, When k = 3, 4 3= 1
II. Divide the answer by the mod When k = 4, 4 =2
III. Record only the remainder in the boxes or When k = 5, 4 =3
cells to complete the table
4 =4 =0
Worked Examples ⇒k = 2
1. Write out the replacement set or elements of The truth set is { k : k = 2}
the following modulo
a) Modulo 2 b) modulo 5 c) modulo 8 2. Construct a table for multiplication ⊗ modulo
6 for the set S = {1, 2, 3, 4, 5.}
Solution Use the table to find:
1. a. Mod 2 = {0, 1} i. 4 ⊗ 2, ii. (3 ⊗1) ⊗(5⊗ 2),
b. Mod 5 = {0, 1, 2, 3, 4} iii. (1 ⊗4) ⊗ (5⊗5) ⊗(2⊗4),
c. Mod 8 = {0, 1, 2, 3, 4, 5, 6, 7} iv. the truth set of m⊗m = m.

2. Draw an addition table for arithmetic modulo Solution

6. Use your table to find;
i. (2 5) (5 2) ii. 4 4 ⊗ 1 2 3 4 5
ii. If 4 k = 0, find the value of k 1 1 2 3 4 5
2 2 4 0 2 4
Solution 3 3 0 3 0 3
4 4 2 0 4 2
0 1 2 3 4 5 5 5 4 3 2 1
0 0 1 2 3 4 5
1 1 2 3 4 5 0 From the table:
2 2 3 4 5 0 1
i. 4 ⊗2 = 2
3 3 4 5 0 1 2
4 4 5 0 1 2 3 ii. (3 ⊗1) = 3 and (5⊗2) = 4
5 5 0 1 2 3 4
Therefore (3 ⊗ 1) ⊗(5⊗2) = 3 ⊗4 = 0

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iii. (1 ⊗4) = 4, (5⊗5) = 1 and (2⊗4) = 2 1⊗4 3
Therefore (1⊗4) ⊗(5⊗5) ⊗(2⊗4)
= (4 ⊗1⊗2) When y = 2
= 4⊗1 = 4 2 ⊗ (2 + 3) = 2 + 2
= 4⊗2 = 2 2⊗5 4
When y > 2, the statement is undefined.
iv. m⊗m = m Therefore y = 2, satisfies the statement.
From the table,
When m = 1, 1 ⊗1 = 1 4. a. Draw the multiplication⊗table for arithmetic
When m = 2, 2 ⊗2 = 4 modulo 7.
When m = 3, 3 ⊗3 = 3 b. Using the table,
When m = 4, 4 ⊗4 = 4 i. state with reasons, whether or not the operation
When m = 5, is commutative,
5 ⊗5 = 1, 1 ⊗1 = 1, 3 ⊗3 = 3 and 4⊗4 = 4 ii. evaluate (4 ⊗6) ⊗(5⊗ 4),
⇒m = 3 or m = 4 iii. find the truth set of n⊗n = n.
Truth set = {m : m = 1, 3 orm = 4}
Solution
3. Draw a multiplication table in modulo 6 and
use your table to find the truth set following ⊗ 0 1 2 3 4 5 6
equations; 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6
i. x⊗x, ii. y⊗ ( y + 3 ) = y + 2, 0 2 4 6 1 3 5
2
3 0 3 6 2 5 1 4
Solution 4 0 4 1 5 2 6 3
5 0 5 3 1 6 4 2
⊗ 0 1 2 3 4 5 6 0 6 5 4 3 2 1
0 0 0 0 0 0 0
1 0 1 2 3 4 5 i. Method 1
2 0 2 4 0 2 4
3 0 3 0 3 0 3 ⊗ 0 1 2 3 4 5 6
4 0 4 2 0 4 2 0 0 0 0 0 0 0 0
5 0 5 4 3 2 1 1 0 1 2 3 4 5 6
2 0 2 4 6 1 3 5
ii.y⊗ ( y + 3 ) = y + 2 3 0 3 6 2 5 1 4
From the table, 4 0 4 1 5 2 6 3
When y = 0 5 0 5 3 1 6 4 2
0 ⊗ (0 + 3) = 0 + 2 6 0 6 5 4 3 2 1
0⊗3 2
The table is symmetrical about the leading
When y = 1 diagonal. Therefore, the operation ⊗ is
1 ⊗ (1 + 3) = 1 + 2 commutative.

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Method 2 When n = 3
i. From the table, 3 ⊗ (3 3) = 3 ⊗ 6 = 3
2 ⊗ 3 = 6 and 3⊗2 = 6 When n = 6
2⊗3=3⊗2=6 6 ⊗ (6 3) = 6 ⊗ 9 = 9
Therefore the operation ⊗ is commutative . When n = 9
9 ⊗ (9 3) = 9 ⊗ 12 = 3
ii. From the table, When n = 12
(4 ⊗6) ⊗(5⊗ 4) 12 ⊗ (12 3) = 12 ⊗ 0 = Not defined
=3 ⊗ 4 = 5 ⇒n⊗ ( n 3) = 9 ⊗ (9 3) = 3
Therefore n = 3 or n = 4
iii. n⊗n = n
From the table, Exercises 14.10
0 ⊗ 0 = 0, 1 ⊗ 1 = 1, 2⊗ 2 = 4, 1. i. Construct a table for multiplication ⊗modulo
3 ⊗ 3 = 2, 4 ⊗ 4 = 2, 5⊗ 5 = 4 9 on the set P = {2, 3, 4, 5,7}
6 ⊗ 6 = 1, ii. Find the truth set of x⊗x = 4, from the table,
Therefore, n = 0 or n = 1 where x∈ P
Truth set = { n : n = 0 or n = 1}
2. i. Make an addition and multiplication table
5. a. Draw an addition and multiplication tables for for „clock arithmetic‟ with a clock showing the
the set K = {3, 6, 9, 12} in arithmetic modulo 15. numbers 0, 1, 2, 3, 4; e.g. 2 + 3 = 0, 4 × 3 = 2
b. From your table: ii. If x is a variable on {0, 1, 2, 3, 4}, with the
i. evaluate (3 ⊗ 3) (6 ⊗ 9), table, solve;
ii. find the truth set of n⊗( n 3) = 3. a. x + 3 = 0 b. x + 2 = 1
c. 3x + 2 = 4 d. 4x + 3 = 1
Solution
a. 3. a. Draw a table for multiplication, ⊗, modulo 7
3 6 9 12 on the set M = { }
3 6 9 12 0 b. Use your table to find on the set M, the truth
6 9 12 0 3 set of r ⊗ (r ⊗ 6) = 3
9 12 0 3 6
12 0 3 6 9 4. a. Draw a table for multiplication ⊗ in modulo
12 on the set {1, 5, 7, 11}

⊗ 3 6 9 12 b. Using the table;

3 9 3 12 6 i. state with reasons whether or not the operation
6 3 6 9 12 ⊗ is commutative,
9 12 9 6 3 ii. evaluate 5 ⊗ (7 ⊗ 11),
12 6 12 3 9 iii. find the truth set of n⊗n = 1.

b. i. (3 ⊗ 3) (6 ⊗ 9) = 9 9=3 5. a. Draw addition and multiplication ⊗ tables

ii. n⊗ ( n 3) = 3 on the set R = {2, 5, 7, 11} in arithmetic modulo 12

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b) From your table: i) (2⊗4) 4 ii)(3 4)⊗2
i) evaluate (5⊗7) (7⊗11),
ii) find r if r ⊗(r 2) =11. 9. a. Copy and complete the addition and
multiplication⊗ tables for arithmetic modulo 7
6. i. Copy and complete the following table for
1 2 3 4 5 6
multiplication modulo 7 on the set { }
1 2 3 4 5 6 0
2 3 5 6 0
⊗ 1 2 3 6
3 4 6 0 2
1
2 4 5 0 2
3 5 6 1 2 3 4
6 6 1 3 5

ii. Use the table to find the truth set of n⊗n⊗n = n ⊗ 1 2 3 4 5 6

1 1 4 5 6
7. i. Copy and complete the table below for 2 2 6 1 3
multiplication modulo 5 on the set {1, 2, 3, 4} 3 3 6 2 5 1 4
4 1 6
⊗ 1 2 3 4 5 5 3 1 4
1 6 5 3 2 1
2
3 4 b. Use your tables to solve for x;
4
i. x 3 = 6 ii. 3 ⊗x = 1 iii. (6 x) ⊗ 4 = 2
ii. Use the table to find the truth set of 3 ⊗n = 1
Modulo and Binary Combined
To solve problems involving a combination of
8. Copy and complete the following table for binary operation and modulo arithmetic:
addition and multiplication⊗ mod 5 I. Identify the binary operation and the given
modulo.
⊗ 1 2 3 4 II. Perform the binary operation and convert the
1
answer to the given modulo.
2
III. Complete the table of values if any, with the
3
4 answers obtained from the conversion.

1 2 3 4 Worked Examples
1 The operation * is defined by a * b = ab + 2
2 in arithmetic modulo 5
3 a. Draw a table for * on the set {1, 2, 3, 4}
4 b. Use the table;
i. evaluate (2 * 1) + (2 * 3)
b) From the tables, find: ii. find the truth set of n* (n * 3) = 3

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Solution 3 * 5 = (3)(5) + 2 = 17(mod 7) = 3
a. 3 * 6 = (3)(6) + 2 = 20 (mod 7) = 6
* 1 2 3 4 5 * 3 = (5)(3) + 2 = 17(mod 7) = 3
1 3 4 0 1 5* 5 = (5)(5) + 2 = 27(mod 7) = 6
2 4 1 3 0 5 * 6 = (5)(6) + 2 = 32(mod 7) = 4
3 0 3 1 4 6 * 1 = (6)(1) + 2 = 8(mod 7) = 1
4 1 0 4 3
* 1 3 5 6
b. From the table
1 3 5 0 1
i. (2 * 1) = 4 and (2 * 3) = 3
3 5 4 3 6
Therefore (2 * 1) + (2 * 3) = 4 + 3 = 7
5 0 3 6 4
But 7 = 2(mod 5)
6 1 6 4 3
ii. n * (n * 3) = 3
ii. From the table,
From the table, a. 3 * n = 3 b. m * m = 4
When n = 1, 3*5=3 3*3=4
1 * (1 * 3) = 1 * 0 (Not defined on the set) ⇒n = 5 ⇒m=3
When n = 2,
2 * (2 * 3) = 2 * 3 = 3 Solved Past Question
When n = 3, 1. The operation* is defined on the set {2, 4, 6}
3 * (3 * 3) = 3 * 1 = 0 by m * n = the unit digit in the product mn.
When n = 4, a. Copy and complete the table.
4 * (4 * 3) = 4 * 4 = 3
⇒n * (n * 3) = 2 * (2 * 3) = 4 * (4 * 3) = 3 * 2 4 6
2 4 8 2
Therefore n = 2 or n = 4 4 6
6
2. An operation * is defined by m * n = mn + 2, in
arithmetic modulo 7. b. Use your table to solve the equations:
i. Copy and complete the table for the operation * i) x * 4 = 8 ii) e * e = e iii) (4 * f) * 4 = f
on the set {1, 3, 5, 6}
Solution
* 1 3 5 6 i.m * n = the unit digit in the product mn.
1 3 5 0 1 4* 2 = 8, 4* 6 = 4, 6* 2 = 2,
3 5 4
6* 4 = 4, 6* 6 = 6
5 0
6 6 4 3
* 2 4 6
2 4 8 2
ii. From the table in (i), find the truth set of:
4 8 6 4
a. 3 * n = 3 b. m * m = 4 6 2 4 6

Solution
b. i. x * 4 =8
i.m * n = mn + 2, in arithmetic modulo 7

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2* 4 = 8 ii. (2 * 4) * (2 * 5)
⇒x = 2 iii. the truth set of (1 * n) * n = 3
iv. why will you say the operation * is
ii.e * e = e commutative?
6*6=6
⇒e = 6 2. The operation * is defined by a * b = a (b + 2)
in arithmetic modulo 8.
iii. (4 * f) * 4 = f a. Prepare a table for * on the set P = {1, 3, 5, 7}
when f = 4, (4 * 4) * 4 b. From the table;
=6*4=4 i. evaluate (3 * 5) * 7
ii. find the truth set of n2 = n
When f = 6, (4 * 6) * 4
= 4* 4 = 6 3. The table below defines the operation ◙ on the
{ f : f = 4 or f = 6} set X = {a, b, c, d}

Exercises 14.11 ◙ a b c d
1. The binary operation * is defined by the a b d a c
statement a * b = 2ab + 1 in arithmetic modulo 6 b d c b a
a. Use this to copy and complete the table below c a b c d
d c a d b
for *on the set {1, 2, 3, 4, 5}

a. Find giving reason(s) whether the following are

* 1 2 3 4 5
correct;
1 3 5 1 3 5
2 i. X is closed with respect to ◙,
3 1 1 1 1 ii. there is an identity element.
4 5 b. Find the where possible, the inverse of the
5 5 1 5 3 elements a, b, c and d
c. Find p and q if:
b. Use the table to evaluate; i. (d ◙ a) ◙ (c ◙ b) = p,
i. 3 * (3 * 5) ii. (q ◙ a) ◙ (c ◙ d) = b.

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15 INDICES AND LOGARITHMS Baffour – Ba Series

Positive Integral Indices I. Arrange the numbers (base) in ascending order.

The task of multiplying a number by itself, a II. Align the base when the problem involves
certain number of times gives rise to the idea of division.
powers of numbers. For example,
can be explained as 2 multiplied by itself 4 times. Worked Examples
This can be written in short form as: 1. Simplify
4

Solution
In the expression, , 2 is called the baseand 4 is
Re-arrange the terms to have
called the index or exponent and it is read as
5–2 7–2
“two exponent four”. = 33 × 55

Generally, in an expression of the form m,a is 2. Evaluate

called the base and m is called the index or
exponent or power and it is read as “a exponent
Solution
m” or “a to the power of m”.
5–2 7–2
= 33 × 55
Laws of indices = 27 × 3125 = 84,375

3. Simplify,
1. × = 5. ( ) =

Solution
2. 6. =1 8–0 10 – 9
= = = = 28 × 9

3. ( = 7. =
4. Evaluate

4. ( = 8. ⁄ ⁄
=( ) Solution
8–0 10 – 9
= =
Application of the Laws of Indices
= 28 × 9 = 256 × 9 = 2,304
1. To simplify an exponential expresson means,
leave the answer in exponent form
2. To evaluate an exponential expression means 5. Simplify
to find the value of the expression.
In dealing with complex exponential expressions, Solution
apply the laws of indices one after the other.
= = × = ×
Also:

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6. Solution
9–9 0

Solution
= = = × 2. 4 × 690

Solution
Exercises 15.1
4 × 690 = 4 × 1 = 4
A. Simplify the following;
1. 2. 3. Exercises 15.2
( ( Evaluate each of the following:
4. 5. 6.
1. 2.

B. Evaluate the following; 3. 4. (811 0

1. 2. 5. Show that; (136 4

) ( 5 5
)=
6 4
3. , 4. ( )
Rational Indices
6 4 4 2
5. ( ) ( ) By the law of indices,
⁄ ⁄
………….(1)
Zero Power of a Natural Number 2 × 2 = 4………………….(2)
Study the pattern below carefully with reference
to the division rule: Comparing eqn (1) and eqn (2)
⁄
am an = am – n =2
⁄
24 But √ = 2. Therefore, we define as the
square root of 4 written as√ = ⁄

⁄
In general,√ =
⁄
For e.g.√ = ⁄ ( =3
⁄ ⁄ ⁄
Similarly, = =a
⁄
We define as the cube root of a written as;
Since the value of = 1, it follows √ = ⁄
that , so (a≠0)
In general, √ = ⁄
⁄ ⁄
Worked Examples For e.g. √ =( =3
Simplify the following; ⁄
Generally, if is taken to be the nth root
⁄
1. of , then =√
⁄ ⁄
For e.g.√ =( =

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 4 5 6
Worked Examples 1. √ 2. √ 3. √
1. Evaluate ⁄
B. Find the values of the following:
⁄
Solution 1. 4. ( 7.
⁄
=( ⁄
= = 3 × 3 × 3 = 27 2. 5. 8.
⁄ ⁄
3. 6. 9.
⁄
2. Evaluate ( ) C. Evaluate the following;
⁄ ⁄ ⁄
1. ( ) 2. ( ) 3. ( )
Solution ⁄ ⁄ ⁄
⁄ ( ⁄ ( ⁄ 4. ( ) 5. ( ) 6. ( )
( ⁄ =( ⁄ =( ⁄ = =
7. 8. (

3. Simplify ( Find x if
1. (x + 1)5 = 243 2. (x – 1)5 =
Solution
⁄ ⁄
( = Negative Indices
⁄ ⁄
= ( ×( =2×7 From the law of indices, = , but
and = 2
4. Evaluate (
By substitution, 1 = = and
Solution therefore, is taken to be equal to
⁄ ⁄ Similarly, = = , where is the
( =
⁄ ⁄
= ( × ( = 3 × 5 = 15 reciprocal of

⁄ ⁄
5. Evaluate In general,
1. , (a ≠ 0) and is the reciprocal of
Solution 2. = (a ≠ 0) and is the
Method 1
⁄ ⁄ ⁄ ⁄ ⁄ reciprocal of
( ( )
= = = = =
Worked Examples
Method II 1. Express as a negative index.
Express each as an exponent with base 2 and
apply the laws of indices. Solution
1. =
Exercises 15.3
m
A. i. Express with signs of the form √
2. Write in the form
1. 2. 3.
Solution
ii. Write the following in exponent form;

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 ⁄ ⁄
= = = 3. ( ) 6. ( ) 9. ( )

3. Simplify C. Find the values of the following;

⁄ ⁄ ⁄
1. 2. 3.
⁄
Solution 4. ⁄
5. ⁄
6. ⁄
= = = =
D. Simplify the following;
⁄ ⁄ ⁄ ⁄ ⁄
4. Find the value of ( ⁄ 1. 5.
⁄ ⁄ ⁄ ⁄
Solution 2. 6.
⁄
( ⁄ ⁄ ⁄ ⁄ ⁄

⁄
3. ⁄ 7. ⁄
⁄ ( )
⁄
=( ⁄ = ⁄ = ⁄
= = ⁄ ⁄ ⁄ ⁄
(
4. ⁄ 8. ⁄ ⁄

⁄
5. Find the value of ( )
Exponential Equations
Solution An equation involving powers of numbers, is
⁄ called an exponential equation.For example,
⁄ ⁄
( ) =( =( =3
, are exponential equations
⁄
6. Find the value of ( ) The steps in solving exponential equations are;
1. Express all the numbers as exponents of the
Solution same base (usually a prime number).
⁄
( ) =( ⁄
=( ⁄
= = 64 2. Express both sides as a power of the same
number, using the laws of indices.
3. Equate the exponents and solve the equation.
Method II
⁄ ⁄ Worked Examples
⁄
=( ) = =( =
⁄ ⁄
1. Find the value of x in 2x= 8

Exercises 15.4
Solution
A. Express as negative exponents;
2x
1. 2. 3.
4. = 3x 5. 6. 2x 3

B. Find the values of the following:

⁄ 2. In , find the value of x
1. ( ) 4. ( ) 7. ( )
⁄ ⁄
2. ( ) 5. ( ) 8. ( ) Solution
,

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 ( )
= (Equating exponents)
3
-3x = -
x–4=3
x–4+4=3+4 -3x× 2 = - × 2 ( Multiply through by 2)
x=7
-6x = -3
3. Solve for x given that = 49 x= =

Solution ⁄
7. Given that = , find the value of y

2 Solution
⁄ ⁄
x+3=2 =( =
x = 2 – 3 = -1 But ⁄
=
= , therefore y = 3
4. If = 256, determine the value of x
Exercises 15.5
Solution
A. Solve the following;
= 256 1. 49x = 7 4. 9x = 27
= 2. = 125 5. 103n = 10000
= 3. 729x = 9 6. 9x = 729
– 5x = 8
x=– B. Solve the following equations:
1. = 27 6. = 128
5. If × , find the value of x 2.
4 x–5
3.
Solution
× 4. = 1000 9. =√
x + 2 + 3x = 0 5. = 81 10. =
4x + 2 = 0
4x = –2 C. Find the value of the variables:
x= = 1. ( )x = 5. =
2. = 6. =
⁄
6. Solve ( ) – ( ) =0
3. 2n = 7. ( ) =

Solution 4.( )x = 125 8.( )x =64

⁄
( ) – ( ) =0
⁄
Logarithms
– =0 From the calculator, it is seen that the log of 100
( )
– =0 to the base 10 is 2. This is stated mathematically
( )
– =0 as log10 100 = 2. This means that 100 = ,

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where 10 is the base and 2 is the index B. Express in exponential form:
(logarithm). Note that log10 is written simply as 1. log 91 = 0 4. –2 = log 3
log.
2. log 4 2 = 5. log 27 3 =
Relating Indices and Logarithms
3. log 5 625 = 4 6. log (0.1 × 108) = 7
In general, ifa = then the logarithm of a is
expressed as: Logba = cread as“the logarithm of a
Laws of Logarithms
to the base b is c”
Logarithms and their operations are govern by
Thus:
some basic properties and laws as shown below
If a = , then =c for a, x andy belonging to the set of real numbers;
1. loga (xy) = logax + logay
e.g. log2 (4 × 2) = log2 4 + log2 2
Any statement in index form has an equivalent = log2 22 + log22
log form, for example, =2+1=3
8 = (index form)
⇒ = 3(log form)
2.loga( ) = logax – logay

Worked Examples e.g. log 2( ) = log 2 4 – log 2 2

1. Express = 16 in logarithmic notation
log 2( ) = log 222 – log 2 2 = 2 – 1 = 1
Solution
In a = log b a = c 3.logaa = 1
If = 16, then log2 16 = 4 e.g. log 3 3 = 1

⁄ 4. logax y= y logax
2. Write = in logarithmic form
e.g. log2 4 = log 2 22 = 2 log 2 2 = 2 × 1 = 2

Solution 5.loga1 = 0
⁄
If = , then log8 = ⁄ e.g. log 10 1 = 0

3. Express in index notation log2 32 = 5 6.loga( ) = logax + logay – logaz

e.g. loga 2 + loga 6 – loga 4
Solution
= loga( ) = loga( ) = loga 3
If log2 32 = 5, then 32 =

Exercises 15.6 7. logax =

A. Write the following in logarithm form:
E.g. log 1000 10 =
1. 125 = 4. 0.01 =
⁄ ⁄ = = = =
2. =7 5. =5
⁄
3. = 64 6. =

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8.loga b = (change of base) Worked Examples
1. Simplify log 2 72

e.g. log 4 64 = = = =3 Solution

9. logax = y, then x = log 2 72 = log 2(8 × 9)
e.g. If log 2 8 = 3, then 8 = = log 28 + log 29
= log 223 + log 232
Exercises 15.7 = 3 + 2 log 23
A. Express as a single logarithm:
1. log 2 + log 5 2. Simplifylog 540
2. log 2 + log 12 – log 6
3. log 3 + log 16 – log 4 – log 6 Solution
4. (log6 4 + log6 9)2 log 540 = log 5(5 × 8)
5. (log525 + log5 15 – log5 75)4 = log 55 + log 58
= 1 + log 523
Common Logarithmic Expressions = 1 + 3 log 52
(Applying the Laws of Logarithms)
A logarithmic expression is a logarithm statement Exercises 15.8
that is not equated or does not contain an equal Simplify;
sign. 1. log372 2. log 363 3. log 5150
4. log 8120 5. log 2√ 6. log √
To evaluate a logarithmic expression is to find the
value of the expression by applying the laws of Type II: Involving Addition and Subtraction
indices. Hence, answers should not contain the 1. Ensure that all the logs are of the same base
word “log”. 2. Express the given expression as a single
logarithm of the form log ab.
To simplify a logarithmic expression is to apply
3. Go throught the processes of type I to complete
the laws to the extend that no law is applicable.
Hence, a simplified answer should contain the simplification, only if a is a factor of b.
word“log” 4. If a is a not a factor of b, then leave your
answer in the form: log ab
Simplifying Logarithms Expressions
Type I: Form log a b Worked Examples
I. Express b as a product of two factors to obtain 1. Simplify 3 log 3 – log 27
log a(c × d)
II. Apply the laws to obtain log ac + log ad Solution
III. Express c and d as exponents, if possible and 3 log 3 – log 27
apply the exponential law of logarithm = log 33 + log 27
IV. Apply other laws if necessary = log 27 – log 27 = log ( ) = log 1

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2. Simplify log 49 +log 4 21– log 47 VI. Make the variable the subject and work out to
obtain the value of the variable. i.e. x=
Solution
log 49 + log 4 21 – log 47
Worked Examples
= log 4( ) = log 427 = log 433 = 3 log 4 3 1. Use tables or calculator to find an approximate
value oflog2 7
3. Simplify log 2 3 – log 215 +log 250
Solution
Solution Letlog27 = x
log 2 3 – log 215 + log 250 ⇒7=
= log 2( ) Taking log on both sides
log 7 = log 2x
= log 210
log 7 = xlog 2
= log 2(2 × 5)
= log 22 + log 2 5 x= = 2.807 (3 d. p.)
= 1 + log 2 5
2. Evaluate log2 64
Exercises 15.9
A. Simplify: Solution
1. log 5 12 + log 510 Let log2 64 = x
2. 2 log 79 – log 781 64 =
= (Express 64 as an index to the base 2)
3. log 45 – log
x=6 (Equating exponents)
4. log 3 24 + log 3 15 – log 3 10
5. log 7 98 + log 7 30 – log 7 15
3. Evaluate log381

Evaluating Logarithmic Expressions Solution

Type I: Form loga b Method 1
Given a logarithmic expression of the form loga b
log3 81 = log334
to determine its value:
I. Equate the expression to any preferred variable, = 4 × × log3 3,
say x. i.e.loga b = x =4× ×1=2
II. Write the logarithmic expression as an
exponential equation. That is:b =
Method 2
III. Express the exponential equation as equation
⁄ ⁄
with a common base, if possible. log381= log3 = log3 ( = log3 9
IV. If not possible, take the log on both sides of
the equation.log b = log Let log3 9 = x
V. Simplify the variable side of the equation. 9=
i.e.log b = x log a ⇒ = x=2

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4. Evaluate log 0.1 10 B. Find the values of the following:
1. log 3 81 4. log 2 64 7. – log 2
Solution
Let log0.1 10 = x 2. log 2 8 5. log 49 7 8. log 2 4
10 = 3. log 27 ( ) 6.log 7 9. log 16 8
10 =( )
10 = ( C. Evaluate to 2 significant figures:
10 = 1. log √ 3. log √
1= - x 2. log √ 4. log 2 2 √
x = -1 (Try other methods)
Type II: Involving Addition and Subtraction
5. Evaluate log3 27√ 1. Ensure that all the logs are of the same base
2. Express the given expression as a single
Solution logarithm of the form: loga b.
Let log3 27√ = x 3. Equate the single log expression to any
27√ = preferred variable to obtain a logarithmic
equation.
× =
4. Write the logarithmic equation as an
= exponential equation.
= 5. Solve for the value of the variable in the
exponential equation as in type I.
x=
Worked Examples
8. Find the value oflog 5 Evaluate to four decimal places;
1. Evaluatelog 5 12 +log 5 10 to four decimal
Solution places
log 5
Solution
= log 5
log 5 12 +log 5 10
= log 5 ( = log 5 (12 × 10) = log 5 120
= log 5
= -2 log 5 5 Let log 5 120= y
= -2 × 1 = -2 (Try an alternative method) 120 =
log 120 = log
Exercises 15.10 log 120 = ylog 5
A. Evaluate:
y= = 2.9746 (4 d.p)
1. log 10 4.log 3 243 7. log 8 2
2 .log 4 1 5. log 27 3 8.
2. Evalutae (log 6 4 + log 6 9)2.
3. log 5 125 6. log 0.01 10 9. log 2 32

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Solution = log 10 ( )
Method 1
log 6 (4 × 9)2 = log 6 (36)2 = log 6 1296 = log 10 ( ) = log 1010 = 1

Let log 6 1296 = y 5. Without using Mathematical tables or

1296 = calculators , simplify;
log1296 = log
log10 ( ) – 2 log10 ( ) + log10( )
log1296 = y
y= =4
Solution
log10 ( ) – 2 log10 ( ) + log10( )
Method 2 ⁄
2 log 6 (4 × 9) log10 ( ) – log10( ) + log10( )
= 2 log 6 36
log10 ( ) – log10 ( ) + log10( )
= 2 log 6 62 = 2 × 2 =2×2×1=4
lg ( )
3.Evaluatelog 4 9 +log 4 21 –log 4 7
lg ( )
Solution lg ( ) = lg ( = lg 10
log 4 9 +log 4 21 –log 4 7
= log 4 ( ) = log 4 27 Exercises 15.11
Find the values of the following:
Let log 4 27 = x 1. log√ + log√ – log√
⇒ 2. log 6 24 + log 6 15 – log 6 10
log 27 = log 3. log 7 98 + log 7 30 – log 715
log 27 = x log 4 4. log 3 – log 15 + log 50
x= = 2.3775 5. log 5 + log 16 – log 4 – log 10

Division of Logarithms
4. Evaluate without using tables or calculators:
To simplify expressions involving division of
log 10( ) – 2 log 10 ( ) + log 10 ( ) logarithms
I. Express each number as an exponent with a
Solution common base
log 10( ) – 2 log 10( ) + log 10( ) II. Apply the law: loga xy = yloga x
III. Simplify by crossing out common factors
= log 10( ) – log 10( ) + log 10( )

= log 10( ) – log 10( ) + log 10( ) Worked Examples

Simplify :
= log 10( )

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Solution = log 10 – log 10 10 2
= x + ylog 10 10 – 2 log 10 10
=x+y–2

Express 125 and 25 as a power of 5

Exercises 15.12
= = Simplify:
1. 2. 3.
2. Simplify √
– 4. 5. 6.
√
√

Solution
Solving Exponential Equations
In an exponential equation, when the exponent of
Make log 6 – log 2 a single log a number at a side of the equation is a variable,
⇒log ( ) I. Take the log on both sides of the equation.
= log 3 II. Simplify the variable part of the equation.
Substitute in place of log 6 – log 2 III. Make the variable the subject and work out
(simplify) to obtain the value of the variable.
⇒

= = Worked Examples
1. Solve the equation = 5, using tables or
calculators to 5 significant figures
3. Simplify
Solution
Solution In = 5,
⁄ Taking log on both sides,
= = = =2
log = log 5
x log 2 = log 5
4. Given that log p = x and log q = y, express log
= = 2.3219 (5 s.f)
in terms of x and y

2. Solve the exponential equation –1=8

Solution
logp = x and log q = y
Solution
p= and q =
–1=8
pq =
=8+1
=
=9
⇒log
log = log 9
= log 3x log 4 = log 9

= log x= = 0.5283 (4 d.p)

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Exercises 15.13 2. Given thatlog 3 5 = 1.465, evaluate without
A. Solve the equations: using tableslog 3 25 +log 3 15
1. =5 2. =2
3. 4. × Solution
5. ( ⁄ =6 6. ( ⁄ = ⁄ log 3 25 + log 3 15
= log 3 52 + log 3 (3 × 5)
B. Solve for n, to 4 significant figures; = 2 log 3 5 + log 3 3 + log 3 5
1. = 20 2. = 2500
3. = 680 4. = 31 Substitute log 3 5 = 1.465
5. = 3460 6. = 4800 ⇒2 × 1.465 + 1 + 1.465
7. = 8. = = 5.395

Substitution ⁄
3. If 6990, find
Sometimes the estimated value of the log of a
number may be given to simplify or evaluate Solution
other logarithmic expressions. In such cases: ⁄
log = log(
I. Express the given log or simplified as an
exponent or a product of the number whose = log( = log 5,
estimated value has been given.
II. Apply the law: log a x y = loga x, if necessary. Substitute log 5 = 0.6990
III. Substitute the estimated value into the ⇒ 0.6990 = 0.4660
expression and simplify
Some Solved Past Questions
Worked Examples 1. If log x 3 = 0.5283 and log x7 = 0.9358, find
Given that log 2 = 0.3010, find without using logarithm to base x of ;
tables or calculators the values of; i. 27 ii. √
i. ii.
Solution
Solution log x3 = 0.5283 and logx7 = 0.9358,
i. log 8 = log = 3 log 2, i. log x 27
= 3 log x33
Substitute log 2 = 0.3010 = 3 log x 3
= 3 × 0.3010 = 0.9030 But logx 3 = 0.5283
⇒ 3 × 0.5283 = 1.5849
ii. log 80 = log ( 8 × 10)
= log 8 + log 10 ii. log x√
= log 23+ log 10 = log x ⁄

= 3 log 2 + log 10 = log x7

= 3 × 0.3010 + 1
= 1.9030 But log x7 = 0.9358

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⇒ 0.9358= 0.4679 = 1 – 3(0.3010 ) = 0.097

5. log 9 = 0.9542, find the value of log 0.009

2. If log x = 0.5 and log y= 1.5, find x + y

Solution
Solution
log 0.009
Log 10x= 0.5 and log y10= 1.5
= log ( 9 × )
x= and y =
= log 9 + log 10
x+y= + = 34.8
= log 9 + (-3) log 10 10
3. Given that log 6 = 0.778, find without using = 0.9542 – 3 (1) = -2.0458
tables or calculators the value of log 600.
6. Given that log 10 2 = 0.3010 and log 10 3 =
Solution 0.4771, evaluatelog 10 0.24
log 600 = log (6 × 100)
Solution
= log 6 + log 100
log 10 0.24
= log 6 + log
= log 6 + 2 log 10 = log 10 ( )
= log 6 + 2(1) = log 10 24 – log 10 100
= log 10 (23 × 3) – log 10 (102)
Substitute log 6 = 0.778 = log 10 23 + log 10 3 – log 10 102
⇒ 0.778 + 2 = 2.778 = 3 log 10 2 + log 10 3 – 2 log 10 10
= 3 (0.3010) + (0.4771) – 2 (1)
4. Given that log 10 2 = 0.3010 and log 10 5 = = - 0.6199
0.6990, evaluate log 10 50 – log 10 40
Exercises 15.14
Solution A. Given that log 2 = 0.3010 and log3 = 0.4771,
log 10 50 – log 10 40 find the values (correct to 4 s.f)
= log 10 ( )= log 10 ( ) = log 10 5 – log 10 4 1. log 12 2. log √ 3. log 18
= log 10 5 – log 10 22 4. log 1.5 5. 2 log 21 – log 98
= log 10 5 – 2 log 10 2
= 0.6990 – 2(0.3010) = 0.097 6. Given that log 5 = 0.6989, find correct to four
places of decimals, the value of log 40
Method 2
log 10 50 – log 10 40 7. Given loga3 0.48 and loga5 1.72, evaluate
= log 10 (5 × 10) – log 10 (5 × 8) loga45 without the use of a calculator
= log 10 (5 × 10) – log 10 (5 × 23)
= log 5 + log 10 – log 5 – log 23 8. If log5 2 = 0.431 and log5 3 = 0.682, find the
= log 5 + log 10 – log 5 – 3 log 2 value of log5 ( ) + 2 log 5( ) – log 5 ( )
= 1 – 3 log 10 2

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C. If log5 2 = 0.431 and log5 3 = 0.682, find the Exercises 15.15
values of the following; A. If f(x) = 2 log 5x , evaluate:
1. log5 6 2.log5 27 3. log5 8 1) f(5) 2) f(25) 3) f( ) 4) f(√ )
4. log5 12 5. √ 6. log5( )
7. log5( ) 8. log5 100 9. log5 1.5 B. Use the change of base method to simplify:
1. log 645 2. log 230 3. log 4128
4. log 1272 5. log 8 2 6. log 30100
Involving Change of Base
Given to simplify using change of base:
Logarithm Equations
I. Find the common factor of the base (a) and the A logarithmic equation is a logarithm statement
number (b) to obtain c that is equated or contains an equal sign. In other
II. Express the logarithm of b to base c to the words, it is an equation involving logarithms.
logarithm of a to base c as a common ratio to
complete the process. i.e.loga b = To solve logarithmic equation is to find the value
III. Simplify where possible or values of the variable that makes the statement
or equation true.
Proof
For y = logb M, There are two methods for solving equations
⇒by = M in exponential form . involving logs:
(Take base – a logarithm of both sides) Method 1
It follows that loga (b y) = logaM 1. Get a single log on both sides of the equation,
y logab = loga M, ( Apply power property of logarithms) II. Equate the expressions and solve for the value
of the variable i.e. write the equation in the form:
y= (Divide by ) log b x = log b y ⇒ x = y and solve

Method 2
But logb M = y
I. Get a single log in the equation,
logb M = II. Change from logarithmic form to exponential
form. i.e. write the equation in the form: log b a =
Worked Examples c as a =
1. Evaluate log9 27
Note:
Solution
1. Make sure that all logs have the same base. If
log9 27= = =
necessary, use the change of base law to write
logs to the same base
2. Simplify log 892 2. Logs are defined for only positive numbers.
Therefore reject any solution that give rise to log
Solution
(negative number)in the original equation
log 892 = = =

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Worked Examples 1. Solve for x in 3 log x + log 3 = log 81
Solve log2 x = 3 – log 2 (x – 2)
Solution
Solution 3 log x + log 3 = log 81
Method 1 log x3 + log 3 = log 81
log 2 x = 3 – log 2 (x – 2) log 3x3 = log 81
log 2 x + log 2 (x – 2) = 3 ⇒ = 81
log 2 x (x – 2 ) = 3 = 27
log 2 (x2 – 2x ) = 3 x = 33
3

= x=3

2. Solve log (8x + 1) – log (2x + 1) = log (x + 2)

( ( Solution
log (8x + 1) – log (2x + 1) = log (x + 2)
log10( ) = log10 (x + 2)
Reject x = – 2. Therefore, x = 4
=x+2
Method 2 = (x + 2) ( )
log 2x = 3 – log 2 (x – 2) 8x + 1 = x(2x + 1) +2(2x + 1)
log 2x + log 2 (x – 2) = 3 8x + 1 = 2x2 + x+ 4x + 2
log 2x(x – 2) = log 2 8 8x + 1 = 2x2 + 5x + 2
log 2( ) = log 28 0 = 2x2 + 5x – 8x + 2 – 1
2x2 – 3x + 1 = 0

( ( a = 2, b = - 3 and c = 1
√
Substitute in x =
( √( ( (
Reject x = – 2. Therefore, x = 4 x= (
√
2 x=
2. Solve log10 x + log10 2 = log10 50
√
x=
Solution √ √
x= or x =
log10x2 + log10 2 = log10 50
= log10(2x2) = log10 50 x = 1 or x =
⇒ = 50
3. Find the value of x if log4 16 = logx 36
x2 = 25
x2 = 52
Solution
x=5
log416 = logx 36
log442 = logx 36
Some Solved Past Questions

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2 log44 = logx 36 Solution
2= log (5x – y ) = log 9………………..(1)
⇒ = 36 log (3x + 2y) = log 8 ………………(2)
⁄ ⁄
x= =( =6
Divide both sides by log
Exercises 15.16 5x – y = 9……………(1)
A. Solve the following equations: 3x + 2y = 8………….(2)
1. log (5x + 2 ) = log 2
eqn (1) × 2
2. log ( 2x – 3 ) = log 3
10x – 2y = 18………(3)
3. log 4 (x + 2) – log 4 3 = log 4 (x – 1 )
4. log 4 (x + 6) – log 3x = log 3 5
eqn (2) + eqn (3)
5. log 2 3 + log2 x = log 2 12
(3x + 10x) + (2y – 2y) = 8 +18
6. log 3 x – log 3 4 = log 3 2
13x = 26
7. log 5 3(x + 5) + log 5 (x – 1 ) = log 5 8
x=2
8. log a (x – 6) + log a ( x – 4 ) = log a x
9. ( – =
Put x = 2 in eqn (1)
5(2) – y = 9
B. Solve the following equations;
10 – y = 9
1. log 2 (x + 2) – log 2x = 3
y = 10 – 9 = 1
2. log (2x – 1) – log 5 = 1
Therefore, x = 2, y = 1
3. log ( x – 7) – log 3 = 2
4. log 2 9 + log 2(x + 3 ) = 3
Exercises 15.17
5. log 5x + log 5 ( x + 4)
Solve the pair of equations;
1. log 2(2x + y) = 3and log 2 (3x – 4y) = 0
C. Find the value of x
2. log 3 (3x – y) = log 3 (y + 1) and log 3 2 +
1. log x – 1 = log (x – 9)
log 3 (x + y) = 2
2. log (2x + 1 ) = 1 – log x
3. log x2 = log y and log (2x + y) – log 3 = 0
3. log 2 (x + 2) + log 2 (x – 2) = 5
4. log 2 2 + log 2 (x + y) = log 2 4 and
4. log 9 x = + log 9 (5x + 18)
log 2 x + log 2y = 2, x > 0, y > 0
5. log 2 (2x – 5) + log 2 (x – 2 ) = 4
The Characteristic and the Mantissa
Involving Simultaneous Equations The logarithm of any number consists of two
Two or more logarithmic equations involving two parts namely: the characteristic which is the
different variables are solved simultaneously whole number and the mantissa which is the
either by method of substitution or elimination. decimal part. For e.g., in = 2.4771, the
whole number at the left of the decimal point, 2 is
Worked Examples the characteristic of and the decimal
1. Solve the simultaneous equations: part which consist of 0.4771 is the mantissa of
log (5x – y ) = log 9 and log (3x + 2y) = log 8

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Finding the Characteristic negating the sum of three and one to get negative
To find the characteristic of any number; four. That is: – (3 + 1) = – 4
Method 1
I. Express the number in standard form/ Worked Examples
II. Identify the power of ten as the characteristic Find the characteristic of the following:
of the number. 1. 0.00574 2. 40322 3.36.19

Method 2 Solution
The following guidelines are used to identify the Method 1
characteristic; 1) 0.00574 = 5.74 ×
I. If the number is equal to or greater than one, The power of 10 = -3
subtract one from the total number of digits to get Characteristics = -3
the characteristic of that number. For e.g. 300
have three digits. Therefore, its characteristic is 3 Method 2
– 1 = 2. Similarly, the characteristic of 71 is 2 – 1 Number of zeros after the decimal point
= 1 and the characteristic of 6 is 1 – 1 = 0 =2+1=3
Negate the sum to get -3
II. If the number is a decimal number greater Characteristics = -3
than or equal to one, subtract one from the total
number of digits before ( or at the right of ) the 2. 40322
decimal point . For e.g. in 300.12, there are three 40322 = 4.0322 ×
digits at the left of the decimal point. Therefore, Characteristics = 4
the characteristic is 3 – 1 = 2. Similarly, the 3. 36.19
characteristic of 7.1 is 1 – 1 = 0 = 3.619 ×
Characteristic = 1
III. If the number is a decimal number less than
one, Exercises 15.18
a. Identify the number of zeros immediately at the Find the characteristic of the following:
right of the decimal point and add one to it 1. 4120 2. 0.000203 3. 6.12
b. Negate the sum in (a) to get the characteristic 4. 211.0045. 78 × 6. 0.344
of the number.
Activities a and b are simplified as; Finding the Mantissa
– (Number of immediate zeros after the decimal The table of logarithm also known as Four
point plus one) For e.g. 0.000304 have three Figure Table is used to find the mantissa (or the
zeroes immediately after the decimal point. decimal part less than one). The four figure table
Therefore, the characteristic is calculated by has three columns as shown below;

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I. The x- column contains the first two significant log 2.5 (equal to 2.50)
figures of a given number. I. At column x of the log table, look for 25
II. The second column, headed 0 – 9 gives the II. At the second column, find the value of 25,
logarithm of the first three significant figures. under the column headed by 0
III. Thus, log 2.5 = 0.3979
III. The third column, also known as “difference
column” headed 1 – 9 is the difference to be 2. log 2.54
added to the value of the log of the first three I. At column x of the log table, look for 25
significant figures in act II. II. At the second column, find the value of 25,
under the column headed by 4
Finding the Log of a Number III. Thus, log 2.54 = 0.4048
a. Identify the numbe.r
b. Find the characteristic of the number. 3. log 2.548
c. Put a decimal point at the right of the I. At column x of the log table, look for 25
characteristic obtained. II. At the second column, find the value of 25,
d. Use the four figure table to determine the under the column headed by 4 to obtain 0. 4048
mantissa (decimal part). III. On the same row, under the difference
column, determine the number under 8, which
is 14.
Worked Examples IV. Interpret 14 as 0.0014 and add it to 0.4048
Use log table to find the following: Thus log 2.548 = 0.4048 + 0.0014
1. log 2.5 2. log 2.54 3. log 2.548 = 0.4072
3. Use tables to find the log of 2548
Solution

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Solution = log 4.069 – log 10000
log 2548 = log 4.069 – log
2.548 × = 0.6095 – 4
Characteristic = 3 = – 4 + 0.6095
25 in the first column, under 4, difference 8 = ̅ .6095
= 0.4048 + 0.0014
= 0.4062 Exercises 15.19
log 2548 = 3 + 0.4022 = 3.4022 Use tables to find the following:
1. log 0.2468 2. log 24.68
4. Use tables to find log 0.0004069 3. log 2468 4. log 0.0004
5. log 3.456 6. log 7.24
Solution
Method 1 The Anti – logarithm of a Number
log 0.0004069 Anti - logarithm is the reverse of logarithm. It
4.069 × implies that if log (a) = b, then a = anti – log (b)
Characteristic = – 4 The anti – logarithm table is used to find the anti
From the log table, find 40,under 6, difference 9 – log of a number. The anti – logarithm table is
= 0.6085 + 0.0010 = 0.6095 similar to that of the logarithm so the process
log 0.0004069 = – 4 + 0.6095 involved in finding the logarithm of a number is
likewise, similar to that of finding the anti-
= ̅ .6095 logarithm of a number. A portion of the four
Method 2 figure table is shown below:
log 0.0004069 = log ( )

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The rule for finding the antilog is as follows: Read 0.07 under 4 difference 9
I. Find the antilog of the decimal part of the log = 1.186 + 0.002
of the given number = 1.188
II. If the integral part (characteristic) is n, shift the ⇒ 1.188 ×
decimal point n places to the right and if it is – n, = 1.188 × 1 = 1.188
shift the decimal point nplaces to the left Anti – log of 0.0749 = 1.188

Exercises 15.20
Worked Examples
Find the antilogarithm:
1. Find the anti – log of 2.5127 1. 0.6914 2. 0.0975 3. 3.5007
4. – 4.2031 5. – 2.4348 6. – 1.8977
Solution
To find the anti-log of 2.5127, Finding the Value of a Variable Given the
I. Identify the characteristic as 2 (meaning ) Value of the Logarithm of the Variable
II. Consider the decimal part, 0.5127 Given the value of the logarithm of avariable, the
III. From the anti–log table, read 0.51 under 2, value of the variable is easily found by the use of
difference 7.That is 0.51 under 2 = 3251 anti – logarithm table.
Interpreted as = 3.251
Difference 7 = 5 Worked Examples
Interpreted as = 0.005 1. If log x = 0.0615, find the value of x
IV. Add the two values to obtain 3.251 + 0.005 =
3.256 Solution
V. Multiply 3.256 by log x = 0.0615
= 3.256 × = 325.6 x = anti – log (0.0615)
Anti – log of 2.5127 = 325.6 From the anti –log table,
I. In the first column, look for 0.06 under 1 in the
2. Find the anti – log of -3.2856 second column to obtain = 1151
II. On the same row as 1151, and under
Solution difference column headed by 5 which is 1
In -3.2856, III. Interpret 1151 as 1.151 and 1 as 0.001
Characteristic = -3 (meaning ) IV. Add to get 1.151 + 0.001 = 1.152
0. 28 under 5, difference 6 = 1. 928 + 0.003 Thus x = 1.152
= 1.931
1.931 × = 0.001931 2. If log x = 1.0607, find the value of x
Anti – log of -3.2856 = 0.001931
Solution
3. Find the anti – log of 0.0749 log x = 1.0607
= 1 + 0.0607
Solution = log 10 + log 1.5
To find the anti – log of 0.0749, = log (10 × 1.15)
Characteristic = 0 (meaning ) = log 11.5

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log x = log 11.5 x = 2755
⇒x = 11.5
3. Let x = √
3. If log x = ̅ .1319, find the value of x ⁄
log x = log (
Solution = log 1.234
log x = ̅ .1319 = × 0.0913
log x = ̅ . + 0.1319
= 0.04565
= - log 100 + log 1.355
Anti – log 0.0456 = 1.111
= log 1.355 – log 100
x = 1.111
= log ( )
= log 0.01355 4. Let x = √
log x = log 0.01355 log x = log ( ⁄
⇒x = 0.01355
= × log 0.4276
Exercises 15.21 = log (4.276 × )
Use tables to find the value of x
1. log x = 0.521 2. log x = 0.0521 = lg 4.276 + lg
3. log x = ̅ .451 = (0.6311 – 1)
= - 0.123
Using Tables to Evaluate Exponential Values,
Anti – log - 0.123 = 0.7534
Values in Roots and Others
x = 0.7534
Worked Examples
Use tables to find the following: Exercises 15.22
1. 1.2342 2. 14.023 3. √ 4. √ A. Use tables to find the value of :
1. 2.4232 2. 3.4563 3. 0.01232
Solutions 4. 12.123 5. 242 6. 0.93
1. Let x = 1.2342
log x = log (1.234) 2 B. Use tables to find the value of :
= 2 log 1.234
1. √ 2. √ 3. √
= 2 × 0.0913 = 0.1826
4. √ 5. √ 6. √
Anti – log 0.1826 = 1.523
x = 1.523
Computations with Logarithms
With the use of the log – tables, computations
2. Let x = 14.023
(like multiplication and division) can be done.
log x = log (
The following examples are illustrations.
= 3 log 14.02
= 3 × (1.1467)
Worked Examples
= 3.4401
Use tables to perform the following:
Anti – log 3.4401 = 2.755 ×
1. 1.345 ×0.0164 2. 0.2352 ÷ 3.342
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3. 4.
No log
√ 0.2352 ̅ .3715
5. 6. 3.342 – 0.5240
√
0.07039 ̅ . 8475
Solutions
Method 1 3.
Let x = 1.345 ×0.0164 No Log
log x = log (1.345 ×0.0164) 1.345 0.1287
0.0164 + ̅ .2148
= log 1.345+ log 0.0164 ̅ .3435
= 0.1287 + ̅ 2148 3.342 - 0.5240
0.006600 ̅ .8195
= 0.1287 + (-2) + 0.2148
= -2 + 0.1287 + 0.2148
= - 2 + 0.3435 = ̅ . 3435 4.
Anti – log ̅ . 3435 = 0.02206
x = 0.02206 No Log
3.342 0.5240
1.345 + 1.287
Method 2
0.6527
0.0164 – ̅ .2148
No Log 2.4379
1.345 0.1287 0.2352 – ̅ .3715
0.0164 + ̅ .2148 1165 3.0664
0.02206 ̅ .3435

5.
2. 0.2352 ÷ 3.342 √

No Log
Method 1 0.0164 ̅ .2148
Let x = 0.2352 ÷ 3.342 13.4 3 × 1.1287 = + 3.3861
= 1.6009
log x = log( ) √ × 0.524 = – 0.2620
= log 0.2352 –log 3.342 21.83 = 1.3389
= ̅ .3715 – 0.5240
= -1 + 0. 3715 – 0.5240 √
= -2 + 1 + 0.3715 – 0.5240 6.
= -2 + 1.3715 – 0.5240
= -2 + 0.8475
No Log
= ̅ .8475
√ × 0.1287 = 0.0429
Anti –log ̅ .8475 = 0.07039 0.23522 2 × ̅ .3715 = + ̅ 7430
x = 0.07039 = ̅ 7859
0.0164 = – ̅ 2148
Method 2 = 0.5711

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 1.2454 4 × 0.0938 = – 0.3752 B. 1. 869 ÷ 432. 6.524 ÷ 2.122
1.571 = 0.1959 3. 3.361 ÷ 5.124 4. 0.0899 ÷ 0.33
5. 4.68 ÷ 4.1 ÷ 32
Exercises 15.22
With the help of four – figure tables, compute C. 1. 2.
the following:
A. 1. 3.142 × 0.123 2. 12.12 × 0.1134 3. 4.
3. 5.23 × 4.86 4. 698 × 62 √ √
5. 3.103 × 4.3 × 0.021 6. 0.002 × 4.34 × 1.23 5. 6.
√ √

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16 SIMULTANEOUS EQUATIONS Baffour – Ba Series

Introduction iii. Substitute the value of the variable obtained in

It involves solving two or more equations at the either eqn (1) or eqn (2) and solve to obtain the
same time. Each of the involving equations value of the other.
consists of two variables of the same identity. vi. Write the answer as ordered pairs (x, y).

Study this case carefully; “when three times a Worked Examples

number is added to another number, the result is 1. Find the values of x and y in the pair of
16 and the sum of the same numbers is 4.” equations; 3x + 2y = 16 and 5x + 2y = 20.
Equations written for cases of this nature are
called simultaneous equations. Solution
3x + 2y = 16 ……………….(1)
Two separate equations are written and solved at 5x + 2y = 20 ………………..(2)
the same time. The set of ordered pairs that y has the same co – efficient in both equations,
satisfy or make both statements true, is called the eliminate y by subtraction.
truth set or solution set of the equations.
eqn (2) – eqn (1)
Solving Simultaneous Equations
(5x – 3x) + (2y – 2y) = (20 – 16)
There are 3 general ways of solving simultaneous
2x = 4,
equations namely;
x = = 2.
1. Elimination method.
2. Substitution method
3. Graphical method Put x = 2 in eqn (1)
Any of the methods can be used to solve 3(2) + 2y = 16
simultaneous equation unless specified. 6 + 2y = 16
2y = 16 – 6
Elimination Method 2y = 10,
In this method, one of the variables, having the y= =5
same co – efficient in both equations is The truth set = {(x, y): x = 2, y = 5}
eliminated (by addition or subtraction).
2. Solve the equations 5x – 6y = 21 and
Type 1: (Equal co-efficients)
3x – 6y = 3 simultaneously.
This involves the situation where a variable has
the same coefficient in both equations. Here:
Solution
i. elimination is done by addition or subtraction.
5x – 6y = 21 ……………(1)
ii. then divide both sides of the equation by the
3x – 6y = 3………………(2)
coefficient of the existing variable to obtain its
value.
Eliminate y by subtraction;

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 eqn (2) – eqn (1); Solution
(3x – 5x) + [- 6y – (- 6y)] = (3 – 21) - 4x + 5y = - 8 ………………..(1)
-2x = -18, - 4x + y = 12……………………(2)
x = 9.
Eliminate x by subtraction;
Put x = 9 in eqn (1) eqn (2) – eqn (1)
5(9) - 6y = 21 (- 4x – (- 4x) + (y – 5y) = [12 – (-8) ]
45 - 6y = 21 - 4y = 20,
- 6y = 21 – 45 y = - 5.
- 6y = - 24
y= =6 Put y = 5 in eqn (1)
- 4x + 5(-5) = -8
The truth set = {(x, y): x = 4, y = }
- 4x – 25 = - 8
- 4x = - 8 + 25
3. Solve the system; - 4x = 17
x - y = 12 and 2x + y = 13
x=

Solution (x, y) = ( , - 5)
x - y = 12 ……..………(1)
2x + y = 13……………(2) 5. Solve the equations; 3x – 7y = - 12 and
3x – 5y = - 6 simultaneously.
Eliminate y by addition;
eqn (2) + eqn (1) Solution
(2x + x) + [y + (-y)] = (13 + 12) 3x – 7y = - 12 …………………(1)
3x = 25, 3x – 5y = - 6 …………………..(2).
x=
Eliminate x by subtraction;
eqn (2) – eqn (1);
Put x = in equation (1) (3x – 3x) + [(-5y – (-7y)] = [- 6 – (-12)]
– y = 12 2y = 6,
y = 3.
- y = 12 –
-y= Put y = 3 in eqn (1)
y= 3x – 7(3) = - 12
3x – 21 = -12
The truth set = {(x, y): x = ,y= } 3x = -12 + 21
3x = 9
4. Find the values of x and y which satisfy the x=3
system - 4x + 5y = - 8 and - 4x + y = 12. (x, y) = (3, 3)

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6. What values of x and y satisfy the system Type 2: (Multiplication by a constant)
- 7x – 2 y = 10 and 7x + 4y = 22? If one of the coefficients of the variables in the
equations is a factor of the other of the same kind
Solution in the other equation, multiply by a constant
- 7x – 2y = 10 …………….(1) number to obtain the same coefficient and name it
7x + 4y = 22………………(2) equation (3). Then carry on with the elimination
process.
Eliminate x by addition;
eqn (2) + eqn (1) Worked Examples
[(7x + (-7x)] + [4y + (-2y) ] = (22 + 10) 1. Find a and b for the pair of equations:
2y = 32, 4a + 4b = 36 and 6a – 2b = 22
y = 16.
Solution
Put y = 16 in eqn (1) 4a + 4b = 36…………….(1)
- 7x - 2(16) = 10 6a – 2b = 22…………….(2)
- 7x – 32 = 10
- 7x = 10 + 32 2 × eqn (2);
- 7x = 42 12a – 4b = 44…………….(3)
x=-6
(x, y) = (- 6, 16) Now, consider eqn (1) and eqn (3)
4a + 4b = 36……………..(1)
7. Solve 4x + y = 10 and - 4x – 3y = 2 12a – 4b = 44…………….(3)
simultaneously.
Eliminate b by addition,
Solution eqn (1) + eqn (3);
4x + y = 10 ……………………(1) 16a = 80
- 4x – 3y = 2 …………………..(2) a=5

Eliminate x by addition; Put a = 5 in eqn (1);

eqn (2) + eqn (2) 4(5) + 4b = 36
[(- 4x + 4x)] + (-3y + y) = (2 + 10) 20 + 4b = 36
- 2y = 12, 4b = 36 – 20
y = - 6. 4b = 16
b=4
Put y = - 6 in eqn (1); (a, b) = (5, 4)
4x – 6 = 10
4x = 10 + 6 2. Solve the pair:
4x = 16 6a + 2b = 18 and 5a + 6b = 28
x= 4
(x, y ) = ( 4, - 6) Solution

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6a + 2b = 18……………...(1) Name the equations as
5a + 6b = 28 ……………..(2) 3x + y = 10 …………………..(1)
2x + 2y = 4 ………………….. (2)
3 × eqn (1);
18a + 6b = 54 ……..…….(3) Obtain equal co efficient of x in both equations;
2 × eqn (1);
eqn (3) – eqn (2); 6x + 2y = 20………………….. (3)
13a = 26
a = 13 3 × eqn (2);
6x + 6y = 12 …………………..(4)
Put a = 13 in eqn (1);
6(13) + 2b = 18 eqn (3) – eqn (4);
78 + 2b = 18 (6x – 6x) + (2y – 6y) = (20 – 12)
2b = - 60 − 4y = 8,
b = -30 y = = −2
(a, b) = 13, - 30)
Put y = −2 in eqn (1);
Type 3: Uncommon coefficient 3x + (−2) = 10
If the co - efficients are unequal, obtain equal 3x – 2 = 10
ones by; 3x = 10 + 2
I. Multiplying the co – efficient of x in eqn (1) by 3x = 12
all the terms of equation (2) and likewise, x=4
multiply the co – efficient of x in eqn (2) by all (x, y) = ( 4, -2)
the terms of equation (1). This gives rise to two
new equivalent equations (3) and eqn (4) 2. Solve 4x + 3y = 6 and 3x + 2y = 3
respectively with common coefficient in one simultaneously;
variable. The coefficient of y can also be used in a
similar fashion to obtain eqns (3) and (4). Solution
4x + 3y = 6 ---- (1)
II. Taking eqn (3) and eqn (4) into consideration, 3x + 2y = 3 ---- (2)
this bring us back to type 1. Therefore, eliminate
the variable with the same co – efficient either by 3 × eqn (1);
adding or subtracting one equation from the other 12x + 9y = 18 …………………(3)
respectively. 4 × eqn (2);
12x + 8y = 12 …………………(4)
Worked Examples
1. Solve the following pair of equations eqn (3) – eqn (4);
simultaneously; 3x + y = 10 and 2x + 2y = 4 (12x – 12x) + (8y – 6y) = 18 – 6
2y = 12,
Solution y= = 6.

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Put y = 6 in eqn (1); 6x + 10y = -12………….(3)
4x + 3(6) = 6
4x + 18 = 6 3 × eqn (2);
4x = 6 – 18 6x – 9y = 45 …………..(4)
4x = −12,
x=− = −3 eqn (3) – eqn (4);
The truth set = {(x, y): x = −3, y = 6} 19y = -57
y = -3
3. If 3x – 7y = 2 and x – 2y = 4, what are the
values of x and y. Put y = -3 in eqn (2);
2x – 3(-3) = 15
Solution 2x + 9 = 15
3x – 7y = 2 ………………..(1) 2x = 15 – 9
x – 2y = 4………………….(2) 2x = 6
x =3
1 × eqn (1); (x, y) = (3, -3)
3x – 7y = 2 ……………….. (3)
Exercises 16.1
3 × eqn (2); A. Solve by elimination method;
3x – 6y = 12 ……………… (4) 1. 3x – 4y = 2 and 7x + 2y = 16
2. 3x – 5y = 19 and 4x – 7y = 26
eqn (4) – eqn (3);
3. 3a – 2b = 5 and a + 4 b = 4
(3x - 3x) – 6y – (−7y) = 12 – 2
4. 3x + 4y = 5 and 5x + 7y = 3
− 6y + 7y = 10
5. 3x – 5y = 8 and x + 2y = 10
y = 10
B. Solve the following by elimination;
Put y = 10 into eqn (2);
1. 5x + 2y = 11 and 3x + 2y = 9
x – 2(10) = 4
2. 3x + 4y = 23 and 2x + 4y = 18
x – 20 = 4
3. 6x + 2y = – 6 and 4x + 3y = 1
x = 4 + 20 = 24
4. – x + 3y = 3 and 2x + 3y = –11
The truth set = {(x, y) : x = 24, y = 10}
5. x + 5y = – 12 and 2x – y = 9
6. 7x – 4y = 40 and 4x + 5y = 43
4. Solve these equations simultaneously;
3x + 5y = - 6 and 2x – 3y = 15
C. Solve the system of equations:
1. 5x + 2y = 14 and 6x + 2y = 16
Solution
2. 7x + 3y = 27 and 6x + 3y = 24
3x + 5y = - 6……………(1)
3. 10x – 2y = 30 and 3x – 2y = 2
2x – 3y = 15……………(2)
4. 9x – 6y = 42 and 6x – 6y = 18
2 × eqn (1); 5. 5x + 3y = 69 and 7x – 3y = 75

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Substitution Method Make y the subject of eqn (1).
Method 1 ⇒y = 10 – 3x
I. Make one variable the subject of any of the
equations and substitute the derived equation into Put/substitute y = in eqn (2);
the other equation. This means that when a ( (y is eliminated).
variable is made the subject of equation (1), the (
derived equation is substituted in equation (2) and
vice – versa.
II. At this stage, one variable is eliminated. The - = - 16
result is then simplified to get the value of the x= =4
existing variable.
III. Find the value of the eliminated variable by Put x = 4 in eqn (1) to obtain the value of y
substituting the value of the known variable in (
any of the equations to obtain the ordered pairs 12 + y = 10
that satisfy both equations. y= = -2
Truth set = {(x, y) : (4, )}
Method 2
I. Choose a preferred variable. Method 2
II. Make that variable the subject of eqn (1) and 3x + y = 10 ………………...(1)
name the derived equation as eqn (3) 2x + 2y = 4………………...(2)
III. Make that same variable the subject of
equation (2) and name the derived equation as Make y the subject of eqn (1);
eqn (4). ⇒ y = 10 – 3x ………….….(3)
IV. At this point, equate equation (3) and
equation (4) and solve to obtain the value of Make y the subject of eqn (2);
2y = 4 – 2x
theinvolving variable.
V. Put the value of the variabke obtained in either y= ……………………(4)
equation (1) or (2) to obtain the value of the other eqn (3) = eqn (4);
variable. 10 – 3x =
VI. Express the answer as ordered pairs (x, y). 2(10 – 3x ) = 4 – 2x
20 – 6x = 4 – 2x
Worked Examples - 6x + 2x = 4 – 20
1. Solve: 3x + y = 10 and 2x + 2y = 4 by - 4x = - 16
method of substitution. x=4

Solution Put x = 4 in eqn (1);

Name the equations as; 3(4) + y = 10
3x + y = 10 ……….(1) 12 + y = 10
2x + 2y = 4………..(2) y = 10 – 12 = - 2
(x, y) = (4, -2)

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2. What are the values of xand y in the following y = 3 – 2x
equations?
3x + 2y = 3 and 4x + 3y = 6 Put y = 3 – 2x into eqn (2);
(
Solution
3x + 2y = 3……... (1)
4x + 3y = 6………(2)
x=
Make y the subject of eqn (1);
2y = 3 – 3x Put x = into eqn (1);
= ( ) =3
y =

Put y = in eqn (2); Truth set = {(x, y) : ( , 2)}

4x + 3 ( )=6
Exercises 16.2
2 × 4x + 2 × 3( )=2×6 A. Solve the following by substitution;
8x + 3(3 – 3x ) = 12 1. 7x – 3y = 40 and 4x + 5y = 43
8x + 9 – 9x = 12 , 2.3x – 2y = -8 and 4x + 4y = -14
8x – 9x = 12 – 9 (Grouping like terms) 3. and
-x=3 4.8x – 2y = - 14 and
x=-3 5. and

Put x= -3 into eqn (1); B. Solve by any preferred method;

3(-3) + 2y = 3 1. 4x + 5y = -8 and 3x + y = 5
- 9 + 2y = 3 2. 2x + 7y = 4 and x + 3y = 1
2y = 3 + 9 3. 6x + 11y = 16 and 2x – 3y = - 8
4. 9x + 7y = 8 and – 4x + 3y = 27
2y = 12
5. 2x - 9 y = 22 and 8x – 11y = 48
y= =6
Truth set = {(x, y) : (- 3, 6)} Solved Past Questions
1. If 3x – y = 5 and 2x + y = 15, evaluate;
3. What are the values of x and y if y + 2x = 3 x2 + 2y
and 3y – 4x = 4 . Use substitution method.
Solution
Solution 3x – y = 5 ……….(1)
y + 2x = 3 ……….. …(1) 2x + y = 15………(2)
3y – 4x = 4….……….. (2)
eqn (1) + eqn (2);
Make y the subject of eqn (1) 5x = 20

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x=4 (2x – x ) + (3y – 3y ) = (8 – 7)
x=1
Put x = 4 in eqn (1);
3(4) – y = 5 Put x = 1 in eqn (1);
12 – y = 5 2(1) + 3y = 8
12 – 5 = y 2 + 3y = 8
3y = 8 – 2
y =7 3y = 6
(x, y) = (4, 7) y=2
(x, y) = (1. 2)
x2 + 2y = (4)2 + 2(7) = 30
2. Find the values of a and b that satisfy the
Involving Fractional Equations
equations; a + b = - 4 and a – b = 1
1. When the equations involve one fraction,
multiply each term of equation by the
denorminator of the fraction to obtain equivalent Solution
equation a + b = - 4……………(i)
2. When the equations involve one or more
fractions, find the L.C.M. of the fractions and a– b = 1…………….(ii)
multiply the L.C.M. by each term of the equation
to obtain equivalent equation(s). Multiply through eqn (1) by L.C.M = 10
(10) a + (10) b = - 4 (10)
Solve the equivalent equations by any preferred 2(6a) + 5(3b) = - 40
method unless stated. 12a + 15b = - 40 …………….(1)

Worked Examples Multiply through eqn (2) by L.C.M = 12

1. Solve the system of equations;
(12) a – (12) b = (12)
x + y = 2 and x + 3y = 7
4a - 3(5b) = 12
4a – 15b = 12 …………….(2)
Solution
1. x + y = 2…………….(i) Solving eqn (1) and eqn (2) simultaneously;
x + 3y = 7 ………………(ii) 12a + 15b = - 40 …………….(1)
4a – 15b = 12 ………………..(2)
Multiply through eqn (1) by L. C. M = 4
( eqn (1) + eqn (2);
x + (4) y = (4) 2
16a = - 28
2x + 3y = 8…………………………….(1)
a=
x + 3y = 7 …………………………..…(2)

eqn (1) – eqn (2); Put a = in eqn (1)

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12( ) + 15b = - 40 3x = 12
x=4
12(-7) + 4(15b) = - 40(4)
(x, y) = (4, 7)
- 84 + 60b = - 160
60b = - 160 + 84
4. Solve the simultaneous equations;
60b = - 76
2x – 3y = 5 and + = – 2
b=
(a, b) = ( ) Solution
2x – 3y = 5……………(1)
3. Solve the simultaneous equations
3x – 2y + 2 = 0 and = + =–2
(6) + (6) = – 2 (6)
Solution 3x + 2y = – 12 ………….(2)
3x – 2y + 2 = 0…………..(i)
Re- arrange to obtain; eqn (1) × 3
3x – 2y = – 2 ……………(1) 6x – 9y = 15……………..(3)
= ……………….(ii)
eqn (2) × 2
Multiply through by L.C.M of 3 and 5 = 15 6x + 4y = – 24……………(4)
15 × = 15 ×
eqn (3) – eqn (4)
5(x – 1 ) = 3(y – 2 )
- 13y = 39
5x – 5 = 3y – 6
5x – 3y = - 6 + 5 y= =–3
5x – 3y = - 1……………….(2)
Put y = – 3 in eqn (1)
eqn (1) × 5 2x – 3(-3) = 5
15x – 10y = – 10…………..(3) 2x + 9 = 5
2x = 5 – 9
eqn (2) × 3 2x = - 4
15x – 9y = - 3……………..(4) x= =–2
(x, y) = (-2, -3)
eqn (3) – eqn (4);
-y=-7
Solved Past Question
y=7
1. If p + q = 1 and p – q = 7, find (p + q)
Put y = 7 in eqn (1)
Solution
3x – 2(7) = – 2
3x – 14 = – 2 p + q = 1 ……………..(1)
3x = – 2 + 14 p – q = 7………………(2)

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2 × eqn (1) Worked Examples
p + 2 q = 2……………..(3) 1. Solve the simultaneous equations
xy = 12 and x + y = 7
eqn (2) – eqn (3)
q=5 Solution
xy = 12……….(1)
–5q = 2 × 5
x + y = 7 …… (2)
–5q = 10
From eqn (1)
q = -2
x=
Put q = -2 in eqn (1)
p + (-2) = 1 Put x = in eqn (2)

p=1+2 + y = 7 (Multiply through by y)

p=3 12 + y2 = 7y
y2 – 7y + 12 = 0
p=2×3=6
(y2 – 4y) – (3y + 12) = 0
(p, q) = (6, -2)
y(y – 4) – 3(y – 4) = 0
(p + q) = 6 + (-2) = 4
(y – 3) (y – 4) = 0
y – 3 = 0 or y – 4 = 0
Exercises 16.3
y = 3 or y = 4
Find the truth set;
1. + y=1 and =7 From eqn (2), when y = 3
2. x + y = 2 and 4x – y = 5 x+3=7
x=7–3 =4
3. (x – 1) + y = 6 and (y – 1) + x = 8
4. y = (x – 2) and x + y = 10 When y = 4,
x+4=7
Challenge Problems x=7–4 =3
Solve the pair of equations simultaneously; Truth set = {(x, y) : (4, 3) and (4, 3) }
1. 2x + y = 3(y – x ) + 7 and = 2 –
2. Solve the equations:
2. – = 4 and x + =
x – y = -2 and xy = 24
Involving Quadratics
Solution
If one of the involving equations is operated by a
x – y = -2 ……………(1)
multiplication sign, or one is a quadratic, the
xy = 24……………….(2)
equations can only be solved by substitution
method, which in turn becomes a quadratic
From eqn (1);
equation, solved by factorization or method of
x = -2 + y
completing squares.

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Put x = -2 + y in eqn (2) y = -3
(-2 + y)y = 24
-2y + y2 = 24 Put x = - 4 in eqn (1);
y2 – 2y – 24 = 0 -4–y=5
(y – 6) (y + 4) = 0 -y=5+4
y = 6 or y = - 4 y=-9
Truth set = {(x, y) : (2, -3) and (- 4, - 9) }
Put y = 6 in eqn (2)
6x = 24 Exercises 16.4
x=4 A. Solve the following systems of equations:
1. y2 = 4x and y = x
Put y = - 4 in eqn (2) 2. xy = 64 and 4x – y = 60
2
- 4x = 24 3. y = 4x + 1 and y = x + 1
x=-6
Truth set = {(x, y) : (4, 6) and (- 6, - 4) } B. Solve the simultaneous equations
1. y + x = 11 and x2 + y2 = 61
3. Find the truth set of the equations 2. y = 2x + 5 and x2 + y2 = 5
x – y = 5 and x2 + 2y = -2 3. y = 2x – 6 and x2 + y2 = 72
4. y = x + 2 and x2 + y2 = 34
Solution 5. x2 + y2 = 40 and y = x – 4
x – y = 5 ……….(1)
x2 + 2y = -2 …….(2) Challenge Problems
Solve the pair of equations;
From eqn (1); 1. + = 3 and + = 7
-y = 5 – x
y = - (5 – x) 2. + = 2 and – = 3
y = -5 + x
3. = 2 and =
Put y = -5 + x in eqn (2); 4. = =
x2 + 2(-5 + x) = -2
x2 – 10 + 2x = -2 5. = =
x2 + 2x – 10 + 2 = 0
Involving Exponential Equations
2
x + 2x – 8 = 0 When given an exponential equation, apply the
(x – 2) (x + 4) = 0 (by factorization) law of indices and reduce it to a linear equation or
⇒x – 2 = 0 or x + 4 = 0 a quadratic equation. Then solve by any preferred
x = 2 or x = - 4 method unless stated

Put x = 2 in eqn (1); Worked Examples

2–y=5 1. Solve the simultaneous equations;
-y = 5 – 2 = 27 and (
=

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Solution eqn (3) + eqn (4);
= 27 3x = 6
= x=2
2x – y = 3……………(1)
Put x = 2 in eqn(1)
(
= 5(2) – 4y = 6
10 – 4y = 6
(
= - 4y = 6 – 10
( - 4y = - 4
=
2(x + y) = - 6 y=1
2x + 2y = - 6 ……….(2) (x, y) = (2, 1)

eqn (1) – eqn (2); Exercises 16.5

-3y = 9 A. Solve the following pair of equations
y= = -3 1. = 25 and

2. 2a + 3b = 3 and =
Put y = -3 in eqn (1) 2
3. y = x – 4 and y = 2x – 1
2x – (-3) = 3
4. x + 6y = 3 and =
2x = 0
x=0 5. = 243 and =
Truth set = {(x, y) : (0, -3) }
Graphical Method
2. Solve simultaneously, the equations: An equation of the form ax + by + c = 0 is a
( linear equation. Its graph can be drawn by finding
5x – 4y = 6 and =
the points where it cuts the x andy – axes. The
coordinates of the points of intersection (x,y) of
Solution
the graph of two linear relations is the truth set of
5x – 4y = 6…………..(1)
the two relations. This is easily found by going
(
= through the following steps:
(
= I. Find where the straight line cut‟s the x and y –
(
= ( axes. To find the intercept on the y – axis, put x =
3(y – x ) = 3(-1) 0, and solve for y in the given relations. To find
3y – 3x = -3 intercept on x– axis, put y = 0 and solve for x in
-3x + 3y = -3…………(2) both relations.
II. Prepare two separate tables of values foreach
3 × eqn (1); relation using x and y values obtained.
15x – 12y = 18……….(3) III. Plot the points on a graph sheet and join them
with a ruler, to obtain two straight lines that
4 × eqn (2); intersects each other.
-12x + 12y = -12………(4 )

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IV. Locate the point of intersection of the two x 6 0
lines. The co-ordinates of this point is the y 0 6
solution set of the equations.
For x – y = 2
Worked Examples Intercept on x – axis, y = 0
1. Find the truth set of x+ y = 6 and x – y = 2 by
graphical method. When y = 0, x – 0 = 2 , x = 2
(x, y) = (2, 0)
Solution
For x + y = 6 Intercept on y – axis, x = 0
Intercept on the x – axis , When x = 0, 0 – y = 2, y = -2;
When y = 0, , x + 0 = 6, x = 6 (x, y) = (0, -2)
(x, y) = (6, 0)
Table of values for x – y = 2
Intercept on the y – axis,
When x = 0, 0 + y = 6, y = 6 x 0 2
(x, y) = (0, 6) y -2 0
Table of values for x + y = 6

y
6
Scale =2cm : 2units on x-axis
and 2cm : 1unit on y 5

2 (4,2)

-10 -8 -6 -4 -2 2 4 6 8 10
-1

-2

-3

-4

-5

-6

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From the graph, the two lines meet at the point Put in eqn (2;)
where x = 4 and y = 2 (
The truth set is: x = 4, y = 2

Exercises16.6
A. Solve by graphical method. x= =8
1. y = 5 – x and y = 2x – 1
2. +2 Put x = 8 in eqn (2);
3.
4.
5. y – =3 and + =5
Truth set = { ( x, y ) : ( 8 , 5 )}
Challenge problems
Method 2 (elimination)
Show that the straight lines with equations;
(
3x + 4y = 2, 5x – 2y = 12 and 2x + 3y = 1, all
(
intersect at the same point. Illustrate in a sketch.
eqn (1) + eqn (2);
Word Problems
( (
Equations involving two variables may be stated
in words. Students are advised to write
mathematical equations to represent the
=8
statements and solve them by any preferred
method unless specified. It is also advisable to
Put x = 8 in eqn (2);
use x and y for the involving variables.
8 + y = 13
Worked Examples y = 13 – 8 = 5
1. The sum of two numbers is 13 and their Truth set = {(x, y) : (8, 5)
difference is 3. Find the two numbers.
2. The sum of two numbers is and their
Solution difference is 11. Find the numbers.
Method 1: (Substitution)
Let x and y be the two numbers. The involving Solution
equations are; Let x andy be the two numbers.
( ……….. (1)
( (
By elimination,
Make y the subject of eqn (1)
eqn (1) + eqn (2)
( (

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 x + y = 40………………(1)
2x = - 8 5x + 50y = 1,280 (changed to pesewas)
x=-4
Divide through by 5
Put x = - 4 in eqn (1) x + 10y = 256…………..(2)
eqn (2) – eqn (1);
y 9y = 216
Truth set = {(x, y) : ( - 4 , - 15) } y= = 24

3. Mr. Brown bought 3 rulers and 6 pencils all for Put y = 24 in eqn (1);
90p. When he purchased 4 rulers and 2 pencils, x + 24 = 40
he paid the same amount. Determine the cost of 1 x = 40 – 24 = 16
ruler and 1 pencil. Therefore, there are 24, 50p coins and 16, 5p
coins.
Solution
Let r represent ruler and p represent pencil 5. A family of three adults and two children paid
3r + 6p = 90……………(1) Gh¢8.00 for a journey. Another family of four
4r + 2p = 90……………(2) adults and three children paid Gh¢11.00 as the
fare for the same journey. Calculate the fare for:
eqn (2) × 3 i. an adult,
12r + 6p = 270…………(3) ii. a child,
iii. a family of four adults and five children.
eqn (3) – eqn (1);
9r = 180 Solution
r = 20 i. Let a represent adult and c represent children
3a + 2c = 8……….(1)
Put r = 20 in eqn (1); 4a + 3c = 11………(2)
3(20) + 6p = 90
6p = 30 4 × eqn (1);
p=5 12a + 8c = 32……….(3)
Therefore the cost of one ruler is 20p and the cost 3 × eqn (2)
of one pencil is 5p. 12a + 9c = 33………(4)

4. A pile of 40 coins consist of 5p coins and 50p eqn (4) – eqn (3);
coins. If the total sum of money is Ghȼ12.80, find c=1
the number of each kind of coin in the pile.
ii. Put c = 1 in eqn (1);
Solution 3a + 2(1) = 8……….(1)
Let x represent the 5p coin and y represent the 3a + 2 = 8
50p coin. 3a = 8 – 2

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3a = 6 7. The product of two positive numbers is 20. The
a=2 sum of squares is 41. Find the numbers.
Therefore, the fare for an adult is Gh¢2.00 and
the fare for a child is Gh¢1.00 Solution
Let the numbers be x and y;
iii. 4a + 5c xy = 20……………………..(1)
= 4(2) + 5(1) x2 + y2 = 41………………...(2)
=8+5
= 13 From eqn (1);
The fare for a family of four adults and five y= ………………(3)
children is Gh¢13.00

Put y = in eqn (2);

6. If the price of a book is reduced by Ghȼ5.00, a
person can buy 5 more books for Ghȼ300.00. + ( ) = 41
Find the original price of the book.
+ = 41
Solution ( + 400 = 41x2
Let the price of a book be x ( – 41x2 + 400 = 0
Number of copies of the book be y
xy = 300……………………..(1) Let x2 = k
(x – 5) (y + 5) = 300…………(2) – 41k + 400 = 0
(k – 16) (k – 25) = 0
From eqn (1); k = 16 or k = 25
y=
Now;
Put y = in eqn (2); x2 = 16 or x2 = 25
x2 – 16 = 0 or x2 – 25 = 0
( – )( ) = 300 x2 – 42 = 0 or x2 – 52 = 0
Applying difference of two squares;
x( )–5( ) = 300
x = 4, x = - 4 or x = 5, x = -5
300 + 5x – – 25 = 300 x = 4 or x = 5 (positive numbers)
5x – = 25
Put x = 4 or x = 5 in eqn (3)
5x2 – 1500 = 25x
When x = 4, y = = 5;
5x2 – 25x – 1500 = 0
When x = 5, y = =4
x2 – 5x – 300 = 0
The numbers are 4 and 5.
(x + 15) (x – 20) = 0 (by factorization)
x = - 15 or x = 20 Some Solved Past Questions
The original price of a book is Ghȼ20.00 1. The cost of a packet of sugar is x cedis and the

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cost of a tin of milk is y cedis. If 3 packets of Put eqn (1) into eqn (2);
sugar and 4 tins of milk cost Gh¢635.00 and 4 120( )t + 15t = 180
packets of sugar and 3 tins of milk cost
60t + 15t = 180
Gh¢695.00, write two equations connecting x and
75t = 180
y. Hence, find x and y.
t= = 2.4
Solution
3x + 4y = 635…………...(1) Put t = 2.4 eqn (1);
4x + 3y = 695…………...(2) c = (2.4) = 1.2
4 × eqn (1);
Therefore each teacher paid Ghȼ2.40 and each
12x + 16y = 2540……….(3)
child paid Ghȼ1.20
3 × eqn (2)
ii. Let A represent the number of students
12x + 9y = 2085………..(4)
required to go on the excursion
A (c) + 20t = 240
eqn (3) – eqn (4);
But c = 1.2 and t = 2.40
7y = 455
1.2A + 20 (2.40) = 240
y= = 65 1.2A + 48 = 240
1.2 A = 240 – 48
Put y = 65 in eqn (1); 1.2A = 192
3x + 4(65) = 635……….(1) A= = 160
3x + 260 = 635
Therefore, 160 students will go on the excursion
3x = 635 – 260
with 20 teachers
3x = 375
x= = 125 Exercise 16.7
x = Gh¢125.00 and y = Gh¢65.00 1. The sum of two numbers is 26 and their
difference is 4. Find the two numbers.
2. A train fare for a school child is half the fare of
a teacher. The total fare for 120 children and 15 2. Two numbers sum up to 30. If the difference
teachers for an excursion is Ghȼ180.00. between the numbers is 8,find the numbers.
i. Find the fare of a child
ii. How many children will go on the excursion 3. There are two numbers x and y. Six times the
with 20 teachers for a total fare of Ghȼ240.00? larger number exceeds twice the smaller number by
156. If the difference of the two numbers is 18, find
Solution the numerical strength of x and y.
Let a child be represented by c and a teacher by t
c = t…………………………(1) 4. A certain number decreased by half of another
120c + 15t = 180 …………….(2) number is -3. When twice the second number is

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subtracted from twice the first number, the result Challenge Problems
is -2. Identify the two numbers. 1. In a pub, brandy is 10p more expensive than
whisky. A round of three brandies and 5
5. I bought 12 pencils and 10 rulers for Gh¢2.10. whiskies cost Ghȼ7.10. Let x be the cost of a
At the same place, I bought 20 pencils and 4 brandy and y the cost of a whisky. Find the cost
rulers for Gh¢1.60.What is the price of one pencil of a whisky.
and that of one ruler?
2. There are five more boys than there are girls in
6. There were 200 people at a concert. Some paid a class. If there were one more girl in the class,
30p each and the rest paid 20p each and the total the ratio of boys to girls would be 5 to 4. How
takings were Ghȼ50.80. How many paid 20p? many boys and girlsare in the class?

7. Mr. Ben saves money by putting every 50p and 4. The expression ax – by has the value 6 when x
every 20p coin he receives in a box. After a = 4 and y = 3. It also has the value 6 when x = 2
while, he finds out that he has 54 coins, and y = . Find the values of a and b.
amounting to Ghȼ17.10. How many 50p coins
does he have?
5. Four years ago, a man was six times as old as
his son, but 5 years‟ time, he will be only three
8. The price of admission to an Obra show was
times as old as his son. What are their ages now?
Ghȼ100.00 for an adult and Ghȼ50.00 for
children. The total amount raised from the sale of
6. If the price of a book is reduced by Ghȼ4.00,
600 tickets was Ghȼ50,000. Find:
Rita can buy 8 more books for Ghȼ192.00. Find
i. the number of adults admitted;
the original price of the book.
ii. the number of children admitted.
7. If 2 is added to the numerator and denominator
9. At a concert 500 tickets were sold: the cheaper
ones cost Ghȼ5.00 and the more expensive one of a certain fraction, it becomes and if 3 is
Ghȼ9.00. The total receipts were Ghȼ3,220. Let x subtracted from the numerator and denominator
and y be the numbers of cheap and expensive of the same fraction, it becomes . Find the
tickets respectively. Form two equations in x and
fractions. Ans
y, and hence find how many cheap tickets were
sold.
8. If twice the age of a son is added to the age of
a father, the sum is 56. But if twice the age of the
10. The value of a fraction expressed as is . If
father is added to the age of son, the sum is 82.
3 is subtracted from the numerator and added to Find the ages of the father and son.
the denominator, its value becomes . Find the
values of p and q. 9. The produt of two positive integers is 45 and
the sum of their square is 106. Find the numbers.

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Applications to a Number and its Reverse Let the reverse of the digit be yx.
Take the two – digit number, 42 from instance. = 10y + x is 27 greater than the given number;
The place value of 4 is tens and the place value of ⇒10y + x = 27 + 10x + y
2 is unit. Thus, 42 can be expanded as: 10y – y + x – 10x = 27
42 = 40 + 2 9y – 9x = 27 (Divide through by 9)
= 4(10) + 2(1) y – x = 3…………(2)

If 42 is reversed, we obtain 24, where 2 is the Put x = in eqn (2);

tens digit and 4 is the unit digit also expanded as:
y– =3
24 = 20 + 4
= 2(10) + 4(1) 6y – 3y = 3 × 6
3y = 18
Similarly, given the two digits number, xy, it can y= =6
be expanded to obtain:
xy = x(10) + y (1) Put y = 6 in eqn (1)
(
= 10x + y x= =3
If xy is reversed, yx is obtained, also expanded as; Therefore, the number xy = 36
yx = y (10) + x (1)
= 10y + x 2. The sum of the digits of a two digit number is
Therefore, to solve problem involving an 9. The number is 27 more than the original
unknown two digit number and its reverse, number with its digit reversed. Find the number.
express them in the expanded form as shown
above and form the appropriate equation(s) Solution
Let the two digit number be xy
Worked Examples x + y = 9…………..(1)
1. A number with two digits is equal to four 10x + y = 27 + 10y + x
times the sum of its digits. The number formed by
reversing the order of the digit is27 greater than 10x – x + y – 10y = 27
the given number. Find the number. 9x – 9y = 27
x – y = 3………….(2)
Solution
Let the two digit number be xy. eqn (1) – eqn (2);
⇒ xy = 10x + y 2y = 6
Four times the sum of the digits = 4(x + y) y=3
⇒ 10x + y = 4(x + y) Put y = 3 in eqn (1);
10x + y = 4x + 4y x+3=9
10x – 4x = 4y – y x=9–3 =6
6x = 3y Therefore, the number xy = 63
x= ………………(1)

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3. A two digit number is 6 more than 4 times the From left to right, the digits form consecutive
sum of its digits. The digits from the left to the integers
right are consecutive even integers. Find the ⇒If one is x, the other (y) is x + 1
number. Substitute y = x + 1 in eqn (1)
10x + y = 5(x + y)
Solution 10x + (x + 1) = 5(x + x + 1)
Let the two digit number be xy 10x + x + 1 = 5(2x + 1)
xy = 6 + 4(x + y) 11x + 1 = 10x + 5
10x + y = 6 + 4(x + y)………..(1) 11x – 10x = 5 – 1
x = 4,
For consecutive even integers, if one is x, the But y = x + 1 = 4 + 1 = 5
other (y) is x + 2 The number is xy = 45
Substitute y = x + 2 into eqn (1)
10x + (x + 2) = 6 + 4(x + x + 2) Exercises 16.8
10x + x + 2 = 6 + 4(x + x + 2) 1. Find a number of two digits which exceeds
11x + 2 = 6 + 4(2x + 2) four times the sum of its digits by 3 and which is
11x + 2 = 6 + 8x + 8 increased by 18 when its digits are interchanged
11x – 8x = 6 + 8 – 2
3x = 12 2. If a certain number with two digits is divided
x= =4 by the sum of the digits, the quotient is 6 and the
remainder is 5. The difference between the given
But y = x + 2 = 4 + 2 = 6
number and the number formed by reversing the
Therefore, the number xy = 46
digit is 18. Find the given number
4. A two digit number is 5 times the sum of its
3. The unit digit of a two digit number is two less
digits. The digits from left to right form
than the tens digit. The number is two more than
consecutive integers. Find the digits.
six times the sum of the digits. Find the number.
Solution
4. A two digit number is five times the sum of its
Let the two digits number be xy
digits. When 9 is added to the number, the results
⇒xy = 5(x + y)
is the original number with its digits reversed.
10x + y = 5(x + y) ……….(1)
Find the number.

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17 PERCENTAGES II Baffour – Ba series

Compound Interest P = Ghȼ300, T = 3 years R = 20% I = ?

It is a kind of interest that accrues at the end of a Interest at the end of 1st year
given period and serves as an element of = 300, T = 1, R = 20%
expanding or strengthening the principal at the = = = Ghȼ60.00
beginning of another or a subsequent period of
transaction. In this situation, an initially earned
Amount at the end of 1st year
interest is added to the initially deposited
= +
principal, thereby creating a higher principal
= Ghȼ300 + Ghȼ60 = Ghȼ360.00
which then earns higher interest at the end of the
stipulated period.
Interest at the end of 2nd year
= 360 T = 1 R = 20%
The Compound Interest Formula
A. Payment without Instalment = = = Ghȼ72.00
Here, compound interest is calculated or payable
at the end of each year. Compound Interest is Amount at the end of 2nd year
equal to the sum of interest at the end of each = +
year over the given number of years. = Ghȼ360 + Ghȼ72 = Ghȼ432.00

Hence, if a principal, p, is deposited or borrowed Interest at the end of 3rd (final) year
at an interest rate, r, and compounded yearly for a = = = Ghȼ86.40
period of t years, the accumulated value, A, is:
A = p(
Amount at the end of 3rd (final) year
= +
The compound interest, I, is calculated by the the
= Ghȼ432 + Ghȼ86.40 = Ghȼ518.40
formula : I = A – P
Compound Interest over the 3 – year period
If an accumulated value, A, is desired after t years
= 3rd year amount – First (initial) Principal
and the money is deposited at an interest rate, r,
= –
and compounded yearly, the present value is
= Ghȼ(518.40 – 300.00) = Ghȼ218.40
P=(
Method 2
Worked Examples P = Ghȼ300, T = 3years, R = 20% , I = ?
1. Find the compound interest on Ghȼ300.00 for Interest at the end of 1st year
3 years at 20% per anum = Ghȼ300, T = 1 and R = 20%

Solution = = = Ghȼ60.00
Method 1

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Amount at the end of 1st year Solution
= + P = Ghȼ800, T = 2, R = 3 % = 3.5% and I = ?
= Ghȼ300 + Ghȼ60 = Ghȼ360.00 Interest at the end of first year
= = = Ghȼ28.00
Interest at the end of 2nd year
= Ghȼ360.00 T = 1 and R = 20%
Amount at the end of first year
= = = Ghȼ72.00
= +
= Ghȼ800 + Ghȼ28 = Ghȼ828.00
Amount at the end of 2nd year
= + Interest at the end of second (final) year
= Ghȼ360 + Ghȼ72 = Ghȼ432.00
= = = Ghȼ28.98

Interest at the end of 3rd (final) year Amount at the end of second (final) year
= +
= = = Ghȼ86.40
= Ghȼ828 + Ghȼ28.98 = Ghȼ856.98

Amount at the end of 3rd (final) year Compound interest

= + = 2nd (final) year amount – 1st (Initial) principal
= Ghȼ(432.00 + ȼ86.40) = Ghȼ518.40 =
= Ghȼ856.98 – Gh¢800.00= Ghȼ56.98
Compound Interest is equal to the sum of interest
at the end of each year over the 3 – year period. Alternatively
That is: Compound Interest
Compound Interest = 1st year Interest + 2nd year = 1st year Interest + 2nd (final) year interest
interest + 3rd (final) year interest = +
= + + = Ghȼ28 + Ghȼ28.98 = Ghȼ56.98
= Ghȼ(60 + 72 + 86.40) = Ghȼ218.40
Exercises 17.1
Method 3(using the formula) A. Find the compound interest on:
P = Ghȼ300, t = 3 years, r = 20%, A = ? 1. Ghȼ3,200.00 for 3 years 12% per anum.
A = p( 2. Ghȼ2,400.00 for 4 years at 17.5% .
A = 300 ( ) (By substitution) 3.Ghȼ50,200.00 for 3 years at 15% .

A = 300 ( = 300 × 1.728 = 518.40

B. 1. A man deposited Ghȼ80,000.00 at a bank at
12% compound interest per anum. Find his total
Compound interest, I = A – P
amount at the end of the third year.
= Ghȼ(518.40 – 300.00) = Ghȼ218.40
2. Mr. Brown borrowed Ghȼ2,500.00 from
2. Find the compound interest on Ghȼ800.00 for
Bayport financial services for 4 years at 28%
2 years at 3 % per anum. compound interest per anum. Calculate the

Baffour – Ba Series Core Maths for Schools and Colleges Page 459
interest and amount to be paid at the end of the Amount at the end of first year;
period. Ghȼ12,000 + Ghȼ1,200 = Ghȼ13,200

3. Mrs. Kaya deposited an amount of 10% of Ghȼ13,200 = Ghȼ1,320

Ghȼ1,320.00 at a bank at 17 % per anum. Find Amount at the end of second year;
Ghȼ13,200 + Ghȼ1,320 = Ghȼ14,520
her compound interest at the end of three years.
10% of Ghȼ14,520 = Ghȼ1,452
4. A district had 75,000 inhabitants at the Amount at the end of third year;
beginning of the year 2007. The population grew Ghȼ14,520 + Ghȼ1,452 = Ghȼ15,972
at a constant rate of 2 % per anum. What was the
population at the beginning of 2010? Exercise 17.1B
Mr. Brown borrowed Ghȼ2,500.00 from a bank at
Calculating the Time 28% compound interest per anum. How many
Identify the given principal(p) , amount (A), and years did it takehis interest to amount to
the rate (r). Ghȼ4,210.89
II. Substitute in the formula : A = p(
III. Take logarithm on both sides of the B. Compound Interest for Instalments
equation and make t the subject. However, other For instalments, payments or compound interest
alternatives can also be used. within a year are payable as follows:
1. Monthly, meaning 12 times a year.
Worked Example 2. Quarterly, meaning 4 times a year.
A man invests Ghȼ12,000.00 at a rate of 10% per 3. Half – yearly, meaning 2 times a year.
anum. At what time will his investment amount
to Ghȼ15,972.00? Method 1
I. Divide the rate, r, by the number of
Solution payments/instalment within a year.
Method 1 II. Use this rate to calculate the interest, I, and the
A = 15,972, p = 12,000,r = 10% = 0.1 t = ? amount, at the end of each instalment.
Substitute in A = p( III. Find the sum of interest for the total number
15,972 = 12,000( of instalments made to obtain the compound
interest.
=(
1.331 = ( Method 2
lg 1.331 = lg( The Compound Interest Formula
lg 1.331 = t lg 1.1 If a principal, p, is deposited or borrowed at an
t= = 3 years interest rate, r, and compounded a certain number
of times, n, within a year for a period of t years,
Method 2 the accumulated value is: A = p( )
10% of Ghȼ12,000 = Ghȼ1,200

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Where A is the amount at the end of the given = Ghȼ2,160 T = 1 and R = 8%
period, p is the principal, r is the rate , n is the = = = Ghȼ186.62
number of times per year interest is compounded
and t is the time in years.
Amount at the end of third quarter
= +
The compound interest, I, is calculated by the the
= Ghȼ2,332.80 + Ghȼ186.62 = Ghȼ2519.42
formula : I = A – P
Interest at the end of 4th quarter (3 months)
Worked Examples
= 2519.42, T = 1 and R = 8%
Find the compound interest on Ghȼ2,000.00 for 1
year at 32% per anum payable every quarter. = = = Ghȼ201.55

Solution Amount at the end of third quarter

Method 1 = +
= 2,000, R = 32%, = Ghȼ2,519.42 + Ghȼ201.55 = Ghȼ2,720.97

T = 1yr = ( ) Compound interest

If rate per 1 year = 32%, then rate per (quarter) a = + + +
= Ghȼ(160 + 172.80 + 186.62 + 201.55)
year = × 32% = 8%. Therefore rate to be used
= Ghȼ720.97
quarterly (3months) = 8% for 4 quarters
Method 2(using the formula)
Interest at the end of 1st quarter (3 months)
P = 2,000, r = 32% = 0.32, n = 4 t = 1
= 2000, R = 32% and T = 1
= = = Ghȼ160.00 Substitute in A = p( )

A = 2,000( )
Amount at the end of first quarter A = 2,000( = Ghȼ2,720.97
= +
= Ghȼ2,000 + Ghȼ160 = Ghȼ2,160.00 I=A–P
I = Ghȼ2,720.97 – Gh¢2,000.00 = Gh¢720.97
Interest at the end of 2nd quarter (3months)
= 2160, T = 1 and R = 8% 2. Find the compound interest on Ghȼ400.00 for
= = = Ghȼ172.80 2 years at 10% per anum, if interest is added half
a year
Amount at the end of second quarter
= + Solution
= Ghȼ2,160 + Ghȼ 172.80 = Ghȼ2.332.80 Method 1
Interest per 1 year = 10%
Interest at the end of 3rd quarter (3 months) Interest per a year = 5%

Baffour – Ba Series Core Maths for Schools and Colleges Page 461
If 1 year is 2 – half years Method 2(using the formula)
Then 2 years is 4 – half years P = 400, r = 10% = 0.1, n = 2, t = 2
Substitute in A = p( )
Interest at the end of first half year
= 400, R = 5% and T = 1 A = 400( )
= = = Ghȼ20.00 A = 400( = Ghȼ486.20
I=A–P
Amount at the end of first half year I = Ghȼ486.20 – Gh¢400.00 = Gh¢86.20
= +
= Ghȼ400 + Ghȼ20 = Ghȼ420.00 Exercises 17.2
A. Find the compound interest:
Interest at the end of second half year 1. Ghȼ2,060.00 for 1 year at 20% per anum,
= 420, T = 1 and R = 5%
payable every quarter
= = = Ghȼ21.00
2. Ghȼ1,540.00 for 2 years at 24% per anum, paid
half a year
Amount at the end of second half year;
= + 3. Ghȼ72,000.00 for a year, payable half a year
= Ghȼ420 + Ghȼ21 = Ghȼ441.00 at a rate of 24% per anum

Interest at the end of third half year;

B. Find the amount at compound interest:
= 441, T = 1 and R = 5% 1. Ghȼ28,000.00 for 1 year at 20% p.a payable
= = = Ghȼ22.05 half yearly.

2. Ghȼ150,000.00 for 1 years at 30% p.a

Amount at the end of third half year;
payable half yearly.
= +
= Ghȼ441 + Ghȼ22.05 = Ghȼ463.05 3. Ghȼ120,000.00 for 1 years at 22% p.a payable
half yearly.
Interest at the end of fourth half year;
= 463.05, T = 1 and R = 5% 4. Ghȼ200,000.00 for 2 years at 28% p.a payable
= = = Ghȼ23.15 quarterly.

5. Ghȼ250,000.00 for 3 years at 30% p.a payable

Amount at the end of third half year; quarterly.
= +
= Ghȼ463.05 + Ghȼ23.15 = Ghȼ486.20 Finding the Outstanding Balance or Refund
I. Calculate the interest and amount at the end of
Compound interest = + + + each period of payment.
= Ghȼ(20.00 + 21.00 + 22.05 + 23.15) II. Deduct the amount of repayment to obtain a
= Ghȼ86.20 reducing balance.

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III. Calculate the interest and amount using on the Interest at 6% for 1st half year 1,200.00
reducing balance and deduct repayment until the Amount at the end of 1st half year 21,200.00
last repayment is made. Deduct 1st repayment 9,000.00
IV. Note that a negative outstanding balance
indicates a refund. Principal for 2nd half year 12,200.00
Interst at 6% for 2nd half year 732.00
nd
Worked Examples Amount at the end of 2 half year 12,932.00
1. A man borrowed Gh¢150,000.00 at 25% per Deduct 2nd repayment 9,000.00
anum compound interest paid half yearly. The
Principal for 3rd half year 3,932.00
man agreed to repay Gh¢50,000.00 at the end of
Interst at 6% for 3rd half year 235.92
each half year. Find the amount of loan rd
Amount at the end of 3 half year 4,167.92
outstanding at the beginning of the fourth half
Deduct 3rd repayment 9,000.00
year
Outstanding Loan - 4,832.08
⇒ Refund = Gh¢4,832.08
Solution
GH¢
Exercises 17.3
Principal for first half year 150,000.00
st 1. A woman borrows Ghȼ50,000.00 at 15% per
Interest at 12.5% for 1 half year = 18,750.00
anum compound interest payable half – yearly.
Amount at the end of 1st half year = 168,750.00
She agrees to repay Ghȼ30,000.00 at the end of
Deduct 1st repayment 50,000.00
each year. Find the amount of loan outstanding at
Principal for 2nd half year 118,750.00 the beginning of the fourth half – year.
Interst at 12.5% for 2nd half year 14,843.75
Amount at the end of 2nd half year 133,593.75 2. At the end of the year, Tony‟s bank increased
Deduct 2nd repayment 50,000.00 the rate of interest from 15% to 18% per anum
payable half – yearly. How much more interest
Principal for 3rd half year 83,593.75 would Tony receive on his savings of
Interst at 12.5% for 3rd half year = 10,449.22 Ghȼ20,000.00 at the end of the year.
Amount at the end of 3rd half year 94,042.97
Deduct 3rd repayment 50,000.00 3. A man borrowed Ghȼ25,000.00 at 24% per
Outstanding Loan 44,042.97 anum compound interest payable monthly for 5
months. He paid back Ghȼ5,000.00 at the end of
2. Mr. Okra borrowed Gh¢20,000.00 at 12% per each of the first 4 months. Calculate how much
anum compound interest paid half yearly. He he paid to clear the debt at the end of the fifth
agreed to repay Gh¢9,000.00 at the end of each month.
half year. Find the amount of outstanding refund
at the end of the third half year. 4. Mr. Adanko borrowed Ghȼ2,500.00 from a
certain financial institution at a rate of 17% per
Solution anum.
GH¢ i. Find the compound interest at the end of 3
Principal for first half year 20,000.00 years.

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ii. If for every month, Mr. Adanko pays Amount at the end of 1st year
Ghȼ155.00 to the financial institution, find his = +
refund at the end of the 3 year period. = Ghȼ180 + Ghȼ27 = Ghȼ207.00

5. If you take a loan of Ghȼ7,000.00 at 21% Interest at the end of 2nd year;
compound interest and as part of the loan = Ghȼ207, T = 1 and R = 15%
agreement, you pay a fixed deposit of = = = Ghȼ31.05
Ghȼ2,400.00 every year; find the outstanding
loan at the end of three years. Amount at the end of 2nd year;
= +
Finding the Rate given the Amount at the End = Ghȼ207 + Ghȼ31.05 = Ghȼ238.05
of a Certain Year
I. Identify the principal (P) and the amount (A) Interest at the end of 3rd (final) year;
II. Find the interest (I) by the formula: = Ghȼ238.05 T = 1, and R = 15%
I=A–P = = Ghȼ35.71
III. Identify the values of P, I and T (time) and
Substitute in R = to obtain the value of R. Amount at the end of 3rd (final) year;
= +
Worked Example = Ghȼ238.05 + Ghȼ35.71 = Ghȼ237.76
1. Mr. Brown invested Ghȼ180.00 at Ghana
Commercial Bank for 3 years at compound Compound Interest over the 3 – year period;
interest. If at the end of the first year, his money = 3rd (final) year amount – Initial Principal
amounted to Ghȼ207.00, calculate: = –
i. the rate per cent per anum = Ghȼ237.76 – Ghȼ180.00 = Ghȼ93.76
ii. the total interest earned by Mr. Brown at the Total interest earned is Ghȼ93.76
end of the three – year period
Method 2
Solution Compound Interest = 1St year Interest + 2nd year
Method 1 interest + 3rd (final) year interest
i. = Ghȼ180, T = 1year, R = ? = + +
I = Ghȼ207 – Ghȼ180 = Ghȼ27.00
= Ghȼ(27 + 31.05 + 35.71) = Ghȼ93.76
Total interest earned at the end of the 3 – year
From I = ,
period is Ghȼ93.76
R= = = 15%
Depreciation
st
ii. Interest at the end of 1 year It is the loss in the value of goods or items. When
= Ghȼ180, T = 1 and R = 15% I = ? items are used for some period of time, its value
decreases (depreciates) as a result of old age. The
= = = Ghȼ27.00
value at which an item depreciates is calculated

Baffour – Ba Series Core Maths for Schools and Colleges Page 464
as a percentage of the cost price. Therefore, if the Method 3:(using the formula)
rate of depreciation is r% then; P = 1,500, r% = 10%= 0.1, n = 1
1. Depreciation = V=p(
V = 1,500( = Ghȼ1,350.00
2. Depreciation = × Cost price
( 2. The value of a T. V. set depreciated by15% of
3. New price = × Cost price. it cost price of Gh¢5,400.00. Find the value of the
T.V after 2 years.
4. Cost/original price = ( × New price
–

Depreciation value or new value,at a constant Solution

rate, at a given period of time is best calculated Method 1
by the formula:V = p ( At the end of the first year,
(
Where V is the depreciated or new value, New value = × C.P
p is the initial value, But r = 15, C.P = Ghȼ5,400
r% is the depreciation rate (
New value = × 5,400
n is the number of calculations
New value = × 5,400 = Ghȼ4,590.00
Depreciation (D) = Initial value – New value
D=P–V
⇒ New value at the end first year
= Ghȼ4,590.00
Worked Examples
1. The value of a sewing machine depreciated by
At the end of the second year;
10% of its value in one year. The cost price of the
r = 15, and C. P = Ghȼ4,590.00
machine was Gh¢1,500.00. Find its new price (
(value) when it is 1 year old. New value = × 4,590 = Ghȼ3,901.50

Solution Method 2
Method 1 At the end of the first year;
(
New price = × C.P Depreciation = × C.P,
But, r = 10, and C.P = Gh¢1,500. But r = 15 and C.P = Ghȼ5,400
(
New price = × 1,500 = Gh¢1,350.00 Depreciation = × 5,400 = Ghȼ810.00

Method 2 New value at the end of first year

Depreciation = × C.P, = Cost price – Depreciation
= Ghȼ5,400.00 – Ghȼ810.00 = Ghȼ4,590.00
But r = 10 and C.P = Ghȼ1,500
Depreciation = × 1,500 = Ghȼ150.00 At the end of the second year;
New Price = Cost price – Depreciation Depreciation = × C.P,
New price = Ghȼ1,500 – Ghȼ150 = Ghȼ1,350.00 But r = 15 and C.P = Ghȼ4,590

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Depreciation = × 4,590 = Ghȼ688.50 Second year = 25%
Depreciation = × C.P,
New value at the end of firt year But r = 25 and C.P = Ghȼ6,000
= Cost price – Depreciation Depreciation = × 6,000 = Ghȼ1,500.00
= Ghȼ4,590.00 – Ghȼ688.00 = Ghȼ3901.50
New Price = Cost price – Depreciation
Method 3:(using the formula)
New price = Ghȼ6,000.00 – Ghȼ1,500.00
P = 54,00, r% = 15%= 0.15, n =2
= Ghȼ4,500.00
V=p(
V = 5,400( = Ghȼ3,901.50
C.P at the end of second year = Ghȼ4,500.00
3. The cost of an item depreciated according to
Third year = 10%
the table below:
Depreciation = × C.P,
First year Nil But r = 10 and C.P = Ghȼ4,500.00
Second year 25% Depreciation = × 4,500 = Ghȼ450.00
Third year 10%
New Price = Cost price – Depreciation
If the cost of the item is Gh¢6,000.00, find its
New price = Ghȼ4,500.00 – Ghȼ450.00
cost at the end of three years
= Ghȼ4,050.00
Solution
4. A business man purchased a copier for
Method 1
Ghȼ8,500.00 and anticipates it will depreciate in
Cost price = Ghȼ6,000.00
value by Ghȼ1,250.00 per year.
First year = Nil, C.P = Ghȼ6,000.00
i. What is the copier‟s value after 4 years?
Second year = 25% ii. How many years will it take the copier‟s value
( to decrease to Ghȼ2,250.00?
New price = × C.P
(
= × 6,000 = Ghȼ4,500 Solution
Cost of machine Ghȼ 8,500
Third year = 10% Depreciation per year Ghȼ 1,250
( Depreciation for 4 years = 4 × Ghȼ1,250
New value = × C.P
= Ghȼ5,000
(
= × 4,500 = Ghȼ4,050.00 Copier‟s value after 4 years;
The cost of the item at the end of three years is = Gh¢8,500 – Gh¢5,000 = Gh¢3,500.00
Ghȼ4,050.00
ii. Value of copier after 5 years;
Method 2 = Gh¢3,500 – Gh¢1,250 = Gh¢2,250.00
First year = Nil Therefore, it will take 5 years for the copier‟s
C. P at the end of first year = Ghȼ6,000.00 value to decrease to Ghȼ2,250.00?

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Exercises 17.4 The original purchase value of a car is
1. The value of a refrigerator depreciated by 15% Gh¢9,000.00. What will be its official valuation
of its value each year. If the value of the during each of the first 4 years?
refrigerator was Gh¢2,040.00, calculate its new
price after 1 year. 7. The cost of an H.P. computer and a B.M.W.
vehicle depreciates with respect to the table
2. The value of a new machine depreciated each below;
year by 10% of its value at the beginning of that
year. The value of the machine when new was Year HP Computer B.M.W
Gh¢200,000.00. Find its value when it was 3 First 14% 15%
years old Second 25% 24%
Third 20% 20%
3. The value of a printer depreciates by 12% of its
value every year. Calculate the new price of the If the cost of the computer is Ghȼ1,850.00 and
calculator after three years, if its value when new the B.M.W is Ghȼ24,400.00
is Gh¢1,250.00. i. Determine the cost of the B. M. W. after 3
years
4. The cost price of a Chinese mobile phone ii. Find the cost of the computer after 3 years
depreciated 5% after one year and 10% after two iii. What is the total cost of the B.M.W. and the
years and 20% after three years. If the value of H.P computer after 3 years?
the mobile phone is Gh¢800.00, find its value
after three years. Partnership in Business
It is the situation whereby two or more people
5. A car which was bought for Gh¢9,000.00 when contribute money to set up business with the aim
new was valued at Gh¢7,500.00 at the end of the of maximizing profit. The profit accrued from
first year. It then depreciated each year by 12 % partnership isusually shared annually according
to the ratio of the contributions, responsibilities
of its value at the beginning of that year.
and other agreed terms and conditions.
Calculate:
i. the depreciation at the end of the first year;
Forms of Business Partnership
ii. the value of the car at the end of the third year.
1. Sole Proprietorship
A Sole Proprietorship is one individual or
6. In a certain country, the official valuation for
married couple in business alone. Sole
cars up to 4 years old is calculated according to
proprietorships are the most common form of
the following table:
business structure. This type of business is simple
Year after Depreciation on original to form and operate, and may enjoy greater
manufacture purchase value flexibility of management, fewer legal controls,
First year Nil and fewer taxes. However, the business owner is
Second year 10% personally liable for all debts incurred by the
Third year 18%
business.
Fourth year 30%

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2. General Partnership Solution
A General Partnership is composed of 2 or more Total profit = Gh¢11,000,000.00
persons (usually not a married couple) who agree Amos share = 30% of the profit
to contribute money, labor, or skill to a business. = × 11,000,000 = Gh¢3,300,000.00
Each partner shares the profits, losses, and
management of the business and each partner is The rest = Gh¢(11,000,000 - 3,300,000)
personally and equally liable for debts of the = Gh¢7,700,000.00
partnership. Formal terms of the partnership are Total ratio = 4 + 4 + 3 = 11
usually contained in a written partnership
agreement.
Ben‟s share = × 7,700,000
3. Limited Partnership = Gh¢2,800,000.00
A Limited Partnership is composed of one or Cudjoe‟s share = × 7,700,000
more general partners and one or more limited = Gh¢2,800,000.00
partners. The general partners manage the
Dadzie‟s share = × 7,700,000
business and share fully in its profits and losses.
= Gh¢2,100,000.00
Limited partners share in the profits of the
business, but their losses are limited to the extent
2. Three friends Ato, Oko and Edem entered into
of their investment. Limited partners are usually
a business partnership. They contributed Ghȼ3.0
not involved in the day – to-day operations of the
million, Ghȼ2.4 million and Ghȼ3.6 million
business. Filing with the state is required.
respectively. It was agreed that profits will be
shared in proportions to their contributions. After
4. Limited Liability Partnership (L.L.P.)
one year of operation, the profit made was 2.7
A Limited Liability Partnership (L.L.P.) is similar
million cedis
to a General Partnership except that normally a
i. Find the amount received by each partner as his
partner doesn‟t have personal liability for the
share of the profit
negligence of another partner. This business
ii. Express Edem‟s share of the profit as a
structure is used most by professionals, such as
percentage of his investment
accountants and lawyers. Filing with the state is
required.
Solution
Total profit = Gh¢2.7 million
Sharing Profit according to Ratio of
= Gh¢2,700,000
Contribution
Worked Examples Ratio of contribution;
1. Four business partners Amos, Ben, Cudjoe and 30 : 24 : 36 = 5 : 4 : 6
Dadzie share a profit of Gh¢11,000,000.00. Amos Total ratio = 5 + 4 + 6 = 15
takes 30% of the profit and the remainder is
Ato‟s share = × Gh¢2,700,000
shared among Ben, Cudjoe and Dadzie in the
ratio 4 : 4 : 3 respectively. Find the share of each = Gh¢900,000.00
person

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OKo‟s share = × Gh¢2,700,000 Babatunde‟s third share
= 3% of the capital he invested
= Gh¢720,000.00
= × 7,000 = Ghȼ210.00

Edem‟s share = × Gh¢2,700,000 Abubakar‟s first share

= Gh¢1,080,000.00 = 3% of the capital he invested
= × 25,000 = Ghȼ750.00
ii. Edem‟s share of the profit as a percentage of
Remaining profit
his investment = × 100 %
= Ghȼ6,400 – Ghȼ(1,000 + 480 + 210 + 750)
= × 100% = 30% = Ghȼ (6,400 – 2,440) = Ghȼ3,960

The remaining Ghȼ3,960 shared in the ratio of

Sharing Profit according to a Given Ratio, their capital contributions
Responsibilities and other conditions ⇒Abubakar : Babatunde = 25,000 : 7,000 = 25 : 7
Worked Examples Total ratio = 25 + 7 = 32
1. Abubakar and Babatunde entered into a Babatunde‟s fourth share
business partnership. The capital from the = × 3,960 = Ghȼ866.25
business is made up of Ghȼ25,000.00 from
Abubakar and Ghȼ7,000.00 from Babatunde.
Babatunde‟s total share
They agreed to share the yearly profit in the
= Ghȼ (1,000 + 480 + 210 + 866.25)
following manner; Babatunde as managing
= Ghȼ 2,556.25
director is paid Ghȼ1,000.00 and additional 7.5%
of the total profit. Each partner is paid a sum Abubakar‟s second share
equal to 3% of the capital he invested. The
= × 3,960 = Ghȼ3,093.75
remainder of the profit is shared between the
partners in the ratio of their contributions to the
capital. If the profit at the end of a certain year Abubakar‟s total share
was Ghȼ6,400.00, calculate the total amount each = Ghȼ (750 + 3,093.75) = Ghȼ 3,843.75
partner received from the profit.
2. Jonas, George and Green are partners in a
Solution business and their contribution to the capital are
Abubakar‟s contribution = Ghȼ25,000 respectively Ghȼ15,000.00, Ghȼ25, 000.00 and
Babatunde‟s contribution = Ghȼ7,000 Ghȼ30,000.00. They agreed to share 40% of any
Profit shared = Gh¢6,400 net profit in the ratio of their contribution to the
Babatunde‟s first share = Ghȼ1,000.00 capital. In 2011, their profit before tax was
Ghȼ16,800.00 and 45% was paid to the
Babatunde‟s second share government as tax.
= 7.5% of the total profit i. Calculate the share of the profit received by
= × 6,400 = Ghȼ480.00 each partner.

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ii. Green invested his share of the profit in 2011 A Partner Joining the Business Later
at 12 % per anum simple interest. Express When a partner does not start the business but
Green‟s share of the profit in 2011 together with joins later, his share of the profit is calculated
interest earned on it in 8 years from the time he joins, but not from the beginning
as a percentage of his initial contribution to the of the year.
capital, giving your answer to two decimal
places. Note:
1. That, we have a calendar year of 12 months so
Solution profits are easily shared according to the number
Total profit = Gh¢16,800.00 of months of joining the business and the capital
Tax = 45% of total profit contributed.
= × Gh¢16,800 = Gh¢7560.00 2. That, the net profit is shared in the ratio of the
product of the capital contributions and time
since each partner joins the company.
The Rest (Net profit) ;
= Gh¢16,800.00 – Gh¢7,560.00 = Gh¢9,240.00
Worked Examples
1. Mr. Blain and Mr. Thomas entered in a
40% of net profit;
partnership with capitals of Ghȼ12,600.00 and
= × Gh¢9,240 = Gh¢3,696.00
Ghȼ1,920.00 respectively. After three months
they were joined by Mr. Ceesay with a capital of
Amount to Shared = Gh¢3,696.00 Ghȼ16,200.00. It was agreed that the profit
Ratio of contribution = 15 : 25 : 30 = 3 : 5 : 6 should be shared in proportion to their capitals.
Total ratio = 3 + 5 + 6 = 14 During the first three month of the year, the
business made a profit of 24% of the working
Jonas‟ share = × Gh¢3,696 = Gh¢792.00 capital and during the remaining nine months, the
profit was 32% of the working capital.
Gorge‟s share = × Gh¢3,696 = Gh¢1320.00
i. Find the amount received by each partner as his
Green‟ share = × Gh¢3,696 = Gh¢1,584.00 share of the profit for the year.
ii. Express Mr. Blain‟s share of the profit as a
percentage of his investment.
ii. P = Gh¢1,584.00 T = 8years, R = 12%
I= = = Gh¢1,520.64
Solution
Working Capital of Mr. Blain and Mr. Thomas;
Green‟s share + Interest = Ghȼ12,600.00 + Ghȼ1,920.00 = Gh¢14,520.00
= Gh¢1,584.00 + Gh¢1,520.64 = Gh¢3,104.64
First three month profit;
Green‟s amount as a percentage of his total = 24% of the working capital
contribution = × 100% = × 14,520 = Gh¢3,484.00
= × 100% = 10.35% (2 d.p)

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Ratio of contribution ii. Mr. Blain‟s share of the profit as a percentage
12,600 : 1,920 = 105 : 16 of his investment;
= × 100%
Total ratio = 105 + 16 = 121
= × 100% = 56%
Mr. Blain‟s share = × Gh¢3,484
= Gh¢3,023.31
2. Yaw started a business with Ghȼ2,400, 000.00.
Mr. Thomas‟ share = × Gh¢3,484 After 6 months he was joined by Esi who
= Gh¢460.69 contributed Ghȼ3,000.00. Two months later, Yaw
and Esi were joined by Kwasi who contributed
After three months capital contribution by Ghȼ3,300.00. They agreed to share the profit as
Mr. Blain, Mr. Thomas and Mr. Ceesay; follows: 20% to Yaw as manager of the company,
= Ghȼ(12,600.00 + 1,920.00 + 16,200.00) 4% to Kwasi as assistant manager. The rest of the
= Gh¢30,720.00 profit was then shared in the ratio of the product
of their capital in the company and the time since
Profit of remaining 9 months; each of them joined the company. If the profit at
= 32% of the working capital the end of the year after Yaw had started the
= × Gh¢30,720 = Gh¢9,830.40 company was Ghȼ1,005,000.00, calculate the
total amount received by each of the three
Share of profit according to ratio of contribution; partners of the company.
= 12,600 :1,920 :16,200 = 105 : 16 : 135
Total ratio = 105 + 16 + 135 = 256 Solution
Profit = Ghȼ1,005,000.00,
Mr. Blain‟s share = × Gh¢9,830.40
= Gh¢4,032.00 Yaw‟s first share of 20% of profit as manager,
Mr. Thomas‟ share = × Gh¢9,830.40 = × Ghȼ1,005,000.00 = Ghȼ201,000.00
= Gh¢614.40
Kwasi‟s first share;
Mr. Ceesay‟s share = × Gh¢9,830.40 = 4% of profit as assistant manager
= Gh¢5,184.00 = × Ghȼ1,005,000.00 = Ghȼ40,200.00

Mr. Blains total share; The rest of the profit;

= Gh¢3,023.31 + Gh¢4,032.00 = Gh¢7,055.31 = Ghȼ(1,005,000 – 201,000 – 40,200)
= Ghȼ763,800.00
Mr. Thomas‟ total share;
= Gh¢460.69 + Gh¢614.40 = Gh¢1,075.09 The rest of the profit shared in the ratio of the
product of their capital and time since each of
Mr. Ceesay‟s total share = Gh¢5,184.00 them joined the company

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Yaw‟s product of capital and time of joining; Gh¢y for 8 months and Gh¢ z for 7 months
= Ghȼ2,400,000 × 12 = Ghȼ28,800,000 respectively.
Then, 14x : 8y : 7z = 5 : 7 : 8.
Esi‟s product of capital and time of joining; ⇒ =
= Ghȼ3,000 × 6 = Ghȼ18,000
7 × 14x = 5 × 8y
Kwasi‟s product of capital and time of joining 98x = 40y
= Ghȼ3,300 × 4 = Ghȼ13,200 y= =
⇒ =
Ratio of product of capital and time of joining
8 × 14x = 5 × 7z
Yaw : Esi : Kwasi
112x = 35z
28,800,000 : 18,000 : 13,200
288,000 : 180 : 132 = 72,000 : 45: 33 z= =
Total ratio = 72,000 + 45 + 33 = 72,078 x:y:z=x: :
Total amount = Ghȼ763, 800.00 x : y : z = 20 : :

Yaw‟s second share = × Ghȼ763,800 Exercises 17.5

= Ghȼ762,973.44 1. David and Goliath contributed Ghȼ21,000.00
Kwasi‟s second share = × Ghȼ763,800 and Ghȼ7,000.00 towards a business. At the end
of one year, the business yielded a profit of
= Ghȼ349.70
Ghȼ42,000.00 which they shared according to the
ratio of their contributions.
Esi‟s share = × Ghȼ763, 800 = Ghȼ476.86
i. How much was received by each partner?
ii. Express the share of each person as a
Total amount received by Yaw; percentage of the total profit.
= Ghȼ201,000.00 + Ghȼ762,973.44
= Gh¢763,973.44 2. A joint business was set up by Summer, Butler
and Winter with respective contributions of
Total amount received by Kwasi ; Ghȼ5million, Ghȼ3.5 million and Ghȼ1.5million.
= Ghȼ40,200.00 + Ghȼ349.70 = Gh¢40,549 The profit was agreed to be shared in proportion
Total amount received by Esi = Ghȼ476.86 to their contributions.
i. Calculate the share of each person if the profit
3.Three partners shared the profit in a business in made after one year was Ghȼ30million. ii. What
the ratio 5 : 7 : 8. They had partnered for 14 percentage of their contributions was received as
months, 8 months and 7 months respectively. profit by the partners?
What was the ratio of their investments?
3. The capital for a joint business is made up of
Solution Ghȼ7,500.00 and Ghȼ12,000.00 contributions by
Let their investments be Gh¢ x for 14 months, Samson and Delilah respectively. They agreed to
share the profit as follows; as the manager,

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Samson will be paid Ghȼ500.00 in addition to 7. Tonton and Sansan entered into partnership
10% of the total profit. Each person will be paid with capitals in the ratio 4 : 5. After 3 months,
5% of the contributions made and the rest of the Tonton withdrew of his capital and Sansan
profit will be shared according to the ratio of their
withdrew of his capital. The gain at the end of
contributions. Calculate the share of each partner
if they made a net profit of Ghȼ8,000.00 at the 10 months was Gh¢7,600.00. Calculate Tonton's
end of the year. share of the profit .

4. Candy, Sandy and Wendy jointly thought of 8. A, B, C rent a pasture. A puts 10 oxen for 7
engaging themselves in a business venture. It was months, B puts 12 oxen for 5 months and C puts
agreed that Candy would invest Gh¢6,500.00 for 15 oxen for 3 months for grazing. If the rent of
6 months, Sandy, Gh¢8,400.00 for 5 months and the pasture is Gh¢175.00, how much must C pay
Wendy, Gh¢10,000.00 for 3 months. Candy as his share of rent?
wants to be the working member for which, he
9. A began a business with Gh¢85,000. He was
was to receive 5% of the profits. The profit
joined afterwards by B with Gh¢42,500.00. For
earned was Gh¢7,400.00. Calculate the share of
how much period does B join, if the profits at the
Sandy in the profit.
end of the year are divided in the ratio of 3 : 1?
5. Esi and Mansah entered into a business
10. Cain and Abel started a business in
partnership. Esi contributed 35% of the capital
partnership investing Gh¢20,000.00 and
while Mansah contributed the rest. At the end of
Gh¢15,000.00 respectively. After six months,
the year, they made a profit of Ghȼ5,600.00. 15
Joseph joined them with Gh¢20,000.00. What
% of the profit was paid into a reserve fund while
will be Abel's share in total profit of Gh¢2,500.00
25% of the remaining profit was paid as income
earned at the end of 2 years from the starting of
tax. They then shared the remaining profit in the
the business?
ratio of their contributions. If Mansah contributed
Ghȼ10,400. 00, find;
Calculating the Profit Given the Ratios, Terms
i. the total contributions of Esi and Mansah;
and Conditions
ii. the total amount paid as income tax,
Sometimes, students are requested to find the
iii. correct to one decimal place, Esi‟s profit as a
total profit shared given the ratio of contributions,
percentage of her contribution.
the share of the profit of one person and other
terms and conditions. In such situations
6. Joel, Martin and Bruce subscribe
I. Calculate the capital contributions of each
Gh¢50,000.00 for a business. Joel subscribes
person.
Gh¢4,000.00 more than Martin and
II. Represent the total profit by any preferred
MartinGh¢5,000.00 more than Bruce. Out of a
variable.
total profit of Gh¢35,000.00, how much will Joel
III. Calculate the share or shares of each person
receive if it is shared according to the ratio of
out of the total profit
their subscription.
IV. Find the total of the shares (total amount used).

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V. Determine the remaining profit using the Let the total profit be x
relation: Keren‟s first share = =
Remaining profit = Profit - total amount used
VI. Determine the one whose share has been
given and find the share of that person out of the Keren‟s second share = × 1,280 = Ghȼ64.00
remaining fraction Kelvin‟s first share = =
VII. Find the total share(s) of the person in
activity VI and equate it to the given total profit Kelvin‟s second share = × 1,920 = Ghȼ96.00
of that person
VIII. Solve for the value of the variable, to get the
Total amount used
total profit shared
= + 64 + + 96
Worked Examples = + + 64 + 96
1. Keren and Kelvin contributed a capital of = + 160
Ghȼ3,200.00 to purchase a vehicle for “trotro”
business. The capital was contributed in the ratio =
2 : 3. The profit for a year was shared under the
following agreed conditions and terms; Keren the Remaining profit;
driver received 15% of the total profit and = Profit – Total amount used
Kelvin, =x–( )
the conductor (mate) received 10% of the profit. (
Each partner is paid additional 5% of the capital = = =
contributed and the rest of the profit was shared
between them in the ratio of their contributions to Ratio for sharing the remaining profit;
the capital. If Keren‟s share of the total profit was Keren : Kelvin = 2 : 3
Ghȼ2,340.00, calculate Total ratio = 2 + 3 = 5
i. the total profit for the year
ii. Kelvin‟s share of the profit as a percentage of Keren‟s share of the remaining profit;
his contribution to the capital. = ×( )
 Keren‟s share of the total profit
Solution = His first share + his second share + his share of
Capital = Ghȼ3,200.00 the remaining profit
Keren : Kelvin = 2 : 3
= + 64 + ( )
Total ratio = 2 + 3 = 5

Capital contributed by each person; But Keren‟s total profit = Ghȼ2,340.00

Keren = × 3,200 = Ghȼ1,280.00 ⇒ + 64 + ( ) = Ghȼ2,340

Kelvin = × 3,200 = Ghȼ1,920.00 + 64 + (3x – 640) = 2,340

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(20) + (20) 64 + (20) (3x – 640) = (20) 2340 Alex : Jimmy = 2 : 1
Total ratio = 2 + 1 = 3
3x + 1280 + 2(3x – 640) = 46,800
3x + 1280 + 6x – 1280 = 46,800
Capital contributed by each person;
3x + 6x = 46,800
9x = 46,800 Alex = × 27,000 = Ghȼ18,000.00
x = 5,200 Jimmy = × 27,000 = Ghȼ9,000.00
The total profit for the year was Ghȼ5,200.00
Let the total profit be x
ii.Kelvin‟s first share;
Alex‟s first share =
= = = Gh¢520.00

Remaining Profit = x – = =
Kelvin‟s second share = Ghȼ96.00

Kelvins total share of the profit; Alex‟s second share = ( )=

= Gh¢520 + Ghȼ96.00 = Gh¢616.00

Kelvin‟s profit as a percentage of his contribution Total share of Alex = +

= × 100% But Alex‟s total share = Gh¢6,850

⇒ + = 6,850
= × 100% = 32%
300 × + 300 × = 300 × 6,850
2. Alex and Jimmy entered into a business 3(5.5x) + 189x = 2,055,000
partnership in January 2013. The total capital was 16.5x + 189x = 2,055,000
Ghȼ27,000.00 which they agreed to contribute in 205.5x = 2,055,000
the ratio 2: 1 respectively. The annual profit from x= = 10,000
2013 was shared as follows:Alex was paid 5 % of The total profit shared is Gh¢10,000.00
the total profit for his service as a manager. The
remainder of the profit was shared between them ii. Jimmy‟s share = Profit – Alex‟s share
in the ratio of their contributions to the capital. If = Ghȼ 10,000 – Ghȼ 6,850 = Ghȼ3,150.00
Alex received a sum of Ghȼ6,850.00out of the
profit. Calculate: Jimmy‟s share of the profit as a percentage of his
i. the total profit for the year; initial contribution to the capital;
ii. Jimmy‟s share of the profit as a percentage of = × 100% = 35%
his initial contribution to the capital;
iii. If Alex had to pay tax at 30% on the amount iii. 30% tax paid by Alex on his share;
he received, how much did he pay?
= × Gh¢6,850
Solution = Gh¢2,055.00
i. Capital = Ghȼ27,000.00

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Exercises 17.6 Bank Transactions and Services
1. Martha and Mercy invest in a business in the Some common transactions and services provided
ratio 3 : 2. If 5% of the total profit goes to charity by the banks are explained below:
and Martha's share is Gh¢855.00. Find the total
profit shared: Savings Account
The savings account allows the customer to
2. Kofi and Yaw entered into business partnership deposit money at the bank and earns interest at a
with a total capital of Ghȼ81 million. They given rate on the deposited amount for a given
agreed to contribute the capital in the ratio 2 : 1 period of time (usually one year). With savings
respectively. The profit was shared as follows; account, banks insist on minimum balance that
Kofi was paid 5% of the total profit for his customers have to leave in their accounts and also
services as the manager. Each partner was paid insist on customers withdrawing money in
3% of the capital he invested. The remainder of person.
the profit was then shared between them in the
ratio of their contributions to the capital. If Kofi‟s Current Accounts
share of the total profit was Ghȼ7.5 million, The current accounts allow customers to deposit
calculate and withdraw money from their accounts at any
a. the total profit for the year to the nearest time within banking hours. With current
thousand cedi accounts, the customer withdraws money by the
b. Yaw‟ share of the profit as a percentage of his use of a cheque and withdrawal can be done
contribution to the capital through a third person. The customer is charged
for all services provided by the bank.
Banking
It is the act of engaging in the business of keeping Fixed Deposit accounts
money for savings or exchange. Institutions Fixed deposit accounts allows a customer to
responsible for these activities are called Banks. deposit money with a bank for a given (fixed)
In Ghana, there is the central bank called Bank of period of time at a given (fixed) rate of interest
Ghana, established to have oversight payable at the due or fixed date. With fixed
responsibilities on all banking activities. Some of deposit, if the customer makes a partial or full
the banks in Ghana are Ghana Commercial Bank, withdrawal of the deposited amount before the
Agricultural Development Bank, SG – SSB, fixed period, it constitutes breach of contract and
National Investment Bank, Eco – Bank, Barclays as such he or she forfeits the interest. It is also
Bank, Standard Charted Bank etc. Added to the called time deposit
list are Rural and Community banks (Micro-
banks) who work under the auspices of the ARB Inward Money Transfer
Apex Bank. This service allows money to be transferred from
The customer of a bank is identified by a unique one person to another from within and outside the
number called account number, required for all country through the banks for a fee. Eg. Money
transactions. gram, western union, bankers draft etc . In the

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case of transfer of foreign currency, the bank pays The pay in slip and payment cheques has the
the customer the cedi equivalent. following empty spaces to be completed
1. Name of customer /depositor
Overdraft 2. Accounts number
This service allows a customer to withdraw more 3. Branch where account is held
money than he or she has in his accounts at an 4. Date
interest. This facility is usually granted for a short 5. Amount deposited (in figures and words)
period of time (one or two months). 6. Signature and others

Loans Value Added Tax (V. A. T)

This service allows customers to take money It is a tax or extra money charged on goods and
from the bank at a given rate of interest for a services. This charge is calculated as the
given period of time. With the loan facility, banks percentage of the basic cost of the good or
usually request from the customer, documents of services.
buildings or any valuable property as collateral
evidence. Others also ask for a third person called The cost of an item without V.A.T is called the
guarantor to offer to pay the loan with its interest basic cost or VAT exclusive cost (always 100%)
to the bank in case the customer defaults and the cost of item with VAT is called VAT
repayment. inclusive cost orsimply, total cost or selling
price.The V.A.T inclusive cost is equal to the
Bank Charges basic cost plus the VAT charge.
For operating an account with a bank or i.e. (100% + r %)
requesting some services or transactions from a
bank, the bank charges the customer some fees. V.A.T is usually calculated as a percentage.
Some of the bank charges or fees are listed
below: Mathematically,
1. Cost of turnover, COT (applicable to current 1. V.A.T = Rate × Basic cost
accounts 2. VAT inclusive cost /selling price
2. Withdrawal charges
3. Overdraft charges = Basic cost + (Rate × Basic cost)
4. Bankers Drat Charges = Basic cost + VAT
5. Telegraphic Transfer charges In Ghana, the VAT rate is 12.5%, subject to
6. Periodic charges change
7. Money Transfer charges
8. Cheque deposit charges Note
9. Card user charges 1. Given the cost price of an item, it simply
implies the basic cost or VAT exclusive cost
Completing a Pay-in-slip and Payment 2. Given the selling price of an item, it simply
Cheques: implies VAT inclusive cost

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Worked Examples But rate = 20% , basic cost = Gh¢620.00
1. The cost of a bracelet is Gh¢1,000.00. VAT = 20% × Ghȼ620
i. How much is the VAT charged at a rate of = × Ghȼ620 = Gh¢124.00
12 %?
ii. What is the VAT inclusive cost of the bracelet? ii. Selling price = Basic cost + VAT charge
= Gh¢620.00 + Gh¢124.00
Solution = Gh¢744.00
i. V. A .T = Rate × Basic cost
Rate = 12 % = % Exercises 17.7A
Basic cost = Gh¢1,000.00 1. Rashid wants to buy a car costing
V. A. T = Rate × Basic cost. Gh¢8,000.00. If VAT is due at a rate of 12 %
V. A. T = × 1,000 = Gh¢125.00 of the cost of the car, how much will he pay for
the car?
ii. VAT inclusive cost = Basic cost + VAT
= Gh¢1,000.00 + Gh¢125.00 = Gh¢1,125.00 2. The cost of a colour television is Gh¢2,200.00.
i. Find the VAT at 17 %.
2. The VAT exclusive charge of a monthly ii. How much will you pay for the television?
electricity bill is Gh¢200.00.Find the V.A.T of
12.5% on the bill and hence, find the total 3. The VAT rate of a certain country is 20%. If
monthly bill. the cost of a Toyota Lancruiser vehicle in that
country is Gh¢36,600.00, calculate:
Solution i. the VAT on the vehicleii. the selling price of
Rate = 12.5%, Basic cost = Gh¢200.00 the car
VAT = Rate × Basic cost
VAT = 12.5% × Ghȼ200 4. At a certain store, the VAT exclusive cost of a
= × Ghȼ200 = Gh¢25.00 set of football jersey is Gh¢375.00, if VAT is
charged at a rate of 12%, find:
i. the VAT on five sets of jerseys
Monthly bill = Basic charge + VAT
ii. the selling price of the five jerseys
= Gh¢(200.00 + 25.00) = Gh¢225.00

Finding the Basic Cost or the VAT Charge

3. The VAT rate of a certain country is 20%.
Given the VAT Inclusive Cost
If the cost of a refrigerator is Gh¢620.00,
From the relation,
calculate:
VAT inclusive cost /selling price/ total cost
i. the VAT charged on the refrigerator.
= Basic cost + (Rate × Basic cost)
ii. the total cost of the refrigerator
Let S represent the VAT inclusive cost, b
represent the basic cost and r the VAT rate
Solution ⇒ S = b + br%
i.VAT = Rate × Basic cost
Baffour – Ba Series Core Maths for Schools and Colleges Page 478
1. b = ( ii. the amount charged as VAT.

2. VAT charge = Inclusive cost – Basic cost

2. A calculator sells at Gh¢800.00 at a VAT
rate of 12.5%. Find:
Worked Examples
i. the basic cost of the calculator.
1. An electricity bill including VAT of 12 % is
ii. the VAT charge on the calculator.
Gh¢450.00.
i. How much is the bill exclusive of VAT? 3. A telephone bill including VAT of 12.5 % is
ii. How much VAT is charged on the bill? Gh¢490.00. How much was the bill before VAT
was added.
Solution
i. VAT inclusive charge, S = Gh¢450.00, VAT 4. A electricity bill including VAT of 12.5% is
rate, r = 12.5% and basic cost (same as VAT Gh¢4,374.00. How much was the bill before
exclusive cost), b = ? VAT was added.
b=( =( = Gh¢400.00
5. The VAT paid on an article amounts to
The bill exclusive of VAT is Gh¢400.00
Gh¢900.00 at a rate of 10%. What is the total
price of the article including VAT?
ii. VAT charge = Inclusive cost – Basic cost
= Gh¢450 .00 – Gh¢400.00 = Gh¢50.00
National Health Insurance Levy (NHIL)
It is a charge in percentage, calculated on the cost
2. A school bag is sold at Gh¢594.00. How much
of goods and services to provide health assistance
is the VAT charged on the bag if the VAT rate is
12.5%? to people. The rate on NHIL in Ghana is 2 %,
subject to changes.
Solution 1. NHIL = Rate × Basic cost
VAT inclusive cost, S = Gh¢594.00, VAT rate, r 2. NHIL inclusive cost / selling price
= 12.5% and basic cost (same as VAT exclusive
= Basic cost + (Rate × Basic cost)
cost), b = ?
= Basic cost + NHIL charge
b=( =( = Gh¢528.00
The basic cost of the bag is Gh 528.00 Note
1. Given the cost price of an item, it simply
ii. VAT charge = Inclusive cost – Basic cost implies the basic cost or NHIL exclusive cost
= Gh¢594 .00 – Gh¢528.00 = Gh¢66.00 2. Given the selling price of an item, it simply
implies NHIL inclusive cost
Exercises 17.7B
1. The selling price of a digital camera is Worked Examples
Gh¢400.00. If VAT is charge at a rate of 12.5%, 1. Find the NHIL on a television that cost Gh¢
calculate: 612.00
i. the basic cost of the camera.

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Solution 3. A telephone bill of Gh¢1,120.00.
NHIL = Rate × Basic cost 4. Twenty bags of rice each at a cost of
where Rate = 2 % = 2.5% Ghȼ120.00

NHIL = × Gh¢ 612 = Gh¢15.30

Finding the Basic Cost or the NHIL Cost Given
the NHIL Inclusive Cost
2. Calculate the NHIL on a refrigerator that cost From the relation, NHIL inclusive cost
Gh¢240.00 = Basic cost + (Rate × Basic cost)
Let C represent NHIL inclusive cost, B represent
Solution basic cost and r represent rate of NHIL
NHIL = Rate × Basic cost ⇒ C = B + (Br)
where Rate = 2 % = 2.5% But r = 2.5% (unless stated otherwise)
Basic cost = Gh¢240.00 C = B + (2.5% B)
C = B (1 + r%)……….(1)
NHIL = × Gh¢240.00 = Gh¢6.00

From eqn (1), given the NHIL inclusive cost, C,

3. A yard of G.T.P. cost Gh¢3.80. Find the NHIL
the basic cost, B, is calculated as:
on 30 yards of G.T.P.
1. B =
Solution 2. NHIL cost = Inclusive Cost – Basic cost
NHIL = Rate × Basic cost
where rate = 2 % = 2.5% Worked Examples
1. A man purchased a set of furniture at an NHIL
Basic cost = Gh¢3.80 × 30 = Ghȼ114.00
inclusive cost of Gh¢6,314.00. If NHIL is
NHIL = × 114 = Ghȼ2.85 charged at a rate of 2.5%, find:
i. the basic cost of the set of furniture
Exercises 17.8 ii. the amount paid as NHIL
A. Calculate the NHIL of 2 % on the
following, if NHIL is exclusive. Solution
1. A car that cost Gh¢8,000.00. i. NHIL inclusive cost, C = Gh¢6,314.00, Basic
2. An electricity bill that cost Gh¢4,600.00 cost, B =? and r = 2.5%
3. A refrigerator that cost Gh¢1,800.00 B= = = Gh¢6,160.00
4. A bag that cost Gh¢1,000.00
5. A bill that cost Gh¢4,700.00 ii. NHIL cost = Inclusive Cost – Basic cost
NHIL cost = Gh¢6,314 – Gh¢6,160 = Gh¢154.00
B. Calculate the NHIL of 5 % on the following
2. What is the NHIL on a car sold for
and determine the NHIL inclusive cost;
1. A car that cost Gh¢400.00 Gh¢20,541.00 at rate of 2 %? Hence, find the
2. An electricity bill of Gh¢2,200.00 NHIL on the car.

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Solution The person who takes the insurance policy is
i. NHIL inclusive cost, C = Gh¢20,541.00, Basic called the policy holder and the company that
cost, B =? and r = 2.5% offers the insurance is called the insurance
Substitute in B = =( company.

= Gh¢20,040.00 The policy holder pays some amount of money at

regular intervals to the insurance company. The
ii. NHIL cost = Inclusive Cost – Basic cost money paid by the policy holder is intended to
NHIL cost = Gh¢20,541 – Gh¢20,040 compensate him in case of occurrence of any of
= Gh¢501.00 the eventualities mentioned above. The amount
paid regularly to the insurance company is called
Exercises 17.9 premium and the compensation or benefit paid to
1. A Plasma television is sold at Gh¢4,220.00. the policy holder is called insuredvalue .The
If NHIL is charged at a rate of 2.5%, calculate; premium is usually a percentage of the insured
i. the basic cost of the plasma television. value.
ii. the NHIL cost.
Mathematically;
2. The cost of a scientific calculator including Premium = Rate × Insured value….. (1)
NHIL is Gh¢738.00. What is the NHIL on the
Insured value = ………….. (2)
calculator, if NHIL is charged at a rate of 2.5%?
Rate = × 100…………. (3)
3. The NHIL inclusive cost of a car is
Gh¢25,584.00. If you purchase this car at the said
cost, what will be your contribution to NHIL, if Worked Examples
the rate of NHIL is 5.5%? 1. Mr. Oppong insured his car against accident.
The value of the car was estimated as
4. At a certain boutique, the NHIL inclusive cost Gh¢9,500.00. If the rate of insurance is 15%,
of a necklace is Gh¢940.00 and that of a bracelet calculate the annual premium paid to the
is Gh¢730.00. If the rate of NHIL is 5.5%; company.
i. find the total basic cost of 15 necklaces and 20
bracelets Solution
ii. what is the total health insurance contribution Premium = Rate × Insured value
on the 15 necklaces and 20bracelets? But rate = 15% =
Insured value = Gh¢9,500
Insurance Premium = 15% × 9,500
In order to take care of future eventualities such
= × 9,500 = Gh¢1,425
as death, illness, accident, fire outbreaks, injuries
and damages to properties, people take insurance The premium paid annually is Gh¢1,425.00
policy.

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2. Mrs. Shirley pays a premium of Gh¢117.00 at Solution
a rate of 13% star insurance company. Calculate Rate = × 100%
the insured value if she contributes towards
insuring her car against accident. Premium = Gh¢144 and Insured value = Gh¢2,400

Solution Rate = × 100 = 6%

Insured value =
Exercises 17.10
But premium = 117 and rate = 13% = 1. The value of a car is Gh¢10,200.00, If it is
Insured value = = Gh¢900.00 insured against accident at rate of 9%. Calculate
the premium paid annually to the insurance
The insured value of the car is Gh¢900.00
company.
3. Kennedy insured his house against fire. If the
2. Mr. Brown insured his hotel against fire. The
insured value of the house is Gh¢2,351.00 at a
insured value of the hotel is Gh¢15,220.00.
rate of 9%, find the yearly premium he pays to
Calculate the yearly premium paidto the
the insurance company.
insurance companyby Mr. Brown at a rate of 15%

Solution
3. King George is a policy holder of a certain
Premium = Rate × Insured value.
insurance company. He insured his vehicle value
But rate = 9% = , at Ghȼ120.00 per month. What is his yearly
Insured value = Gh¢2,351 premium?
Premium = 9% × 2,351
= × 2,351= Gh¢211.59 4. Mrs. Pomaa contributes a yearly premium of
Gh¢189.00 to state insurance company as
4. Calculate the insured value of Mr. Peter‟s car if commitment to his insured yutong bus against
he contributes a premium of Gh¢144.00 a year at accident at a rate of 20%. Find the insured value
a rate of 6% of the Yutong bus.

Solution 5. Calculate the insured value of a property which

Insured value = attracts a yearly premium of Gh¢204.00 at a rate
of 17%.
Premium = Gh¢144, Rate = 6% =
Insured value = = = Gh¢2,400 6. Maame Duku pays Gh¢156.00 yearly premium
as a property insurance to Enterprise insurance
The insured value is Gh¢2,400.00 company at a rate of 12%. Calculate the insured
value of her property.
5. Find the rate of insurance given the insured
value as Gh¢2400, 00 and the premium as 7. The premium paid on an insured car valued at
Gh¢144.00.

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Gh¢5,000.00 is Gh¢250.00, find the rate per Note that words like “NIL” “NO” FREE
anum. ALLOWANCE” against any amount of money
represent tax free (allowance) income.
8. The value of a ship is estimated at
Gh¢3,600.00. If the owner contributes a premium Worked Examples
of Gh¢324.00 a year; calculate the insurance rate. 1. Suppose income tax regulations are as follows:
Allowances:
Income Tax Personal allowance: Gh¢45.00
It is the portion of a person‟s salary or wage paid Marriage allowance: Gh¢30.00
to the government. Income tax levies are used by Child allowance: Gh¢25.00
the government for developmental projects such (For each child under 18years)
as schools, roads, hospitals, transport and others. Disability Allowance: = Ghȼ20.00
Income tax is deducted from workers salary
before it is paid to them. Tax on Taxable income:
5% for first Gh¢60.00
In Ghana the agency responsible for collecting 10% for second Gh¢60.00
income tax is Internal Revenue Service (I.R.S). 20% for the remainder

Note the following terms: a. Calculate the tax paid by Mr. Peter, a
Gross Income: The total income before tax is physically challenged person, whose salary is
deducted. Gh¢3,800.00 per annum, if he has a wife and
Tax Allowance: The part of the income that is three of his children under 18years.
not taxed b. Calculate Mr. Peter‟s net salary.
Taxable Income: It is the part of the income that
is taxed Solution
Net Income:It is the part of the salary left after all Income Gh¢3,800.00
taxes has been taken. That is: Allowances Gh¢
Net salary = Gross salary – Total tax Personal 45.00
a. Tax free income Marriage 30.00
= sum of all tax allowances Disability 20.00
b. Taxable income Child 25 × 3 = 75.00
= Annual Income – tax allowances Total allowance 170.00 Gh¢170.00
c. Annual tax paid Taxable income Gh¢3,630.00
= Tax rate × Taxable amount
d. Monthly tax paid = Taxes
5% tax on first Gh¢60.00
e. Net annual income
= × 60 = Gh¢3.00
= Annual gross income – Annual tax paid
f. Net monthly income = 10% tax on second Gh¢60.00

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= × 60 = Gh¢6.00 Total allowance 65.00 Gh¢65.00
Taxable income Gh¢5,735.00
Remainder = Ghȼ3,630 – (60 + 60)
Taxes
= Ghȼ(3,630 – 120) = Gh¢3,510.00
10% tax on first Gh¢50.00 = × 50 = Gh¢5.00
20% tax on the remainder, 15% tax on next Gh¢50.00 = × 50 = Gh¢7.50
= × Ghȼ3,510 = Gh¢702.00
Remainder = Ghȼ5,735 – Ghȼ(50 + 50)
Total Tax = Gh¢(3 + 6 + 702) = Gh¢711.00 = Gh¢5,735 − Gh¢ 100 = Gh¢5,635.00

Net Salary = Gross salary – Total tax 20% tax on the remainder;
Net salary = Gh¢3,800 – Ghȼ711 = × 5,635 = Gh¢1,127.00
= Gh¢3,089.00
Total tax = Gh¢(5 + 7.50 + 1,127)
2. Mr. Green earns a salary of Gh¢5,800.00 per = Gh¢1,139.50
annum. He has three children whose ages are 19 Mr. Green paid a total tax of Gh¢1,139.50
years, 21 years and 23 years. If Mr. Green is
visually impaired but has devoiced his wife, Net salary = Gross income – Total tax
calculate the tax paid by him and his net salary = Gh¢5,800.00 − Gh¢1,139.50
using the following regulations; = Gh¢4,660.50

Allowances: 3. Suppose the income tax regulation of a country

Personal allowance =Gh¢40.00 is as follows.
Marriage allowance =Gh¢ 30.00
Allowances Amount (Ghȼ)
Child allowance = Gh¢20.00, but for each
Personal 100
child under 18, Marriage 400
Disability allowance = Gh¢25.00 Disability 500
Car maintenance 600
Tax on taxable income: Child (under 18) 250
10% on first Gh¢50.00 Dependant relative 200

15% on next Gh¢50.00

20% on the remainder Taxable income Tax Rate
Gh¢ Gh¢
First 800.00 Nil
Solution Next 1,200.00 5.5%
Gross income Gh¢ 5,800.00 Next 1,200.00 10.5%
Allowances: Gh¢ Next 600.00 15%
Personal 40.00 Remainder 5%
Disability 25.00

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Mr. Green is a widower with 2 children aged 20 3. Darling earns a monthly salary of
and 21 years. Recently, he lost one of his legs Gh¢6,500.00. He is allowed a tax free of
through an accident, which did not involve his Ghȼ500.00. If he pays 17% tax in his taxable
own car. If his salary per anum is Gh¢4,850.00, income, calculate;
calculate: i.his taxable income,
i. his annual taxable income and how much he ii.his total tax paid,
pays as income tax. iii.his net salary.
ii. how much he receives as net annual income.
Solution
Solution i. Gross monthly salary = Gh¢6,500.00
Tax allowance = Gh¢500.00
Income (Gh¢) 4,850 Taxable income = ?
Personal 100
Disability 500 Taxable income
Car maintenance 600 = Gross monthly salary – Tax allowance
Tax free 800 2,000 = Gh¢6,500.00 – Gh¢500.00 = Ghȼ6,000.00
Taxable income 2,850
ii. Tax paid is 17% of taxable income
Taxes = × 6,000 = Gh¢1,020.00
5.5% on Gh¢1,200
× 1,200 = 66 iii. His net salary = Gross salary – Total tax
Net salary = Gh¢6,500 − Gh¢1,020
Remaining = 2,850 – 1,200 = 1650
Net salary = Gh¢5,480.00
10.5% on Gh¢1,200 4. Mrs. Akushika‟s monthly taxable income is
× 1,200 = 126 Gh¢9,850.00. Use the table below to calculate:
Remaining = 1,650 – 1,200 = 450 i.how much she pays as income tax
ii.her net monthly income.
15% on Gh¢450 (not Gh600)
× 450 = 67.5 Taxable Tax
income Rate
Gh¢ Gh¢
Total taxes First 600.00 Nil
= 66 + 126 + 67.5 = 259.5 Next 600.00 50%
Total tax paid is Gh¢259.5 Next 600.00 10%
Next 1,200.00 15%
ii. Net Salary = Gross – taxes Next 1,200.00 25%
= 4,850 – 259.5 = 4,590.50 Next 5,400.00 35%
Remainder 55%
Net annual income = Gh¢4,590.50

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Solution For every cedi of the next 480.00 7 p
i. Gross income Gh¢9,850.00
For every cedi of the next 480.00 10p
Allowances Gh¢
For every cedi of the next 960.00 12 p
First 600.00
Total allowance 600.00 Gh¢600.00
Taxable income Gh¢9,250.00 For every cedi of the next 1,140.00 15p
a. Calculate the income tax payable by Agbedefu
Taxes who earned Ghȼ3,000.00 per anum
b. What percentage of Agbedefu‟s annual income
50% of Gh¢600.00 = × 600 = Gh¢300.00
was payable as income tax?
10% of Gh¢600.00 = × 600 = Gh¢60.00
Solution
15% of Gh¢1,200.00 = × 1,200 = Gh¢180.00 Gross income Gh¢3,000
Allowances Gh¢
25% of Gh¢1,200.00 = × 1,200 = Gh¢300.00
First 300
35% of Gh¢5,400.00 = × 5,400 = Gh¢1,890.00 Total allowance 300 Gh¢300
Taxable income Gh¢2,700.00
Remainder
= Ghȼ9,250 – Ghȼ(600 + 600 + 1,200 + 1,200 + Taxes
5,400) 5% of Gh¢240.00 = × 240 = Gh¢12.00
= Gh¢9,250 − Gh¢9,000 = Gh¢ 250.00
Remaining = Gh¢2,700 – Gh¢240 = Gh¢2,460.00
55% tax on remainder = × 250 = Gh¢137.50

7.5% of Gh¢480.00 = × 480 = Gh¢36.00

Total tax
Remaining = Gh¢2,460 – Gh¢480 = Gh¢1,980.00
= Gh¢(300 + 60 + 180 + 300 + 1,890 + 137.50)
= Ghȼ2,867.50
10% of Gh¢480.00 = × 480 = Gh¢48.00
ii. Net income = Gross income – Total tax. Remaining = Gh¢1,980 – Gh¢480 = Gh¢1,500.00
Net income = Gh¢ 9,850.00 − Ghȼ2,867.50
Net income = Gh¢6,982.50 12.5% of Gh¢960.00 = × 960 = Gh¢120.00
Therefore, her net income is Gh¢6,982.50. Remaining = Gh¢1500 – Gh¢960 = Gh¢540.00

Some Solved Past Questions 15% of Gh¢540 (Not Gh¢1,140)

1. In Ghana, the annual income tax payable by an
= × 540 = Gh¢81.00
individual in a certain year was assessed at the
following rate:
Total tax paid (Income tax);
Rate of Tax = Gh¢(12 + 36 + 48.00 + 120 + 81)
For every cedi of the first 300.00 Nil = Gh¢297.00
For every cedi of the next 240.00 5p

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b. × 100% Personal allowance = Gh¢1,200.00
Wife allowance = Gh¢600.00
= × 100% = 9.9%
Children allowance = Gh¢250.00 per child.

Exercises 17.11 Tax on taxable income:

A. In each of the following cases, calculate the 20% on the first Gh¢2,000.00
income tax. 15% in the next Gh¢4,000.00
Calculate the officer‟s:
Annual Salary Allowance Tax Rate
(Ghȼ) (Ghȼ) (%) i. taxable income,
800.00 45.00 10% ii. total tax paid a month,
1500.00 500.00 15% iii. net monthly salary.
11,070.00 3,550.00 20%
25,000.00 5,200.00 30% 5. A company‟s director married with 6 children
is on an annual salary of Ghȼ12,000.00. His tax
B.1. Madam Fausty receives Gh¢5,000.00 a free allowances are as follows;
year. She is allowed a tax free of Gh¢300.00. If Personal allowance Ghȼ1,200.00 plus
she contributes 12% to social security and 2% p. 10 % of excess of his salary over Ghȼ10,000.00
a to health insurance, calculate: Wife allowance Ghȼ600.00
i. her taxable income, Children allowance Ghȼ250.00 per child
ii. her total tax, for the first four children
iii. her net salary. Dependent relatives Ghȼ550.00

2. Mama Jane‟s monthly salary is Gh¢4,500.00. On the taxable income, the rates of tax are as
She is allowed a tax free of 12% of her monthly follows:
salary. She contributes 17 % to social security 10p in the Ghȼ on the first Ghȼ2,000.00
and 2 % per month to health insurance. 15p in the Ghȼ on the next Ghȼ4,000.00

Determine: 22 p in the Ghȼ on the next Ghȼ5,000.00

i. her tax allowance, 30p in the Ghȼ on the next Ghȼ10,000.00
ii. her total tax payment per month. 37 p in the Ghȼ on the rest. Calculate:
iii. her net monthly salary. a. the director‟s taxable incomeb. the average
monthly tax he pays to the nearest pesewa,
3. A man earns Gh¢38,400.00 per anum and pays
b. the percentage of his monthly salary he pays as
an income tax of Gh¢480.00 per month. What
tax, correct to one decimal place.
percentage of his salary does he pay as income
tax? 6. A man‟s annual salary is Ghȼ36,900.00. He is
entitled to of this amount free of tax and a
4. The Chief Executive Officer of Down – Below
Football Club is married with two children. He is further tax free allowance of Ghȼ700.00. On his
paid a yearly salary of Gh¢18,000.00. His tax remaining salary after these allowances, he pays
free allowances are as follows: 10p in the cedi on the first Ghȼ6,000.00, 40p in

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the cedi on the next Ghȼ4,500.00 and 17p in the into the country as well as articles that are
cedi on the balance if any. produced locally. Taxes paid on articles brought
a. How much tax does he pay annually? from other countries are import duties and sale
b. Calculate his annual net salary tax. These are collectively called custom duties,
for e.g., cars, clothing, shoes, and other goods
Challenge Problem brought from abroad.
1. Mr. White receives a legacy of Gh¢100,000.00.
He considers two ways of investing it, as shown Taxes are also paid on goods produced, sold or
below: Calculate his net used within a country. This is called excise
income in each case. duties. For e.g., soft drinks, cement, textiles and
i. In a savings bank, he can invest Ghȼ5,000.00 in other made in Ghana goods.
the ordinary branch, bearing 4% p. a. interest and
the rest in the special branch, bearing 7% p. a. The tax or duty amount depends on the value of
interest. Interest received in the ordinary branch the article. Thus, the higher the value of the
is tax free: interest received in the special branch article, the bigger the tax amount.
is subject to tax at a standard rate of Ghȼ300.00
ii. He can buy oil shares costing Ghȼ20.50 each Mathematically, Duty = Rate × Item value.
which pay a dividend annually of Gh¢1.20 per
share, all of which is liable to income tax at the Worked Examples
standard rate. 1. An import duty of 25% is charged on some
goods. How much duty must be paid on a
2. Suppose the following allowances were computer valued at Gh¢12,500.00?
granted free of income tax from an employer‟s
annual salary. Solution
Employees personal allowance: Gh¢30,000.00 Duty on computer = Rate × Item value
Employees wife allowance: Gh¢24,000.00 But rate = 25% = , value = Gh¢12,500
Allowance for each child under 18: Gh¢12,
000.00. Find : Duty = × 12,500 = Gh¢3,125.00
a. the annual taxable income,
b. the net income of the following employees; 2. The import duty of 15% is charge on a
i. Mr. Benson whose income is Gh¢195,000.00 p. calculator. If the calculator is valued at
a and who is married but has no child, Gh¢300.00 Calculate the duty that must be paid
ii. Mr. Okoro who also earns Gh¢195,000.00 and on the calculator
married with two children under the age of 18,
iii. Mr. Keji whose salary is Gh¢195,000.00 per Solution
anum and who has 4 children, two of whom are Duty = Rate × value
above the age of 18, but a widower. Rate = 15% = , value = Ghȼ300.00

Custom Duties Duty = × 300 = Gh¢45.00

Taxes are charged on goods or items imported

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3. The exercise duty on a bag of cement is 5%. barrel of crude oil. How much duty will be paid
How much will a factory pay for producing 270 on 600 barrels of crude oil?
bags of cement each costing ¢ per
bag? 4. The exercise duty on a bag of cement is 12%.
How much will a man pay for importing 240 bags
Solution of cement each at a cost of Gh¢350.00
Method 1
Duty on 1 bag of cement Household Bills (Tariffs)
= Rate × value of bag Tariff is defined the system of prices which
But rate = 5% = and value = Gh¢300.00 companies charge for the services they provide.
Duty = = Gh¢15.00
In Ghana, some of the common household tariffs
are electricity, water and telephone bills. These
Duty on 1 bag is Gh¢15.00.
services are provided by the Electricity Company
Duty on 270 bags = 270 Gh¢15.00
of Ghana, Ghana Water Company and Ghana
= Gh¢4,050.00 Telecommunications respectively. Each service is
charged monthly according to a given schedule or
Method 2 pattern in accordance with the amount of units
1 bag of cement cost Gh¢300.00 consumed.
270 bags will cost 270
= Gh¢81,000.00 Worked Examples
1. In a household, the meter reading for water at
5% duty on Gh¢81,000.00 the end of October 2009 was 7848 thousand
= = Gh¢4,050.00 litres. The meter reading at the end of November,
2009 was 7908 thousand litres. The household
Exercises 17.12 was charged for the consumption at the following
1. An import duty of 25% is charged on a rates:
Toyota car valued at Gh¢55,000.00. How much The first 10 thousand litres at Gh¢500.00 per
duty will you pay if you own the car? thousand litres
The next 30 thousand litres at Gh¢1,300.00 per
2. If the import duty on a bag of flour is 5%. thousand litres
How much duty is paid on 200 bags, if each bag The next 40 thousand litres at Gh¢1,820.00 per
cost Gh¢400.00 thousand litres. Calculate:
3. An import duty of Gh¢15.00 is charged on a a. the consumption at the end of November
b. the total charge for the consumption.

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Solution thousand litres = 30 × Gh¢1,300
a.The consumption at the end of November = Gh¢39,000,000
= End of November - end of October Remaining liters= 60,000 – 10,000 – 30,000
= 7908 thousand litres – 7848 thousand = 20,000 liters
= 60 thousand liters = 60,000 liters
20 thousand litres at Gh¢1,820.00 per thousand
b. Units consumed = 60,000 units litres = 20 × Gh¢1,820 = Gh¢36,400.00
First 10 thousand litres at Gh¢500.00 per Total amount;
thousand litres = 10 × Gh¢500 = Gh¢5,000 = Gh¢(5,000 + 39,000 + 36,400) = Gh¢80,400.00

The next 30 thousand litres at Gh¢1,300.00 per

Method 2
a. Units consumed = 7908 thousand litres – 7848 thousand
= 60 thousand liters
= 60,000 liters
b.
Charges (Ghȼ) Total Charge (Ghȼ) Amount (Ghȼ) Remaining Liters
10,000l at 500/1000l 10 × 500 5,000 60,000 –10,000 = 50,000l
30,000l at 1300 /1000l 30 × 1300 39,000 50,000 –30,000 = 20,000l
40,000l at 1820 /1000l 40 × 1820 72,800 20,000 – 40,000 = - 20,000l
20,000l at 1820/1000l 20 × 1820 36400 0

Total charge = Ghȼ(50,000 + 39,000 + 36400)= Ghȼ 80,400,000.00

Note:
The negative sign (on the table) indicates that iii. Find correct to 2 decimal places, the
there is no remaining liters. Therefore, all figures percentage change in Kwaku‟s consumption
on that row are not used in the calculation. of electricity from July to August that year

2. In a certain year, Kwaku paid for electricity Solution

consumed each month as follows; the cost of the i. Total units consumed = 420
first 30 units was Gh¢1.50 per unit; the cost of 30 units was 1.50 per unit
the next 30 units was Gh¢1.70 per unit; the cost = 30 × Gh¢1.50= Gh¢45.00
of each additional unit is 50p
i. If Kwaku used 420 units of electricity in July The next 30 units was 1.70 per unit
that year, calculate the amount he paid. = 30 × Gh¢1.70 = Gh¢51.00
ii. If kwaku paid Ghȼ194.00 in August last year,
calculate the number of units of electricity he Remaining units = 360 units
used. 360 units at 50p per unit
= 360 × 50p = Gh¢180.00

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Total amount But cost of each additional units;
= Gh¢45 + Gh¢51 + Gh¢180 = Gh¢276.00 = 50p = Gh¢ 0.5
Units consumed for Gh¢98.00 at 50p per unit
ii. Amount paid in August last year = Ghȼ194.00 = = 196 units
Less cost of first 30 units = Gh¢45.00
Less cost of next 30 units = Gh¢51.00
Total units consumed last year august
Remaining = Gh¢98.00
= 30 units + 30 units + 196 units = 256 units
Total cost of each additional units;
= Gh¢98.00
Method 2
Units consumed = 420

Charges (Ghȼ) Total Charge (Ghȼ) Amount (Ghȼ) Remaining Units

30 at 1.5 each 30 × 1.5 45 420 – 30 = 390
30 at 1.7 each 30 × 1.7 51 390 – 30 = 360
Additional at 0.5 each 360 × 0.5 180

Total amount = Ghȼ(45 + 51 + 180) = Ghȼ276.00

b. Amount Charged = Ghȼ194.00

Charges (Ghȼ) Amount (Ghȼ) Less Amount (Ghȼ) Units

30 at 1.5 each 30 × 1.5 = 45 194 – 45 = 149 30
30 at 1.5 each 30 × 1.7 = 51 149 – 51 = 98 30
Remaining at 50p each 196

Total units consumed = 30 + 30 + 196 = 256 units

3. In a chargeable period, a household consumed Solution

700 units of electricity. Use the table below to First 50 units = Gh¢270.00
find the cost. Next 150 units = Gh¢(150 × 4.35) = Gh¢652.50
Electricity Company of Ghana Next 400 units = Gh¢(400 × 6.86) = Gh¢2,744.00
Power Tariff Schedule Additional 100 units = Gh¢(100 × 14.72)
Domestic 50 units or less @ Gh270.00
= Gh¢1,472.00
Next 150 units @ Gh4.35 each
Next 400 units @ Gh6.86 each
Additional units @ Gh14.72 each Total units = 700
Total amount;
Commercial Each unit supplied @ Gh17.00
Service charge Gh115.00 = Gh¢(270 + 652.50 + 2,744 + 1,472)
= Gh¢5,138.50

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4. The monthly electricity charges in a Units consumed for Gh¢34,760.00 at Gh¢220 per
country are calculated as follows;
unit = = 158 units
First 50 units = Ghȼ4,000.00
Next 100 units = Ghȼ120.00 per unit
Next 150 units = Ghȼ150.00 per unit Total units consumed by the man;
Next 300 units = Ghȼ220.00 per unit = 50 + 100 + 150 + 158
Remaining unit = Ghȼ350 per unit = 458 units
a. How much did Mr. Owusu pay for using 720
units in a month? Exercises 17.13
b. A man paid Ghȼ73,260.00 for electricity 1. The monthly electricity bill for a household is
consumed in a month. How many units of calculated by adding a fixed charge
electricity did he consume? Ghȼ15,500.00 to the cost of the number of units
of electricity used. If the cost per unit is
Solution Ghȼ500.00, what is the bill for a household that
Total units consumed = 720 units uses 111 units in a month?
Remaining units
= 720 – 50 – 100 – 150 – 300 2. Consider the table below and use it to answer
= 120 units the questions that follow:

First 50 units = Ghȼ4,000 Electricity Company of Ghana

Next 100 units = 100 × Ghȼ120.00 Power Tariff Schedule
= Ghȼ12,000 From July
Domestic 50 units or less @ Gh270.00
Next 150 units = 150 × Ghȼ150.00 Next 150 units @ Gh4.35 each
= Ghȼ22,500 Next 400 units @ Gh6.86 each
Additional units @ Gh14.72 each
Next 300 units = 300 × Ghȼ220.00
= Ghȼ66,000.00 Commercial Each unit supplied @ Gh17.00
Remaining unit = 120 × Ghȼ350 = Ghȼ42,000 Service charge Gh115.00
Total amount;
= Ghȼ4,000 + Ghȼ12,000 + Ghȼ22,500 + Ghȼ i. How much does a man pay for using 900 units
66,000 + Ghȼ42,000 of electricity at home?
= Ghȼ146,500.00 ii. A clerk made use of 35 units in his office, how
much did he pay just for the units he used?
ii. Amount paid = Ghȼ73,260.00 iii. If the bill for a commercial office is
Less first 50 units = Ghȼ4,000 Gh¢15,415.00. Find the number of units
Less next 100 units = 100 × Ghȼ120.00 consumed.
= Ghȼ12,000
Less next 150 units = 150 × Ghȼ150.00 3. In a certain year, Mr. Harry paid for domestic
= Ghȼ22,500 electricity consumed each month as
Remaining = 34,760 at 220 per unit follows:

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The cost of the first 30 units used is 100p per First 10 liters Ghȼ 0.50 per liter
unit; the cost of the next 30 units used is 70p per Next 10 liters Ghȼ 0.75 per liter
unit and the cost of each additional unit used is Next 10 liters Ghȼ 1.00 per liter
50p per unit, Next 10 liters Ghȼ 1.25 per liter
i. Mr. Harry used 430 units of electricity in Remaining liters Ghȼ 1.50 per liter
august that year, calculate the amount he paid
If Mr Brown used 200 liters of water in April of
ii. if he paid Gh¢301.00 the next month, find the that particular year, how much did he pay?
units of electricity he consumed.
Challenge Problem
4. The Ghana Water Company charges the Below is the tarrif schedule for electricity
following rates for water consumed each month: consumption in a particular country;
First 20,000 liters at Gh¢400.00 per 1,000 liters, First 50 units or less = Gh¢0.27 per unit
Next 30,000 litres at Gh¢520 .00 per 1,000 litres, Next 50 units = Gh¢0.31 per unit
Remaining liters at Gh¢900 per 1,000 liters Next 50 units = Gh¢0.68 per unit
i. If Mr Boadi used 62,000 liters in a particular Next 30 units = Gh¢0.80 per unit
month, calculate his total bill for the month, Additional units = Gh¢0.95 per unit
ii. If he paid Gh¢28,000.00 in a particular month, Service Charge = Gh¢2.00
calculate the liters of water he consumed to the Street light = 5% of additional units
nearest liter.
If a household initial and final meter readings for
5. The Ghana water company tariff for a certain a particular month is 4122kw and 4473kw,
year is given below: calculate the total bill for that month.

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18 VARIATIONS Baffour – Ba Series

Meaning of Variations From y = k ,

In mathematical sense, variation is concerned k = = , where and are corresponding
with ways and means by which one variable
values of x and y
relates or depends on other variable(s). The type
of variations includes;
3. “y varies as the square root of x”, is expressed
1. Direct variation
mathematically as:y √
2. Inverse variation
3. Joint or Combined variation y = k√ , where k is the constant
4. Partial variations or variation in parts From y = k√ ,
k= = , where √ and y1 are corresponding
√ √
Direct Variation
values of √ and y
It is a kind of relation that exists between two
quantities such that an increase in one quantity
4. “y varies as the cube of x”, is expressed
causes an increase in the other quantity and vice –
mathematically as: y
versa. For e.g. If 4 pens (x)cost Ghȼ2.00 (y), then
y= , where k is the constant of variation
an increase in the number of pens means the cost
From y = ,
will be more than Ghȼ2.00 and a decrease in the
number of pens means the cost will be less than k = = , where and are the
Ghȼ2.00. In this instance, the cost of pens (y) is corresponding values of y and
said to be in a direct proportion to the number of
pens(x). Simply put “y is (directly) proportional 5. “y varies as the cube root of x”,is expressed
to x”.This kind of relationship is called a direct mathematically as: y √
variation. ⇒ = k √ , where is the constant
From y = k √ ,
In variations, the appropriate phrase to use
= = , where √ and are
instead and their mathematical expressions are √ √
explained below: corresponding values of √ and y
1. “y varies directly as x”,is expressed
mathematically as:y x, Worked Examples
y = kx…….. (1), where k is called the constant of 1. Given that p varies directly as the square of q
variation and p = 4 when q = 2. Find q when p = ⁄
From (1), k = = , where and are
Solution
corresponding values of x and y
p varies directly as the square of q
2. “y varies as the square of x”is expressed p
mathematically as:y x2, ⇒ p = k …………….(1)
y= , where k is the constant From eqn (1)

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k= y= x
Whenp = 4 and q = 2 y = × 15 = 5
k= = = =1 Therefore, a = 5, and b = 6

Put k = 1 in equation (1) to obtain the variation 3. P varies as the square of (Q + 1) and P is 2
equation as p = when Q is 3.
i. Write an equation connecting P and Q.
When p= ⁄ , q = ?andk = 1 ii. Find the possible value of Q, when P = 8.
From the variation equation;
q2 = Solution
i. P varies as the square of (Q + 1)
q =√ = √ = P (Q + 1)2
⇒P (Q + 1)2 ………..(1)
2. In the table below, y varies directly as x
changes. Find the values of a and b From Eqn (1)
k=(
y 1 2 a
x 3 b 15 When P = 2 and Q = 3,
k=( = = =
(
Solution
y varies directly as x
Put k = in eqn (1)
y x
⇒y = kx ……….(1) P (Q + 1)2
From eqn (1) The equation connecting P and Q is:
k= P (Q + 1)2
From the table when y = 1, x = 3
k= = ii. When p = 8, Q = ?andk =
8 (Q + 1)2
Put k = in eqn (1) to obtain the variation Multiply both sides by 8
equation asy = x 64 = (Q + 1)2

From the table, when y = 2, x = b and k = √ = √(

8=Q+1
From the variation equation,
Q=8–1=7
x=
⇒when p = 8, Q = 7
x = × 2= 6
4. Given that y varies directly as x2 and that y = 20
From the table, when y = a, x = 15 and k = when x = 2, find the value of y when x = 3
From the variation equation

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Solution 6. If A varies directly as the squere of r and V
y varies directly as x2 varies directly as the cube of r, find the
y x2 percentage increase in A and V if r is increased by
⇒y = kx2……….(1) 20 percent.
From eqn (1)
k= Solution
A r2
When y = 20, x = 2 and k= ? A= kr2

k= = =5
When r is increased by 20%,
=r+( )
Put k = 5 in eqn (1) to obtain the variation
equation as y = 5x2 =r+ = =

When x = 3, y= ?andk = 5
A = k( ) =
From the variation equation
y = 5x2 = 5 × 32 = 5 × 9 = 45
Therefore, when x = 3, y = 45 Increase in area;
= – kr2
5. The resistance to an aeroplane varies directly
as the squareof its speed. If the resistance is 1260 =
ohms when the air craft is travelling 45 m/s, what =
is the resistance at 65m/s?
Percentage increses in A;
Solution
Let R represent the resistance and s represent the = × 100%
speed
⇒R s2 =( ) × 100%
R ks2 =( ) × 100%
= × 100 % = 44%
When R = 1,260, s = 45
1260 k (45)2 Workout the % increase in V Ans : (72%)

k=
Exercises 18.1
Hence, R s2 A. Express the following as an equation:
1. u varies as the square root of v
When s = 65, R = ? and k = 2. The cube of a varies as the square of b
3. If y x2, state the factor by which y is
R (65)2 = = 2,630 multiplied when x is multiplied by -3

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B. 1. v varies as w and w = 5 when v = 40. Find: 10. P varies as the square of the sum of x and y
i. v when w = 8 ii. w when v = 4 when P is 4, x is 3 and y = 5. Find P when x = 6
and y = 8
2. If y varies asx3 and y = 54 when x = 3, find y
when x = 1.5and obtain an expression for y in 11. Find, √ from the table below leaving
terms of x your answer in surd form if p varies directly as q2

3.u varies directly as the square of v. If u = 14 P 2 x 32

whenv = 2, find the value of uwhen v = 3 q 1 2 y

4. It is known that y varies as x and y = 9 when x 12. For the entries in the table below, verify that
= 3. Find y in terms of xand the value S2 W and hence find the equation connecting W
of y when x = 4 and

5. The area, A of a sector, containing a specified S 1 2 5

angle, varies as the square of its W 2 8 50
radius, r. When r is 2cm, A is 3.6cm2
a. Find A when r is 7cm 13. The breaking weight, W tonnes of a rope of
b. Find r when A is 8.1cm2 diameter d mm is proportional to the square of its
diameter. Given that W = 6 when d = 40, find W
6. Find the values of x and y in the table below: when d = 24/

T 1 2 3 x 5 14. The amount, in litres of petrol consumed

S y 4 6 8 10 varies directly as the number of kilometres
travelled by a car. It is known that a certain
7. y varies as the square root of x, and y = 3 when saloon car used up 4.5 litres of petrol when it
x = 1.44. Find: travelled a distance of 53km. Find how much
i. y when x = 0.36 ii. x when y = 10 petrol is needed for the car to cover 275km.

8. In the table below, find the missing entry 15. The distance, d meters travelled by a falling
assuming that V object in t seconds is given by the formula d =
k . When t = 4, d = 80
h 125 275 i. Find d when t = 8, and t when d = 1125
V 25 ii. Change the subject of the formula to t, and
copy and complete the statement: „The time in
9. Given that the mass of a sphere varies directly seconds an object takes to fall varies ……….‟
as the cube of its radius, complete the table.
16. The area of a circular sector containing a
Mass 54 given anglevaries as the square root of the radius
Radius 3 4 of the circle. If the area of the sector is 2cm2

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when the radius is 1.6cm, find the area of the From y = , k=y = , where, are
sector containing the same angle when the radius
corresponding values of x and y respectively
of the circle is 2.7cm.
4. “y varies inversely as the cube root of x”
Inverse Variations
is expressed mathematically as y ,, therefore
It is a kind of relationship that exists between two √
quantities such that an increase in one quantity y = , where k is the constant of variation.
√
causes a decrease in the other quantity and vice
versa at standard conditions. For e.g. If six Fromy= , k = y√ = √ , where, √ are
√
girls(x) can sweep a classroom in 10 minutes(y), corresponding values of x and y respectively.
then an increase in the number of girls implies
that the work can be completed in less than 10 Worked Examples
minutes and a decrease in the number of girls 1. Find the variation constant and an equation of
implies that the work can be completed in more variation if y varies inversely as x, and y = 32
than 10 minutes. In this instance, the time is said when x = 0.2.
to be in an inverse proportion to the number of
girls. Simply put, “y is inversely proportional to Solution
x”. This kind of relationship is called inverse “y varies inversely as the x”
variation, y
In variations, the appropriate phrase to use
y ……..(1)
instead and their mathematical expressions are
explained below:
1. “y varies inversely as x” is expressed From eqn (1)
k = xy
mathematically as y ,, therefore y = , where k
When x = 0.2, y = 32 and k = ?
is the constant of variation. k = 0.2 × 32 = 6.4
From y= ,k = yx = , where , are The variation constant k = 6.4
corresponding values of x and y respectively Hence, the equation of variation isy

2. “y varies inversely as the square of x” 2. y varies inversely as the square of x. When x =

is expressed mathematically as y ,, therefore 3, y = 100. Find:
y= , where k is the constant of variation. From a. the equation connecting x and y
b. the value of x when y = 25
y = , k = y = , where, , are c. the value of y when x = 15
corresponding values of x and y respectively
Solution
3. “y varies inversely as the cube of x”is a. y varies inversely as the square of x
expressed mathematically as y ,therefore y
y= , where k is the constant of variation. y ……….(1)

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⇒k = x2y Solution
2
k = 3 × 100 = 900 i. R varies inversely as the square of (3q – 2)
⇒R (
Substitute k =100 in eqn (1)
R= ……….(1)
y (
k = R(3q – 2)2
Hence,the equation connecting x and yis y
When R = 4, q = 2 and k= ?
b. When y = 25, x = ?andk = 900 k = 4[3(2) – 2]2 = 4(6 – 2)2= 64
From eqn (1)
Hence, the variation equation is R = (
x2 =

x=√ =√ =6 When q = 1, R = ? and k = 64

From eqn (1)
R= = = 64
c. When x = 15, y = ?and k = 900 ( ( (

From eqn (1)

ii. When R = 16, q = ? andk = 64
y = =4
From eqn (1)
R(3q – 2)2 = k
3. R varies inversely as the cube of S. If R = 9
(3q – 2)2=
when S = 3, find S when R =
3q – 2 = √
Solution
3q = √ + 2
R varies inversely as the cube of S
R √ (√ )
q= = =
R …..(1)
3
k= y= × 9 = 243 5. If y is inversely proportional to x + 2 and
Hence, the variation equation is R y = 48 when x = 10, find x when y = 30

Solution
When R = , S = ? andk = 243
“y is inversely proportional to (x + 2)”
From eqn (1)
y
3
S =
y= ……….(1)
S= √ = √ =√ =√ =4 k = y(x + 2)
⁄

4. If R varies inversely as the square of (3q – 2) Wheny = 48, x = 10

and R = 4 when q = 2, find: k = 48 (10 + 2) = 48(12) = 576
i. R when q =1 ii. q when R = 16 Hence, the variation equation is y =

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When x = ?, y = 30
Q = √ – 1= √ –1=√ –1=2–1 =1
30 =
30(x + 2) = 576
2. A garrison of 400 men had food for 40 days.
30x + 60 = 576
After 10 days, 200 more men joinedthem. How
30x = 576 – 60
long will the food last now? Assume that the
30x = 516
amount of food taken by each man is almost the
x= = 17.2 same.

Some Solved Past Questions Solution

1. P varies inversely as the square of (Q + 1) and 400 men ate the food for 40 days. Hence, they
P is 2 when Q is 3 can now eat the remaining food for 40 – 10 = 30
a. Write an equation connecting P and Q days.
b. Find the possible values of Q when P = 8
Let the remaining food last for x days. When 200
Solution men joined them, then the number of men
P varies inversely as the square of (Q + 1) becomes 400 + 200 = 600
P (
No of men 400 600
P= ……….(1)
( No of days 30 x
k = P(Q + 1)2
From the table above,
When P = 2, Q = 3 and k= ? 400 × 30 = 600x
k = 2(3 + 1)2 ⇒x = 20 days
k = 2(4)2= 2 16 = 32
3.The following table shows the safe exposure
Substitute k = 32 in P = to obtain time for people with less sensitive skin.
(

P= UV Index Safe Exposure Time

(
The equation connecting P and Q is ( in mins)
2 150
P= 4 75
(
6 50
8 37.5
b. When P = 8, Q = ?andk = 32 10 30
From eqn (1)
P(Q + 1)2 = k a. Determine whether the table indicates direct
2
(Q + 1) = variation or inverse variation.
b. Find an equation of variation that approximates
Q+1=√ the data. Use the data point (6, 50)

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c. Use the equation to predict the safe exposure 4. Given that y is inversely proportional to x and
time for a person with less sensitive skin when that y = 24 when x= , find the value of y when x
the UV rating is 3
= 16
5. Given that y is inversely proportional to the
Solution
a. The table indicates an inverse variation square of x and that y = 24 when x= ,
i. Find the equation connecting x and y
b. Index, I = 6 and time, T = 50 ii. Find the value of y when x = 3
From the table, I
6. Given that y is inverselyproportional to x and
I = ……….(1)
that y = 4 when x= , find the value of y when x =
From eqn (1)
k = I × T = 6 × 50 = 300 2

Substitute k= 300 in eqn (1) 7. Given that N is inversely proportional to r2 and

I= that N = 3, when r = 5, find the value of N when r
= 10
c. When I = 3, K = 300 and T = ?
8. If s varies inversely as t3 and s = 4 when t = 2,
From eqn(1) find s in terms of t and the value of s when t = 7
T= = = 100
9. Given that y is inversely proportional to (x – 2)
and that y = 6 when x = 5, calculate the value of y
Exercises
when x = 7
A. Express the following as an equation:
1. p varies inversely as
10. It is given that the force between two particles
2. p varies inversely as v
is inversely proportional to the square of the
3. H varies inversely as the square of d
distance between them. If the force is F when the
4. T varies inversely as the square root of g
distance between them is r and cF. When the
B. 1. y varies inversely as x and y = 6 when x = 6. distance is 5r, write down the value of c. Ans
Find: i. y when x = 24 ii.x when y = 32 0.04

2. A quantity y varies inversely as another Joint or Combined Variation

quantity x. If y = when x = 9, express y in terms It is a kind of relationship that exists between
three (or more) quantities such that one quantity
of x
relates to the others in a direct proportion. In
other words, joint variation is a type of variation
3. Given that y is inversely proportional to x2 and
where a quantity varies directly as the product of
that x = 10, when y = 5, find the value of y when
two or more other quantities. When one variable
x =15
varies directly as two or more others, the word

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jointly is sometimes used. For example, in V = π k= =5
V varies jointly as h and
Put k = 5 in eqn (1)
S = 5qr
The joint variation is of three types. Its
mathematical expressions are explained below:
The equation connecting s, q and r is
S = 5qr
I. The variation is direct for all the variables
E.g. The variable z varies directly as x and ii. S= ?q = 6, r = 8 and k = 5
directly as y. That is z = kxy, where k is the
S = 5 × 6 × 8 = 240
constant of proportionality
iii. q= ? S = 60, r = 0.5 and k = 5
II. The variation is inverse for all the variables. s = 5qr
E.g. The variable z varies inversely as x and
60 = 5 × q × 0.5
inversely as y. That is z = , where k is the q= = 24
constant of proportionality
2. z varies directly as x and inversely as the
III. The variation is a combination of direct square of y. Find z in terms of x and y , given that
and inverse variations z = 10 when x = 8 and y = 2. Find;
E. g. z varies directly as x and inversely as y are i. the value of z when x = 12, y = 4
expressed as z x and z respectively. By ii. the value of x when y = 5, z = 4
combining the two, z
Solution
⇒ z= “z varies directly as x and inversely as y2”
z x and z
Worked Examples
z
1. If s varies directly as q and directly as r and
s = 40 when q = 2 and r = 4, z ……………(1)
i. Write an equation connecting s, q and r
ii. Find the value of S, when q = 6 and r = 8 Whenz = 10, x = 8, y = 2 and k= ?
iii. Find the value of q when S = 60 and r = 0.5
10

Solution 10 × 22 = 8k
“S varies directly as q and directly as r” 10 × 4 = 8k
S q and S r k= =5
S qr
S = kqr……………….(1) Put k = 5 in eqn (1)
k= Hence the variation equation is z
When s = 40, q = 2 and r = 4,

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i. When z = ?x = 12, y = 4 and k = 5 √
( ⇒v = √ = = 3.2
z = = 3.75
4. x varies jointly as the square of m and the cube
ii. Whenx= ?y = 5, z = 4 and k = 5 of n. When x = 9, m = ⁄ and n = ⁄
4 a. determine the relationship between X, m and n
4 × 52 = 5x b. calculate, correct to three significant figures the
4 × 25 = 5x values of the following:
x= = 20 i. x when m = and n
ii.m when x = 5 and n =
3. A quantity f varies jointly as the square of v
and the inverse of r. If f = 12 when v = 2 and r = Solution
3, find: a. “x varies jointly as the square of m and the
a. the constant of variation cube of n”
b. the value of v when r = 4 and f = 23 x m2 and x n3
x m2n3
Solution x = km2n3 ……….(1)
a. “f varies jointly as the square of v and the
inverse of r” Whenx = 9, m = ⁄ and n = ⁄ , k = ?
f v2 and f
k= = = = 128
( ) ( ) ( )( )
f
f …………….(1)
Put k = 128 in eqn (1)
x = 128m2n3 (variation equation)
If f = 12 when v = 2 and r = 3,
(
The relationship between x, m and n is:
⇒12 x = 128m2n3
12 × 3 = 4k
k= =9 b. i. Whenx= ?, m = , n and k = 128
But x = 128m2n3 (variation equation)
Hence, the constant of variation, k = 9
Put k = 9 in eqn (1) x = 128 ( ) ( )
f x = 128 ( ) ( ) = 128( ) = 0.455 (3 s. f)
Hence, the variation equation is f
ii. When m = ?, x = 5 , n =
b.v = ? , r = 4, f = 23 and k = 9
From x = 128m2n3 (variation equation)
23
5 = 128m2( )
4 × 23 = 9v2
= v2 5 = 128m2( )

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5 × 512 = 128m2 9
2
( (
m =
9
⇒m = √ = 4.47 9 × 21 = k
k = 189
5. Find the equation of variation if y varies jointly
Put k = 189in eqn (1)
as x and z and inversely as the square of w, and y
= 105 when x = 3, z = 20 and w = 2 t
The equation connecting t, v and w is t
Solution
“y varies jointly as x and z and inversely as the 7. y varies directly as (x – 2) and varies inversely
square of w”, as x2. When x = 12, y = 8.
⇒y x z and y i. Find the relation between x and y
y ii. Find the value of y when x = 4

y …………..(1) Solution
“y varies directly as (x – 2) and varies inversely
Wheny = 105, x = 3, z = 20 and w = 2 as x2 ‟
( ( ⇒y ( and y
105
(
105 = y
(
105 × 4 = 60k y ……………(1)
k= =7
When x = 12, y = 8
(
Put k = 7 in eqn (1) 8=
y (variation equation) 8=
8 × 144 = 10k
6. t varies inversely as v and inversely as w. If v =
k= = 35.2
7, w = 3 and t = 9. Find an equation connecting t,
v and w
Put k = 35.2 in eqn (1)
Solution (
y
“t varies inversely as v and inversely as w”
t and t When y = ? , x = 4
(
t y= = = 4.4
t ……………(1)
8. x varies directly as y2 and y varies inversely as
When v = 7, w = 3 and t = 9 z. If y = 3 and z = 2 when x = 12, find:

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i. an expression for x in terms of z alone. When z = 4, x = ( = =3
ii. the values of x and y when z = 4
Hence, when z = 4, x = 3 and y = 1.5
Solution
9. Given that p varies directly as q whiles qvaries
“x varies directly as y2 and y varies inversely as
inversely as r, how does p varies with r
z”
x and y Solution
x ……………(1) “p varies directly as q whiles qvaries inversely as r”
y ………………(2) p q
⇒p = k1q……….(1)
Where c and k are variation constants
Wherek1 is the constant of variation
From eqn (1),
Also, q
Whenx = 12, y = 3 substitute in x
12 ( ⇒q = …….(2)
12 = 9c Where k2 is the constant of variation
c= =
Substitute eqn (2) into eqn (1)
P= where k1 × k2 = k
Put c = in eqn (1)
P=
x
⇒P , P varies inversely as r
Hence, variation equation is x …(i)
From eqn (2),
Some Solved Past Questions
When z = 2, y = 3 substitute in y 1. x varies directly as the square root of t and
3 inversely as S. When x = 4, t = 9 and s = 18
i. Express x in terms of s and t.
2×3=k
ii. Find x when t = 81 and s = 27.
k=6
Hence, variation equation isy …(ii)
Solution
Now put y into x i.“x varies directly as the square root of t and
inversely as S”
x ( ) =
⇒x √ and x
The expression for x in terms of z is x = √
x
√
b. From eqn (ii),y x …………..(1)

when z = 4, y = = 1.5 When x = 4, t= 9 and s = 18

From the expression, x = √
4

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4 × 18 = √ 30 =
√
k= = = 24
√ 30 × √ =5×9
60 √ = 45
Put k = 24 in eqn (1)
√ √ 0.75
Hence, the variation equation is x
Q=( = 0.5625

ii. When x=? t = 81 and s = 27 4. P varies directly as the square of S and

√
x = =8 inversely as R. When S = 6 and R = 3, P = 46.
Find the value of P when S = 10 and R = 5.
2. Three quantities P, Q and R are connected so
that P varies directly as R and inversely as the Solution
square root of Q. If P = 6 when R = 12 and Q = “P varies directly as the square of S and inversely
25. as R”
a. Find an expression for P in terms of Q and R P and P
b. Find the value of Q when P = 30 and R = 9
P
Solution P ……………(1)
a. “P varies directly as the as R and inversely as
the square root of R”. When S = 6 and R = 3, P = 46, k = ?
P and P (
√ 46
P 46 × 3 = 36k
√
k= =
P ……………(1)
√

Put k = in eqn (1)

If P = 6, R = 12 and Q = 25
6 P
√
6 Hence, the equation of variation is P
6 × 5 = 12k
30 = 12k When P = ?, S = 10, R = 5 and k =
k= (
P = = 76.67
(

Put k = in eqn (1)

3. x, y and z are such that x varies directly as z
Hence, the variation equation is P and inversely as the cube root of y. If x = 8, y =
√
27 and z = 4. Find:
a. an expression for x in terms of y and z
When Q = ?, P = 30 and R = 9 and k =

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b. the value of y when x= 12 and z = 10 P ………………(1)
√

Solution
When P = 3, Q = 4, R = 25 and k= ?
a. “x varies directly asz and inversely as the cube
root of y” P
√
(
x and x 3
√ √
x 3 × √ = 4k
√
3 × 5 = 4k
x ……………(1)
√ k= =

Whenx = 8, y = 27 and z = 4
( Put k = in eqn (1)
8
√
P
8 × √ = 4k √

8 × 3 = 4k The equation connecting P, Q and R is P

√
k= =6
Put k = 6 in eqn (1) b. When R = ? P = 6, Q = 7 and k =
Hence, the variation equation is x = P
√ √
(
6
√
b. When y = ?, x = 12, z = 10 and k = 6
√ =
x=
√ √ = = 4.375
(
12 =
√ R=√ =2
(
√ = = =5
y= = 125 5. The electrical resistant R ohms of a wire varies
directly as the length, Lcm and inversely as the
4. P varies directly as Q and inversely as the square of the diameter dcm
square root of R. If P = 3 when Q = 4 and R = 25 a. Express d in terms of L, R and the constant of
a. Write an equation connecting P, Q and R variation,k
b. Find R when P = 6 and Q = 7 (correct to the b. Find the value of d, correct to 2 decimal places,
nearest whole number) when L = 15cm, R = 0.23ohms and k = 1.25 ×

Solution
a. “P varies directly as Q and inversely as the Solution
square root of R” a. “R varies directly as Lcm and inversely as the
square of d”
P and P
√ R and R
P
√ R

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R b. whend = ?, F = 30, M = 7.5 and m = 4
F=
d2 =
( (
30 =
d=√
= =2

b. When d= ?, L = 15cm, R = 0.23 and k = 1.25 × d=√

Exercises 18.3
(
d=√ =√ = 0.08cm A. 1.Express the following as an equation,
where k is a variation constant
6. The force of attraction, F, between two bodies, i. Z varies jointly as x2 and y
varies directly as the product of their masses M ii. V varies directly as h and r2 jointly
and m and inversely as the square of the distance iii. R varies directly as the square root of x and
d, between them. Given that F =20N when M = varies inversely as the square of y
25kg, m = 10kg and d = 5 meters. Find:
a. an expression for F in terms of M, m and d, B. 1. z varies directly as the square of x, and
b. the distance d, when F = 30N, M = 7.5kg and inversely as √ . Find z in terms of x and y, given
m = 4kg. that z = 8 when x = 6 and y = 9. Find the value of
z when x = 2 and y = 4.
Solution
a. “F varies directly as the product of M and m 2. q varies as x and inversely as , and q = 10
and inversely as the square of d”. when x = 30 and z = 14. Find:
F Mm andF i. the formula connecting q, x and z,
ii. q when x = 36 and z = 42.
F
F= ……………(1) 3. The volume V of a given mass of gas varies
directly as the absolute temperature T and
When F = 20, M = 25, m = 10, d = 5, k= ? inversely as the pressure P, When V = 490, T =
350 and P = 750. Calculate V when T = 400 and
F=
P = 560.
( (
20 =
20 × = 250k 4. If y varies jointly as xand (x + 2), when x = 4, y
= 72. Find the value of y if x = 7.
k= =2

5. If c varies directly as the cube of m and

Put k = 2 in eqn (1)
inversely as n, and c = 8 when m = 4 and n = 12,
F= find c in terms of m and n and find the value of k
when m = 2 and n = 16.
Hence, the variation equation is F =

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6. If y varies jointly as x2 and z3, and y = 12 when according to the number required. Some of its
x = 2 and z = 3, find y in terms of xand z and the mathematical expressions are explained below:
value of y when x = 3 and z = 4 I. y is partly fixed (constant) and partly varies as
x is expressed asy = C andy x,
7. Given that C varies as the square of x and y = C and y= kx,
inversely as the cube of y, and that C = 18 when x By putting them together, y = C+ (kx), where C
= 3 and y = 2. Find the value of: and k are variation constants
i. C when x = 2 and y = 4
ii. x when C = 48 and y = 2 II. y is partly fixed (constant) and partly varies
as the inverse of x is expressed as
8. In the table below, W , where W, R y = C and y ,
and Q positive integers. Solve for W2 and r3 y = C and y = ,

W R Q By putting them together, y = C + ( ), where C

3 4 4 and k are variation constants
W2 1 2
8 r3 6 III. ypartly varies as xand partly varies as z
expressed asy x andy z,
Challenge Problems y = Cx andy= kz
1. It is given that x varies directly as y, and that y By puttingthem together,y = Cx+ (kz), where C
varies inversely as the square root of z. and k are variation constants
If x = 3 and y = 12 when z = 4
i. Calculate the values of x and y when z = 9 Solving Partial Variation Problems
ii. Express x in terms of z Depending on the given variation phrase,
I. Form two equations by substituting the set of
2. The force between two magnetic poles varies values of the given quantities
inversely as the square of their distance apart. If II. Solve the equations simultaneously to obtain
the distance apart varies as the square of the time, the values of the constants, C and k
a. Find how the force varies with the time, t III. Substitute the values of C and kobtained in
b. If the force is 0.25 N, when t = 3 the variation equation to establish the actual
i. Find the force when t = 4 relationship between the given quantities
ii. Find the time when the force is 0.001 N
Worked Examples
Partial Variation or Variation in Parts 1. n is partly constant and partly varies as m.
It is a type of variation which consist of two parts When m = 7, n = 11 and when m = 10, n = 23.
namely a fixed (constant) part and a variable part. Find:
For instance, when a sound system and chairs are i. the equation connecting n and m
hired for an activity, the sound system is hired at ii. the value of n when m = 25
fixed cost whilst the cost of the chairs varies iii. the value of m when n = 43

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Solution i. Find the equation connecting x and y
n is partly constant and partly varies as m. ii. What is the value of y when x = 24
n = c and n m iii. Find the value of x when y = 12
n = c and n = km
⇒ n = c + km Solution
“y is partly constant and partly varies as the
When m = 7, n = 11, square of x”
11 = c + k(7) y = c and y x2
11 = c + 7k……………(1) y = c and y = kx2
⇒y = c + kx2
When m = 10, n = 22,
23= c + k(10) When x = 2, y = 6 substitute in y = c + kx2
23 = c + 10k……………(2) 6 = c + k(2)2
6 = c + 4k ……………(1)
eqn (2) – eqn (1)
12 = 3k When x = 6, y = 10 substitute iny = c + kx2
k= =4 10 = c + k(6)2
10 = c + 36k ……………(2)
Put k = 4 in eqn (1)
11 = c + 7(4) Solving eqn (1) and eqn (2) simultaneously
11 = c + 28 6 = c + 4k ………………(1)
c = 11 – 28 = -17 10 = c + 36k ……………(2)

Put k = 4 and c = -17 in n = c + km to obtain the eqn (2) – eqn (1)

equation of variation as n = -17 + 4m 4 = 32k
k=
ii. When n = ? m = 25, Substitute in n = -17 + 4m
n = -17 + 4(25) = 83
Put k = in eqn (1)
iii. When m = ?n = 43, 6 = c + 4( )
Substitute in n = -17 + 4m
6=c+
43 = -17 + 4(m)
43 + 17 = 4m 6×8=8×c+8× h
60 = 4m 48 = 8c + 4
m= = 15 48 – 4 = 8c
44 = 8c
2. A variable y is partly constant and partly varies c=
as the square of x. When x = 2, y = 6
and when x = 6, y = 10
Substitute c = and k = in y = c + kx2

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y= + x2 13 = c + 4k …………..(2)

The equation connecting x and y is y = + x2

eqn (2) – eqn (1)
-27 = 3k
ii. When y = ? , x = 24 k= 9
y= + x2
y= + (24)2 Put k = in eqn (1)
40 = c + (
y= + (576) = + 72 = 77.5 c = 40 + 9 = 49

iii. When y = 12, x = ? Putc = and k = in Q = c + kR2

12 = + x2 Q = 49+ (-9)R2
The equation connecting x and y is Q = 49 – 9R2
8 × 12 = 8 × + 8 × x2
96 = 44 + x2 When Q = ?, R = 3, put in Q = 49 – 9R2
96 – 44 = x2 Q = 49 – 9(3)2 = 49 – 81 = -32
52 = x2
x = √ = 7 (nearest whole number) 4. The cost of producing a radio component is
partly constant and partly varies inversely with
3. Q is partly constant and partly varies as the the number made per day. If 100 components are
square of R. If Q is 40, R is 1 and if Q is13, R is made per day, the cost is Ghȼ2.00 per article; If
2. Find the value of Q when R is 3. 200 components are made per day the cost is
reduced to Ghȼ1.50. What would be cost of a
Solution component if 500 were made per day?
Q is partly constant and partly varies as the
square of R Solution
Q = c and Q R2 Let c represent the cost of producing a radio
Q = c and Q = kR2 component and n represent the number made per
⇒Q = c + kR2 day
⇒c is partly constant and partly varies inversely
When Q = 40, R = 1 substitute in Q = c + kR2 with n
40 = c + k(1)2 c 1 and c

40 = c + k ……………(1) c = k1 and c =
c = k1 +
When Q = 13, R = 2, put inQ = c + kR2
13 = c + k(2)2 When c = 2, n = 100
13 = c + 4k ……………(2) 2 = k1 +

Solving eqn (1) and eqn (2) simultaneously 100 × 2 = 100k1 + 100 ×
40 = c + k ……………(1) 200 = 100k1 + k2 ……….(1)

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When c = 1.5, n = 200 c is partly constant and partly varies inversely as
1.5 = k1 + n
c 1 and c
200 × 1.5 = 200k1 + 200 ×
300 = 200k1 + k2 ……….(2) c = k1 and c =
c = k1 +
Solving eqn (1) and eqn (2) simultaneously
When c = 1600, n = 4
eqn (2) – eqn (1)
1600 = k1 +
100 = 100k1
k1 = 1 1600 × 4 = 4× k1 + 4 ×
6400 = 4k1 + k2 ……….(1)
Put k1 = 1 in eqn (1)
200 = 100(1) + k2 When c = 1420, n = 5
200 = 100 +k2
1420 = k1 +
200 – 100 = k2
k2 = 100 1420 ×5 = 5 ×k1 + 5 ×
7100 = 5k1 + k2 ……….(2)
Put k1 = 1 and k2 = 100 in c = k1 +
Solving eqn (1) and eqn (2) simultaneously
⇒c = 1 +
eqn (2) – eqn (1)
When c = ?, n = 500, put in c = 1 + 700 = k1
c=1+ k1 = 700

c=1+ Put k1 = 700 in eqn (1)

5 × c = 5(1) + 5( ) 6,400 = 4(700) + k2 ……….(1)
6,400 = 2,800 +k2
5c = 5 + 1
k2 = 6,400 – 2,800 = 3,600
5c = 6
c = = 1.2
Put k1 = 700 and k2 = 3,600 in c = k1 +
The cost of a component would be Ghȼ1.20
⇒c = 700 +
5. The cost, c, of maintaining a motor car in a The relation between c and n is c = 700 +
factory is partly constant and partly varies
inversely as the number, n, of cars produced per 6. The cost of maintaining a school is partly
day. The cost of producing 4 cars per day is constant and partly varies as the number of
Gh¢1,600.00 and that of producing 5 cars per day students. With 50 students, the cost is
is Gh¢1,420.00 Find the relation between c and n. Ghȼ15,705.00 and with 40 students, it is
Ghȼ13,305.00. Find the cost when there are 44
Solution students.

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Solution C 1and c
Let C = Cost of maintaining a school C = k1 and c = 2LB
n = number of students C= k1 + 2 LB
C = a + bn, where a, and b are constants
When c = 15,705 and n = 50 When C = 85,000, L = 50 and B = 20 put in
15,705 = a + 50b……….(1) C = k1 + 2 LB
85,000 = k1 + 2 (50) (20)
When c = 13,305 and n = 40 85,000 = k1 + 2 ……….(1)
13,305 = a + 40b……….(2)
When c = 100,000, L = 40 and B = 30 put in C=
eqn (2) – eqn (1);
k1 + 2 LB
2400 = 10b
100,000 = k1 + 2 (40) (30)
b = 240
100,000 = k1 + 2 ……….(2)

Put b = 240 in eqn (1)

Solving eqn (1) and eqn (2) simultaneously;
15,705 = a + 50(240)
eqn (2) – eqn (1)
15,705= a + 12,000
15,000 = 200k2
a = 15,705 – 12,000 = 3,705
k2 = 75
⇒The variation equation is C = 3,705 + 240n
Put k2 = 75 in eqn (1)
Now, put a = 3705, b = 240 and n = 44 in 85,000 = k1 + ( ……….(1)
C = a + bn 85,000 = k1 +
C = 3,705 + (240) (44) 85,000 – 75,000 = k1
C = 3,705 + 10,560 = Ghȼ14,265.00 k1 = 10,000

7. The cost, C, of weeding a rectangular plot of Put k1 = 10,000 and k2 = 75 in

land is partly constant and partly varies jointly as C = 10,000 + 75 LB
the length, L, and the breadth B, of the plot. For a The relationship between C L and B is
plot of length 50m and breadth 20m, the cost of C= 10,000 + 75 LB
weeding is Ghȼ85,000.00 and for a plot of length
ii. When C = ? L = 50 and B = 40
40m and breadth 30m, the cost of weeding is
C = 10,000 + 75 LB
Ghȼ100, 000.00
C = 10,000 + 75 (50) (40)
i. Find the relationship between C, L and B.
C = 10,000 + 150,000 = 160,000
ii. Calculate the cost of weeding a plot of length
50 m and breadth 40m.
8.The cost (c) of producing n bricks is the sum of
a fixed amount (h) and a variable (y)
Solution
a. y varies directly as n, write (in symbols) the
i. C is partly constant and partly varies jointly as
relation between;
L and B
i. y and n

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ii. c, h and n iv. The variable V varies partly as k and partly as
b. If it cost Ghȼ520.00 to produce 200 bricks and the inverse of the cube of r
Ghȼ1,030.00 to produce 400 bricks, calculate the v. The variable z varies partly as the square root
cost of producing 300 bricks. of x and partly as the constant

Solution B. 1. T varies directly as r and partly constant. If

a. i. C = h + y T = 35, r = 2 and if T = 55, r = 4
y = kn
i. Write down an equation connecting T and r
ii. C = h + kn ii. Find the value of:
a. r when T = 75b. T when r = 1.5
b. When C = Ghȼ520.00, and n = 200
substitute in C = h + Kn 2. y is partly constant and partly varies as x. If y =
520 = h + k(200)……….(1) 2 when x = 4 and y = 6 when x = 6, find y when x
=9
When C = Ghȼ1030.00, and n = 400 substitute in
C = h + Kn 3. x and y are connected by the relation of the
1030 = h + k(400)……….(2) form y = ax + ⁄ . When x = 1, y = 2 and when x
= 4, y = 15 . Find the relation and also the value
eqn (2) – eqn (1);
510 = 200k of y when x = 4
k= = 2.55
4. y varies partly as x and is partly constant .
When x = 0, y = 1 and when x = 1, y = 0.
Put k = 2.55 in eqn (1); a. Find the constant of variation and the relation
520 = h + (2.55) (200) between x and y
520 = h + 510 b. calculate the value of y when x = 2
h = 520 – 510 = 10
5.The annual cost C, of running a certain car is
Put k = 2.55, h = 10 and c = 300 in made up of two parts, one of which is fixed and
C = h + kn the other varies as the distance, d, run by the car
C = 10 + (2.55) (300) = 10 + 765 = 775 in the year. In one year, it ran 6,000 km at a cost
The cost of producing 300 bricks is Ghȼ775. 00 of Ghȼ9,000.00; in the next year, it ran 7,200km
at a total cost of Ghȼ9,500.00. Find:
Exercises 18.4 i. the relationship between C and d,
A. Write the partial variations equations: ii. how much it cost to run the car in a year during
i.y is partly constant and partly varies as x2 which it ran 1200km.
ii. y varies partly as x and partly as x3
iii. The variable y varies partly as the square of x 6. The cost (C) of paying electricity bills per
and partly as the constant month is partly constant and varies directly as the

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square of the number of appliances (n) used in the cost Ghȼ125.00 per day to run, and a school of
household. If Mr Brown uses 5 appliances, he 400 costs Ghȼ160.00 per day, find the formula
pays Ghȼ82,000.00 and if he uses 7 appliances he giving the expected cost in terms of the number
pays Ghȼ250,000.00 How much will Mr. White of pupils, and hence find the expected daily cost
pay, if he uses 13 appliances? of running a school with 500 pupils.

7. A super market pays its sales personnel on 10. The cost, C, of running a training course is
weekly basis. At the end of each week, each sales partly constant and partly varies as the number of
person receives a basic wage plus a candidates (n) and the number of weeks (w) that
bonus which varies directly as the number of the course lasts. When 110 candidates attended
complete weeks the particular person has worked the course for 10 weeks, the running cost was
in the shop. At the end of her fourth week, a sales Ghȼ120,000.00 and when 150 candidates
girl received a pay package containing attended the course for 6 weeks, the running cost
Gh¢2,060.00. Six weeks later her pay had jumped was Ghȼ100,000.00. Find:
to Gh¢2,150.00. Find the exact relation for i. the equation connecting c, n and w
determining how much the shop sales personnel ii. the cost of running the course for 100
are paid every week. candidates for 12 weeks

8. The cost per mile, p, of running mycar in 11. The cost (c) of producing n bricks is the sum
Kumasi is partly constant and partly inversely of a fixed amount, h, and a variable amount y,
proportional to n, the number of miles travelled where y varies directly as n. If it costs Gh¢950.00
per month. When n = 500, c = 10, and when n = to produce 600 bricks and Gh¢1,030 to produce
1,000, c = 8. 1000 bricks,
i. Find the formula giving c in terms of n. i. Find the relationship between c, h and n.
ii. What would be the cost per mile at a travelling ii. Calculate the cost of producing 500 bricks
distance of 1500 miles per month?
12. V is the sum of two parts. One part varies
directly as t, one part varies directly as t2. When
9. The daily cost of running a school is assumed t = 1, V = 14 and when t = 2, V = 7
to be partly constant and partly proportional to i. Find the relationship between t and V.
the number of pupils. If a school of 300 pupils ii. When t = 3, find the value of V.

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19 STATISTICS II Baffour – Ba Series

Measures of Positions Let the lower quartile be Q1, the median be Q2

Measures of positions are techniques that divide a and the upper quartile be Q3
set of data into equal groups. The different Ordering the data: 2, 2, 3, 4, 5, 6, 9
measures of positions are quartiles, deciles and n=7
percentiles (
Q1 = = = 2nd position
2, 2, 3, 4, 5, 6, 9
To determine measurement of position, the data
↓
must be sorted from the lowest to the highest.
Q1 = 2
Quartiles
(
The quartiles are the three values of the variable Q2 = = = 4th position
that divide an ordered data set into four equal 2, 2, 3, 4, 5, 6, 9
parts.Q1(lower quartile), Q2(middle quartile) and ↓
Q3(upper quartile) determine the values for 25%, Q2 = 4
50% and 75% of the data. Q2 coincides with the
median. (
Q3 = = = 6th position
Determining Quartiles 2, 2, 3, 4, 5, 6, 9
I. Order or arrange the data from smallest to ↓
largest Q3 = 6
II. Find the place that occupies every quartile,
Quartiles of Even Number of Data
using the expression, , where k = 1, 2, 3, for
Quartiles of even number of data, for eample
Q1, Q2, and Q3 respectively a, b, c, d, e, f, g, h, are determined by the
following procedure:
Quartiles of Odd Number of Data I. Order the data and divide the data into two
Quartiles of odd number of data are determined equal parts as shown below:
(
by the formula, , where k = 1, 2, 3 for Q1,
Q2, and Q3 respectively a, b, c, d, e, f, g, h, k, l
( ( (
Q1= , Q2 = , Q3 = , II. Identify the average of e and f as the median or Q2

a, b, c, d, e f g, h, k, l
Worked Example
Find the lower, middle and upper quartiles of Q2 =
the data 2, 5, 9, 3, 2, 6, 4

III. If the data from a to e is an odd number,

Solution

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identify the middle number of a to e as the first Solution
quartile, Q1 Method 1
Ascending order:
a, b, c, d, e f, g, h, k, l
2 2 3 3 4 5 6 7 7 8 99
Q1
IV. If the data from f to l is an odd number, Q1 Q2 Q3
identify the middle number of f to l as the third Q1= =3 Q2 = = 5.5 Q3 = = 7.5
quartile, Q3
Method 2
a, b, c, d, e, f, g, h, k, l Ascending order :
2 2 3 3 4 5 6 7 7 8 99
Q1 Q3
n = 12
Q1 = (12)th and (12)th + 1
V. If the data from a to e and f to l are even
numbers, identify the average of the two middle = 3rd and 4th positions
numbers from ato e as Q1 and the average of the Q1= =3
two middle numbers from f to l as Q3.

a, b,c, d e ,f, g, h, k, l Q2 = (12)th and (12)th + 1

= 6th and 7th positions
Q2 = = 5.5
Q1 = Q2 = Q1 =

Using the Formula Q3 = (12)th and (12)th + 1

For even number of data, two positions are = 9th and 10th positions
required for each quartile. Thus:
Q3 = = 7.5
Q1 = (n)th and (n)th + 1

Q2 = (n)th and (n)th + 1 2. Find Q1, Q2 and Q3 of the data; 2, 7, 6, 14, 7,

11, 9, 3, 4, 4, 8
Q3 = (n)th and (n)th + 1
Solution
2, 3, 4, 4, 6, 7, 8, 9, 11, 14
Use the positions to identify the two numbers and
n = 10
divide the sum by two to obtain the value of the
quartile. Q1 position = (10)th and (10)th + 1
Q1 position = 2.5th and 3.5thpositions
Worked Examples Q1 position = ( ) = 3rd position
1. Determine the quartiles of the data; 2, 2, 5, 6,
Q1 = 4
7, 7, 9, 3, 3, 4, 8, 9

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Q2 position = (10)th and (10)th + 1 ⇒ 4 = = 5.5

= 5th and 6th positions

(
= 6 and 7 D5 = = = 7th position
Q2 = = 6.5 ⇒D5 = 6

Q3 position = (10)th and (10)th + 1 D7 =

(
= = 9.8th position
Q3 position = 7.5th and 8.5th
It implies that D7 lies between the 9th and 10th
Q3 position = ( ) positions. But 9th position = 8 and 10th position = 9

Q3 position = 8th position ⇒D7 = = 8.5

Q3 = 8
2. From the data; 4, 1, 3, 12, 3, 2, 11, 6, 11, 10, 9, 10,
Deciles 8, 7, determine the values of D3, D5 and D8
The deciles (D) are the nine values of the variable
that divide an ordered set of data into ten equal Solution
parts. The deciles determine the values for 10%, 1, 2, 3, 3, 4, 6, 7, 8, 9, 10, 10, 11, 11, 12
20%, 30%...90%. The median corresponds with N = 14
D5. (
D3 = = = 4.5th position
⇒D3 lies between the 4th and 5th positions
Calculating Deciles of Raw Data
I. Re- arrange the data in ascending order D3 = = 3.5
II. Identify the number of entries as N
III. Find the place or position that occupies every (
D5 = = = 7.5th position
(
decile, using the relation, D = , where k = ⇒D5 lies between the 7th and 8th positions
1, 2, 3…9 for D1, D2, D3...D9 respectively. D5 = = 7.5
(
Worked Examples D8 = = = 12th position
1. Find D4, D5 and D7 of the data : 6, 4, 7, 6, 11, ⇒D8 = 11
8, 5, 3, 2, 9, 10, 12, 3
Percentiles
Solution Percentiles are the 99 values of the variable that
2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 12 divide an ordered data set into 100 equal parts.
N = 13
( The percentiles determine the values for 1%, 2%,
D4 = = = 5.6th position
3%...99% of the data. D1 = P10, D2 = P20… The
It implies that D4 lies between the 5th and 6th median corresponds to P50
positions. But 5th position = 5 and 6th position = 6

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Determining Percentiles 2. 8, 9, 1, 2, 7, 3, 2, 4, 7, 8.
I. Re- arrange the data in ascending order. 3. 7, 11, 9, 2, 4, 2, 10, 4, 6.
II. Identify the number of entries as n. 4. 10, 11, 14, 11, 18, 19, 11, 12, 13, 15, 16.
III. Find the place or position that occupies every
( Cumulative Frequency Table
percentile, using the relation, P = , where
Consider the table below:
k =1, 2, 3…99 for P1, P2, P3 … P99 respectively.
Marks (x) Frequency (f)
Worked Examples 11 – 20 7
1. Find P15 and P84of the data : 6, 4, 7, 6, 11, 8, 5, 21 – 30 3
3, 2, 9, 10, 12, 3 31 – 40 4
41 – 50 6
Solution 51 – 60 5
2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 12
To draw up a cumulative frequency table,identify
N = 13
the upper class boundaries of the class intervals.
(
P15 = = 2.1th position Thus, for the class interval 11 – 20, with
⇒P15 lies between the second and third positions. frequency 7, it means that 7 students had a mark
But second position = 3 and third position = 3 of 20.5(upper class boundary) or less. For the
P15 = =3 class interval 21 – 30, we consider students who
had marks less than 30.5, and this is found by
adding the frequency of the previous class to that
(
P84 = = 11.76 th position of itself. i.e. 7 + 3 = 10
⇒P84 lies on the 11th and 12th positions The cumulative frequency table for the above
P84 = = 10.5 information is drawn as shown below:

Marks Frequency Cumulative

Exercises 19.1 less than (f) frequency (f)
A. Find Q1, Q2 and Q3 from the data: 20.5 7 7
1. 2, 5, 8, 8, 3, 4, 5, 1. 30.5 3 7 + 3 = 10
2. 8, 9, 1, 2, 7, 3, 2, 4, 7, 8. 40.5 4 10 + 4 = 14
3. 7, 11, 9, 2, 4, 2, 10, 4, 6. 50.5 6 14 + 6 = 20
4. 10, 11, 14, 11, 18, 19, 11, 12, 13, 15, 16. 60.5 5 20 + 5 = 25

B. Find D3, D5 and D8 of the following data: Worked Examples

1. 2, 5, 8, 8, 3, 4, 5, 1 1. The heights of 100 girls were measured to the
2. 7, 11, 9, 2, 4, 2, 10, 4, 6 nearest centimetre. The results were as follows
3. 10, 11, 14, 11, 18, 19, 11, 12, 13, 15, 16
Height (Cm) Frequency
C. Find P35, P50, and P70 130 – 134 4
1. 2, 5, 8, 8, 3, 4, 5, 1. 135 – 139 7
140 – 144 14

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 145 – 149 20 a.
150 – 154 24 Marks less Frequency Cumulative
155 – 159 16 than Frequency
160 – 164 9 29.5 1 1
165 – 169 6 39.5 3 4
49.5 3 7
a. Draw upa cumulative frequency table 59.5 15 22
b. How many girls were 149.5cm tall or more? 69.5 4 26
79.5 5 31
Solution 89.5 1 32

Height Frequency Cumulative b.Students who obtained 49.5% or less = 7

less than Frequency
134.5 4 4 c. Students who obtained 69.5% or more?
139.5 7 11 32 – 26 = 6 students
144.5 14 25
149.5 20 45 Exercises 19.2
154.5 24 69 1. The followingtable gives the masses of 100
159.5 16 85
clients of a company to the nearest kilogram.
164.5 9 94
169.5 6 100
Mass Mass
(Nearest kg) Freq (Nearest kg) Freq
b. Number of girls who were 149.5cm tall or
55 – 59 3 85 – 89 11
more 60 – 64 5 90 – 94 7
= Total girls – Number of girls who were not 65 – 69 10 95 – 99 4
149.5cm tall or more = 100 – 25 = 75 70 – 74 18 100 – 104 2
75 – 79 20 105 – 109 2
2. The following is the records of marks(%) 80 – 84 17 110 – 114 1
obtained by some students in a test
a. Draw up a cumulative frequency table.
59 59 64 50 74 79 33 57 57 53 67 49
b. Using the frequency table, find how many
80 57 32 57 76 48 74 24 56 50 56 clients had mass more than 79.5kg.
50 60 39 52 57 30 61 73 40
a. Using the class intervals of 20 – 29, 30 – 39, 40 2. The following table gives the masses of 160
babies born in a maternity home to the nearest
– 49 …construct a cumulative frequency table for
tenth of a kilometre.
the data
b. How many students obtained 49.5% or less? Mass Frequency
1.5 – 1.9 5
c. How many students obtained 69.5% or more? 2.0 – 2.4 15
2.5 – 2.9 60
Solution 3.0 – 3.4 54

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 3.5 – 3.9 25
4.0 – 4.4 1 Marks Frequency
11 – 20 1
a. Draw up a cumulative frequency table 21 - 30 1
b. Using the cumulative frequency table, find 31 – 40 2
how many babies had mass of less than 29.5kg at 41 – 50 4
51 – 60 6
birth.
61 – 70 7
c. What proportion of the total number of births is 71 – 80 7
this? 81 – 90 4
91 – 100 1
Cumulative Frequency Curve (Ogive)
Steps to follow when drawing the cumulative Draw a cumulative frequency curve for the data.
frequency curve:
I. Prepare a cumulative frequency table. Solution
II. Draw two perpendicular axes on a graph sheet.
III. Label the two axes, (horizontal axis for upper Marks
class boundaries and vertical axis for cumulative Marks Less Frequency Cumulative
than Frequency
frequencies) 11 – 20 20.5 1 1
IV. Choose a suitable/ given scale and mark of 21 - 30 30.5 1 2
the values. 31 – 40 40.5 2 4
V. Plot the points from the cumulative frequency 41 – 50 50.5 4 8
table drawn up in step I 51 – 60 60.5 6 14
61 – 70 70.5 7 21
VI. Join the points with a smooth curve, but not a 71 – 80 80.5 7 28
ruler. 81 – 90 90.5 4 32
91 – 100 100.5 1 33
Worked Examples
1.The table below gives the distribution of marks
obtained by 33 candidates in an examination.

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 35

30

Cumulative frequency
25

20

15

10

0
10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5 110.5
Marks

2. A class of students obtained the following 60 39 52 57 30 61 73 40

marks (%) in a test Use this information to draw a cumulative
59 59 64 50 74 79 33 57 frequency table and a cumulative frequency
57 53 67 49 80 57 52 57 curve, using the intervals 20 – 29, 30 – 39…
76 48 74 24 56 50 56 50

Solution

Marks Marks less than Frequency Cumulative frequency

20 – 29 29.5 1 1
30 – 39 39.5 3 4
40 – 49 49.5 3 7
50 – 59 59.5 15 22
60 – 69 69.5 4 26
70 – 79 79.5 5 31
80 – 89 89.5 1 32

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 35

Cumulative frequency 30

25

20

15

10

0
19.5 29.5 39.5 49.5 59.5 69.5 79.5 89.5
Marks

Exercises 19.3
1. The marks scored by 30 students in a 3. The marks obtained by 40 students in an
mathematic test are given below: examination are as follows:
38 31 50 18 51 63 10 34 42 89 73 11 51 46 38 68 21 51 58 64 72 33 86 48
33 31 41 25 76 13 26 23 29 30 51 91 67 93 71 63 44 50 22 91 78 66 52 81
37 64 19 86 09 20 43 64 53 82 45 58 57 72 62 77 61 74
a. Using class intervals 1 – 20, 21 – 40 …Prepare 88 35 43 56
for the distribution, a cumulative frequency table a. Using class intervals of 20 – 29 , 30 – 39 , 40 –
b. Use your table to draw a cumulative frequency 49 … construct a frequency distribution table for
curve the data.
b. Draw a cumulative frequency curve for the
2. The height of 40 citrus plants in a school farm distribution.
are recorded, to the nearest centimetre as follows:
103 116 127 101 118 125 119 127 114 117 Estimating the Quartiles from a Cumulative
120 117 114 128 112 118 119 129 125 130 Frequency Curve
117 110 121 109 115 118 113 126 123 131 The positions of quartiles on a cumulative
109 117 105 122 124 114 124 121 123 115 frequency curve involving N observations are
a. Form a group frequency table using the determined as follows:
intervals 100 – 104, 105 – 109 … Q1 = , Q2 = and Q3 =
b. Calculate the mean height of the plant
c. Form a cumulative frequency table and draw
the ogive
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The median Q2 is estimated on the cumulative Worked Examples
frequency curve as follows: 1. The following are the marks obtained by some
students in an achievement test.
I. Find ( )
90 25 31 35 52 50 48 15
II. Look for this value on the vertical axis (the 19 40 60 83 23 38 40 70
cumulative frequency axis) 55 65 43 68 58 57 62 83
III. Draw a line perpendicular to the vertical axis 46 33 09 75 59 46 71 05
at mark and extend it till it meets the 63 19 50 47 42 47 28 21
cumulative frequency curve a. Construct a cumulative frequency table for the
IV. Draw a line from the point of intersection distribution using the class intervals 1 – 10, 11 –
with the curve perpendicular to the horizontal 20, 21 – 30 …
axis b. Draw a cumulative frequency curve for the
V. Read off the corresponding value on the distribution
horizontal axis as Q2 c. Use your graph to find:
This same procedure is used to estimate the i. the median;
values of Q1 and Q3 ii. the lower quartile;
Note that Q1 corresponds to the 25th percentile ii. the upper quartile;
and Q3 corresponds to the 75th percentile iv. the marks obtained by a student who was 15th
in the class.

Solution
a.
Marks Marks less than Frequency Cumulative frequency
1 – 10 10.5 2 2
11 – 20 20.5 3 5
21 – 30 30.5 4 9
31 – 40 40.5 6 15
41 – 50 50.5 9 24
51 – 60 60.5 6 30
61 – 70 70.5 5 35
71 – 80 80.5 2 37
81 – 90 90.5 3 40

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b.
45

40

35
Cumulative frequency

30

25

20

Upper quartile
15
Lower quartile
10

Median
5

0
0.5 10.5 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 100.5
Marks

c. From the cumulative frequency curve, 75 59 53 71 73 79 57 73 60 93

72 78 71 95 61 65 97 67 74 62
i. Median (Q2) = ( )
60 74 83 68 66 78 61 89 94 77
But n = 40 71 79 68 60 96 78 76 82 75 95
Q1 = ( ) = 20th position = 46.5 Using class intervals 50 – 54, 55 – 59 …
construct a cumulative frequency table for the
distribution
ii. Lower quartile (Q1) = ( ) b. Draw the cumulative frequency curve.
Q1 = ( ) = 10th position = 32.5 c. Use your graph to estimate;
i. the median mark,
ii. the lower quartile,
iii. Upper quartile (Q3) = ( ) iii. the probability that a student selected at
( random scored a mark between 80 and 89.
Q1 = ( ) = 30th position = 60.5
2. The marks obtained by 40 students in an
iv. From the graph, the marks obtained by a examination are as follows:
student who was 15th in the class is 52.5 51 46 38 68 21 51 58 64 72 33 86 48 67 93
71 63 44 50 22 91 78 66 52 81 43 64 53 82
Exercises 19.4 45 58 57 72 62 77 61 74 88 35 43 56
1. The marks scored in Mathematics by 50 a. Using class intervals of 20 – 29, 30 – 39 , 40 –
students in an examination are recorded below 49 … construct a frequency distribution table for
93 88 85 62 68 84 73 81 90 68 the data

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b. Draw a cumulative frequency curve for the
distribution Marks Marks Freq. Cumulative
c. Use your curve to estimate; less than frequency
i. the upper quartile 0–9 9.5 4
10 – 19 19.5 7
ii. the pass mark if 31 students passed
20 – 29 29.5 5
30 – 39 10
3. The following are the marks obtained by a 40 – 49 13
number of students in an examination; 50 – 59 20
79 80 49 68 70 51 41 10 18 28 60 – 69 15
19 29 30 36 33 50 43 49 41 45 70 – 79 13
47 55 55 60 50 40 31 32 10 20 80 – 89 3
90 – 99 1
44 11 21 29 30 25 26 35 33 48
42 70 51 20 26 34 35 34 67 71
a. Copy and complete the table.
a. Construct a cumulative frequency table for the
b. Draw a cumulative frequency curve for the
distribution using the intervals 1 – 10, 11 – 20…
distribution.
b. Draw a cumulative frequency curve for the
distribution. 6. The table below gives the distribution of the
c. Using your cumulative frequency curve: ages of all people (the nearest thousands) in a
i. determine the pass mark if one – fourth of the certain village
class passed.
ii. Find the probability that a student selected at Age 0–4 5–9 10 – 15 – 20 –
random had distinction if the minimum mark for (yrs) 14 19 24
distinction is 75% Freq. 2 3 6 15 12
Marks 25 – 30 – 35 –
4. The table below shows the distribution of (%) 29 34 39
marks of 800 candidates in an examination.
Freq 7 4 1
Marks 0–9 10 – 20 – 30 – 40 –
Draw a cumulative frequency table and hence,
(%) 19 29 39 49
draw a cumulative frequency curve for the
Freq. 10 40 80 140 170
Marks 50 – 60 – 70 – 80 – 90 – distribution
(%) 59 69 99 89 99
Freq 130 100 70 40 20 Mean Deviation of a Raw Data
a. i. Construct a cumulative frequency table. The mean deviation or average deviation is the
ii. Draw the ogive. arithmetic mean of the absolute deviations and it
iii. Use your ogive to determine the 50th is denoted by ̅ . Thus:
̅ ̅ ̅
percentile ̅ =

5. The table below shows the distribution of The steps involved in finding the mean
marks of candidates in an examination, deviation about the mean are as follows:

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I. Find the mean of the given data = 3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22
II. Subtract the mean from each of the Mean deviation about the mean = = 2.75
observations and record your results
III. Find absolute values of the deviations Exercises 19.5
obtained. 1. Calculate the mean deviation of the following
IV. Add the absolute values obtained. distribution: 9, 3, 8, 8, 9, 8, 9, 18
V. Divide the results of the last step by the
number of observations to obtain the mean 2. Find the mean deviation of the following set of
deviation of the data. numbers: 3, 4, 4, 4, 5, 8, 8, 9

Clas x f fx d = fd 3. For the set of numbers 11, 14, 14, 13, 15, 20,
s ̅ 17, determine the mean deviation

Mean Deviation of a Grouped Data

∑ ∑ To find the mean deviation ( ̅ of a grouped
= ∑ = data,
=
I. Prepare and complete a tablesuch as the one
below:
Worked Examples
II. From the table, determine the mean by the
Find the mean deviation about the mean for the ∑
following set of data: 6, 7, 10, 12, 13, 4, 8, 12 formula: Mean ( ̅ ) = ∑
III. Calculate the mean deviation by the formula:
Solution ∑
̅ = ∑
Data = 6, 7, 10, 12, 13, 4, 8, 12
IV. Leave the answer to two decimal places
Mean = = =9
unless stated.

Deviations from the mean; Worked Example

= 6 – 9, 7 – 9 , 10 – 9 , 12 – 9 , 13 – 9 , 4 – 9 , The marks obtained by 20 students in a test
8 – 9 , 12 – 9 (marked out of 20) were recorded as follows;
= -3, -2, 1, 3, 4 -5, -1, 3 12 9 14 13 2 9 14 5 12 7
18 13 6 10 8 10 15 3 12 16
Absolute deviations; i. Construct a group frequency table for the data,
= /-3/, /-2/, /1/, /3/, /4/, /-5/, /-1/, / 3/ using the class intervals 1 – 5, 6 – 10, 11 – 15 …
= 3, 2, 1, 3, 4, 5, 1, 3 ii. Use your table to calculate the mean deviation
of the distribution.
Sum of absolute deviations;

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Solution

Class x f fx d= ̅ fd
1–5 3 3 9 7.25 21.75
6 – 10 8 7 56 2.25 15.75
11 – 15 13 8 104 2.75 22
16 – 20 18 2 36 7.75 15.5
∑ = 20 ∑ = 205 ∑ 75

∑
( ̅)= ∑
= = 10.25
∑
Mean deviation, ̅ = ∑
= = 3.75(2.d.p)

Exercises 19.6 161 – 165 7

1. The scores obtained by 42 pupils in a test were: 166 – 170 5
25 20 18 34 26 10 17 15 15 35 30 10 16 171 – 175 1
19 20 11 10 10 05 07 25 10 23 08 05 15
05 13 20 10 05 13 07 22 06 05 10 09 12 Calculate the mean deviation
15 35 26
i. Group the data as 5 – 9, 10 – 14 … and prepare 4. The following is the records of the marks of 40
a frequency table for the data candidates in an examination
ii. From the table, determine the mean deviation. 65 84 91 58 43 86 73 33 76 80
57 33 53 29 40 27 72 19 52 67
2. The scores out of 100 in a science test were: 37 14 18 92 13 45 61 39 23 22
95 81 78 67 78 84 60 72 66 60 33 36 60 22 41 27 51 63 47 19 35 39 76
72 27 63 42 18 42 36 27 66 54 27 42 63 a. Using class intervals of 11 – 20, 21 – 30 etc,
27 18 33 30 24 39 45 24 39 30 33 30 30 prepare for the distribution a frequency table and
33 45 9 use your table to find the mean deviation of the
a. Draw a frequency table for the intervals: 0 – 9, distribution.
10 – 19, 20 – 29 …
b. From the table, determine: b. Find the modal class and the median class.
i. the mean deviation.
ii. the modal class and the median class. Measures of Dispersion
Measures of dispersion indicate the degree of
3. The heights of 25 boys are measured to the „spread‟ of the data. The most common statistics
nearest cm and are then grouped as follows: used as measures of dispersion are the range, the
interquartile range, variance and standard
Height(cm) Frequency deviation
151 – 155 4
156 – 160 8

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The Range (
Q2 = = = 8th position = 143cm
The range of the numbers in a group of data is the
(
difference between the greatest number in the Q3 = = = 12th position = 147cm
data and the least number in the data. For
example, given the list 11, 10, 5, 13, 21, the range iii. Interquartile range
of the numbers is 21 – 5 = 16 = Q3 – Q1 = 147 – 137 = 10

The Interquartile Range iv. Semi – interquartile range

The interquartile range is defined as the
= = = =5
difference between the third quartile and the first
quartile. i. e. Q3 – Q1. Thus, the interquartile
Exercises 19.7
range measures the spread of the middle half of
1. For the data: 11, 11, 13, 15, 17, 19, 23, 31, 45,
the data
47, 49, find the range and the inter quartile
range
Semi - Interquartile Range
The semi – interquartile range is another measure
2. Find the range, the quartiles and the
of dispersion. To find the semi –interquartile
interquartile range of the data 30, 64, 49, 45, 30,
range, divide the inter quartile range by 2.
55, 47, 49
i.e.
4. Find the interquartile range and the semi –
Worked Examples interquartile range of the following data: 43, 34,
The height of 15 boys to the nearest cm is 45, 31, 40, 22, 30, 17, 25, 18, 27
given as: 138, 136, 143, 143, 147, 158, 146, 145,
137, 135, 143, 140, 129, 156, 149 Variance
From these data, find: The variance is defined as a measure of how the
i. the range data distributes itself about the mean. It is
ii. the quartiles, computed as the average of the squared
iii. the inter quartile range, differences from the mean.
iv. the semi – interquartile range.
To calculate the variance, follow the steps below:
Solution I. Work out the mean.
i. Arranging the data in order of size, II. Then for each number subtract the mean and
129, 135, 136, 137, 138, 140, 143, 143, 143, 145, square the results(the squared difference)
146, 147, 149, 156, 158 III. Then work out the averages of those squared
The range 158 – 129 = 29 differences.

ii. n = 15 Note:
( When you have n data values that are:
Q1 = = = 4th position = 137cm

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1. The population: divide by n when calculating dispersed around the mean. It is the square root of
∑( ̅ the variance.
the variance. That is σ2 = , where ̅ is the
mean and n is the number of scores.
The steps to follow in computing the standard
deviation are as follows:
2. A sample (a selection taken from a bigger
I. Find the mean ( ̅ ) of the distribution.
population): divide by n – 1 when calculating
∑( ̅
II. For each given value, find the deviation from
the variance. That is: σ2 = the mean. i.e. x - ̅ .
III. Find the square of each deviation.
Worked Examples i. e. ( ̅
Find the variance of the following scores in a test: IV. Find the sum of all the squared deviations.
17, 12, 14, 13, 19 i. e. ∑( ̅
V. Divide this sum by the number of values in the
Solution ∑( ̅
17, 12, 14, 13, 19, n=5 data. i. e.
Mean ( ̅ ) = = = 15 VI. Find the positive square root of the answer
∑( ̅
obtained. i. e. √
Variance
(σ2) =
( ( ( ( ( Generally, the standard deviation, S, is calculated
( ( ( ( ( ∑( ̅
σ2 = by the formula: S = √
σ2 = = = 6.8
Standard Deviation of a Raw Data
Properties of Variance I. Find the mean ( ̅ ) of the distribution by the
1. The variance is always positive or in the event ∑
formula: ̅= =
that the values are equal, the variance is
II. Prepare and complete a table of values as
zero.
shown below:
2. If all the values of the variables are added by
the same number, the variance does not change.
x x- ̅ ( ̅
3. If all the values of the variables are multiplied
by the same number, the variance is multiplied by ∑( ̅ =
the square of that number.
II. Carefully identify the values of and
Exercises 19.8 ∑( ̅ and substitute in the formula:
Determine the variance:
∑( ̅
1. 30, 64, 49, 45, 30, 2. 22, 30, 17, 25, 18, 27 S=√ to obtain the standard deviation of
the data, ususally to two decimal places.
Standard Deviation
Standard deviation is the measure of the extent to Worked Examples
which the given values or data are spread or

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1. Calculate the standard deviation for the data: 2, 3. The scores obtained by students in a test are:
4, 4, 7, 8 21, 25, 27, 25, 27, 21, 24, 23, 23 and 24.
Calculate the standard deviation.

Solution
Solution
Mean ( ̅ ) = = =5 ̅= = 24

x x- ̅ ( ̅ x x- ̅ ( ̅
2 2 – 5 = -3 ( =9 21 -3 9
4 4 – 5 = -1 ( =1 21 -3 9
4 4 – 5 = -1 ( =1 23 -1 1
7 7–5=2 ( =4 23 -1 1
8 8–5=3 ( =9 24 0 0
∑( ̅ = 24 24 0 0
25 1 1
25 1 1
∑( ̅
But S = √ , where ∑( ̅ = 24, n = 5 27 3 9
27 3 9
S=√ = 2.1909 = 2.19 (2 d. p) ∑( ̅ = 40

∑( ̅
2. Find the standard deviation for the data: But S =√ , where ∑( ̅ = 40 and n = 10
3, 4, 5, 6, 8, 10
S=√ = 2.00 (2 d. p)

Solution
Exercises 19.9
Mean ̅ = = =6 1. The marks scored by 8 pupils in a science test
are 3, 7, 8, 8, 5, 8, 4, 8.What is the standard
x x- ̅ ( ̅ deviation?
3 3 – 6 = -3 ( =9
4 4 – 6 = -2 ( =4 2. The scores of 10 students in an examination are
5 5 – 6 = -1 ( =1 given as follows; 45, 12, 75, 81, 54, 51, 24, 67,
6 6–6=0 ( =0 19 and 39. What is the standard deviation of the
8 8–6=2 ( =4 score?
10 10 – 6 = 4 ( = 16
∑( ̅ = 34
3. The price of 12 commodities in cedi was
recorded as: 5, 2, 5, 3, 3, 5, 3, 2, 4, 3, 3, 2
∑( ̅
S =√ , where ∑( ̅ = 34 and n = 6 Calculate the standard deviation.

S=√ = 2.3805 = 2.38(2 d. p) 4. The ages in years of 8 boys are: 14, 14, 15, 15,
12, 11 13, 10. Calculate the standard deviation.

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Standard Deviation of Ungrouped Data ∑
S=√ ∑
( ̅ =√ ( = 3.03(2d.p)
To find the standard deviation of an ungrouped
data, prepare and complete a table as shown
below: 2. The table below gives the distribution of the
number of letters per word in the first 50 words of
an easy.
x f fx fx2
No. of
letters 1 2 3 4 5 6 7 8 9
∑ ∑ ∑
No. of
II. From the table, calculate the mean by the words 1 3 x 10 12 y 6 4 1
∑
formula: ̅ = ∑
a. If the average number of letters per word is 5,
III. Carefully substitute the values of ∑ , ∑
find the values of x and y
∑
and ̅ in S = √ ∑
(̅ and simplify to obtain b. Calculate correct to two decimal places, the
standard deviation of the distribution
the value of the standard deviation.

Solution
Worked Examples
1.The marks scored by 15 pupils in a test are as
x f fx fx2
follows; 14, 14, 11, 13, 17, 14, 11, 13, 20, 19, 17,
1 1 1 1
11, 20, 14, 17. Find:
2 3 6 12
a. the mean mark,
b. the standard deviation, correct to 2 decimal 3 x 3x 9x
places. 4 10 40 160
5 12 60 300
Solution 6 y 6y 36y
7 6 42 294
x f fx fx2 8 4 32 256
11 3 33 363 9 1 9 81
13 2 26 338 ∑ = ∑ = ∑
14 4 56 784 (37 + x (190 + 3x (1104 +
17 3 51 867 + y) + 6y) 9x + 36y)
19 1 19 361
20 2 40 800
a. From the table,
∑ 15 ∑ ∑
225 3513
∑ = (37 + x + y) =50
From the table, 37 + x + y = 50
∑ x + y = 50 – 37
̅= = = 15
∑ x + y = 13……….(1)
Substitute the values of ̅ , ∑ and ∑ in

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 ∑ 5 52 260
̅= ∑
=
6 80 480
But ̅ = 5 7 59 413
8 56 448
⇒ =5 9 21 189
= 50 × 5 10 16 160
= 250
3x + 6y = 250 – 190 Find correct to one decimal place;
3x + 6y = 60 a. the mean mark of the distribution
x + 2y = 20……….(2) b. the standard deviation of the distribution

eqn (2) – eqn(1) Solution

(x – x) + (2y – y) = (20 – 13)
y=7 Marks Frequency fx
(x) (f)
1 14 14 14
Put y = 7 in eqn (1)
2 30 60 120
x + 7= 13……….(1) 3 32 96 288
x = 13 – 7 = 6 4 40 160 640
(x, y) = (6, 7) 5 52 260 1300
6 80 480 2880
ii. ∑ = (190 + 3x + 6y) 7 59 413 2891
= 190 + 3(6) + 6(7) 8 56 448 3584
= 190 + 18 + 42 =250 9 21 189 1701
10 16 160 1600
∑ 400 ∑ = ∑ =
∑ (1104 + 9x + 36y)
1880 15018
= 1104 + 9(6) + 36(7)
= 1104 + 54 + 252 = 1410 ∑
a. ̅ = ∑
= = 4.7
Substitute the values of ̅, ∑ and ∑ in in S
∑ b. Substitute the values of ̅, ∑ and ∑ in S
=√ ∑
( ̅ =√ ( = 1.79 (2 d. p)
∑
=√ ∑
( ̅ =√ ( . 5 (2 d. p)

3. The table below gives the frequency

distribution of marks scored by 400 students in a Exercises 19.10
test. 1. Calculate the standard deviation of the
frequency distribution below:
Marks(x) Frequency (f) fx
Marks 1 2 3 4 5 6
1 14 14
Freq 4 2 1 2 1 2
2 30 60
3 32 96
4 40 160
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2. The following is a record of scores obtained by I. Prepare a table frequency as shown below:
30 J.H.S. 2 pupils in a test marked out of 5:
5 3 2 4 5 2 4 3 1 3 3 4 2 3 4 class Mid- f fx fx2
5 3 4 3 2 4 3 1 2 3 3 2 4 2 1 point x
Construct a frequency table for the data and a–c
d–f
calculate the standard deviation.
∑ ∑ ∑
= = =
3. The marks obtained by 16 pupils in a test were
recorded as follows: 4, 3, 8, 7, 7, 6, 6, 4, 5, 1, 4, II. From the table, calculate the mean by the
7, 8, 4, 3, 2. ∑
formula: ̅ = ∑
i. Construct a frequency distribution table for this
data. III. Carefully substitute the values of ∑ , ∑
ii. Calculate the mean mark. ∑
and ̅ in S = √ ∑
(̅ and simplify to obtain
iii. Calculate the standard deviation.
the value of the standard deviation to two decimal
4. The table below shows the marks obtained by places
some students in a test. Note: x is the class mid-point of each class.

Marks 5 6 7 8 9 10 Worked Examples

No. of student 1. Calculate the standard deviation of the
5 4 2 3 7 7 frequency distribution below:

i. How many students took part in the test?

Age 1–3 4–6 7–9 10 –12 13 – 15
ii. Find from the table, the median score. Freq 4 10 20 11 5
iii. Calculate the mean score.
iv. What is the standard deviation of the Solution
distribution?
class x f fx fx2
5. The table gives the frequency distribution of 1–3 2 4 8 16
marks obtained by a number of students in a test. 4–6 5 10 50 250
7–9 8 20 160 1280
Marks 3 4 5 6 7 8 10 – 12 11 11 121 1331
Frequency 5 9 4 1 13 – 15 14 5 70 980
∑ ∑
If the mean mark is 5, calculate the; = = ∑
i. value of 50 409 = 3857
ii. mode and median
∑
From the table, ̅= ∑
= = 8.18
Standard Deviation of Grouped data
Put ̅ = 8.18, ∑ = 3857 and ∑ = 409 in the
For a grouped data, the standard deviation is
formula;
calculated as follows:

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 ∑ 174 – 183 11
S=√ (̅ =√ ( = 3.20
∑ 184 – 193 14

Exercises 19.11 Calculate the mean and the standard deviation of

1. Find the standard deviation for the data the distribution of heights.
below: Ans ̅ = 8.24 S = 3.27
Age 1–3 4–6 7–9 10 – 12 13 – 15 Using the Assumed Mean to Calculate the
Freq 2 5 10 5 3 Standard Deviation
Using an assumed mean, A, to calculate the
2. Calculate the standard deviation of the standard deviation, prepare and complete a
frequency distribution below: S = 6.87 frequency table similar to the one below:

Age 1– 6 – 10 11 – 16 – 21 –
5 15 20 25
Freq 2 5 10 5 3 class x d=x–A f fd

3. The following is the marks scored by 50

students in a test: ∑ ∑ ∑
21 35 52 70 55 42 09 48 57 36
46 15 35 12 29 61 48 22 43 58 The standard deviation is then calculated by the
25 42 01 60 44 38 54 47 69 30 formula:
47 18 35 32 21 50 11 24 41 50 ∑ ∑
S =√ ∑
(∑ )
53 30 54 47 34 48 60 45 16 33
i. Make a frequency table for these data using
equal class intervals of 1 – 10, 11 – 20, … Worked Example
ii. Calculate the standard deviation from the table 1. The table below shows the marks obtained by
iii. Draw a histogram for the data students in an examination

4. The following table gives the distribution of Marks 0 – 9 10 – 20 – 30 – 40 –

height (in cm) of 60 boys 19 29 39 49
Freq 2 3 15 10 10
Height (Cm) Frequency
144 – 153 4 Assuming the mean is 15, calculate the standard
154 – 163 15 deviation.
164 – 173 16

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Solution
A = 15
class x d= x–A f fd
0–9 4.5 -10.5 2 -21 220.5
10 – 19 14.5 -0.5 3 -1.5 0.75
20 – 29 24.5 9.5 15 142.5 1353.75
30 – 39 34.5 19.5 10 195 3802.5
40 – 49 44.5 29.5 10 295 8702.5
∑ = 40 ∑ = 610 ∑ = 14,080

∑ ∑
S =√ ∑
( ∑ ) =√ ( ) =10.93 (2 .d.p)

Exercises 19.12 41 – 50 4
1. Consider the test results of a form three class in 51 – 60 3
Mathematics 61 – 70 4

marks 40 – 50 – 60 – 70 – 80 – The Range and Interquartile Range from

49 59 69 79 89 Frequency Distribution
Freq. 1 4 19 15 3 The Range
The range can be determined from the frequency
Taking the assumed mean as 64.5, find the distribution table in two ways
standard deviation of the distribution 1. Finding the difference between the mid points
of the highest class and the lowest class.
2. The test results of 50 students were recorded as 2. Finding the difference between the highest
follows: value and the lowest value

Marks 50 – 60 – 70 – 80 – 90 – Worked Example

59 69 79 89 99
From the frequency distribution table below, find
Freq. 6 11 19 9 5
the range.
Choose a suitable working mean and use it to
calculate the standard deviation Marks 1 – 11 – 21 – 31 – 41 – 51 –
10 20 30 40 50 60
3. Calculate the standard deviation of the Freq. 2 6 7 14 20 35
following data using the assumed mean method.
Solution
class Frequency Method 1
10 – 20 4 Midpoint of the highest class
21 – 30 5 = = 55.5
31 – 40 4 Midpoint of the lowest class

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= = 5.5 distribution is still the difference between the
third quartile (Q3) and the first quartile (Q1). i. e.
Range = 55.5 – 5.5 = 50
Q3 – Q1
But the first quartile is determined by the
Method 2
From the table, the highest value = 60 and the formula: Q1 = L + i, where:
lowest value = 1
L is the lower boundary of the class containing
Range = 60 – 1 = 59
the first quartile,
n is the number of frequencies,
Exercise 19.13
CF is the cumulative frequency of all the classes
Determine the range of the frequency distribution
preceding the class in which the first quartile lies,
table below:
f is the frequency in which the first quartile falls.
Marks 11 – 21 – 31 – 41 – 51 – 61 –
20 30 40 50 60 70 The Third quartile Q3= L + i where all
Frequency 5 21 15 10 7 3
variable carry the same meaning except L, CF, f
The Interquartile Range and i which refer to the values needed from the
The inter quartile range for data in a frequency third quartile

Worked Example
The following table shows the distribution of the masses of parcels in a local post Office

Marks (kg)20 – 24 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59
Frequency 2 3 7 26 29 25 6 2

Find the interquartile range for the data.

Solution

Classes Class boundaries Frequency Cumulative frequency

20 – 24 19 .5 – 24.5 2 2
25 – 29 24.5 – 29.5 3 5
30 – 34 29.5 – 34.5 7 12
35 – 39 34.5 – 39.5 26 38
40 – 44 39.5 – 44.5 29 67
45 – 49 44.5 – 49 .5 25 92
50 – 54 49.5 – 54.5 6 98
55 – 59 54.5 – 59 .5 2 100

Determine the first quartile as follows:

Q1 = L + i

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But = = 25 75 > 67 but 75 < 92⇒ Q3 lies in sixth class.
Therefore L = 44.5
25 > 12 but 25 < 38 ⇒ Q1 lies in fourth class.
Cumulative frequencies of all the classes
Therefore L = 34.5
preceding the class containing Q3 is 67, so CF =
Cumulative frequencies of all the classes
preceding the class containing Q1 is 12, so CF 67
=12 The frequency of the class containing Q3 is 25.
The frequency of the class containing Q1 is 26. Therefore f = 25
Therefore f = 26 The class interval of the class containing Q1 is
The class interval of the class containing Q1 is 49.5 – 44.5 = 5. Therefore i = 5
39.5 – 34.5 = 5. Therefore i = 5 In all, there are 100 observations, so n = 100
In all, there are 100 observations, so n = 100 By substitution,
By substitution,
Q3= L + i
Q1 = 34.5 + ×5
= 34.5 + × 5 = 37.0 (

Q3 = 44.5 + ×5
Determine the third quartile as follows: = 44.5 + × 5 = 46.1

Q3= L + i
The interquartile range;
(
But = = 75 = Q3 – Q1 = 46 .1 – 37.0 = 9.1

Exercises 19.14
Find the range and interquartile range of each of the data below;
1.

Weight 116 – 118 119 – 121 122 – 124 125 – 127 128 – 130
Frequency 7 19 28 16 5

2. Marks 40 – 44 45 – 49 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74
Frequency 4 11 20 31 19 11 4

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20 PROBABILITY Baffour – Ba Series

Definition there are only two outcomes when a 20p coin is

Probability is defined as the ratio of the number tossed namely,
of successful outcomes of a random experiment 1. Coat of arms 2. Cocoa pod
to the total number of possible outcomes.
Likewise, in the game of football, only three
Mathematically, outcomes are expected namely,
Probability = 1. Win 2. Draw 3. Lose
The outcome of an experiment deals with
numbers.
It may also be defined as the ratio of event to
sample space. In this case,
(
Sample Space of an Experiment
Probability(P) = = It is the set of all possible outcomes of an
(
experiment. It is denoted by S; For instance, if
Measures of probability are fractions between 0 one side of a coin is represented by H and the
and 1 i.e 0 ( 1. The probability of zero other side by T, then the sample space, for a coin
(0) means an event will not occur and a is S = { }. Similarly, the sample space for a
probability of one means the event will always Ludo die is S ={ }.
occur.
Event
Random Experiment An event is the occurrence with which a
It is an experiment whose results depend on particular interest is attached. It is always the
chance; meaning the result of the experiment subset of the sample space. For example, in a die
cannot be predetermined. For instance, the toss of experiment, the sample space, S = {1, 2, 3, 4, 5,
a coin is a random experiment because it cannot 6}. If one is only interested in the occurrence of
be determine with surety and certainty, the face an even number, then the event, E = {2, 4, 6}
that will show up. Also, in the same experiment, if one is interested
in the occurrence of a number more than 4, then
Similarly, the game of football is a random the event, E = {5, 6}
experiment because; one cannot predetermine the
outcome (whether win, draw or lose) with The first letter of an event is usually used to
certainty. Again, in boxing bout, the winner represent the probability of that event. Thus, the
cannot be predetermined so it is a random probability of an even number can be written as
experiment. P(E), the probability of an odd number can be as
P(O) and that of a number less than 5 as P(<5).
Outcome of an Experiment
It is the result we get from an experiment. In Exercises 20.1
tossing the 20p coin for instance, either the coat List all the possible outcomes of the following;
of arms or the cocoa pod will show up. Therefore, 1. Days of the week on which a baby is born.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 539
2. Months of the year in which workers are paid. ii. S = { }, E= { }
3. Time of the day in which greetings are offered. n(S) = 6, n(E) = 3
4. Observing the sex of a baby. (
P(M2) = = =
(
5. Playing the game of football.
6. Taking a penalty kick.
iii. S = { }, E={ }
7. Tossing a die?
n(S) = 6, n(E) = 3
(
Worked Examples P(P) = = =
(
1. A ludo die is thrown once, what is the
probability of scoring; 3. In a class of 24 students, 8 of them are girls
i. an even number and 16 of them are boys. Find the probability of
ii. an odd number picking at random,
iii. a number less than 5 i. a boy ii. a girl

Solution Solution
i. Sample space S = { } Let G represent girls and B represent boys
n(S) = 6, E ={ }, n(E) = 3 n(S) = 24, n(B) = 16, n(G) = 8
(
But P =
(
, P(E) =
(
= = i. P(B) = = =
(
( (

ii. Sample space S ={ }, (

ii. P(G) = = =
(
n(S) = 6 E ={ }, n(E) = 3
( (
But P = , P(O) = = = 4. A bag contains 12 red pens and 16 blue pens. If
( (
a pen is randomly selected, what is probability
iii. Sample space S ={ }, that it is of: i. Blue color ii. Red color
n(S) = 6 , E ={ } n(E) = 4
( Solution
But P = , Let R represent red pens and B represent blue
(
( pens
P(<5) = (
= =
n(S) = n(B) + n(R) = 16 +12 =28
n(S) = 28, n(R) = 12, n(B) = 16
2. A fair die is thrown once. (
i. Write down the set of all possible outcomes i. P (B) = = =
(
ii. Find the probability of obtaining a multiple of 2.
iii. What is the probability of obtaining a prime (
ii. P (R) = = =
(
number?
5. A bag contains 12 mangoes of which 4 are not
Solution ripe. What is the chance of picking at random a
i. The set of all possible outcomes
ripe mango from the bag?
S={ }

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Solution i. either a prime or a multiple of 3
n(S) = 12, n(not ripe) = 4, ii. not a prime
n(ripe) = n(S) – n(not ripe)
= 12 – 4 = 8 The Probability of a Letter in a Word
( Sometimes, a word may be given to determine
P(ripe) = = =
(
the probability of selecting a letter from it.
Learners must therefore take note of the
Exercises 20.2
following:
A. 1. What is the probability of an event that
1. that the set of English alphabets, A = {a, b,
which is certain to happen?
c…z} and the number of alphabets, n(A) = 26
2. What is the probability of an event that is 2. that the set of English vowels, V = {a, e, i, o,
impossible? u} and the number of vowels, n(V) = 5
3. the rest of the alphabets excluding the vowels
3. Given that the probability that an event will are consonants and therefore, the number of
happen is a(0 , what is the probability consonants, n(C) = 21
that the event will not happen?
For instance, in the word “HIPPOPOTAMUS,”
B. 1. A number is selected at random from the the chance of selecting a vowel at random, if the
set of whole number from 1 to 9 inclusive. What letters are written on a card and placed in a box,
is the probability that the number is even? is determine as follows;
S ={ } n(S) = 12,
2. A number is chosen at random from the set W
V={ } ( =5
= {11, 12, 13,…20}. Find the probability that the
(
number is at least 15. P(V) = =
(

3. A bag contains 12 good and 9 bad oranges. If

Worked Examples
an orange is picked at random from the bag, what
1. A letter is selected from the letters of the
is the probability that it is a good orange?
English alphabets. What is the probability that the
letter is selected from the word
4. A box contains 40 pens of equal size. 10 of
MATHEMATICS?
them are green and 18 red. If a pen is chosen at
random from the box, what is the probability that
Solution
it is neither green nor red?
Alphabets, S = {a, b, c…z} n(S) = 26
Word, W = {MATHEICS}, n(W) = 8
5. There are 100 cars in park. 28 of them are blue (
and red 34. If a car is selected at random, what P(W) = = =
(
the probability that it neither blue nor red?
2. The letters of the word “examination” are
6. Find the probability that a number chosen at written on a card and placed in a box. If Mr.
random from the natural numbers 2 to 12 Greenpicks a letter at random from the box, what
inclusive is: is the probability that he picks:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 541
i. a vowel? ii. the letter “I”? 4. A letter is chosen at random from the letters of
iii. the letter “O”? iv. the letter “M”? the word “STATISTICS”. Find the probability
that the letter is:
Solution i. S ii. not a vowel.
i. Let S represent sample space and
V represents vowels Solution
S={ }, n(S) = 11, i. S = { }, E = {S, S, S}
V={ }, n(V) = 6 n(S) = 10, n(E) = 3
( (
P(V) = = P(S) = =
( (

ii. n(S) = 11, n(I) = 2 ii.S = { }, ̅ = {STTSTCS}

( n(S) = 10, n( ̅ ) = 7
P(I) = = (
(
P( ̅ ) = ( = = 0.7

iii. n(S) = 11, n(O) = 1

( Exercises 20.3
P(O) = =
( 1. In the word “encyclopedia”, what is the
probability of randomly selecting; ?
iv. n(S) = 11, n(M) = 1 i) a vowel ii) the letter e iii) a consonant
(
P(M) = =
(
2. The letters of the word “Photosynthesis” is
written on a card and placed in a box. Whatis the
3. Determine the probability of picking the
probability of selecting at random,
following at random when the letters of the word
i. a letter that is not a vowel from the box?
“Length” is place in a bag.
ii. a letter that is S
i. a vowel, ii. a consonant, iii. the letter “g.”
3. A letter is selected from the letters of the
English alphabets. What is the probability that the
Solution
letter is selected from the word OPPOSITION?
S={ }, V = { }
n(S) = 6, n(V) = 1 Equally Likely Events
(
i. P(V) = = Equally likely events are events of the same
(
experiment that have equal chance of occurring.
ii. S = { }, C = { } In other words, they are two or more events of the
n(S) = 6, n(C) = 5 same experiment that have the same probability.
( For e.g, if a die is tossed once:
P(C) = =
( S = {1, 2, 3, 4, 5, 6} n(s) = 6,
Even numbers, E ={ }, n(E) = 3
iii. n(S) = 6, n(g) = 1
(
P(g) = = Odd numbers, O = { } ( .
(

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The probability of scoring an even number, Solution
(
P(E) = = = n(S) = 20, n(E) = ? P(Red) =
(
Likewise, the probability of scoring an odd But n(E) = P(R) n(S).
( n(E) = 20 = 5
number, P(O) = = =
(
The number of red beads in the bag is 5
These two events, P(E) and P(O), are said to have
equally likely outcomes because they have the 2. A box contains 25 balls some of which are blue
same chance of occurrence or probability which and others green. If the probability of picking a
is . Two or more of such events are called green ball randomly from the box is .

equally likely events and they are said to have i. Determine the number of green balls in the box.
equally likely outcomes. ii. How many blue balls are in the box?

Solution
Exercises 20.4
i. n(S) = 25, n(E) = ?, P(Green) =
1. A number is chosen from the set S = {1, 2, 3,
But n(E) = P(Green) n(S).
4, 5, 6} show by working, four events whose
outcome is equally likely. n(E) = 25 = 10
There are 10 green balls in the box
2. In the word “existentialism,” identify any four
events of the letters that have equal outcomes. ii. n(Green) + n(Blue) = n(S)
3. Show by working, any three events of the But n(G) = 10, n(S) = 25 and n(B) =?
letters that are likely to have equal outcomes in n(B) = n(S) – n(G)
the word “preposition” n(B) = 25 – 10 = 15
Therefore, there are 15 blue balls in the box
Finding the Number of an Event, n(E), Given
3. The probability of selecting a girl at random
the Probability P, and Number of Sample
Space, n(S) from a class of 27 students is .
To find the n(E), given P and n(S), make n(E), i. How many girls are in the class?
( ii. Find the number of boys in the class.
the subject of the formula, P(E) = to obtain,
(
n(E) = P n(S) where, Solution
n(E) = number of event,
i. n(S) = 27, n(E) = ? P(Girls) =
n(S) = number of sample space
P = probability of obtaining that event. But n(E) = P(Girls) n(S)
n(E) = 27 = 9
Worked Examples There are 9 girls in the class
1. There are 20 beads in a bag. The probability
that a red beads is taken at random from the bag ii. n(Girls) + n(Boys) = n(S)
is . Find the number of red beads in the bags. But n(G) = 9, n(S) = 27 and n(B) =?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 543
n(B) = n(S) – n(G) ⇒The total number of balls in the bag is 50
= 27 – 9 = 18
There are 18 boys in the class n(S) = 50, n(B) = 20
(
P(B) = = =
4. A school bus has a capacity of 40 people, some (

of which are boys and others girls. The

Some Solved Past Questions
probability of picking a girl at random is . How
1. The probability of picking a footballer from a
many girls are in the bus? group of 120 sportsmen is How many
sportsmen were not footballers?
Solution
n(S) = 40, n(E) = ?, P(Girls) = Solution
But n(E) = P(R) × n(S) Method 1
Let the number of footballers be n(F) and the
n(E) = 40 = 30 girls
number of sportsmen who were not footballers be
n( ̅
5. There are 16 white, 20 blue and a number of n(F) = × 120 = 75
green identical tennis balls in a box. The
( ̅ = 120 – 75 = 45
probability of picking a green tennis ball from the
box is . Find the probability of picking a blue
tennis ball from the same box. Method 2
P(F) + ( ̅ = 1
Solution ( ̅ = 1 - P(F)
Let W represents white tennis ball,
But P(F) =
B represents blue tennis ball,
G represents green tennis ball, (̅ = 1 -
x represents total number of balls in the bag. P (̅ = – = =
n(W) = 16, n(B) = 20, n (x) = ?
n(F) = × 120 = 45
P(G) = , P(W) = , P(B) =
But P(W) + P(B) + P(G) = 1
2. A bag contains some balls of which are red.
⇒ + + =1
Forty more balls of which five are red are added.
+ =1– If of all the balls are red, how many balls were
there originally?
+ = –

= Solution
Let the number of balls be x
18x = 36 × 25 n (R) =
x= = 50

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If forty more balls are added, total number of ball P(G) = =
= x + 40
Exercises 20.5
Additional 5 red balls;
1. Nissan and Benz cars numbered 48 are parked
n (R) = +5 for sale. If the likelihood of Mr. White buying a
of all the balls are red; Nissan car is ,
i. how many Nissan cars are on the park?
( = +5 ii. find the number of cars that are Benz.
20 × ( = 20 × + (20)5
2. A box contains 50 pens. The probability of
4( = 5x + 100
4x + 160 = 5x + 100 picking a blue pen at random from the box is .
60 = 5x – 4x i. How many blue pens are in the box?
60 = x ii. If the rest of the pens are red, how many are
Therefore, 60 balls were there originally they in the box?

3. Three blue balls, five green balls and a number 3. There are 54 identical balls of white and green
of red balls are put together ina sack. One ball is colors in a bag. The probability of picking a green
picked at random from the sack. If the probability ball from the bag at random is . Find the number
of picking a blue ball is , find: of green balls in the bag.
i. the number of red balls in the sack,
ii. the probability of picking a green ball. 4. A bag contains a number of balls, 30 of which
are red and the remainder blue. If a ball is chosen
Solution at random, the probability that it is red is 0.60.
i.n(B) = 3, n(G) = 5 and n(R) = x Find the number of blue balls in the bag. Ans = 20
n(S) = 3 + 5 + x
=8+x 5. A box contains30 white, 25 blue and a number
of green identical balls. The probability of
(
P(B) = (
= , but P(B) = picking a green ball from the box is . Find:
= i. the total number of balls in the bag
3×6=8+x ii. the number of green balls in the bag
18 = 8 + x iii. the probability of picking a blue ball from the
x = 18 – 8 = 10 box.

But n(R) = x 6. A bag contains 60 balls, some of which are red

n(R) = 10 and some blue. The probability of a ball drawn at
random, being red is . Find:
(
ii. P(G) = = , but x = 10 i. the number of blue balls in the bag,
(

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ii. the number of red balls that should be added to (
⇒ =
the bag to change the probability to .
n(M) = = 25
There are 25 males in the class
Rules of Probability
1. If the probability that an event E, will occur,
2. A box contains a number of green and yellow
denoted by P(E) = a, then the probability that the
beads. If the probability of choosing a yellow
event will not occur denoted by:
P( ̅ = 1 – P(E) = 1 – a. bead is , find the probability of choosing a
For e.g. If the probability that George will go to green bead.
school today is , then the probability that he will
Solution
not go to school is :1 – = – = P(G) + P(Y) = 1
But P(Y) = , P(G) = ?
2. For any two events A and B,
P(G) = 1 – P(Y)
P (A) + P (B) = 1…… (1)
P(G) = 1– = – = =
From eqn (1)
P (A) = 1 – P (B)…. .. (2)
3. A class consists of some boys and girls. The
P (B) = 1 – P (A)… …(3)
probability of picking a girl for a quiz is . Find
For e.g if the probability of obtaining a “head”
the probability of picking a boy.
when a coin is tossed is , the probability of
obtaining a “tail” is: Solution
1– = – = = P(B) + P(G) = 1
P(G) = , P(B) = ?
Worked examples P(B) = 1 – P(G)
1. There are 30 males and females in a class. The P(B) = 1 – = – = =
probability of selecting a female at random is ,
i. what is the probability of selecting a male? 4. The probability that it will rain on a particular
ii. how many males are in the class? day is . What is the probability that it will not
rain on that day?
Solution
i. P(M) + P(F) = 1, But P(F) = , and P(M) = ? Solution
P(M) = 1 – P(F) = 1– = – = = Let the probability that it will rain be P(R) = ,
the probability that it will not rain ,
P( ̅ = 1 – P(R)
ii. n(S) = 30, P(M) = n(M) = ?
( =1– = – = =
But P(M) = (

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5. A bag contains a number of red, blue and green Challenge Problems
pens. The probability of picking at random a red 1. All possible two – digit numbers are formed
pen is and the probability of picking at random from the digits 1, 2, 3, 4, 5, 6. Find the
probability that one of these numbers chosen at
a green pen is . What is the probability of
random will be divisible;
picking a blue pen at random from the bag? i. by 5 ii. by 6

Solution Compound Events

P(R) = , P(G) = and P(B) = ? Events can be combined by the words “or” and
P(R) + P(B) + P(G) = 1 “and”. Events which are combined as such are
called combined events. The word “or”
+ + P(G) = 1
corresponds to union (U) and the word “and”
+ P(G) = 1 corresponds to Intersection(∩). These are further
P(G) = 1 – = explained below:
I. AUB is the event that occurs if A or B (or both)
occur. In probability, “or” means addition. i.e
Exercises 20.6
P(A or B) = P(A) + P(B)
1. A carton consists of Pepsi and Fanta drinks.
⇒P(A ∪ B) = P(A) + P(B)
The probability of taking a Fanta at random is ,
what is the probability of taking Pepsi? II. A B is the event that occurs if A and B occur
together.In probability, “and” means
2. The probability of picking a yellow T‟ shirt at multiplication. That is:
random from a bag containing yellow and red
T‟shirts is . What is the probability of picking a P(A and B) = P(A) × P(B)
red T‟shirt at random from the bag? ⇒P(A B) = P(A) × P(B)

3. A bag contains a collection of Nokia and Worked Examples

Samsung mobile phones. If the probability of 1. A fair die is thrown once, what is the
picking a Nokia phone at random from the bag is probability of obtaininga factor of 4 or number
greater than 4?
, what is the probability of not picking a Nokia
mobile phone from the bag? Solution
i. S = {1, 2, 3, 4, 5, 6}
4. A bag contains a number of red, black and n(S) = 6,
white tenisballs. The probability of picking at Fcators of 4 = {1, 2, 4} n(F4) = 3
random a black tenis ball from the bag is and Numbers of greater than 4 = {5, 6} n(G4) = 2
(
the probability of picking at random a white tenis P(F4) = =
(
ball from the bag is . What is the probability of (
P(G4) = =
(
picking a red tenis ball at random from the bag?

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P(F4) or P(G4) = P(F4) + P(G4) Solution
⇒P(F4∪G4) = + = i. P(S) = , P(E) = , P(J) =
Probability that only Sammy will hit the target
2. A number is chosen at random from the set means Eddy and Jimmy will not hit the target;
X = {1, 2, 3…10). What is the probability that the ⇒ ( and P( ̅ = and P( ̅ =
number chosen is either a factor of 24 or multiple
= P(S and ̅ and )̅
of 5?
= P(S ̅ )̅
= P(S) × P( ̅ × P( ̅
Solution
Let the probability of choosing a factor of 24 be = × × =
P(F24) and the probability of choosing a multiple
of 5 be P(M5) ii. Probability that only Sammy and Eddy will not
X = {1, 2, 3…10}, n(S) = 10 hit the target;
⇒P( ̅ and ̅ and J)
Factors of 24 = {1, 2, 3, 4, 6, 8}, n(F24) = 6 = P( ̅) × P( ̅ ) × P( J)
Multiples of 5 = {5, 10}, n(M5) = 2 = × × =
(
P(F24) = = =
(
( 5. The probabilities that Randy, Sandy and
P(M5) = = =
(
Wendy will win the U. S. lottery are , and
⇒P(F24 or M5) = P(A) + P(B) respectively. Find the probability that;
= + = = i. Randy or Sandy or Wendy will not win,
ii. Only Randy and Sandy or only Randy and
3. The probabilities that Manful, John and Ernest Wendy will win.

will pass an examination are , and

Solution
respectively. Find the probability that all three
i. P(R) = , P(S) = , P(W) =
will pass the examination.
The probability that Randy or Sandy or Wendy
Solution will not win;
P(M and J and E), ⇒P( ̅ or ̅ or ̅ )
= P(M) × P(J) × P(E), P( ̅ ) + P( ̅ + P( ̅
= × × = = + + =

4. The probabilities that Sammy, Eddy and ii. Probability that only Randy and Sandy or only
Randy and Wendy will win,
Jimmy will hit a target are , and respectively.
⇒ P(R and S and ̅ ) or P( and and ̅ )
Find the probability that;
P(R) × P(S) × P( ̅ ) or P(R) × P(W) × P( ̅)
i. only Sammy will hit the target,
ii. only Sammy and Eddy will not hit the target. = ( ) or ( )

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= ( ) or ( ) = + = = b. either P or Q or R will win,
c. None of these teams will win.
Exercises 20.7
The Coin and Die Experiments
1. The probability of picking a red pen from a bag
A. Tossing a Coin a Number of Times
is and that of picking a green pen from a box is 1. If a fair coin is tossed once, the possible
. Find the probability of picking; outcomes are, S = {H,T}, and the number of
i. both pens, possible outcomes, n(S) = 2
ii. only one color,
2. If a fair coin is tossed twice or two different
iii. none of the pens.
coins are tossed once, the possible outcomes are
S = {HT, TH, TT, HH}, the number of possible
2. The probabilities that Sam, Dan and Jake will
outcomes, n(S) = 4. This is illustarated as :
go to school on monday are , and
respectively. Find the probability that: H T
i. all of them will go to school on Monday. H HH HT
ii. all of them will not go to school on monday T TH TT

3. The probabilities that Mr. Brown will wear a 3. If a fair coin is tossed three times, the possible
white shirt to school is , the probability that he outcomes, S = {HHH, HHT, HTH, HTT, THH,
THT, TTH, TTT}, the number of possible
will wear a black trousers to school is and the outcomes, n(S) = 8. This is illustrated as :
probability that he will wear a yellow shoe is .
What is the probability that: HH HT TH TT
H HHH HHT HTH HTT
i. he will wear only a white shirt and a black
T THH THT TTH TTT
trousers to school?
ii. he will wear neither a white shirt nor a black The number of possible outcomes of the above
trousers nor a yellow shoe to school? experiments is summarized in the table
iii. he will wear all the three to school? below;

4. The probabilities that A, B, C will occur are , No. of times a No. of possible
and respectively. What is the probability that; coin is tossed outcomes
1 2=
i. A or B or C will not occur? 2 4=
ii. Only A and B or only A and C will occur? 3 8=
n
5. The probabilities of three teams P, Q, R
winning a football competition are , and From the table if a coin is tossed n times, the
number of possible outcomes =
respectively. Calculate the probability that;
a. either P or Q will win,

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Worked Examples b. What is the probability of obtaining?
A fair coin is tossed twice. i. Exactly two heads?
i. Write down the set of possible outcomes. ii. No tail?
ii. What is the probability of obtaining? iii. Not more than one head?
a. exactly two heads, c. at least one head,
b. no head, d. a head and a tail, Solution
a. The set of possible outcomes,
Solution S = {HHH, HHT, HTH, HTT, THH, THT, TTH,
i. S = {HH, HT, TH, TT}, n(S) = 4 TTT},

ii. a. S = {HH, HT, TH, TT}, n(S) = 4 b. i. Let A represent the event of obtaining
Let A denote the event of obtaining two heads exactly two heads;
A = {HH}, n(A) = 1 S = {HHH, HHT, HTH, HTT, THH, THT, TTH,
P(A) =
(
= TTT},n(S) = 8,
( A = {HHT, HTH, THH}, n(A) = 3
The probability of obtaining two heads = P(A) =
(
=
(

b. Let B be the event of obtaining no head;

ii. Let B represent the event of obtaining no tail;
B = {TT}, n(B) = 1
( S = {HHH, HHT, HTH, HTT, THH, THT, TTH,
P(B) = = TTT}, n(S) = 8,
(

The probability of obtaining no head = B = {HHH}, n(B) = 1

(
P(B) = = =
(
c. Let C be the event of obtaining at least one
head; iii. Let C represent the event of obtaining not
C = {HH, HT, TH}, n(C) = 3 more than one head;
(
P(C) = = S = {HHH, HHT, HTH, HTT, THH, THT, TTH,
(
TTT},n(S) = 8,
The probability of obtaining at least one head = C = {HTT, THT, TTH, TTT}, n(C) = 4
(
P(C) = ( = =
d. Let D denote the event of obtaining a head and
a tail
B. Throwing a Ludo Die
D = {HT, TH}, n(D) = 2 1. If a fair die is thrown once, the sample space, S
(
P(B) = = = = {1, 2, 3, 4, 5, 6} and the number of sample
(

The probability of obtaining a head and a tail = space, n(S) = 6

2. If a fair die is thrown twice or two dice are

2. Three fair coins are tossed once. thrown together, the sample space, S, is obtained
a. Write down the set of all possible outcomes in a tabular form shown below:

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 Die 1
1 2 3 4 5 6 From the table, the total number of possible
1 1,1 1,2 1,3 1,4 1,5 1,6
outcomes, n(S) = 36
2 2,1 2,2 2,3 2,4 2,5 2,6
3 3,1 3,2 3,3 3,4 3,5 3,6
Die 2

4 4,1 4,2 4,3 4,4 4,5 4,6 ii. a. The pair of numbers for which the sum is 5
5 5,1 5,2 5,3 5,4 5,5 5,6 = (1, 4), (2, 3), (3, 2), (4, 1)
6 6,1 6,2 6,3 6,4 6,5 6,6
b. The pair of numbers for which the sum is 10 =
Number of possible outcomes, n(S) = 36 (4, 6), (5, 5), (6, 4)

Note: c. The pair of numbers for which the sum is more

than 10 = (5, 6), (6, 5) and (6, 6)
1. The pair of numbers through which the
diagonal drawn from n to n passes sum up ton d. The pair of numbers for which the sum is at
2. The pair of numbers at the left (up) of a least 10 = (4, 6), (5, 5), (6, 4), (5, 6), (6, 5) and (6,
diagonal drawn from n to nare less than n 6)
3. The pair of numbers at the right (down) of a
diagonal drawn from n to n are greater or more iii.a. Let A represent the pair of numbers that
than n show different scores ;
n(A) = 30 and n(S) = 36
(
Worked Examples P(A) = = =
(
1. Two fair dice, A and B, each with faces
numbered 1 to 6are thrown together.
b. Let B represent the pair of numbers that show
i. Construct a table showing all the equally likely
the same number;
outcomes
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)
ii. From your table, list the pair of numbers on the
n(B) = 6 and n(S) = 36
two dice for which the sum is (
a. 5 b. 10 c. more than 10 d. at least 10 P(B) = = =
(
iii. Find the probability that the two dice show:
a. Different scores b. the same scores iv.a. Let C represent the pair of numbers that sum
iv. Find the probability that the sum of the up to 5 = (1, 4), (2, 3), (3, 2), (4, 1)
numbers on the two dice is: n(C) = 4 and n(S) = 36
(
a. 5 b. 10 c. more than 10 d. at least 10 P(C) = = =
(

Solution
b. Let D represent the pair of numbers that sum
i. Die 1
up to 10 = (4, 6), (5, 5), (6, 4)
1 2 3 4 5 6
n(D) = 3 and n(S) = 36
1 1,1 1,2 1,3 1,4 1,5 1,6
(
2 2,1 2,2 2,3 2,4 2,5 2,6 P(D) = = =
(
Die 2

3 3,1 3,2 3,3 3,4 3,5 3,6

4 4,1 4,2 4,3 4,4 4,5 4,6
5 5,1 5,2 5,3 5,4 5,5 5,6 c. Let E represent the pair of numbers that sum up
6 6,1 6,2 6,3 6,4 6,5 6,6 more than 10 = (5, 6), (6, 5), (6, 6)

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n(E) = 6 and n(S) = 36 E = [(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)]
( n (E) = 6 n(s) = 36
P(E) = = =
(
P(same number) = =
d. Let F represent the pair of numbers that sum up
to at least 10 iii. A total of not less than 5
= (4, 6), (5, 5), (6, 4), (5, 6), (6, 5) and (6, 6) n (E) = 30 n(s) = 36
n(F) = 6 and n(S) = 36 P(not less than 5) = =
(
P(F) = = =
(
3.In the throw of two fair dice, what is
2. Two fair dice are thrown at the same time. the probability of throwing?
a. Draw the sample space for the possible i. a pair of even numbers;
outcomes. ii. a pair of prime numbers;
b. Find the probability of obtaining: iii. a total score that is at most 4.

i. a total score of 6 or 8, Solution

ii. the same number on the two dice, i. Die 1

iii. a total not less than 5. 1 2 3 4 5 6

1 1,1 1,2 1,3 1,4 1,5 1,6
Solution 2 2,1 2,2 2,3 2,4 2,5 2,6
Die 2

a. Die 1 3 3,1 3,2 3,3 3,4 3,5 3,6

1 2 3 4 5 6 4 4,1 4,2 4,3 4,4 4,5 4,6
1 1,1 1,2 1,3 1,4 1,5 1,6 5 5,1 5,2 5,3 5,4 5,5 5,6
2 2,1 2,2 2,3 2,4 2,5 2,6 6 6,1 6,2 6,3 6,4 6,5 6,6
3 3,1 3,2 3,3 3,4 3,5 3,6
Die 2

4 4,1 4,2 4,3 4,4 4,5 4,6 n(S) = 36

5 5,1 5,2 5,3 5,4 5,5 5,6 A pair of even numbers, E;
6 6,1 6,2 6,3 6,4 6,5 6,6 (2, 2), (2, 4), (2, 6), (4,2), (4, 4), (4, 6), (6, 2), (6,
4), (6, 6) n(E) = 9
b. E (6) = [(5, 1) (4, 2), (3, 3), (2, 4), (1, 5)] P(E) = =
n(E6) = 5
E (8) = [(5, 3) (4, 4), (3, 5), (2, 6)]
ii. a pair of prime numbers, P
n(E8) = 4
= (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3),
From the table, the total number of possible
(5, 5) n(P) = 8
outcomes, n(S) = 36
P(P) = =
i. The probability of a total score of 6 or 8
( ( iii. A total score that is atmost 4,
P(6 or 8) = + = + =
( (
E = (1,1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)
n(E) = 6
ii. The probability of the same number on the two
dice P(E) = =

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4. Two dice are thrown together and the scores Worked Examples
are added. Copy and complete the following table 1. In a game, a player throws a die and tosses a
of total scores; coin. If the coin lands heads, he scores twice the
number on the die. If the coin lands tails, he
1 2 3 4 5 6 scores three times the number on the die. What is
1 2 the probability of scoring more than 6?
2
3 7
4 Solution
5
6 12 Dice
1 2 3 4 5 6
i. What is the probability of scoring exactly 9? ii. Coin H 2 4 6 8 10 12
What is the probability of scoring an even? T 3 6 9 12 15 18
iii. What is the probability of scoring either 7 or 11?
n(S) = 12,
Solution More than 12 = { 8, 9, 10, 12, 12, 15, 18}
n( More than 12) = 7
1 2 3 4 5 6 P( More than 12) =
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9 Other Related Experiments
4 5 6 7 8 9 10 In some related instances, two setsof elements
5 6 7 8 9 10 11 may be provided for an experiment. For example,
6 7 8 9 10 11 12 given A = {1, 2, 3} and B = {t, h, e}, if A and B
occur at the same time, then the set of possible
n(S) = 36 outcomes or sample space is represented as
i. Probability of scoring exactly 9 = shown below:

ii. Probability of scoring an even number =

1 2 3
iii. Probability of scoring either 7 or 11 = t 1t 2t 3t
h 1h 2h 3h
e 1e 2e 3e
C. Throwing a Coin and a Die Together
If a coin and a die are thrown together or one Number of sample space, n(S) = 9
after the other, the outcomes are obtained as
shown in the table below; Worked Examples
1. A number is selected from each of the sets {2,
Dice 3, 4} and {1, 3, 5}. Find the probability that:
1 2 3 4 5 6
Coin
i. the sum of the two numbers is less than 7 and
H
T greater than 3;
ii. the sum of the numbers is a prime number;

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Solution 3.A two digit numeral (base ten) is formed by
choosing both digits at random from the set {6, 7,
2 3 4 8, 9}. The same digit may be chosen twice. Find
1 (1, 2) (1, 3) (1, 4) the probability that the number is;
3 (3, 2) (3, 3) (3, 4) a. even, b. divisible by 4, c. prime.
5 (5, 2) (5, 3) (5, 4)
Solution
n(S) = 9
i.
Sum of the two numbers less than 7 and greater
6 7 8 9
than 3 = (1, 3), (1, 4), (3, 2), (3, 3)
6 66 67 68 69
n(E) = 4
( 7 76 77 78 79
P= =
( 8 86 87 88 89
9 96 97 98 99
ii.sum of the numbers that is a prime number
E = (1, 2), (1, 4), (3, 2), (3, 4), (5, 2), n(E) = 5 n(S) = 16
P(P) = Even numbers, E = {66, 68, 76, 78, 86, 88, 96, 98}
n(E) = 8
2. If two letters are selected from the sets {G, O, P(E) = =
L, D} and {C, O, A, L}, what is the probability
that: ii. Numbers divisible by 4,
i. both letters are vowels, = {68, 76, 88, 96}
ii. the letters contain at least one vowel. n( numbers divisible by 4) = 4
P( numbers divisible by 4) = =
Solution
G O L D iii. Prime numbers = {67, 79, 89, 97}
C CG CO CL CD n(prime) = 4
O OG OO OL OD P(Prime) = =
A AG AO AL AD
L LG LO LL LD
Exercises 20.8
n(S) = 16 1.The following is an incomplete table of possible
V = {OO, AO}, n(V) = 2 outcomes when a die is thrown twice
( Die 1
P(V) = = =
( 1 2 3 4 5 6
1 1,1 1,2 1,3
ii. Let X represent the set that contains at least 2
Die 2

one vowel 3
⇒ X = { CO, OG, OO, OL, OD, AG, AO, AL, 4
5
AD, LO }, n(X) = 10
(
6
P(X) = = =
(

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i. Copy and complete the table ii. a double is thrown,
ii. Use your table to find the probability of iii. there is a multiple of three on each die,
throwing: iv. the total score is a prime number.
a. no six, b. at least one five, c. two sixes.
7. A boy tosses a coin and throws a die. If the
2.a. A fair coin is tossed three times,write down coin lands heads, he scores the number shown on
the set of possible outcomes. the die. If the coin lands tail, he scores double the
b. What is the probability of obtaining? die number. What is the probability that he will
i. no tail, score an odd number?
ii. exactly two heads,
iii. at most two heads, 8. A number is selected from each of the sets {1,
iv. at least one head. 2, 3, 4} and {3, 5, 7}. Find the probability that:
i. the sum of the two numbers is greater than 5
3. A pair of fair dice each numbered 1 to 6 are and less than 8,
tossed together. Find the probability of scoring: ii. the sum of the numbers is an even number,
i. a sum of at least 9 iii. the sum ofthenumbers is at most 8 or atleast 11.
ii. a sum of 7
iii. a sum that is atmost 10 9. If two letters are selected from the sets M = {C,
A, T} and N = {D, U, C, K}, what is the
4. Show as an array of ordered pairs all possible probability that:
outcomes in throwing a pair of dice. Hence find i. both letters are vowels,
the probability of getting: ii. the letters contain at least one vowel,
i. two sixes, iii. none is a vowel.
ii. at least one six,
iii. a total score of six, 10. Two dice are thrown together and the scores
iv. a total score of not six, multiplied. Tabulate all the possible results, and
v. a total score of eleven, find the probability that the resulting score is:
vi. a total score which is a multiple of three. a. odd b. less than 20 c. a factor of 36

5. I throw two dice, one after the other. Make an 11. A two digit number in base ten is formed by
array of possible outcomes, and find the choosing the tens digit from the set {4, 5, 6, 7, 8}
probability that I score: and the units digits from the set {1, 2, 3, 9). Find
i. a sum that is exactly 8, the probability that the resulting number is:
ii. a sum that is at least 8, a. even,
b. greater than 50,
6. Draw a possibility space to show all the c. both even and greater than 50,
outcomes when thrown together a normal die and d. either even or greater than 50.
a die with its sides numbered 1, 1, 1, 2, 2, 3.Use
this probability space to find the probability that: 12. A two digit numeral (base ten) is formed by
i. the total score is more than 4, choosing both digits at random from the set {6, 7,

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8, 9}. The same digit may be chosen twice. Find ii. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} n(S) = 10
the probability that the number is; Let P(1) denote the probability of selecting 1
a. odd, b. prime . P(10) denote the probability of selecting 10
n(1) = 1 and n(10) = 1
13. A = {1, 2, 3}; B = {4, 5, 6, 7}; C = {(a, b) : (
P(1) = =
a∈ A, b∈ B}. (
P = {8, 9}; B = {10, 11, 12, 13, 14}; R = {(p, q) : (
P(10) = =
p∈ P, q∈ Q}. Find: (
a. n (C)
b. the probability that a member of C ∪ R, chosen ⇒P(1 or 10) = P(1∪10)
at random , is a member of C = P(1) + P(10)
= + =
Mutually Exclusive Events
Any two events that cannot occur at the same
time are said to be mutually exclusive. For 2. In a football match, the probability of Kotoko
example, in throwing the ludo die once, the scoring a goal is 0.25 and that of Hearts scoring is
sample spaces, S = {1, 2, 3, 4, 5, 6,}. Suppose 0.41. What is the probability of Kotoko or Hearts
one is interested in playing the number 2 or 5. It Scoring?
is clear that 2 occurs once on the die and 5 also
occurs once on the die; hence these two numbers Solution
cannot show up at the same time in a single Let P(K) represent the probability of Kotoko
throw. The event of playing 2 and 5 are said to be scoring and P(H) represent the probability of
mutually exclusive events in this instance. Hearts scoring
P(K) = 0.25 and P(H) = 0.41
In general, if A and B are mutually exclusive P(K or H) = P(KUH) = P(K) + P(H)
events, then the probability of A or B occurring in = 0.25 + 0.41 = 0.66
a single draw is given by:
( ( 3. Given that X and Y are mutually exclusive
P(A or B) = +
( ( events such that P(X) = 0.5, P(Y) = n and
P(A or B) = P(A∪B) = P(A) + P(B) P(X or Y) = 0.8,
This is called the addition rule of probability. i. find the value of n;
ii. what is the probability of X or Y occurring in
For mutually exclusive events, terms of X and Y?
P(A and B) = P(A B) = 0
Solution
Worked Examples
P(X or Y) = P(X) + P(Y)
1. i. Write down the sample space for the set of
But P(X or Y) = 0.8, P(X) = 0.5, P(Y) = n
first ten positive integers.
By substitution,
ii. What is the probability of selecting either 1 or 10?
P(X or Y) = P(X) + P(Y)
Solution 0.8 = 0.5 + n
i. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 0.8 – 0.5 = n

Baffour – Ba Series, Core Maths for Schools and Colleges Page 556
0.3 = n
A is independent of B. Likewise B is independent
ii. P(X or Y) = P(X) + P(Y) of A
1. AUB = A +B = {a, b, m, n}
4. Two mutually exclusive events A and B are
such that P(A) = and P(B) = . What is the In terms of probability,
P(A∪B) = P(A) + P(B)
probability of either A or B occurring?
This is the addition rule of probability for

Solution independent events. That is P( ) = P(A)

P(A or B) = P(A) + P(B) 2. If A and B are two independent events then the
But P(A) = and P(B) = probability that both occur is the product of their
separate probabilities.
⇒P(A∪B) = + = i.e. P(A and B) = P(A B) = P(A) × P(B)
The probability of A or B occurring is This relation is called the multiplication rule of
probability.
Exercises 20.9
Similarly, for three independent events A, B and C,
1. If A and B are mutually exclusive events and
P(A B C) = P(A) × P(B) × P(C)
P(A) = and P(B) = , find the probability of A or
B occurring. Two events that are not independent are said to
be dependent. In other words, an event A is said
2. In a certain study, pupils were classified as to be dependent on event B if the occurrence of
being underweight or overweight or having event A affects the occurrence of event B and vice
normal weight. The probability that a pupil is - versa
underweight is 0.3 and the probability that a pupil
has a normal weight is 0.5 and the probability that A B U
a pupil is overweight is 0.2. Find the probability z
x y,
that a pupil selected at random is either
,
underweight or over weight.

Dependent and Independent Events AUB = A + B = {x, z} + {z, y}

An event , A is said to be independent on an AUB = {x, y, z, z} – {z}, but A∩B = z
event B, if the occurrence or non– occurrence of AUB = A + B – (A∩B)
event B does not affect the occurrence of event A. In terms of probability,
The diagram below further explains independent P(AUB) = P(A) + P(B) – P(A∩B)
events.
A B U Worked Examples
A. 1. Use the given probabilities to determine
m, a b, n
whether or not the events A and B are
independent events:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 557
P(A) = , P(B) = , P(A∩B) =
4. i. X and Y are two independent events such

Solution that P(X) = and P(X∩Y) = , find P(Y) .

If two events A and B are independent events, ii. Find P(XUY) if the events X and Y are not
then P(A∩B) = P(A) × P(B) independent.
P(A∩B) =
Solution
P(A) × P(B) = × = = If X and Y are two independent events, then
P(A B) = P(A) × P(B) = P(X∩Y) = P(X) × P(Y)
Events A and B are independent. P(X) = and P(X∩Y) =
By substitution,
2. Use the probabilities given to determine = P(Y)
whether or not, events X and Y are independent:
P(X) = 0.3, P(Y) = 0.7 and P(X Y) = 0.14 P(Y) = = × =

Solution ii. P(XUY) = P(X) + P(Y) – P(X∩Y)

If two events X and Y are independent events, = – =
then P(X Y) = P(X) × P(Y)
P(X Y) = 0.14 B. 1. A number is chosen at random from the set
P(X) × P(Y) = = 0.21 of factors of 30. What is the probability that the
P(X Y) P(X) × P(Y) number chosen is a prime number or a multiple of
Therefore events X and Y are not independent 3?

3. Two independent events A and B are such that Solution

P(A) = and P(A B) = Find: S = {1, 2, 3, 5, 6, 10, 15,30}, n(S) = 8
i. P(B) ii. P(A∪B) P = {2, 3, 5}, n(P) = 3
M = {3, 6, 15, 30}, n(M) = 4
Solution P M = {3}, n(P O} = 1
i. If A and B are two independent events, then ( (
P(P) = ( = P(M) = ( =
P(A∩B) = P(A) × P(B)
P(P M} =
But P(A) = and P(A B) =
P(P or O) = P(P) + P(O) – P(P∩O)
By substitution, P(P ∪ O) = P(P) + P(O) – P(P∩O)
= P(B) = + – = =
P(B) = = × = =
2. A number is chosen at random from the
ii. P(AUB) = P(A) + P(B) integers 1 to 12 inclusive. Find the probability
= = = = that the number chosen is :

Baffour – Ba Series, Core Maths for Schools and Colleges Page 558
i. a multiple of 2 or a multiple of 3, Solution
ii. a factor of 18 or a greater than 9. i. Let P( represent the probability that Romeo
passes,
Solution ⇒ the probability that Romeo fails is P( ̅ ,
i. S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, Let P( represent the probability that Juliet
n(S) = 12 passes the exams.
Let represent the set of multiples of 2 and ⇒P( ̅ represent the probability that she fails
represent the set of multiples of 3 The probability that both Romeo and Juliet pass
the examination,
= {2, 4, 6, 8, 10, 12}, n( =6 P(R∩J) = P(R) × P(J)
= {3, 6, 9, 12}n( =4 But P(R) = and P(J) =
= {6, 12) n( )=2 By substitution,
P( P( + P( – P( P(R∩J) = P(R) × P(J)
P( P( + P( – P( P(R∩J) = =
( ii. The probability that both Romeo and Juliet fail
ButP( = = =
( the examination is given by:
P( =
(
= = P( ̅ ̅ = P( ̅ × P( ̅
(
P( + P( ̅ = 1, but P( =
P( = + P( ̅ = 1
P( P( + P( – P( P( ̅ = 1 – = – =
= + – = P( + P( ̅ = 1, but P( =
⇒ + P( ̅ = 1
ii. Factors of 18, F = {1, 2, 3, 9}, n(F) = 4 P( ̅ = 1 – = – =
Numbers greater than 9, G = {10, 11, 12}
n(G) = 3 P( ̅ ̅ = P( ̅ × P( ̅ = × =
P(F G) = { }⇒ (Independent event)
⇒P(FUG) = P(F) + P(G) iii. The probability that at least one of them
= + = passes the examination is given by:

Method I
3. The probability that Romeo passes his end of P(RUJ) = P(R) + P(J) – P(R∩Y)
term examination is and the probability that By substitution,
Juliet passes the same examination is . Find the P(RUJ) = ( )–
probability that: =( )– = =
i. both Romeo and Juliet pass the examination
ii. both fail the examination, Method II
iii. at least one of them passes the examination. P(R) P( ̅ + P(J) P( ̅ + P(R) P(J)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 559
= × + × + Method 2
= + = P(MUF) = P(M) + P(F) – P(M ∩F)
But P(M ∩F) = P(M) × P(F)
4. The probability that Maame goes to school on = =
a market day is and the probability that her
brother, Frank also attends school on the same P(MUF) = ( )– =
day is . What is the probability that:
Method 3
i. both will be absent?
P(MUF) = 1 – P( ̅ ∩ ̅ )
ii. at least one will be present?
=1– = – =
iii. exactly one of them will be present?

Solution iii. The probability that exactly one of them is

present:
Let P( represent the probability of Maame
going to school; P(M) × P( ̅ + P(F)× ( ̅
P( ̅ represent the probability ofMaame not = × + × = + =
going to school.
P( represent the probability of Frank going to 5. Bag X contains 10 balls, ofwhich 3 are red and
school; 7 are blue. Bag Y contains 10 balls of which 4 are
P( ̅ represent the probability of Frank not going red and 6 blue. A ball is drawn at random from
to school. each bag. Find the probability that:
P( ( ̅ = 1, but P( i. both are red, ii. at least one is blue.
+ (̅ =1
Solution
(̅ =1– = – = i. The probability that both are red;
⇒Bag X is red and Bag Y is red
P( ( ̅ = 1, but P( n(X) = 10, n(R) = 3, P(Rx) =
(̅ = 1 – = – = n(Y) = 10, n(R) = 4 , P(Ry) =
P(Rx and Ry) = P(Rx∩ Ry)
i. The probability that both were absent
= P(Rx) × P(Ry) = × = =
P( ̅ ̅ = P( ̅ × P( ̅
× = =
ii. The probability that at least one is blue;
ii. The probability that at least one was present; = P(Bx∩ By) or P(Bx∩ Ry) or P(Rx ∩ By)
Method I = P(Bx∩ By) + P(Bx∩ Ry) + P(Rx ∩ By)
P(M)P( ̅ ) or P(F)P( ̅ )orP(M)P(F) = ( )( ) ( )( ) ( )( )
= × + × + ×
= + = =
= + + = =

Baffour – Ba Series, Core Maths for Schools and Colleges Page 560
Exercises 20.10 the probability that:
1. If X and Y are two independent events such i. both will break the record;
that P(X) = and P(X∩Y) = , find: ii. neither breaks the record;
iii. exactly one of them breaks the record.
i. P(Y) ii. P(XUY)

8. Two proof readers, M and N, read the proofs of

2. From the given probabilities, determine
a book, and working independently, correct
whether or not events A and B are independent
misprints. The probability of M detecting a
events:
misprint is P(M) = 0.8 and N is P(N) = 0.75, find:
i. P(A) = , P(B) = and P(A∩B) =
i. the probability that a misprint is detected by
ii. P(A) = , P(B) = and P(A∩B) = both M and N,
ii. the probability that a misprint is detected by
3. The probability that Mugu will go to church on neither M nor N,
iii. the probable number of undetected misprints
if M finds 212.
sunday is and that of his brother, Yaaro is .
What is the probability that: 9. Statistics show that the probability of a baby
i. both will go to church? being a boy is 0.55. Find the probability that the
ii. at least one of them will go to church? first three children born in the family include at
iii. exactly one of them will go to church? least two boys.

4. Mrs. Brown is expecting a baby. The 10. Two men, Sugri and Dabre play the game of
probability that the baby will be a boy is and the „oware‟. Sugri usually wins two games to every
probability that the baby will be fair in one won by Dabre. If they play three games, what
complexion is . What is the probability that the is the probability that:
i. Sugri wins all of them?
baby will be fair complexioned boy?
ii. Dabre wins twice and Sugri once?
5. A number is selected at random from the set S
Selection with Replacement
= {1, 2, 3…15}. What is the probability that the
“Selection with replacement” means selecting an
number is less than 10 or even?
item from a lot and putting it back from where it
was taken. The addition and multiplication rules
6. A number is selected from the set of factors of
of probability stated above are used to determine
30. What is the probability that the number is
the probability of A or B (two or more items of
either a prime number or a multiple of 3?
different or the same kind) from a lot with
replacement.
7. The probability that an athlete A breaks a
record of a race is ; the probability that an athlete Generally, if a selection is done with
B breaks the record in the same race is . Find replacement;

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1. The number of items in the various groups b.Let the probability that the first ball is white be
does not change. P( and the probability that the second ball is
2. The total number of items in the lot (Sample white be P(
space) does not change. P( = P( (

Worked Examples But P( = , and P( =

1. A bag contains 4 red balls and 6 black balls. A P( = P( (
ball is selected from the bag after which it is = × =
replaced and a second ball selected and replaced.
What is the probability that both balls are red? c. Let the probability that the first ball is black be
P( and the probability that the second ball is
Solution black be P(
Both balls are red means, the first ball is red and P( = P( (
the second ball is also red;
But P( = , and P( =
S = 4red balls and 6black balls,
n(S) = 10 balls P( = P( (
n(R) = 4 and n(B) = 6 = × =
Let P(R) represents the probability of selecting a
red ball; P( or P(
P(R∩R) = P(R) × P(R) = P( ( + P( (
= × = = =( )+( )
= +
2. A box contains 5 white balls, 3 black balls and
2 red balls of the same size. A ball is selected at
random from the bag and replaced. Find the Exercises 20.11
probability of obtaining: 1. A bag contains 3 red pens and 7 blue pens.
a. two red balls, What is the probability that the first pen drawn is
b. two white balls, red and the second is blue, if the selection is done
c. two white balls or two black balls. with replacement?

Solution 2. A box contains 4 black balls and 8 white balls

a. Let the probability that the first ball is red be of the same size. What is the probability that the
P( and the probability that the second ball is first ball drawn is white and the second ball is
red be P( black if the selection is with replacement?
P( = P( (
3. Assming that a child is equally likely to be a
But P( = , and P( = boy or a girl. A family contains 3 childred. State
P( = P( ( the probability that :
= × = i. they are all boys,
ii. they are all girls,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 562
iii. the family contains both boys and girls. random and is not replaced, a second ball is then
selected at random, find the probability that:
4. If two numbers are selected at randon, one after i. both bulbs are red,
the other, with replacement from the set A = {5, ii. the first ball is green and the second is red.
6, 7, 8, 9}, find the probability of selecting at
least one prime number. Solution
i.“Both balls are red” means, the first ball is red
Solution and the second is also red;
Let P( the probability of the first red ball
5 6 7 8 9 and P( be the probability of the second red
5 5, 5 5, 6 5, 7 5, 8 5, 9 ball;
6 6, 5 6, 6 6, 7 6, 8 6, 9
P( = P( = P( (
7 7, 5 7, 6 7, 7 7, 8 7, 9
8 8, 5 8, 6 8, 7 8, 8 8, 9 But P( = = and P( = =
9 9, 5 9, 6 9, 7 9, 8 9, 9 By substitution,
P( = P( (
E = (5, 5), (5, 6), (5, 7), (5, 8), (5, 9), (6, 5), (6, 7)
(7, 5), (7, 6), (7, 7), (7, 8), (7, 9) (8, 5), (8, 7), (9, = × = =
5), (9, 7)
n(E) = 16 n(S) = 25 ii.Let P( the probability of the first green
ball and P( the probability of the second
red ball;
P(at least one prime number) =
P( = P(
= P( (
Selection without Replacement
But P( = = and P( = =
“Selection without replacement” means selecting
an item from a lot without putting it back from By substitution,
where it was taken. P( = P( (
= × = =
Generally, if a selection is done without
replacement, 2. A box contains 10 marbles, 7 of which are
1. The number of balls or items in the various black and the rest red. Two marbles are drawn,
groups reduces one after the other without replacement. Find the
2. The total number of items in the lot (Sample probability of getting:
space) reduces i. a red and a black marble,
3. The probability of subsequent draws depends ii. two black marbles,
on the previous selection iii. twomarbles of the same color.

Worked examples Solution

1. A box contains 10 red bulbs and 15 green balls i. n(S) = 10
of the same type and size. A bulb is selected at Number of black marbles = 7

Baffour – Ba Series, Core Maths for Schools and Colleges Page 563
Number of red marbles = 10 – 7 = 3 P(R1) =
Let P( the probability of the first red
marble and P( the probability of the Without replacement,
second black marble; n( R) = 5 and n (S) = 23
P(R2) =
P( = P( (
P(R1 and R2) =P(R1 R2)
But P( = and P( = =
= P(R1) × P(R2)
By substitution,
= × = =
P( = P( (

= × = = b. P(R1 and Y2) = P(R1 Y2)

= P(R1) × P(Y2)
ii.Let P( be the probability of the first black = × = =
marble and P( the probability of the
second black marble; 2. Three bags P, Q and R contains red, blue and
P( = P( ( white balls respectively of equal sizes. The raio
But P( = and P( = = of the ball in the bag are P : Q = 2 : 3, and Q : R
By substitution, = 4 : 5. All the balls are removed into a big bag
P( = P( ( and properly mixed together.
= × = = a. Find the probability of picking a:
i. red ball ii. blue ball iii. white ball
b. If two balls are picked at random one after the
iii. Two marbles of the same color;
other with replacement, find the probability of
= P( or P(
picking:
= [P( ( ] + [P( ( ]
i. a white ball and a blue ball
=( )+( ) ii. blue ball first, then a red ball.
= + =
Solution
Some Solved Past Questions a. = and =
1. A bag contains 6 red, 8 black and 10 yellow = × =
identical beads. Two beads are picked at random,
one after the other without replacement. Find the = × =
probability that:
a. both are red, ⇒ P = 8, Q = 12 and R = 15
b. one is black and the other yellow. n(R) = 8, n(B) = 12 and n(W) = 15
n(S) = 8 + 12 + 15 = 35
Solution i. P(R) = , P (B) = and P (W) =
a. n (R) = 6 , n(B) = 8 and n(Y) 10
n(S) = 6 + 8 + 10 = 24

Baffour – Ba Series, Core Maths for Schools and Colleges Page 564
b. i. P(white and blue) ; 5. There are 5 bulbs in a box, 2 of which are
= P(W and B) defective. Two bulbs are selected at random, one
= × = = after the other without replacement. Find the
probability that:
i. both bulbs are defective,
ii. P(Blue and Red) ;
ii. the first is defective and the second is not.
P(B and R) = × =
6. A bag of sweets contains 7 toffees, 3 chocolate
Exercises 20.12 and 5 chewing gums, all wrapped identically.
1. A bag contains 3 red pens and 7 bluepens. Sweets are drawn out one at a time and not
What is the probability that the first pen drawn is replaced. Find the probability that the first:
red and the second pen is blue if selection is done i. drawn is a toffee,
without replacement? ii. drawn is a toffee and the second a chocolate,
iii. and second drawn are both toffees.
2. A box contains 5 red, 4 white and 3 blue balls.
If three balls are drawn from the box one after the 7. Calculate the probability of an event of
other without replacing them, find the probability drawing two red balls from a set of 5 white and 5
that the first ball is red, the second is white and red balls in a bag when the ball drawn first:
the third is blue. i. is replaced before the second draw,
ii. is not replaced before the second draw.
3. A man has 9 identical balls in a bag. Out of
these, 3 are black, 2 are blue and the remaining is 8. Calculate the probability of drawing three
red. prime numbers in succession from a set of 10
i. If a ball is drawn at random, what is the cards numbered 1 – 10, where the drawn cards
probability that it is: are not replaced before the next draw.
a. not blue,b. not red.
ii. If 2 balls are drawn at random one after the 9. A bag contains three green, 2 white and 4 black
other, what is the probability that both of them balls. If I draw two balls in succession, without
will be: replacingthe first, what is the probability that;
a. black, if there is no replacement? I. I have a green and a black ball.
b. blue if there is replacement? ii. I draw two balls of the same color.

4. A bag of mixed toffees contains 20 cream, 15 Probability and Tree Diagrams

milo and 10 nuts. A tree diagram called probability tree diagram,
i. If I select one toffee at random, what is the can be used to solve problems involving
probability that it is a nut? combined or compound events
ii. If the first toffee is not a nut and I eat it, what Steps:
is the probability that the next one I pick I. For each action in the problem, show the
will be a nut? possible outcomes (or events) at the end of a
branch of the tree

Baffour – Ba Series, Core Maths for Schools and Colleges Page 565
II. Each action in the problem is shown by a stage Solution
rd
in the diagram. For example, for two actions, 3 Coin
nd
there will be two actions as shown below: 2 Coin H
st
H
1 Coin T
2 H
H
T
1 T
H
H
T
T
T H

III. At the end of each route along the branches of T

Possible outcomes = {HHH, HHT, HTH, HTT,
the tree, the final outcome is written.
THH, THT, TTH, TTT}
IV. Work out the probability of each outcome
shown by a branch on the diagram. Write each
3. One element is drawn at random from each of
probability on it branch of the tree, not forgetting
the three sets, A = {a, b}, B = {c, d, e}, C = {f,
the fact that probabilities on adjacent branches
g}. Use tree diagram to list all possible outcomes
add up to 1.
V. Determine the probability of the final outcome
Solution f
by finding the route that leads to the final
c g
outcome. Multiply together the probabilities from
f
any branch as you go along.
d g
a f
Worked Examples e g
1. A fair coin is tossed twice. List all the possible f
c g
outcomes using a tree diagram.
f
b d
Solution nd
2 Coin g
H
st
1 Coin Possible outcomes f
e
H = {acf, acg, adf, adg, aef, aeg,
T
bcf, bcg, bdf, bdg, bef, beg} g
H
T 4. There are 5 red pens and 2 red pens in a
T bag. A pen is taken at random from the bag with
replacement. A second pen is then taken from the
Possible outcomes = {HH, HT, TH, TT}
bag. What is the probability that:
i. both pens are red?
2. With the aid of a tree diagram, list all the
ii. both pens are of the same color?
possible outcomes when a coin is tossed three
times

Baffour – Ba Series, Core Maths for Schools and Colleges Page 566
Solution Solution
i. Number of red pens (R) = 5 i.
R
Number of blue pens (B) = 2
Total number of pens = 5 + 2 = 7 B
R
P(R) = and P(B) = nd
2 action G
R R
st
1 action
B
R B
B
G
B R
G
B B

R
G
i. Probability that both marbles are red;
P(R∩R)
The probability that both pens are red;
P(R∩R) = P(R) × P(R) = × = =
= =
ii. Probability that both marbles are of the same
ii. The probability that both pens are of the same color;
color; P(R∩R) or P(B∩B) or P(G∩G)
P(R∩R) or P(B∩B) =( )+( ) +( )
= P(R) × P(R) + P(B) × P(B)
= + + = =
=( )+( )
= + =
Exercises 20.13
5. A bag contains 4 red, 5 blue and 7 green 1. Three coins are tossed once. Use the tree
marbles, which are all identical except for color. diagram to list all the possible outcomes.
A marble is selected at random, the color is noted
and it is not replaced in the bag. A second marble 2. In two independent events, neither of the two
is selected at random from those remaining in the has an influence on the other. A game is played
bag. Find the probability that the marbles selected by rolling a die and then tossing a coin. You win
are: if the die shows 3 or a 4 and the coins shows
i. both red, ii. of the same color ? heads.
i. Use tree diagram to find the sample space.
ii. What is your probability of winning?

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3. Two balls are drawn without replacement from ii. both pens are of the same color.
a bag containing 12 similar balls, 4 red, 2 yellow
and 6 blue. 5. A coin is tossed three times in succession.
i. Represent the possible outcomes on a tree Draw the probability tree diagram for the various
diagram. outcomes and hence write down the probability
ii. From the tree diagram, find the probability of of obtaining ;
selecting two blue balls. i. two heads only,
iii. From the tree diagram, find the probability of ii. one head only,
selecting one red and one yellow ball. iii. no heads.

4. There are 5 red pens and 3 blue pens in a box. 6. A bag contains 5 white balls and 3 black balls.
A pen is chosen at random from the box. Use tree One ball is drawn from the bag and a second is
diagram to find the probability that: then drawn. Find the probability of drawing one
i. both pens are blue, ball of each color.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 568
21 QUADRATIC EQUATIONS Baffour – Ba Series
Describing a Quadratic Equation Factorize the following equations;
Any equation which can be put in the form ax + 2 1. x2 + 6x + 8 = 0 2. x2 − 13x + 36 = 0
bx+ c = 0, where a ≠ 0 is called a quadratic 3. x2 − 2x – 15 = 0 4. x2 + 4x − 12 = 0
equation. In ax2 +bx + c = 0,
a is called the co-efficient of x2, Solutions
bis called the co-efficient of x and 1.x2 + 6x + 8 = 0
cis called the constant term. (x2 + 4x) + (2x + 8) = 0
x(x + 4) +2(x + 4) = 0
Whena = 1, the quadratic equation becomes x2 + (x + 2) (x+4) = 0
bx + c = 0.
For all quadraticequations, the forms;ax2 + bx + c 2.x2 −13x + 36 = 0
= 0 and y = ax2 + bx + c areequivalent equations (x2 − 4x) − (9x + 36) = 0
and as such y = 0 x(x − 4) − 9(x − 4) = 0
(x −4) (x − 9) = 0
Factors of Quadratic Equations
Make use of the knowledge of factorization of 3. x2 −2x − 15 = 0
quadratic expressions to find the factors of a (x2 − 5x) + (3x + 15) = 0
quadratic equation. Make sure the quadratic x(x − 5) + 3(x − 5) = 0
equation is equated to zero to assume its standard (x +3) (x − 5) = 0
form.
i.e. x2 + b x + c = 0. By illustration; 4. x2 − 4x − 12 = 0
x2 - (sum of roots)x + (product of roots)= 0 (x2 − 6x) + (2x − 12) = 0
Thus, the factors of x2 + b x + c = 0 is found as x(x − 6) + 2(x − 6) = 0
follows; (x + 2) (x − 6) = 0

x2 + b x + c = 0… (1) Exercises 21.1`

Find the factors of the following;
m+n m×n 1. x2 + 11x + 18 = 0 2. x2 + 2x – 3 = 0
3. x2 + x – 30 = 0 4. y2 + 2y – 8 = 0
By substitution,
5. 3x2– 13x – 10 = 0 6. 7x2 + 11x – 6 = 0
x2 + mx + nx + mn = 0
(x2 + mx) + (nx + mn) = 0
x(x + m) + n(x + m) = 0 Truth set of a Quadratic Equation
(x + n) (x + m) = 0. The truth set of a quadratic equation is the set of
Thus (x + n) (x + m) are the factors of values that satisfy the equation or makes the
x2 + b x + c = 0 equation true. It is also called the solution set or
the roots of the equation.
Worked Examples

Baffour – Ba Series, Core Maths for Schools and Colleges Page 569
To solve a quadratic equation is to find the values (x + 2) (x – 3) = 0
of x the make the equation true (or equal to zero). x + 2 = 0 or x – 3 = 0
This is done by applying any of the following x = – 2 or x = 3
methods: Truth set = { }
1. Factorization
2. Completing the square 2. Solve x2 − 13x + 36 = 0
3. Quadratic formula
4. The graphical method Solution
x2 – 9x – 4x + 36 = 0
Method of Factorization (x2 – 9x) − (4x + 36) = 0
To solve quadratic equations of the form: x ( x − 9 ) – 4( x − 9) = 0
x2 + b x + c = 0 , (x – 4) (x – 9) = 0
I. Find two factors of the constant term, c, that (x − 4) = 0 or (x − 9) = 0
sum up to the coefficient of x, as illustrated below x = 4 or x = 9
The truth set = {x : x = 4 or x = 9}
x2 + b x + c = 0… (1)
3. Solve 2x2 + 5x − 3 = 0
m+n m×n
Solution
II. Substitute mx + nx = bx in eqn (1) to obtain
2x2 + 5x - 3 = 0
four terms as:
2x2 – x + 6 x - 3 = 0
x2 + mx + nx + mn = 0
(2x2 – x) + (6x - 3) = 0
III. Group the terms and factorize completely as
x(2x – 1) + 3 (2 x – 1) = 0
shown below;
(2 x – 1) = 0 or x = - 3
(x2 + mx) + (nx + mn) = 0
2x = 1 or x = -3
x(x + m) + n(x + m) = 0
(x + n) (x + m) = 0. x= or x = − 3
IV. At this point, the implication is that either x + The truth set = { }
n = 0 or x + m = 0
x = − n or x = − m Solving Related Problems
Therefore, the truth set or solution set of the Worked Examples
equation is { }
1. Write the equation x – 11 + = 0 in the form

Worked Examples ax2 + bx + c = 0, and hence find the truth set

1. Find the truth set of x2 – x − 6 = 0
Solution
Solution x – 11 + =0
x2 – x − 6 = 0, Multiply through by x
x2+ 2x – 3x – 6 = 0 (By factorization) x2– 11x + 24 = 0
(x2+ 2x) – (3x – 6) = 0 (x – 3) (x – 8) = 0
x (x + 2) – 3(x + 2) = 0 x – 3 = 0 or x – 8 = 0
Baffour – Ba Series, Core Maths for Schools and Colleges Page 570
x = 3 or x = 8 ( = 25 + +

2. If (p, q) is the truth set of the equation Solution

x2 + 10x = 96, evaluate (p + q) Comparing the statements
( = 25 + + and
Solution ( = + 2ab + , it is seen that:
x2 + 10x = 96 i. = 25
x2 + 10x – 96 = 0 a=√ = 5x
x2 + 16x – 6x – 96 = 0
(x2 + 16x) – (6x – 96) = 0 ii. 2ab = 70x
x(x + 16) – 6( x + 16) = 0 But a = 5x
(x – 6) (x + 16) = 0 2(5x)b = 70x
x – 6 = 0 or x + 16 = 0 10bx = 70x
x = 6 or x = -16 b = 7 so = = 49
Truth set (6, -16) = (p, q) Therefore the complete statement is
⇒( p + q ) = 6 + (-16) = - 10 ( = 25 + + 49

Exercises 21.2 2. Copy and complete the statement

A. Solve the following equations;
( = – +
1. x2 − 3x– 4 = - 6 5. x2 + x – 12 = 0
2. x2 − 6x − 16 = 0 6. x2 + 16x + 63 = 0
3. x2 + 13x + 25 = -5 7. 24 + 11x + x2 = 0 Solution
4. 14 – 5x – x2 = 0 8. 44 + 15x + x2 = 0 Comparing the statements:
( = – + and
B. Solve the following equations; ( = – 2ab + it is seen that
1. x2 + 7x − 5x − 35 = 0 3. 17 − 15x – 2x2 = 0 i. ( =(
2. 2x2 – 13x + 20 = 5 4. 3x2 + 2x – 8 = 0
Therefore a = and a2 =
C. Use factors to solve these equations
1. x2 – 6x+ 5 = 0 2. x2 + 9x + 20 = 0 ii – 2ab = – x, but a =
3. 4x2 + 4x + 1 = 0 3. x ( x – 8 ) = 33
– 2( b=–x
Method of Completing Squares –b=–1
The method of completing the squares depend on b = 1, so =1
the identities: Therefore the complete statement is:
1. ( = + 2ab + ( = – +1
2.( = – 2ab +
3. If m, n∈z, find m and n such that
Worked Examples x2 + 12x + m = (
1. Complete the squares in the following;
Baffour – Ba Series, Core Maths for Schools and Colleges Page 571
Solution
( ) =– c + ( )
Comparing the statements:
x2 + 12x + m = ( and V. Introduce a square root sign on both sides of
2 the equation
a + 2ab + = (
i.x + 12x + m = a2 + 2ab +
2
i.e( )= √ ( )
Therefore a = x
6. Make x the subject of the equation
ii. 2ab = 12x, but a = x x= √ ( )
2bx = 12x
Find the value or values of x that satisfies the
b= =6
equation
But b = n and =m
6 = n and =m Worked Examples
1. Find the truth set of x2 + 8x+ 15 = 0
By substitution,
x2 + 12x + m = ( Solution
x2+ 12x + 36 = ( x2 + 8x + 15 = 0
Therefore m = 36 and n = 6 x2 + 8x = – 15
x2 + 8x + ( ) = – 15 + ( )
Exercises 21.3
1. The truth set of the equation ax2 + bx = 4 is {- x2 + 8x + = – 15 +
3, 2}. Find the numerical values of the constants ( = – 15 + 16
a and b ( =1
x+4= √
Quadratic Equations of the Form: x = – 4 + √ or x = – 4 – √
x2 + b x + c = 0 x = – 4 + 1 or x = – 4 – 1
To complete the squares of quadratic equations of x = – 3 or x = – 5
the form: x2 + bx + c = 0 Truth set = { }
I. Transpose the constant, c,to the right side of
the equation to assume the opposite sign i.e. x2 + 2. Solve x2 + 3x – 28 = 0
bx = – c
II. Divide the coefficient of x (number attached to Solution
x) by 2. i.e. x2 + 3x – 28 = 0
x2 + 3x = 28
III. Square to get ( ) and add ( ) to both
x2 + 3x + ( ) = 28 + ( )
sides of the equation
( ) = 28 +
i.e. x2 + bx + ( ) = – c + ( )
IV. Complete the squares at the left side of the ( ) =
equation to get

Baffour – Ba Series, Core Maths for Schools and Colleges Page 572
x+ = √ = + =–
III. Find of i.e. × =
x+ =
x=– + or x = – – IV. Square to get ( ) and add ( ) to both
sides of the equation
x= or x =
x = 4 or x = - 7 i.e. x2 + +( ) = –c+( )
Truth set = { } V. Complete the squares at the left side of the
equation to get
3. Find the truth set of x2 – 5x – 14 = 0
( ) =–c+( )
VI.Introduce the square root sign on both sides of
Solution
x2 – 5x – 14 = 0 the equation
x2 – 5x = 14 √–
i.e. ( )= ( ) ,
2
x – 5x + ( ) = 14 + ( ) VII. Make x the subject to get the value or values
( ) = 14 + of x that satisfies the equation;

x=– √– ( )
( ) =

x– = √ Worked Examples
x– = 1. Find the truth set of 3x2 + 8x + 5 = 0

x= Solution
x= or – 3x2 + 8x + 5 = 0
3x2 + 8x = – 5
x= or x = –
x2 + =–
x = 7 or x = – 2
Truth set = { } x2 + +( ) = – +( )

Quadratic Equations of the Form: x2 + +( ) = – +( )

ax2 + b x + c = 0, a > 1
x2 + +( ) = – +( )
To complete the squares of quadratic equations of
the form: ax2 + bx + c = 0, ( ) =– +
I. Transpose the constant, c, to the right side of
the equation to assume the opposite sign i.e. ax2 + ( ) =
bx = – c
x+ = √
II. Divide through the equation by a, which is the
coefficient of , i.e. + =– x+ =

Baffour – Ba Series, Core Maths for Schools and Colleges Page 573
x=– Solution
3x2 –2x – 5 = 0
x=– or x = –
3x2 – 2x = 5
x= orx = x2 – =
x=– or x = –
x2– +( ) = +( )
x = – 1 or x = –
x2– +( ) = +( )
Truth set = { – }
x2 – +( ) = +( )
2. Find the truth set of 3x2 + 2x – 5 = 0 ( – ) = +

Solution ( – ) =
3x2 + 2x – 5 = 0
3x2 + 2x = 5 x– = √
x2 + = x– =
x2 + +( ) = +( ) x=

x2 + +( ) = +( ) x= or x =
x= or x =
x2 + +( ) = +( )
x= or x = –
( ) = +
x = or x = – 1
( ) =
Truth set = { – }
x+ = √
Exercises 21.4
x+ = A. Complete the squares in the following;
x=– 1. ( = + x+9
2. ( = – +
x=– or x = –
3. ( = – +
x= or x = 4. ( = – +
x= or x = – 5. ( = – +4
6. ( = – +
x = 1 or x = –
Truth set = { – } B. Solve by completing the square;
1. x2 + 7x – 3 = 0 4. x2 + 9x + 20 = 0
3. Find the truth set of 3x2 –2x – 5 = 0 2. 10 + 3x – 2 = 0 5. x2 + 4x – 21 = 0
3. 4x2 – 6x – 1= 0 6. 3x2 + 12x + 6 = 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 574
The Quadratic Formula Solution
The quadratic formula is derived from the 3x2 + 8x + 5 = 0 and ax2 + bx + c = 0 compared,
quadratic equation ax2 + bx + c = 0 as follows: a = 3, b = 8 and c = 5
In ax2 + bx + c = 0 √
Substitute in x =
I. Subtract c from both sides
√ ( (
ax2 + bx = – c x= (
II. Divide through by a √
x=
+ ( )x = – ( )
√
x=
III. Add ( ) to both sides of the equation
x=
+ ( )x + ( ) = ( ) – ( )
x= or x =
+ ( )x + ( ) = – x= orx =
IV. Factorize the left – hand side
x = – 1 or x = –
( ) =
Truth set = { – }
Take the square root of both sides

x+ = √( ) 2. Find the truth set of 3x2 + 2x – 5 = 0

√(
x+ = Solution
Subtract from both sides 3x2 + 2x – 5 = 0 and ax2 + bx + c = 0
compared, a = 3, b = 2 and c= – 5
√( √
x=– Substitute in x =
So if ax2 + bx + c = 0, then x=
√ ( (
(
√
x= √
x=
This formula is used to solve quadratic equations √
x=
which cannot readily be solved by factorization.
In other words, to solve the quadratic equation x=
ax2 + bx + c = 0 x= or x =
I. Attempt to factorize the quadratic expression
on the left - hand side x= or x =
II. If factorization is not possible, use the x = 1 or x = –
√
formula: x = Truth set = { – }

Worked Examples 3. Find the truth set of 3x2 – 2x – 5 = 0

1. Find the truth set of 3x2 + 8x + 5 = 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 575
Solution Exercises 21.5
3x2 – 2x – 5 = 0 and ax2 + bx + c = 0 A. Solve by using the formula:
compared, a = 3, b = – 2 and c= – 5 1. 3x2 – 7x –1= 0 4. 5x2 – 3x – 7 = 0

Substitute in x =
√ 2. 4 – 2x – 2 =0 5. 3x2 = 7x + 2
( √( ( ( 3. x2 – 6x + 5 = 0 6. x2 – 3x = 0
x= (
√
x= B. Solve by any suitable method;
√ 1. 2x2 + 5x – 12 = 0 3. 3x2 + 14x + 8 = 0
x=
2. 6x2 – x = 2 4. 8x2 = 9 – 6x
x=
C. Find the truth set of the equations:
x= or x =
1. = 4. =
x = or x = – 1
2. = 5. x – 1 = 1 +
Truth set = { }
3. = 6. + =1
3. Find the roots of – to one
decimal place Challenge Problems
Solve the equations, wherex∈ R
Solution
1. ( = 81 2. ( ) =9
–
( ( 3. ( = 4. ( = 64
(
=1
(

=1 Factors of Equations of the Form:

a2 − b2 = 0
=1
To find the factors of equations of the form;a2 −
- 2x – 1 = x2 + x b2 = 0, use the method of difference of two
x2 + x + 2x + 1 = 0 squares as shown below;
x2 + 3x + 1 = 0 a2 − b2 =(a + b) (a – b) = 0. This is the factors of
a2 − b2 = 0
Let a = 1, b = 3 and c = 1 and substitute in the
√
quadratic formula: x = Worked Example
x=
√( ( ( Factorize x2 – 9 = 0
(
√
x= Solution
x=
√
or x =
√ x2 – 9 = 0
x2 – 32 = 0
x = - 0.4 or x = - 2.6
The roots are x = - 0.4 or x = -2.6 (x – 3) (x + 3) = 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 576
Solving Equations of the Form: a2 −b2 = 0 (√2x − √3) (√2x + √3) = 0
To solve equation of the form a2 − b2 = 0 is to (√2x - √3) = 0 or √2x + √3 = 0
find the values of a andb that make the √2x = √3 or √2x = − √3
statement true. This is easily done by using the x=
√
or x =
√
method of difference of two squares which is √ √
√ √
illustrated by the statement: x= or x =
√ √
a2 − b2 = (a + b) (a – b) = 0
= (a + b) = 0 or (a – b) = 0 4. If a2 – b2 = (a + b) (a – b), evaluate
⇒a = − b or a = b. 9.322 – 0.682
The truth set is a = − b or a = b.
Solution
Worked Examples If a2 – b2 = (a + b) (a – b), then
1. Solve x2 – 9 = 0 (9.32)2 – (0.68)2
= (9.32 + 0.68)(9.32 − 0.68)
Solution = (10) (8.64)
x2 – 9 = 0 = 86.4
x2 – 32 = 0
x2 − 32 = ( x + 3 ) ( x – 3) = 0 Exercises 21.6
(x + 3) = 0 or (x – 3) = 0 A. Factorize the following;
x = -3 or x = 3 1. x2 −16 = 0 2. 9x2 = 144
3. x2 = 81 4. 9x2 – 25 = 0
5. 64x2 – 169 = 0 6. 4x2 – 25 = 0
2. Solve 4x2 – 25 = 0
B. Find the truth set :
Solution 1. x2−16 = 0 2. 9x2 = 144
4x2 – 25 = 0 3. x2 = 81 4. 9x2 – 25 = 0
= (2x)2 – (5)2 5. 64x2 – 169 = 0 6. 4x2 – 25 = 0
= (2x + 5) (2x −5) = 0
= (2x + 5) = 0 or (2x −5) = 0 Challenge problems
2x = -5 or 2x = 5 1. Solve 2x2 – 1 = 0
x= or x =
2. Find the truth set of the equation;
Truth set is x = or x = 16 (x + 1)2 − 2 √5 = 0

3) Solve 2x2 = 3 3. Given – = 77 and x + y = 11, find the

values of: i. x – y (Ans7) ii. x and y (Ans9, 2)
Solution
2x2 = 3 4. Factorize the right – hand side of the formula
2x2 – 3 = 0 A = π( . Then calculate A when 3.14 is
(√2x)2 – (√3)2 = 0 used as an approximation for π and

Baffour – Ba Series, Core Maths for Schools and Colleges Page 577
i. x = 5.2, y = 4.8 ii. x = 65, y = 35 ⇒ 2(L + B) = 48
L+B=
Word Problems
L + B = 24cm
In solving word problems involving quadratic
⇒B = 24 – L……….(1)
equations, write the mathematical equation for
the problem and solve it, taking note of the fact
But area of rectangle = L × B = 128cm2
that the problem does not end with solving the
L × B = 128cm2…….(2)
quadratic equation. You should therefore go back
to the word problem and answer the question it
Put eqn (1) into eqn (2)
demands.
⇒L(24 – L) = 128cm2
Worked Examples 24L – L2 = 128
1. The sum of two numbers is 18. The sum of the L2 – 24 L + 128 = 0
squares of the numbers is 194. Find the numbers. L2 − 8L – 16L + 128 = 0
(L2− 8L) – (16L+ 128) = 0
Solution L(L – 8) – 16 (L – 8) = 0
Let x be the number (L – 16) (L– 8) = 0
Then the other number is (18 – x). L− 16 = 0 or L− 8 = 0
Sum of squares = x2 + (18 – x)2. L = 16 or L = 8
But this is given as 194
x2 + (18 – x)2 = 194 When L =16, B = 24 – 16 = 8
x2 + x2 – 36x + 324 = 194 When L = 8, B = 24 – 8 = 16
2x2 − 36x + 130 = 0 Therefore, the length and breadth of the rectangle
x2 − 18x + 65 = 0 are 16cm and 8cm respectively.
x2 – 5x – 13x + 5 = 0
(x2 − 5x) – (13x + 65) = 0 3. The breadth of a rectangle is 1cm less than its
x ( x − 5) – 13 ( x − 5) = 0 length. If the area is 42cm2, find the breadth.
(x −13) (x − 5) = 0
Solution
x − 13 = 0 or x – 5 = 0
Let the breadth be xcm. It means thatthe length is
x = 13 or x = 5
(x +1)cm
Therefore, the numbers are 5 and 13.
But area = length × Breadth
2. A certain rectangle has perimeter of 48cm and 42cm2 = x (x + 1)
area of 128cm2. Find the length and breadth of
42cm2 = x (x + 1)
the rectangle. 41= x2 + x
x2 + x − 42 = 0
Solution x2 + 7 x – 6 x − 42 = 0
Let the length be L cm and breadth be B
(x2 + 7x) – (6 x – 42) = 0
P = 2(L + B) „
x2( x + 7) – 6 ( x + 7) = 0
But P = 48 (x – 6) (x + 7) = 0
Baffour – Ba Series, Core Maths for Schools and Colleges Page 578
x – 6 = 0 or x + 7 = 0 y = 0 , y – 1 = 0 or y – 4 = 0
x = 6 or x = -7 ⇔ y (y – 1) (y – 4) = 0
Since the breadth of a rectangle cannot be ⇔ y(y2 – 5y + 4)= 0
negative, the breadth of the rectangle is 6cm. ⇔y3 – 5y2 + 4z = 0

4. The present ages of a man and his son are 44 6. A man cycles 20km from C to B. If he
and 13 years. How many years ago was the increases his speed by 3km/h, he saves 30
product of their ages 140? minutes on the journey. Find his original speed in
km/h.
Solution
Let the number of years ago be x
Solution
(44 – x ) (13 – x ) = 140
44(13 – x ) – x (13 – x ) = 140 Let x km/h be the original speed
572 – 44x – 13x + x2 = 140 The time taken for the journey at this speed is
x2 – 57x + 572 = 140 hours
x2 – 57x + 572 – 140 = 0 Time taken for the journey at this increased speed
x2 – 57x + 432 = 0
would be = ( ) hours
a = 1, b = -57 and c = 432
√ Thus we have – =
Substitute in x =
20(x + 3) – 20(x) = x (x + 3) Multiply through by x(x + 3)
( √( – ( (
Substitute in x = 20x + 60 – 20x =
(

√ – x2 + 3x – 120 = 0
x= ⇒ a = 1, b = 3 and c = -120
√
√ – Substitute in x =
x= √ ( (
x=
√
x= √
x=
x= √
x=
x= or x =
x=
x = 48 or x = 9
x = -1.5 11.06
Since the ages must be positive number of years,
x = -1.5 11.06 or x = -1.5 11.06
x = 9 years ago.
x = 9.56 or x = -12.56
Therefore the original speed is 9.56km/h
5. It is known that a number y is either 0, 1 or 4.
Form an open sentence or equation in which y is
Exercises 21.7
the variable.
1. The sum of a number and its square is 6. Find
the number.
Solution
2. The square of a number is 17 times that
y = 0 or y = 1 or y = 4
number, what is the number?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 579
3. The area of a rectangle is 84cm2. If the length received four more shirts. Calcualte the number
is 5cm greater than the width, find the length of of shirts bought.
the rectangle.
4. A farmer encloses a rectangular piece of land Solution
of area 2,500cm2 with fencing of total length Let x be the cost of a shirt and y be the number of
250m. Find the breadth of the rectangular land. shirts bought.
xy = 720…………………..(1)
5. Asaase is 7 years younger than Atikopo. If the
product of their ages is 78, find Asaase‟s age. If each shirt was Ghȼ2.00 cheaper he would have
6. 18 added to the square of a number is equal to received four more shirts.
11 eleven timesthe number. Find the number. (x – 2) (y + 4) = 720……..(2)

7. My brother‟s age is 7 less than mine. If I From eqn (1);

square his age and subtract one, I get the same x=
results as when I multiply my age by 5. How old
am I?
Put x = in eqn (2)
8. Squaring a number gives the same result as
multiplying it by 10 and adding 24 to the result. ( – ) (y + 4) = 720
Find the possible values for the number. (Ans 12, -2 ) +
(
– 2y – 8 = 720
Challenge Problems 720 + – 2y – 8 = 720
1. A number on base ten has two digits, the
– 2y – 8 = 720 – 720
second one being two more than the first. The
number itself is 22 more than the product of it‟s – 2y – 8 = 0 (multiply through by y )
digits. Find the number.
2,880 – 2y2 – 8y = 0
2y2 + 8y – 2880 = 0
2. Due to a fault, the speed of a train was reduced
(y – 36) (y + 40) = 0
by 10km/h over a journey of 100km. The journey
y = 36 or y = - 40 (ignore y = - 40 )
took 20 minutes longer. Find the usual speed of
The number of shirts bought is 36
the train. A

x+8 2. Each term a house master was given

3. What value must be
2x+ 2 Ghȼ21.00 to share equally among his students as
chosenfor x so that < B
pocket money. One term, an extra five students
is a rightangle?
B x C were assigned to his house, but the total pockets
money was unchanged. As a result, each student
Word Problems Type 2 recieved 10p less that term. Find the original
1. A man bought some shirts for Ghȼ720.00. If number of students in his house.
each shirt was Ghȼ2.00 cheaper, he would have

Baffour – Ba Series, Core Maths for Schools and Colleges Page 580
Solution 4. If the price of a radio is reduced by Ghȼ4.00 a
Let x be the number of students and y be the a man can buy 8 more radio sets for Ghȼ192.00.
amount received by each student. Find the original price of the radio.
xy = 2100
y= …………………..(1) 5. A firm employed a number of workers who
were paid minimum daily wage. The total wages
bill for these labourers was Ghȼ36.00. When the
Extra 5 student assigned to the house and each
miminum dialy wage was increased by 10p, the
student received 10p less that term
firm reduced the number of labourers by 5 and
y – 10 = …………….(2)
the total wage bill was still Ghȼ36.00. Find the
new minimum daily wage.
Put eqn (1) in eqn (2);
( ) – 10 = (Multiply through by x ) 6. A shop sells only one type of radio. Every
week, there is a total income of about Ghȼ450.00
2100 – 10x = (Multiply through by x + 5 )
from sales of these radio. When the shop had
2100 (x + 5) – 10x (x + 5) = 2100x reduction sales, the owner calculated that if he
2100 x + 10500 – 10x2 – 50x = 2100x reduced the price of the radio by Ghȼ5.00 and
10500 – 10x2 – 50x = 2100x – 2100x sold an extra 5 radio per week, he would obtain
10500 – 10x2 – 50x = 0 an extra Ghȼ50.00 per week. What was the
10x2 + 50x – 10500 = 0 original piece of each radio?
(x – 30) (x + 35) = 0
x = 30 or x = - 35 (ignore x = -35) 7. A piece of copper wire cost Ghȼ240.00. If it
The number of students in the house is 30 was 4 meter longer and price of each meter of
copper wire was Ghȼ3.00 less, the total cost of
Exercises the piece would remain unchanged. Find the
1. A shop keeper buys some books for Gh ȼ80.00 length of the copper wire.
If he had bought four more books for the same
amount, each would have cost Ghȼ1.00 less. Find 8. A group of student went to a restaurant for a
the total number of books he bought . meal. When the bill of Ghȼ175.00 was brought
by a waiter, two of the cheeky ones from the
2. A shop keeper buys some books for Ghȼ group just sneaked off before the bill was paid,
1,200.00. If he had bought ten more books for the which resulted in the payment of extra Ghȼ10 by
same amount, each book would have cost him each remaining student. How many students were
Ghȼ20.00 less. How many books did he buy? in the group as well? Ans 7
Ans 20

Graphical Solution of Quadratic Equations

3. Alice buys a number of books for Ghȼ40.00. If
The graph of ax2 + bx + c = 0, a > 0 is a parabola
he had bougth two more books for the same
of U – shape and the graph of ax2 + bx + c, a < 0
amount, each book would have cost Ghȼ1.00 less.
is a parabola of ∩ – shape. The points at which
How many books did he buy?
the parabola cuts the x – axis is the roots of the

Baffour – Ba Series, Core Maths for Schools and Colleges Page 581
equation or the truth set or the solution set or the the quadratic equation are also called the zeros of
zeros of the equation. the equation.
y

Line of symmetry
From the graph, the roots of a function is
ax2 + bx + c = 0 determine with cognizance to the nature of the
parabola in relation to the x – axis ;
x I. Whether U or ∩ – shaped, the points at which
-m n the parabola cuts the x – axis is the roots or zeros
Minimum point of the equation as shown below.
y

Fig. I
From the diagram above (Fig. I), the intercept on x
the x – axis is -m and n. Therefore the truth set is a
b
x = - m or x = n y

Maximum point The zeros or truth set of the function is x = a or

x=b

x
Line of symmetry

e f
-ax2 + bx + c = 0 a b
x

Fig. II

From the diagram above (Fig. II), the intercept on The zeros or truth set of the function is x = a or x = b
the x – axis is -e and f. Therefore the truth set is
x = e or x = f II. Whether U or ∩ – shaped, if the parabola does
not cut the x – axis, the equation is said
Drawing the Graph of ax2 + bx + c = 0, to have no roots or no zeros as shown below
To draw the graph of ax2 + bx + c = 0, a ≠ 0 I. y y
Prepare a table of values for a given range of
values of x.
x
II. Plot the points (x, y) on a graph sheet, using a x
given or a convenience scale;
III. Join the points to make a free hand sketch of
the required parabola/curve; In both diagrams, the functions have no roots or
no zeros.
The Roots or Zeros of QuadraticEquation
The roots of a quadratic equation is the value(s) Equation of Axes or Line of Symmetry
of x for which the equation, y = 0. The roots of Parabolas can be described as being symmetrical,
meaning that a line can be drawn through a

Baffour – Ba Series, Core Maths for Schools and Colleges Page 582
parabola, to divide it into two equal parts, Worked Examples
creating mirror images of each other. The straight 1. What is the value of x and y at the turning point
line bisecting the parabola is calleda line of of y = 2x2 – 8x + 3
symmetry.
Solution
For all quadratic equations of the form,y = ax2 + In y = 2x2 – 8x + 3, a = 2 and b = – 8
bx +c, a ≠ 0, the line of symmetry has the At the turning point, x = – =–
(
=4
equation, x = – Put x = 4 in y= 2x2 – 8x + 3
Put x = 4 in y = 2x2 – 8x + 3
Likewise, on the graph, if the curve cuts the x – y = 2(4)2 – 8(4) + 3 = 3
axis at say x1 and x2, the equation of the axis of At the turning point of 2x2 – 8x + 3 = 0,
symmetry is calculated as,x = x = 4 and y = 3

The Minimum and Maximum Points Worked Examples

The vertex of a parabola is the point of 1. Draw the graph ofy = x2 – x – 6, for the values
intersection of the line of symmetry and the –3 , for the scales 2cm to 2 units on both
parabola itself. The vertex is the turning point: axes. Find from the graph:
either maximum (highest) or minimum (lowest) i. the equation of the line of symmetry,
point of the parabola. ii. the minimum point of the parabola,
iii. the minimum value of the curve,
When the parabola is U – shaped, it is said to iv. the truth set of x2 – x – 6 = 0
have a minimum or least turning pointand when
it is ∩ – shaped, it is said to have a maximum or Solution
greatest turning point. Method 1
x2 – x – 6 = 0
Values of x and y at the Turning Point Let y = x2 – x – 6
For all equations of the form: When x = -3,y = (-3)2 – (-3) – 6 = 6
y = ax2 + bx + c, a at the turning point, When x = -2, y = (-2)2 – (–2) – 6 = 0
x=– When x = -1, y =(-1)2 – (-1) – 6 = – 4
When x = 0, y = (0)2 – (0) – 6 = – 6
When x = 1, y = (1)2 – (1) – 6 = – 6
To get the value of y at the turning point,
When x = 2, y = (2)2 – (2) – 6 = – 4
substitute x = – in y = ax2 + bx + c When x = 3, y = (3)2 – (3) – 6 = 0
The value of yobtained is called the maximum or Table of values
minimum value, depending on the nature of the
parabola x -3 -2 -1 0 1 2 3 4
y 6 0 -4 -6 -6 -4 0 6

Baffour – Ba Series, Core Maths for Schools and Colleges Page 583
 Method 2
x -3 -2 -1 0 1 2 3 4
9 4 1 0 1 4 9 16
-x -3 -2 -1 -0 -1 -2 -3 -4
-6 -6 -6 -6 -6 -6 -6 -6 -6
y 6 0 -4 -6 -6 -4 0 6

0
-8 -6 -4 -2 0 2 4 6 8
-2

-4

-6

-8

i.Equation of the axis of symmetry,

(x, y) = (0.5, -6.25)
x=– . The minimum point (x, y) = (0.5, - 6.25)
But fromy = x2 – x – 6, a = 1, b = -1
⇒x= - = = 0.5 iii. The minimum value of the curve is the y value
(
of the minimum point. Hence, y = - 6.25
The equation of the axis of equation is x = 0.5
(Shown on the graph)
iv. The truth set of x2 – x – 6 = 0, is the point
where the curve cuts the axis.
ii. To find the minimum point,
From the graph, x = -2 or x = 3.
substitute x = 0.5 in x2 – x – 6 = 0
⇒when x = 0.5, y = (0.5)2 – 0.5 – 6 = - 6.25

2. The following is an incomplete table for, y = 4x2 – 8x – 21 for – 2.0 x 4.0

x -2.0 -1.5 -1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
y 11 -9 -21 -21 -9 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 584
a. Copy and complete the table above. Solution
b. Draw the graph of y = 4x2 – 8x – 21, using a a. In y = 4x2 – 8x – 21
scale of 2cm to 1 unit on x axis and 2 cm to 5
units on the y - axis. When x = - 1.5, y = 4(-1.5)2 – 8(-1.5) – 21 = 0
c. Find from the graph: When x = 0.5, y = 4(0.5)2 – 8(0.5) – 21 = -24
i. the equation of the line of symmetry, When x =1, y = 4(1)2 – 8(1) – 21 = -25
ii. the minimumpoint of the parabola, When x = 1.5, y = 4(1.5)2 – 8(1.5) – 21 = -24
iii. the minimum value of the curve, When x = 2.5, y = 4(2.5)2 – 8(2.5) – 21 = -16
iv. the truth set of y = 4x2 – 8x – 21. When x = 4, y = 4(4)2 – 8(4) – 21 = 11

Completed table of values

x -2.0 -1.5 -1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
y 11 0 -9 -21 -24 -25 -24 -21 -16 -9 0 11

b.
15

10

0
-3 -2 -1 0 1 2 3 4 5
-5

-10

-15

-20

-25

-30

c. i. Equation of the axis of symmetry, ii. The minimum point of the parabola;
x= Substitute x = 1 in y = 4x2 – 8x – 21
When x = 1, y = 4(1)2 – 8(1) – 21 = -25
But from 4x2 – 8x – 21, a = 4 and b = - 8
(x, y) = (1, -25)
x= = = =1 The minimum point (x, y) = (1, -25)OR
(
x = 1 (shown on the graph) Read directly from the graph to obtain;
(x, y) = (1, -25)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 585
iii. The minimum value is the value of y at the iv. The truth set of y = 4x2 – 8x – 21
minimum point. Hence, y = -25 Since the curve cuts the x – axis at the points
-1.5 and 3.5, the truth set is x = -1.5 or x = 3.5

3. Copy and complete the table for the relation y = 3 – 2x – x2, for the interval -5 ≤ x ≤ 3

x -5 -4 -3 -2 -1 0 1 2 3
y -12 0 4

b. Using a scale of 2 cm to 1 unit onx – axis and 2 Solution

cm to 2 units on y – axis, draw the graph of the a. In y = 3 – 2x – x2
relationy = 3 – 2x – x2 When x = -4, y = 3 – 2(-4) – (-4)2 = -5
c. Use your graph to find: When x = -2, y = 3 – 2(-2) – (-2)2 = 3
i. the equation of the axis of symmetry When x =0, y = 3 – 2(0) – (0)2 = 3
ii. the truth set of y = 3 – 2x – x2 When x = 1, y = 3 – 2(1) – (1)2 = 0
iii. themaximum point of the curve When x = 2, y = 3 – 2(2) – (2)2 = -5
iv. the maximum value of the curve When x = 3, y = 3 – 2(3) – (3)2 = -12

Completed table of values

x -5 -4 -3 -2 -1 0 1 2 3
y -12 -5 0 3 4 3 0 -5 -12

b.
6
4
2
0
-6 -5 -4 -3 -2 -1 0 1 2 3 4
-2
-4
-6
-8
-10
-12
-14

Baffour – Ba Series, Core Maths for Schools and Colleges Page 586
c. i. Equation of the axis of symmetry, (x, y) = (-1, 4)
x= The maximum point (x, y) = (-1, 4)
OR read directly from the graph to obtain
But fromy = 3 – 2x – x2, a = -1 and b = - 2
(x, y) = (-1, 4)
x= = - = -1
(
⇒The equation of axis of symmetry is x = -1 iii. The maximum value is the value of y at the
(Shown on the graph) maximum point. Hence, y = 4

ii. To find themaximum point of the parabola, iv. The truth set of y = 3 – 2x – x2
substitute x = -1 in y = 3 – 2x – x2 Since the curve cuts the x – axis at the points;
-3 and 1, the truth set is {x :x = -3 or x = 1}
When x = -1, y = 3 – 2(-1) – (-1)2 = 4

4. Copy and complete the following table of values for the relation y = 4 + 5x – 2x2, for the interval
-3 x 5
x -3 -2 -1 0 1 2 3 4 5
y -14 -3 6 -21

b. Using 2 cm to 1 unit on the x – axis and 2 cm a. In y = 4 + 5x – 2x2

to 5 units on the y – axis, draw the graph of y = 4 When x = - 3, y = 4 + 5(-3) – 2(-3)2 = -29
+ 5x – 2x2 When x = 0, y = 4 + 5(0) – 2(0)2 = 4
When x = 1, y = 4 + 5(1) – 2(1)2 = 7
c. From your graph, find:
i. the value of x for which y is maximum, When x = 3, y = 4 + 5(3) – 2(3)2 = 1
ii. the value of x for which 1 + 5x – 2x2 = 0 When x = 4, y = 4 + 5(4) – 2(4)2 = -8

Solution

Completed table of values

x -3 -2 -1 0 1 2 3 4 5
y -29 -14 -3 4 7 6 4 -8 -21

Baffour – Ba Series, Core Maths for Schools and Colleges Page 587
 10

0
-4 -3 -2 -1 0 1 2 3 4 5 6
-5

-10

-15

-20

-25

-30

c. i. Equation of the axis of symmetry, b. i using a scale of 2cm to 1 unit on x – axes and
x= 2cm to 2 units on y – axis, draw the graph
of y = x2 – 4x + 3
But from y = 4 + 5x – 2x2,a = -2 and b = 5
c. Use your graph to find;
x= = 1.3 i. the equation of the axis of symmetry of the
(
The equation of axis of symmetry is x = 1.3 curve,
(Shown on the graph) ii. the zeros of y = x2 – 4x + 3,
iii. the least point of the curve,
ii. The maximum point of the parabola; iv. the least value of the curve.
Substitute x = 1.25 in y = 4 + 5x – 2x2
When x = 1.25, y = 4 + 5(1.25) – 2(1.25)2 = 7 2. a. Copy and complete the following table for y
The maximum point (x, y) = (1.25, 7) = x2 – x – 2 for – 3 x 4

iii. The maximum value is the value of y at the x -3 -2 -1 0 1 2 3 4

maximum point. Hence, y = 7 y 10 -2

iv. The truth set of y = 4 + 5x – 2x2 ; b. i. using a scale of 2cm to 2units on both axes,
Since the curve cuts the x – axis at the points draw the graph of y = x2 –x – 2
-0.6 and 3.4, the truth set is x = - 0.6 or x = 3.4 c. Use your graph to find:
i. the zeros of y = x2 –x– 2,
Exercises 21.8 ii. the equation of the axis of symmetry,
1. a. Copy and complete the table below for the iii. the minimum point of the curve,
relation y = x2 – 4x + 3 for – 2 x 6 iv. the minimum value of the curve.

x -2 -1 0 1 2 3 4 5 6 3. Copy and complete the following table for y =

y 15 3 2x2 – 3x – 9 for – 2 x 3

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x -2 -1 0 1 2 3 II. Identify the x value of the curve at the extreme
y 5 -9 left as a
III. Substitute the values of a and b obtained from
b. Use the table to draw the graph of y = 2x2 – 3x– the graph in a x b to get the range of values
9, for the scales 2cm : 2 unit on x – axis, and 2 cm of x for which y decreases as x increases.This
: 2 units on x – axis. shown in the diagram below:
c. Use your graph to find; y
i. the equation of the axis of symmetry,
ii. the zeros of y = 2x2 – 3x – 9 , y decreases
as x increases
iii. the least point of the curve, a x< b
iv. the least value of the curve. x
a b c

4. a. Using x values of -2, -1, 0, 0.5, 1, 2, 3,

draw the graph of y = x2 – x using a scale of 2
cm to1 unit on both axis.
b. i. What is the equation of the line about which
the graph is symmetrical? Explanation
ii. What is the minimum value of the curve? Each point of the curve is represented by a pair of
points (x, y). From the extreme left, as the curve
5. a. Copy and complete the table below for the moves downward, the yvalues descend
relation y = 2 – x – x2 in the interval – 5 x 4 representing a decrease inyvalues. At the same
time, its corresponding x values move to the right
x -5 -4 -3 -2 -1 0 1 2 3 4
or ascend representing an increase in the x values.
y 0 0
This occurs until the curve reaches its turning
b. Use your table to draw the graph of point. Within this range (from the extreme left to
y = 2 – x – x2 for the scales 2cm to 1 unit on x – the turning point), we say the y values decreases
axis and 2 cm to 5units on y – axis as the x values increases
c. From the graph, find:
i. the values of x for which y = 2 – x – x2 = 0, B. When the curve is ∩ - shaped
ii. the equation of the axis of symmetry, I. Identify the x value of the turning point of the
iii. the maximum point of the curve, curve as b;
iv. themaximum value of the curve. II. On the x – axis, identify the extreme value of
the curve at the right as c;
Range of Values of x for which y Decreases as III. Substitute the values of b and c obtained from
x Increases the graph in b x c to get the range of values
A. When the curve is ∪ – shaped of x for which y decreases as x increases.
I. Identify the x value of the turning point of the
curve as b This is shown in the diagram below:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 589
 This is shown in the diagram below:
y

y increases
as x increases
b x< c
x
a b c
a b c x
n y decreases
as x increases
b<x c

Explanation Explanation
From the turning point, as the curve moves down From the turning point, as the curve moves up the
the right, the y values descend representing a right, the y values ascend representing an increase
decrease inyvalues. At the same time, its inyvalues. At the same time, its corresponding x
corresponding x values move to the right or values move to the right or ascend representing
ascend representing an increase in the x values. an increase in the x values. This occurs till the
This occurs till theend of the curve at the extreme end of the curve at the extreme right. Within this
right. Within this range (from the turning point to range (from the turning point to the extreme right
the extreme right of the curve), we say the y of the curve), we say the y values increase as the
values decrease as the x values increase. x values increase

Range of Values of x for which y Increases as x B. For ∩ – shaped

Increases I. Identify the x value of the turning point of the
A. For U – shaped curve as b
I. Identify the x value of the turning point of the II. On the x – axis, identify the extreme value of
curve as b the curve at the right as a
II. On the x – axis, identify the extreme value of III. Substitute the values of a and b obtained from
the curve at the right as c the graph in a x b to get the range of values
III. Substitute the values of b and c obtained from of x for which yincreases as x increases.This
the graph in b x < c to get the range of values shown in the diagram below:
of x for which yincreases as x increases.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 590
 y y 4 -8 -6

b. Using a scale of 2 cm to 1 unit on the x – axis

and 2 cm to 2 units on the y – axis, draw

the graph of the relation
c. From your graph, find:
a b c x i. the equation of axis of symmetry of y,
na< x b ii. the turning point of y,
y increases
as xincreases iii. the minimum value of y,
iv. the truth set of x2 – 5x– 2 = 0,
v. the range of values of x for which y decreases
as x increases,
Explanation vi. the range of values of x for which y increases
From the extreme left, as the curve moves up, the as x increases.
y values ascend representing anincrease in y
values. At the same time, its corresponding x Solution
values move to the right or ascend representing In y = x2 – 5x– 2,
an increase in the xvalues. This occurs till the When x = 0, y = (0)2 – 5(0) – 2 = -2
curve reaches its turning point. Within this range When x = 1, y = (1)2 – 5(1) – 2 = -6
(from the extreme left to the turning point), we When x = 3, y = (3)2 – 5(3) – 2 = -8
say the y values increase as the x values When x = 5, y = (5)2 – 5(5) – 2 = -2
increase. When x = 1, y = (6)2 – 5(6) – 2 = 4

Worked Examples Completed table of values

1. a. Copy and complete the following table for
y = x2 – 5x – 2 for -1 x 6 x -1 0 1 2 3 4 5 6
y 4 -2 -6 -8 -8 -6 -2 4
x -1 0 1 2 3 4 5 6

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b.

5
4
3
2
1
0
-2 -1 -1 0 1 2 3 4 5 6 7
-2
-3
-4
-5
-6
-7
-8
-9

c. i. Equation of the axis of symmetry,

x=– .
v. The range of values of x for which y decreases
But from y = x2 – 5x – 2, a = 1, b = -5
as x increases = a x< b.
⇒x=- (
= = 2.5 From the graph, a = -1 and b = 2.5substitute in a
x = 2.5 (show the line x = 2.5 on the graph) x < b = -1 x < 2.5

ii. To find the minimum point of the parabola, vi. The range of values of x for which y increases
substitute x = 2.5 in y = x2 – 5x– 2 as x increases = b<x c.
When x = 2.5, y = (2.5)2 – 5(2.5) – 2 = -8.3 From the graph, b = 2.5 and c = 6, substitute in b
(x, y) = (2.5, -8.3) < x c = 2.5 < x 6
The minimum point (x, y) = (2.5, -8.3) OR read
directly from the graph to obtain (x, y) = (2.5, -8.3) 2. a. Copy and complete the following table of
values for the relation y = 10 + 6x – 3x2
iii. The minimum value is the value of y at the for -3 x 5
minimum point. Hence, the minimum value is
x -3 -2 -1 0 1 2 3 4 5
y = -8.3
y 10 13 1 -14

iv. The truth set of y = x2 – 5x – 2 ;

b. Using a scale of 2 cm to 1 unit on the x – axis
Since the curve cuts the x – axis at the points –
and 2 cm to 5units on the y – axis, draw the graph
0.2 and 5.4, the truth set is {x : x = - 0.2 or x =
of the relation.
5.2}
Baffour – Ba Series, Core Maths for Schools and Colleges Page 592
c. Find the following from the graph: c. i. The curve cuts the x – axis at x = -1.1 and
i. the truth set of 10 + 6x – 3x2 = 0, x = 3.1. Therefore, the truth set is
ii. the equation of the axis of symmetry, {x : x = -1.1 or x = -3.1}
iii.the turning point of y = 10 + 6x – 3x2.
iv. the maximum value of y = 10 + 6x – 3x2 ii. From y = 10 + 6x – 3x2, equation of the axis of
v. the range of values of x for which y decreases symmetry, x = = = =1
(
as x increases;
vi. the range of values of x for which y increases iii. Put x = 1 in y = 10 + 6x – 3x2
as x increases. When x = 1, y = 10 + 6(1) – 3(1)2 = 13.
The maximum point, (x, y) = (1, 13)
Solution
a. y = 10 + 6x – 3x2 iv. The maximum value is the value of y at the
When x = -3, y = 10 + 6(-3) – 3(-3)2 = -35 maximum point. Therefore, y = 13.
When x = -2, y = 10 + 6(-2) – 3(-2)2 = -14
When x = -1, y = 10 + 6(-1) – 3(-1)2 = 1 v. The range of values of x for which y decreases
When x = 2, y = 10 + 6(2) – 3(2)2 =10 as x increases = 1 x < 5.
When x = 5, y = 10 + 6(5) – 3(5)2 = -35
vi. The range of values of x for which y increases
Completed table of values as x increases = -3 < x 1.

x -3 -2 -1 0 1 2 3 4 5
y -35 -14 1 10 13 10 1 -14 -35

20
15
10
5
0
-4 -3 -2 -1 0 1 2 3 4 5 6
-5
-10
-15
-20
-25
-30
-35
-40

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3. The following is an incomplete table for the relation y = 2x (4x – 7) – 9, where – 2 x 3

x -2 -1 0 0.5 1 1.5 2 3 4
y 51 -9 -14 63

a. Copy and complete the table Solution

b. Using a scale of 2cm to 1 unit on the x – axis a.y = 2x (4x – 7) – 9
and 1 cm to 5 units on the y – axis, draw the y = 8x2 – 14x – 9
graph of the relation When x = -1, y = 8(-1)2– 14(-1) – 9 = 13
c. Estimate, correct to one decimal place; When x = 1, y = 8(1)2– 14(1) – 9 = -15
i. the x – coordinates of the point where y starts When x = 1.5, y = 8(1.5)2–14(1.5) – 9 = -12
increasing with respect to x When x = 2, y = 8(2)2– 14(2) – 9 = -5
ii. the values of x for which 2x(4x – 7) + 4 = 0 When x = 3, y = 8(3)2– 14(3) – 9 = 21

x -2 -1 0 0.5 1 1.5 2 3 4
y 51 13 -9 -14 -15 -12 -5 21 63

b.

C.i. The x – coordinates of the point where ystarts The curve cuts the x – axis at x = - 0.5 and x =
increasing with respect to x= 0.9 x < 4 2.2. Therefore, {x :x = 0.5 or x = 2.2}

ii. Values of x for which 2x(4x – 7) + 4 = 0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 594
 Range of Values of x for which y is Negative or The values of x at the left of a and at the right
y<0 bproduce negative values of y respectively.
a. When the curve is U – shaped Therefore, the values of x for which y < 0 or y is
y negative for an – shaped parabola is x < a and x
> b and the range is a > x > b or b < x < a
Positive area Positive area
Worked Examples
x 1. Copy and complete the following table for the
a Negative area b
relation y = ( (

x -3.5 -3 -2 -1 -0.5 0 1 2
y 6 3.5 0 6

Explanation
b. Using a scale of 2 cm to 1 unit on each axis,
From the diagram above, when y = 0, x = a or x =
b. This means that y < 0 or „y is negative‟ is draw the graph of the relation y =
the region below the x – axis. ( ( for -3.5 x 2
The values of a to b and b to a on the x axis c. Use your graph to find;
produce negative values of y respectively. i. the value of x for which y is least,
Therefore, the values of x for which y < 0 or y is ii. the truth set of ( ( = 4,
negative for a ∪ – shaped parabola is x > a and x iii. the values of x for which y is negative.
< b and the range is a < x < b or b > x > a
Solution
b. When the curve is ∩ - shaped
In y = ( (
y When x = -1,
y= [ ( ][( ]= -1.5
When x = -0.5,
Positive area
x y= [ ( ][( ]= -1.5
a b
Negative area Negative area When x = 0,
y= [ ( ][( ]= -1
When x = 1,
Explanation y= [ ( ][( ]= 1.5
From the diagram above, when y = 0, x = a orx =
b. This means that y < 0 or y is negative is the x -3.5 -3 -2 -1 -0.5 0 1 2
region below the x – axis, y 6 3.5 0 -1.5 -1.5 -1 1.5 6

Baffour – Ba Series, Core Maths for Schools and Colleges Page 595
 7

0
-4 -3 -2 -1 0 1 2 3
-1

-2

-3

c. From the graph c. From your graph, find:

i. the value of x for which y is least = -0.8 i. the equation of the line of symmetry of the
curve
ii. the truth set of ( ( =0 ii. the values of x for which the relation:
y = 3 – 2x – x2 is less than zero
= {x :x = -2 or x = 0.5}

Solution
iii. the values of x for which y is negative is
a. From y = 3 – 2x – x2
x > -2 and x < 0.5
When x = - 4, y = 3 – 2(-4) – (-4)2 = -5
Range = -2 < x < 0.5
When x = - 2, y = 3 – 2(-2) – (-2)2 =3
When x = - 1, y = 3 – 2(-1) – (-1)2 = 4
2. Copy and complete the table of values for the
When x = 1, y = 3 – 2(1) – (1)2 = 0
relation y = 3 – 2x – x2 for the interval – 5 x 3
When x = 2, y = 3 – 2(2) – (2)2 = -5
x -5 -4 -3 -2 -1 0 1 2 3
y -12 0 3 -12 x -5 -4 -3 -2 -1 0 1 2 3
y -12 -5 0 3 4 3 0 -5 -12

b. Using a scale of 2cm to 1 unit on the x – axis

and 2 cm to 2 units on the y – axis, draw the
graph of the relation

Baffour – Ba Series, Core Maths for Schools and Colleges Page 596
b.
6

0
-6 -5 -4 -3 -2 -1 0 1 2 3 4
-2

-4

-6

-8

-10

-12

-14

x = b. This means that y > 0 or y is positive is the

ii. The values of x for which the relation, region above the x – axis,
3 – 2x – x2 is less than zero is x < -3 and x > 1
Range = -3 > x >1 The values of x at the left of aand at the right b
produce positive values of yrespectively.
Range of Values of x for which y is Positive or Therefore, the values of x for which y > 0 or y is
y> 0 positive for an ∩ – shaped parabola is x < a and x
a. When the curve is U – shaped > b and the range is b < x < a.
y
b. When the curve is ∩ - shaped

Positive area Positive area y

x
a Negative area b

Positive area
a b x
Negative area Negative area
Explanation
From the diagram above, when y = 0, x = a or

Baffour – Ba Series, Core Maths for Schools and Colleges Page 597
Explanation y= ( ( for -3.5 ≤ x ≤ 2
From the diagram above, when y = 0, x = a or x =
c. Use your graph to find:
b. This means that y > 0 or „y is positive‟ is the
i. the value of x for which y is least,
region above the x – axis.
ii. the truth set of ( ( = 4,
The values of a to b and b to a on the x axis
produce negative values of yrespectively. iii. the values of x for which y is positive.
Therefore, the values of x for which y > 0 or y is
positive for an ∩ – shaped parabola is x > a and x Solution
< b and the range is a < x < b In y = ( (
When x = -1,
Worked Examples
y= [ ( ][( ]= -1.5
1. Copy and complete the following table for the
When x = -0.5,
relation y = ( (
y= [ ( ][( ]= -1.5
When x = 0,
x -3.5 -3 -2 -1 -0.5 0 1 2
y 6 3.5 0 6 y= [ ( ][( ]= -1
When x = 1,
b. Using a scale of 2 cm to 1 unit on each axis, y= [ ( ][( ]= 1.5
draw the graph of the relation:

x -3.5 -3 -2 -1 -0.5 0 1 2
y 6 3.5 0 -1.5 -1.5 -1 1.5 6

0
-4 -3 -2 -1 0 1 2 3
-1

-2

-3

Baffour – Ba Series, Core Maths for Schools and Colleges Page 598
c. From the graph: b. Using a scale of 2cm to 1 unit on the x – axis
i. the value of x for which y is least = -0.8 and 2 cm to 2 units on the y – axis, draw the
graph of the relation.
ii. the truth set of ( ( =0 c. From your graph, find:
= {x :x = -2 or x = 0.5} i. the equation of the line of symmetry of the
curve,
iii. the values of x for which y is positive is ii. the values of x for which the relation,
x< -2 and x > 0.5 3 – 2x – x2 is greater than zero.
Range = -2 > x > 0.5 OR 0.5 < x < -2
Solution
2. Copy and complete the table of values for the a. From y = 3 – 2x – x2
relation y = 3 – 2x – x2 for the interval When x = - 4, y = 3 – 2(-4) – (-4)2 = -5
– 5 x 3. When x = - 2, y = 3 – 2(-2) – (-2)2 =3
When x = - 1, y = 3 – 2(-1) – (-1)2 = 4
x -5 -4 -3 -2 -1 0 1 2 3 When x = 1, y = 3 – 2(1) – (1)2 = 0
y -12 0 3 -12
When x = 2, y = 3 – 2(2) – (2)2 = -5

x -5 -4 -3 -2 -1 0 1 2 3
y -12 -5 0 3 4 3 0 -5 -12

b.
6

0
-6 -5 -4 -3 -2 -1 0 1 2 3 4
-2

-4

-6

-8

-10

-12

-14

Baffour – Ba Series, Core Maths for Schools and Colleges Page 599
ii. The values of x for which the relation y = 3 – iv. the range of values of x for which y increases
2x – x2 is greater than zero is x > -3 and x < 1 as x increases.
Range = -3 < x <1
4. a.Construct a table of values for y = 3 + 5x –
Exercises 21.9 2x2 in the interval - 4 x 5
1. Copy and complete the table of values for the b. i. Draw the graph of y = 3 + 5x – 2x2 using the
relation y = x2 – 2x - 3 for the interval scale 2cm to 1 unit on x – axis and 2cm to 5 units
–2 x 4 on y – axis.
c. Find:
x -2 -1.5 -1 0 1 2 2.5 3 3.5 4
y -5 0 -3 0
i. the range of values of x for which y decreases
as x increases.
b. Using a scale of 2cm to 1 unit on both axes, ii. the range of values of x for which y increases
draw the graph of the relation as x increases.
c. Use your graph to find: iii. the values of x for which y is positive
i. the equation of the line of symmetry of the iv. the values of x for which y is negative
curve,
ii. the solution set ofx2 – 2x - 3 = 0, 5. The following is an incomplete table for the
iii. the range of values of x for which y is relation y = (x+ 1) (3 – x) for – 3.0 x 5.0
negative.
x -2 -1 0 1 1.5 2 2.5 3 4 5
y 0 3 1.75
2. Construct a table of values and draw the
graph of y = x2 + 2x + 3 for the interval -5 x 3,
a. Copy and complete the table.
using a scale of 2cm to 1 unit on x – axis and 2
b. Taking 2cm as 1 unit on x – axis and 2cm to 2
cm to 2 units on y axis
units on y – axis , draw the graph of the relation
b. From the graph, find:
for the given interval.
i. the axis of symmetry,
c. Use the graph to find;
ii. the least value of the curve,
iii. the range of values of x for which y decreases
i. the truth set of ( x + 1) (3 – x) = 0,
as x increases,
ii. the greatest value of y,
iv. the range of values of x for which y increases
iii. the value of x for which y is the greatest
as x increases.
iv. in the given range, the value of x for which
y decreases as x increases
3. a. Draw the graph of y = x2 + 2x – 3 for the
values of x from -5 to 3 for the scales 2cm :1 unit
Solving Quadratic and a Linear Equation
on x – axis and 2cm : 2 units on y – axis
Simultaneously on the Same Graph
b. From the graph, find:
To solve a quadratic and a linear equation on the
i. the equation of the axis of symmetry,
same graph:
ii. the minimum point of the curve,
I. Prepare or copy and complete the table of
iii. the range of values of x for which y decreases
values for each of the equations.
as x increases,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 600
II. Draw the graph of the curve (quadratic y
equation) and the straight line (linear equation)
on the same graph sheet.

y y2

y1
x1 x
x2

x
Worked Examples
III. Identify the point of intersection of the curve 1. a. Copy and complete the following table for y
and the straight line as the solution set of the two = 3 + 2x – x2 for the interval – 2 x 4
equations as shown in the diagrams below:

x -2 -1 0 1 1.5 2 2.5 3 3.5 4

y 0 3 1.75 -5

b. Taking 2cm as 1 unit on both axes draw the a. In y = 3 + 2x – x2

graph of the relation for the given interval When x = - 2, y = 3 + 2(-2) – (-2)2 = -5
When x = 1, y = 3 + 2(1) – (1)2 = 4
c. Draw on the same axes, the graph of x – y = 0 When x = 1.5, y = 3 + 2(1.5) – (1.5)2 = 3.75
d. Using your graphs solve 3+ 2x – x2 = x When x = 2, y = 3 + 2(2) – (2)2 = 3
When x = 3, y = 3 + 2(3) – (3)2 = 0
Solution When x = 3.5, y = 3 + 2(3.5) – (3.5)2 = -2.25

Completed table of values for y = 3 + 2x – x2

x -2 -1 0 1 1.5 2 2.5 3 3.5 4

y -5 0 3 4 3.75 3 1.75 0 -2.25 -5

From the relation x – y = 0, x = y

Completed table of values for x - y = 0

x -2 -1 0 1 1.5 2 2.5 3 3.5 4

y -2 -1 0 1 1.5 2 2.5 3 3.5 4

Baffour – Ba Series, Core Maths for Schools and Colleges Page 601
 5
4
3
2
1
0
-3 -2 -1 -1 0 1 2 3 4 5

-2
-3
-4
-5
-6

d. From the graph the truth set of the equation 3 + b. Use your graph to find;
2x – x2 = x is {x : x = 2.3 or x = -1.3} i. the greatest value of 1 + 6x – x2 = 0
ii. the truth set of the simultaneous equation y = 1
2. a. Using a scale of 2cm to represent 1 unit on + 6x – x2 and x + y = 11
the x – axis and 2cm to 2 units on the y –
axis, draw the graph of the following relations for Solution
the interval 0 x 6 a. i. Table of values fory = 1 + 6x – x2
i. y = 1 + 6x – x2
ii. y = 11 – x
x 0 1 2 3 4 5 6
1 1 1 1 1 1 1 1
6x 0 6 12 18 24 30 36
-x2 0 -1 -4 -9 -16 -25 -36
y 1 6 9 10 9 6 1

ii. Table of values forx + y = 11

⇒y = 11 – x
x 0 1 2 3 4 5 6
11 11 11 11 11 11 11 11
-x -0 -1 -2 -3 -4 -5 -6
y 11 10 9 8 7 6 5

Baffour – Ba Series, Core Maths for Schools and Colleges Page 602
 12

10

0
0 1 2 3 4 5 6 7

b. i. From the graph, the greatest value of y is 10. i. solve the equation 3 + 2x – x2 = x,
ii. The truth set of the simultaneous equation; ii. find the values of x for which 3 + 2x – x2 = 2.
y = 1 + 6x – x2 and x + y = 11 is x = 2 or x = 5.
3.a. Draw a table of values for the relation:
Exercises 21.10 (x – 1) (x – 1) = 6 for - 4 x 7
A. 1. On the same set of axes and taking values of b. Using a scale of 2 cm to represent 2 units on x
x from – 4 to 4, draw the graphs of y = x2 – 2 and – axis and 2cm to 5 units on yaxis, draw the
y = 4 – x, using a scale of 2cm to 1 unit on x – graphs;
axis and 2 cm to 2 units on y axis . Write down i. (x – 1) (x – 1) = 6,
the coordinates of the two positions where the ii. x + y = 8 within the same interval.
graphs cross.
c. Use your graphs to find, correct to two
2. 1. a. Copy and complete the following table for significant figures, the truth set of the
y = 3 + 2x – x2 for the interval – 2 x 4 simultaneous equations (x – 1) (x – 1) = 6 and
x+y=8
x -2 -1 0 1 1.5 2 2.5 3 3.5 4
0 3 0
B. Using a suitable scale, solve the following
pair of equations graphically, for the range
b. Taking 2cm as 1 unit on x – axis and 2 cm to 2 - 4 x 4.
units on y - axis, draw the graph of the relation 1. y = x2 – 5x 2. y = 6x - x2
for the given interval. y=x+4 y=x–2
c. Draw on the same axes, the graph of x – y= 0 2
3. y = x + 4x – 5x 4. y = x2 + x – 1
d. Using your graphs to;
y=x+6 y = 2x

Baffour – Ba Series, Core Maths for Schools and Colleges Page 603
5. y = x2 – 8x - 7 6. x2 – y – 5 = 0 range -3 x 5
2y = x – 17 4x + 3y + 6 = 0
x -3 -2 -1 0 1 2 3 4 5
Using the Graph of One equation to find the y 10 13 1 -14
Truth Set of Related Equations
I. Name the equation of the curve as eqn (1) b. Using a scale of 2cm to1 unit on the x – axis
II. Name the given or related equation as equation and 2 cm to 5 units on the y – axis, draw the
(2) graph of the relation for the given intervals.
III. Ensure the two given equations have the same c. From the graph, find the truth set of the
coefficients of x and x2 and each is equated to equations;
zero. i. 10 + 6x – 3x2 = 0
IV. Subtract equation (2) from equation (1) to ii. 5 + 2x – x2 = 0
obtain the value of the line y.
V. Draw line y to touch the curve. Solution
VI. Identify the value(s) of x at which line y a. y = 10 + 6x – 3x2
intersects the curve as the truth set of the When x = -3, y = 10+ 6(-3) – 3(-3)2 = -35
related equation or truth set of eqn (2)
When x = -2, y = 10+ 6(-2) – 3(-2)2 = -14

Worked Examples When x = -1, y = 10+ 6(-1) – 3(-1)2 = 1

1. a. Copy and complete the table of values below When x = 2, y = 10+ 6(2) – 3(2)2 =10
for the relation y = 10 + 6x – 3x2 for the When x = 5, y = 10+ 6(5) – 3(5)2 = -35

Completed table of values

x -3 -2 -1 0 1 2 3 4 5
y -35 -14 1 10 13 10 1 -14 -35

b.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 604
 20
15
10
5
0
-4 -3 -2 -1 0 1 2 3 4 5 6
-5
-10
-15
-20
-25
-30
-35
-40

c. i. The curve cuts the x – axis at x = -1.1 and -3 x 5

x = 3.1. Therefore, the truth set is
{x :x = -1.1 or x = -3.1} x -3 -2 -1 0 1 2 3 4 5
y -29 -3 7 -8
ii. From 10 + 6x – 3x2 = 0
y = 10 + 6x – 3x2 ……………(1) b. Using 2 cm to 1 unit on the x – axis and 2 cm
to 5 units on the y – axis, draw the graph of
From 5 + 2x – x2 = 0 y = 4 + 5x – 2x2
y = 5 + 2x – x2 ………………(2) c. From your graph, find the values of x for which
3 × eqn (2) 1 + 5x – 2x2 = 0
y = 15 + 6x – 3x2………….....(3)
Solution
eqn (1) – eqn (3) In y = 4 + 5x – 2x2,
y = -5 (shown on the graph) Whenx = -2, y = 4 + 5(-2) – 2(-2)2= -14
From the graph, when y = -5, x = -1.5 or x= 3.5. Whenx = 0, y = 4 + 5(0) – 2(0)2 = 4
Therefore the truth set of y = 5 + 2x – x2 is Whenx = 2, y = 4 + 5(2) – 2(2)2 = 6
{x : x = -1.5 or x = 3.5}
Whenx = 3, y = 4 + 5(3) – 2(3)2 = 1
2. Copy and complete the following table of Whenx = 5, y = 4 + 5(5) – 2(5)2 = -21
values for the relation y = 4 + 5x – 2x2 for

Baffour – Ba Series, Core Maths for Schools and Colleges Page 605
 x -3 -2 -1 0 1 2 3 4 5
y -29 -14 -3 4 7 6 1 -8 -21

b.

c. From 4 + 5x – 2x2 = 0
y = 4 + 5x – 2x2………eqn (1)
From 1 + 5x – 2x2 = 0
y = 1 + 5x – 2x2……….eqn (2)

eqn (1) – eqn (2)

y = (4 – 1) + (5x – 5x) – 2x2 – ( -2x2)
y=3
From the graph, when y = 3, x= – 0.2 or x = 2.7.
Therefore, the values of x for which 1 + 5x – 2x2 = 0 is x = – 0.2 or x = 2.7

Exercises 21.11
1. a. Copy and complete the table of values below for the relation y = x2 – 4x – 21

x -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
y 11 0 -21 -24 -21 -16 -9 0 11

Baffour – Ba Series, Core Maths for Schools and Colleges Page 606
b. Using a scale of 2 cm to 2 units on the x – axis and 2 cm to 5 units on the y – axis, draw the graph of
the relation
c. Use your graph to find:
i. the truth set of x2 – 4x – 16 = 0
ii. the range of values of x for which y > 0

2. a. Copy and complete the following table for y = 7 + 4x – 3x2, for the interval -3 x 4

x -3 -2 -1 0 0.5 1 1.5 2 2.5 3 3.5 4

y -32 7 8.25 3 -8 -15.75

b. Taking 2cm as 1 unit on x – axis and 2 cm to 5 units on the y – axis, draw the graph of y = 7 + 4x –
3x2, for the given interval.
c. Draw on the same axes, the graph of the relation y + 2x + 2 = 0
d. Using your graphs:
i. solve the equation 9 + 6x – 3x2 = 0
ii. find the values of x for which 7 + 4x – 3x2 = 0
iii. the range of values of x for which y = 7 + 4x – 3x2 is negative.

3.The following is an incomplete table for the b. Using a scale of 2 cm to 1 unit on the x – axis
relation y = 2x (4x – 7) – 9, where -2 x 4 and 2 cm to 2 units on the y – axis, draw on the
same graph sheet, the graphs of the relations:
x -2 -1 0 0.5 1 1.5 2 3 4 i. y = 4 + 3x – x2,
y 51 -9 63 ii. 2y + 3x = 6.
c. Use your graph to find:
a. Copy and complete the table. i. 4 + 3x – x2 = 0,
b. Using a scale of 2cm to 1 unit on the x – axis
ii. 4 + 3x – x2 = 3 – x,
and 2 cm : 10cm on the y – axis, draw the graph
iii. the maximum value of y = 4 + 3x – x2.
of the relation.
c. Estimate, correct to one decimal place;
i. the x – coordinates of the point starts increasing 5. a. Copy and complete the following table of
with respect to y, values for the relation y = x2 – 2x + 5 for the
ii. the truth set of 2x (4x – 7) + 4 = 0. interval – 2 x 5

x -2 -1 0 1 2 3 4 5
4. Copy and complete the following table for y =
y 8 8 2
4 + 3x – x2, for the interval -2 x 5
b. Using a scale of 2cm to 1 unit on the x – axes
x -2 -1 0 1 2 3 4 5 and 2 cm to 2 units on the y – axis, draw the
y -6 4 0 graph of the relation for the interval – 2 x 5

Baffour – Ba Series, Core Maths for Schools and Colleges Page 607
c. Use your graph to determine: ⇒- 6 = (1)2+ m(1) + n
i. the truth set of x2 – 2x – 2 = 0 - 6 = 1 +m + n……………(1)
ii. the coordinates of the lowest point on the
graph. Whenx = 4.5, y = 2.8
iii. the range of values of x for which y increases Substitute in y = x2 + mx + n
as x increases. ⇒2.8 = (4.5)2+ m(4.5) + n
2.8 = 20.25 + 4.5m + n……………(2)
6. Draw the graph of y = x2 - 2x + 2 for the
interval -2 , using a suitable scale. Use
your graph to find the solution of the equation: eqn (2) – eqn (1)
x2 – 6x + 8 = 0 2.8 – (- 6) = (20.25 – 1) + (4.5m – m) + (n – n)
8.8 = 19.25 + 3.5m
2
7. Draw the graph of y = - 1 + 3x – x for the 8.8 – 19.25 = 3.5m
interval - 4 , using a suitable scale. Use -10.45 = 3.5m
your graph to find the solution of the equation – m= = -3
3+ 4x – x2 = 0
Put m = -3 into eqn (1)
Other Applications
- 6 = 1– 3 + n
Worked Examples
-6–1+3=n
1. Below is an incomplete table of values for the
-4=n
relation y = x2+ mx + n, where m and n are
⇒n = - 4
constants for the interval -1.5 x 4.5

x -1.5 -1.0 0 1 2 3 4 4.5 Put m = -3 and n = -4 in y = x2+ mx + n

y -6 2.8 ⇒ y = x2+ (-3)x + (-4)
y = x2 –3x – 4
Use the table to find the values of m and n and The required equation is y = x2 –3x – 4
use it to complete the table
b.Using a scale of 2 cm to 1 unit on both axes, When x = -1.5, y = (-1.5)2 –3(-1.5) – 4 = 2.75
draw the graph of the relation for the given When x = -1, y = (-1)2 –3(-1)– 4 = 0
intervals. When x = 0, y = (0)2 –3(0) – 4 = - 4
When x = 2, y = (2)2 –3(2) – 4 = - 6
Solution When x = 3, y = (3)2 –3(3) – 4 = - 4
From the table, when x = 1, y = -6 When x = 4, y = (4)2 –3(4) – 4 = 0
Substitute in y = x2+ mx + n

Completed table of values

x -1.5 -1.0 0 1 2 3 4 4.5

y 2.8 -1 0 -6 -6 -4 0 2.8

Baffour – Ba Series, Core Maths for Schools and Colleges Page 608
 4
3
2
1
0
-2 -1 -1 0 1 2 3 4 5

-2
-3
-4
-5
-6
-7

2. The roots of a quadratic equation mx2 + nx+ r = Solution

0 are and -3 a.x2– (sum of roots)x + (product of roots) = 0

a. Find the values of the constants m, n and r x = or x = -3

b.Using a scale of 2 cm to 1 unit on x – axes and x2– ( )x + ( )=0
2 cm to 5 units on y - axis, draw the graph of the
relation y = mx2 + nx + r for the intervals -5 x x2– ( )x + ( )=0
4 x2+ x – =0
c. Use your graph to find the truth set of 2x2+ 5x – 3 = 0 compared to mx2 + nx+ r = 0
2x2 + 5x – 18 = 0 ⇒m =2, n = 5 and r = -3

b. In 2x2+ 5x – 3 = 0

x -5 -4 -3 -2 -1 0 1 2 3 4
2x2 50 32 18 8 2 0 2 8 18 32
+5x -25 -20 -15 -10 -5 0 5 10 15 20
-3 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
y 22 9 0 -5 -6 -3 4 15 30 49

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 55
50
45
40
35
30
25
20
15
10
5
0
-6 -5 -4 -3 -2 -1 -5 0 1 2 3 4 5

-10

c. 2x2+ 5x – 3 = y …………..(1) ii. the solution set of 5 + 3x – x2 = 0,

2x2 + 5x – 18 = y………….(2) iii. the maximum value of y.

eqn (1) – eqn (2) 2. The table is for the relation y = px2 – 5x + q
y = 15 (Shown on the graph)
x -3 -2 -1 0 1 2 3 4 5
y 21 6 -12 0 13
The line y = 15 cuts the curve at x = - 4.5 and
x = 2. The truth set of 2x2 + 5x – 18 = 0 is
a. i. Use the table to find dthe values of p and q
{x : x = - 4.5 or x = 2}
ii. Copy and complete the table.
b. Using scales of 2 cm to 1 unit on x – axis and 2
Exercises 21.12
cm to 5 units on the y – axis draw the graph of the
1.Copy and complete the following table for
relation for -3 x 5.
y = 7 + 3x – x2, for the range -3 x 5
2. Copy and complete the table for the relation y
x -3 -2 -1 0 1 2 3 4 5
y -11 7
= 2 + 2x – x2, for the interval -1 x 3

x -1 -0.5 0 0.5 1 1.5 2 2.5 3

b. Using a scale of 2 cm to 1 unit on the x – axis
y -1 0.75 2
and 2 cm to 2 units on the y – axis, draw
the graph of the relation y = 7 + 3x – x2 for the
b. Using a scale of 4 cm to 1 unit on both axes,
given interval
draw the graph of the relation
c. Use your graph to find:
c. Draw on the axes, the graph of x + y = 3
i. the solution set of 7 + 3x – x2 = 0,
d. Using your graphs;

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i. solve the equation 2 + 2x – x2 = 3 – 1, placing a ruler so that it just touches the graph at
ii. find the truth set of 2 + 2x – x2 = 1. the required point, and then adjust the ruler so
that it just touches the graph at the given point.
The Gradient of a Curve at a Given Point
All straight lines have a fixed slope or gradient, Worked Examples
known as “m” in the equation y = mx + c. 1. a. Copy and complete the table below for
Quadratic graphs on the other hand, have no such y = x2 – 4x + 3 for -2 to -6
fixed or constant gradient. As the graph is curved,
the value of the gradient changes all the time x -2 -1 0 1 2 3 4 5 6
based on its position on the graph. y 15 3 0 15

The gradient of a curve at a given point is b. Using scale of 2cm to 1 unit on x – axis and
determined as follows: 2cm to 2 units on y – axis, draw the graph of
I. Identify the given point on the curve y = x2 – 4x + 3
II. Draw a tangent to the curve at that point c. From the graph, determine the gradient of the
curve the points x = 4
III. Use the tangent to construct a right angled
triangle
Solution
IV. From the triangle, calculate the gradient by
a. Iny = x2 – 4x + 3
the formula: m = When x = -1, y = (-1)2 – 4(-1) + 3 = 8
When x = 1, y = (1)2 – 4(1) + 3 = 0
Drawing the Tangent When x = 2, y = (2)2 – 4(2) + 3 = -2
A tangent is a straight line which touches the When x = 4, y = (4)2 – 4(4) + 3 = 3
curve at one point. The tangent is drawn by
When x = 5, y = (5)2 – 4(5) + 3 = 8

x -2 -1 0 1 2 3 4 5 6
y 15 8 3 0 -1 0 3 8 15

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 y
16

14

12

10

0 x
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
-2

c. From the graph (triangle): c. From your graph , find;

= 3, = 5, = -1, , i. the value of x for which y is maximum,
ii. the gradient at x = 0.
m= = = =4

a. Solution
2. a. Copy and complete the following table of In y = 4 + 5x – 2x2,
values for the relation y = 4 + 5x – 2x2 for the When x = -2, y = 4 + 5(-2) – 2(-2)2 = -14
intervals -3 When x = 0, y = 4 + 5(0) – 2(0)2 = 4
x -3 -2 -1 0 1 2 3 4 5
When x = 2, y = 4 + 5(2) – 2(2)2 = 6
When x = 3, y = 4 + 5(3) – 2(3)2 = 1
y -14 4 7 -21
When x = 5, y = 4 + 5(5) – 2(5)2 = -21

x -3 -2 -1 0 1 2 3 4 5
b. Using 2cm to 1 unit and on the x – axis and
y -29 -14 -3 4 7 -14 1 -8 -21
2cm to 5 units on the y – axis, draw the graph of y
= 4 + 5x – 2x2 for -3

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 20

15

10

0
-5 -4 -3 -2 -1 0 1 2 3 4 5 6
A -5
B
-10

-15

-20

-25

-30

-35

c. From the graph: ii. Find the gradient of the curve at the point (3,
i. the value of x for which y is maximum is 2)
x = 1.3
3. i. Draw the graph of y = x (6 – x ) from x = 0 to
ii. Gradient at x = 0 x = 6, taking 2cm to 1 unit on the x – axis, and 1
From the graph (triangle): cm to 1 unit on the y – axis, and plotting points
for which x = 1, 2, 2.5, 3, 3.5, 4, 5, 6
= -1.8, = 1, = -5, ,
ii. Find the gradient of the curve at the point (4,
m= = = = 8) Ans: -2
iii. state the coordinates of the points at which the
gradient of the curve is zero Ans : (3, 9)
Exercises 21.13
1. Draw the graph of y = x2, using a scale of 2cm
4. i. Draw the graph of y = 8x – x2 from x = 0 to x
to1 unit on x – axis and 2cm to 2 units on y – axis
= 8, taking 2 cm to 1 unit on the x – axis, and 1
. Determine gradient of the curve at the point
cm to 5 units on the y – axis.
(9,3).
ii. Find the gradient of the curve at the points
where x = 3 Ans: 2
2. Using a scale of 2cm to1 unit on x – axis and
iii. Use the answers in ii to give the gradients
2cm to 10 units on y – axis, draw the graph of x2 –
where x = 5, 6 and 7 . Ans : - 2, - 4, - 6
3x + 2 = 0 for the interval, -2 to 4

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5. Copy and complete the table below for the The roots of a quadratic equation of the form
curve y = 2x2 – x – 15 + bx + c = 0 is the values of x that satisfy the
equation. It is also called the truth set or solution
x -4 -3 -2 -1 0 1 2 3 4
set.
y -21 -15 13

i. Draw the graph of y = 2x2 – x – 15, to a scale of Quadratic equations of the form: + bx + c = 0
2 cm to 1 unit on x axis and 2 cm to 5 units on y – is related to its roots as shown below;
axis. x2 – (sum of roots)x + (product of roots)= 0
ii. Find the gradient of the curve at the point ⇒ sum of roots = b
wherex = - 1 Ans: - 4.83 II. product of roots = c

6. Taking a scale of 2 cm to 1 unit on each axis, Forming an Equation from its Roots
plot the points given by the following table on a Given the roots of the equation as m and n,
graph sheet, and draw a smooth curve through I. Find the sum of roots asb = m + n and the
them. product of roots asc = m × n.
II. Substitute b = m + nandc = m × n in the
x 0 1 2 3 4 4.6 4.9 5 equation:
y 0 0.1 0.4 1 2 3 4 5 x2– (sum of roots)x + (product of roots) = 0
to obtain – (m + n)x + (m × n) = 0
i. Draw the tangents to the curve at the points III. Simplify to obtain an equation in the form:
where x = 2 and x = 4
ax2 + bx + c = 0
ii. Find the gradient of the curve at each of these
point to 1 decimal place Ans: 0.3, 1.4
Worked Examples
7. a. Copy and complete the following table of 1. Find the quadratic equation whose roots are 5
values for the relation y = - x2 + x + 2 for the and 3
interval– 3 x 3
Solution
x -3 -2 -1 0 1 2 3 Method 1
y -4 -4 Roots 5 and 3
Sum of roots, b = 5 + 3 = 8
b. using scales of 2cm to 1 units on the x – axis Product of roots, c = 5 × 3 = 15
and 2cm to 2 units on the y – axis draw a graph of By substitution,
the relation y = - x2 + x + 2 x2 – (sum of roots)x + (product of roots)= 0
c. From the graph, find the ; – 8x + 15 = 0
i. the maximum value of y The quadratic equation is – 8x + 15 = 0
ii. roots of the equation
iii. gradient of the curve at x = – 0.5 Method 2
Roots of the equation are 5 and 3
Quadratic Equation and its Roots Let the roots of the equation be x

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x = 5 or x = 3 Roots of the equation are and
x – 5 = 0 or x – 3 = 0
Sum of roots, b = + = =
(x – 5) (x – 3) = 0
x( x – 3) – 5 (x – 3) = 0 Product of roots, c = × =
– 3x – 5x + 15 = 0 By substitution,
–8x + 15 = 0 x2 – (sum of roots)x + (product of roots)= 0
– x+ =0
2. Find the quadratic equation whose roots are
Multiply through by 20
– 4 and 7
–13x + 2 = 0
Solution The quadratic equation is – 13x + 2 = 0
Roots of the equation are – 4 and 7
Sum of roots, b = – 4 + 7 = 3 6. A quadratic equation has roots – 3 and , find
Product of roots, c = – 4 × 7 = – 28 the equation.
By substitution,
x2– (sum of roots)x + (product of roots) = 0 Solution
– 3x – 28 = 0 Roots of the equation are and –
The quadratic equation is – 3x – 28 = 0
Sum of roots, b = 3 + ( ) = =
3. What quadratic equation has the roots -3 and - Product of roots, c = 3 × – = –
4? By substitution,
x2– (sum of roots)x + (product of roots)= 0
Solution
– x– =0
Roots of the equation = – 3 and – 4
Sum of roots = – 3 + – 4 = – 7 Multiply through by 2
Product of the roots = – 3 × – 4 = 12 – 5x – 3 = 0
By substitution, The quadratic equation is – 5x – 3 = 0
x2 – (sum of roots)x + (product of roots)= 0
– (–7)x + 12 = 0 Exercises 21.14
The quadratic equation is +7x + 12 = 0 Find equation with the following roots:
1. 2 and 7 2. – 3 and 8
3. – 4 and – 2 4. – 7 and 2
4. If the roots of a quadratic equation are and
5. – 5 and 6 6. 4 and – 7
, find its equation 7. – 2 and 8. 4 and –
Solution

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22 MENSURATION I Baffour – Ba Series

The Idea of a Circle Length of an Arc, Area of a Sector and

A circle is a set of points in a plane which are at Perimeter of a Sector
the same distance from a fixed point. The fixed
point is called the centre of the circle and the set Major arc
of points form the circumference of the circle. O
r θ r =
It can also be defined as the path of s point which
moves so that it is always equidistance from a A B
Minor arc
fixed point called thecenter. It is also said to be L
the path around a circular region
If the central angle subtended by the minor arc
AOB of a circle of radius r is θ, then by
Parts of a Circle
proportion =
Segment
Chord
The length of the arc is given by:
L= × 2πr

If the central angle subtended by the major arc

Diameter
AOB of a circle of radius r is 3600 – θ, then the
length of the arc is given by:

Circumference Sector L= × 2πr

Circumference: It is the distance Round a circular Similarly, if the central angle of a sector of a
region. It is also the length or perimeter of a circle circle of radius r is θ, then by proportion:
Diameter: It is a straight line divides a circle into =
two equal halves
Semi - circle: It is half of a circle The area of the sector is given by: A = × πr2
Chord: It is a straight line that connects any two
points on a circle The perimeter of the sector AOB, formed by the
: It is a any portion of the circle radii and the minor arc AB is given as:
Segment: It is the area bounded by an arc and a P = r + r + length of arc
chord P=2r+L
Radius: It is a line drawn from the center of a circle
to touch any part of the circumference. The plural is Worked Examples
radii 1. The area of a circle with centre O is 120 cm2,
Sector: It is the area bounded by two radii and an angle AOB is 60º. Find :
arc i. the area of the sector AOB.
ii. the length of the minor arc AB

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iii. the perimeter of the monir sector L= × 2πr

O But θ = 72º, r = 14cm and π

0
60
L= × 14 = 17.6cm.
A B
ii. Area of the shaded sector AOB,
Solution A= × πr2
i. Area of sector AOB
A= × πr2 A= × × (14)2 = 123.2cm2

But θ = 60º, πr 2 = 120cm2

iii. P = 2r + L.
A= × 120cm = 20cm2 But r = 14cm and L = 17.6cm
P = 2(14cm) + 17.6cm = 45.6cm
ii. Length of the minor arc AOB =
L= × 2πr 3. In the diagram below, O is the centre of the
2
But πr = 120cm 2 circle and r is its radius. Calculate the area of the
non-shaded region.
r=√ ⁄
= 6cm

L= ×2× × (6) = 6.29 cm

O
0
60 r
iii. The perimeter of the sector
P = 2r + L A B
P = 2(6) + 6.29 = 18.29 cm Solution
Area of the non-shaded region
2.
= × πr2 = × πr2 = πr2
O
14cm 0
72
4. Calculate the perimeter 10m
A B of the figure below. 0
45
The diagram above shows a circle with centre O
Solution 10m
and radius 14cm. The shaded region AOB is a
θ = 45º, r = 10m
sector with angle AOB = 72º. ( π = ). Find:
P=2r+L
i. the length of the minor arc AB
L= = = 7.86 (2 d. p)
ii. the area of the shaded sector AOB.
iii. the perimeter of the shaded region
P = 2 r + arc length
Solution P = 2 × (10) + 7.86 = 27.86m.
Length of arc AB,

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5. In the diagram below, O is the center of the straight edges coincides to form a cone, find the
circle and PR is a tangent to the circle at R. If radius of the base of the cone. π =
= 12cm and = 8cm, calculate:
i. the radius of the circle;
Solution
ii. length of the minor arc QR, correct to two
2. a. 130 = × × 12 × 12
decimal places ( π =
327600 = 3168θ
θ = 1030
P
8cm
Length of arc of sector,
Q
12cm
L= × × 2 × 12 = 21.7cm

O R Base of the cone = Length of arc of sector

2π r = 21.7
r= = 3.45 cm
Solution
Let the radius of the circle be x,
Exercises 22.1
OR= OQ = x.
1. The arc PQ subtends an angle of 400 at the
center of a circle of radius 3cm. What is:
By Pythagoras theorem, i. the length of the arc PQ?
122+ x2 = (8 + x)2 ii. the area of the sector the angle subtends?
144 + x2 = 64 + 16x + x2 (expansion of bracket) iii. the perimeter of the sector the angle subtends?
x2 – x2 – 16x = 64 – 144
– 16x = – 80
2. The angle of a sector of a circle of diameter
x= P
8cm is 135o. Find;
x = 5 cm i. the area of the sector,
13 cm
12 cm ii. the length of the minor arc,
Length of minor arc QR, θ iii. the perimeter of the sector.
L= × 2πr = O R
3. A circle of radius 6m subtends an angle of 1050
But angle, θ , subtended by QR
at the center of the circle. Calculate:
sin θ = =
i. the perimeter of the sector,
θ= ( ) = 670 ii. the area of the sector (л = )

Length of the minor arc QR, 4. A sector of a circle subtends an angle of 135o at
L= × 2 × ( ) × 5 = 5.85cm the center of a circle. If the radius of the circle is
14cm, calculate;
6. The area of a sector of a circle of radius 12 cm, i. the length of the minor arc,
is 130cm2. If the sector is folded such that is ii. the area of the sector,

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iii the perimeter of the sector. 3. In the diagram below, OAB is a sector of circle
with center O and radius 8cm. < BOA is in
5. A garden plot is in the form of a sector of a degrees. OAC is a semi-circle with diameter OA.
circle as shown below. The boundary is an arc of The area of the semicircle OAC is twice the area
a circle of radius and two radii each of length of the sector OAB B
10m, the central angle is 45o. What is the
8 cm
perimeter of the plot?
12m
O θ A
55o 8 cm

12m
D
Challenge Problems i. Find the value of θ
1. In the diagram below, triangle XYZ has XY = ii. Find the perimeter of the complete figure in
6cm, YZ = 9 cm XZ = 4 cm. The point W lies on terms of π.
the line XY. The circular arc ZW is the major arc
of the circle with centre X and radius 4cm. Finding the Central Angle
i. find < ZXY; Given the radius, r, and the arc length,l, the
ii. the area of the major sector XZWX; central angle subtended by the arc, , is
iii. the area of the whole figure; calculated by the relation: =
iv. the perimeter XWYZ.
Given the radius of a circle, r, and the area of a
X sector, A, formed by the radii, the central angle is

W
calculated by the relation: =
Z Y
9cm
2. A sector AB is to be cut from a circle of Worked Examples
cardboard with radius 25cm, and then folded so 1. The diameter of a circle is 14cm. Find the
that radii OA and OB are joined to form a cone, angle which an arc of length 16cm subtends at the
with a slant height 25cm. center of the circle

Solution

α O =
A
O
= ⁄
16cm
B = = 1310

If the vertical height of the cone is to be 20 cm,

2. The area of the sector of a circle is 66cm2. If
what must the angle α be? the radius of the the circle is 9cm, calculate the
angle of the sector (π = 3.14)

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Solution between the points of movement is calculated as:
Area of sector , A = 66, r = 9cm, and =? θ= × 3600
Substitute in =
Note that the minute arm forms the radius of the
= (
= = 930 circular face of the clock. Hence, given the length
of the minute arm, the area, A, covered by
Exercises 22.2 moving from one point to another is calculated by
1. The area of the sector of a circle with radius the formula: A = × πr2
6cm is 35.4cm2. Calculate the angle of the sector
(π = 3.14)
The distance covered by the minute arm in
moving from say A to B is equivalent to the
2. The area of the sector of a circle is 52.7cm2. If
length, L, of the arc AB and is calculated by the
the radius of the the circle is 8cm, calculate the
angle of the sector (π = 3.14) formula:
L= × 2πr
3. The length of an arc of a circle is 33 cm.
Calculate the angle it substends at the center of Worked Examples
the circle if its radius is 24cm. (π = 3.14) 1.If the minute hand of a clock with length 3 ⁄ cm
moves from 9:20 a.m to 10:00 a.m, find: i. the
Application to the Clock area covered by the minute hand;
The hands (minute and hour hands) of the analog ii. the distance covered by the minute hand (π = )
clock engage in a circular motion, in the
clockwise direction.
Solution
Consider that face of the clock below.
i. 12
0 11 1
360
12
11 1
2
3.5cm

10
0 2
10 30
9 θ
3
270
0
9  3 0
90
8
4
8
4
7 5
7 5 6
60
180
Angle in a complete revolution = 3600
It implies that the angle, θ, turned by moving from Minutes in a complete revolution = 60
one unit (hour) to another unit (hour) is 300.
60 minutes = 3600
Hence, the following conclusions can be made:
Minutes between 9 : 20 and 10 : 00 = 40 min
If the minute hand moves from say A to B in x Angle, θ, between 9 : 30 and 10 : 00
minutes, the angle, θ, formed at the center θ= × 3600 = 2400

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Area covered by the minute hand, Angle between the hour and minute hand;
A= × πr 2 = 2400 – 1400 = 1000

A= × × (3.5)2 = 25.6cm2 Exercises 22.3

1. Complete the following table (give the smaller
ii. Distance covered by the minute hand, angle in each case)
L= × 2πr
Time 1:00 2:30 7:00 8:45 10:30 11:20
L= ×2× × 3.5 = 14.67cm Angle 300 750

Angle between the Minute and Hour Hands 2. Determine the area and distance covered by the
For one complete revolution of the hour hand; minute hand of length 5cm as it moves from 6 :
12 hours = 3600 00 to 7 : 15.
1 hour = 300
B. Find the angle formed by the hour and
For one complete revolution of the minute hand; minute hands at the following time.
60 min = 3600 1. 7 : 15 pm 2. 4 : 20 pm 3. 3 : 30 am
1 min = 60 4. 2 : 20 am 3. 11 : 50 pm 4. 10 : 55 am

To find the angle between the minute and hour Area and Perimeter of Segments
hand; The segment of a circle is the region bounded by
1. Convert the time to hours and multiply by 300 a chord and the arc subtended by the chord
to obtain the angle covered by the hour hand.
2. Multiply the minutes by 60 to obtain the angle
covered by the minute hand.
3. Find the difference between the two values. 

Segment
Worked Examples
Find the angle between the hour and minute hand
when it is 4 : 40 pm? To find the area of a segment,
I. Find the area of the sector by the formula:
Solution A= × πr2
Change 4:40 hours; II. Find the area of the triangle by the formula: A
4 =4 = hours = b h or A = ab sin
III. Subtract the area of the triangle from the area
Angle covered by hour hand;
of the sector to obtain the area of the segment.
× 300 = 1400
Angle coverd by minute hand; Asegment = Asector – Atriangle
40 × 60 = 2400

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The perimeter of a segment is calculated as: A= × πr2
= length of the arc + length of chord P
2
A= × ×7 7cm 600
Worked Examples A = 25.66 cm2 0

7cm
M 60
1. Find the area of the segment of a circle with 0
60
central angle of 1200 and a radius of 8cm, to the 7cm
Q
nearest whole number
Area of triangle, A = ab sin ;
Solution
Area of sector subtended by 1200 A = (7) (7) sin = 21.23cm2
A= × πr2
8 cm
0
Asegment = Asector – Atriangle
A= × × 82 = 67.05 cm2 120 Asegment = 25.66 – 21.23 = 4.44cm2

For the two equal circles, the total area of

Area of triangle, A = ab sin segment = 2(4.44cm2) = 9cm0
A = (8) (8) sin (Nearest whole number)
A = 27.71cm2
3. In the diagram below, O is the center of the
Asegment = Asector – Atriangle circle radius r cm and < XOY = 900. If the area of
the shaded part is 504cm2, calculate the value of r
Asegment = 67.05 – 27.71
K
Asegment = 39cm2 (Nearest whole number)
X Y

2. In the diagram, M and N are the centres of two r

r
circles of equal radii 7cm. The circles intercept at
P and Q. If < PMQ = <PNQ = 600. Calculate
correct to the nearest whole number, the area of
the shaded portion (л = ) Solution
Area of sector subtended by 900
P
A= × πr2 = × ×r2 = r2cm2
7cm 7cm
0 0
M 60 60 N
Area of triangle,
A = ab sin
Q
A = r2 sin = r2cm2

Solution Asegment = Asector – Atriangle

r = 7cm and л = 504cm2 = r2 – r2
Area of sector subtended by 600; 14,112 = 22r2 – 14r2

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14,112 = 8r2 bisector of WY.
2
r = = 1,764 If /XZ/ = 8 cm and /WY/ = 64cm, calculate the
radius of the circle
r=√ = 42 cm
Solution
4. PQ is a chord of a circle, center O. The radius
Draw a complete circle as shown below;
of the circle is 8cm. /PQ/ = 12 cm. Find:
i. the angle subtended at the center by the minor X
From the diagram,
arc PQ.
the radius of the circle , 8
ii. the perimeter of the minor sector OPQ
r = x + 8 cm cm32cm
iii. the perimeter of the minor segment. (π ) 32cm
Y
W
x
x+8
Solution 
i. ∆ PQR isisosceles
By Pythagoras
̅̅̅̅ bisects ̅̅̅̅ theorem,
O
 x2 + 322 = (x + 8)2
< POQ =
8cm x2 + 322 = (x + 8) (x + 8)
x2 + 1024 = x2 + 16x + 64
Sin = = 0.75
P 6cm N 6cm Q 1024 – 64 = 16x
= (0.75) 16x = 960
= 48.590 x= = 60
= 97.180 ⇒r = 60cm + 8 cm = 68cm
The angle substended at the centre is 97.180
Exercises 22.4
ii. the perimeter of the minor sector OPQ; 1. Calculate the area of segment of a circle whose
P=2r+l angle is 200 and its radius is 3 cm.
But r = 8 and L = × 2πr
2. Calculate the area of a segment of a circle
L= × 2 ( ) × 8 = 13.57cm whose angle is 1250 and its radius is 18 cm.

P = 2 (8) + 13.57 = 16 + 13.57 = 29.57 3. ABC is a semi circle, center O, such that <BOC
= 600 and arc BC = 15cm. Find :
iii. the perimeter of the minor segment; i. the radius of the circle,
= length of minor arc PQ + length of chord PQ, ii. the length of the arc AB,
= 13.57 + 12 = 25.57 X iii. the perimeter of the semi circle.

Solved Past Question 8cm 4. In the figure below,

In the diagram, WXYZ the diameter of the 0
W Y
isa segment of a circle circle is 14cm.
64cm
such that XZ is the The length of the A B
perpendicular C

Baffour – Ba Series, Core Maths for Schools and Colleges Page 623
chord ̅̅̅̅ is 7cm. Calculate; Solution
i. the length of the minor arc of a sector OAB, P = 20 cm + 16 cm + 17 cm = 53cm
ii. the perimeter of the sector OAB,
iii. the perimeter of the segment ABC. 2. Find the perimeter of the figure below.

5. Determine the area of the shaded region in the 5cm

figure below: 7cm

12cm

0 Solution
10 cm 700
P = 5cm + 7 cm + 12 cm = 24 cm
A B
C
Perimeter (P) of a Square
The perimeter ofa square is calculated by adding
6. In the diagram below, ABCDEO is two – thirds all the lengths of the sides of the square. Consider
of a circle O. The radius AO is 7cm and /AB/ = the square below with side, Lcm.
/BC/ = /CD/ = /DE. Calculate correct to the P = Lcm + Lcm + Lcm + Lcm
nearest whole number , the area of the shaded P = 4 Lcm
portion
Worked Examples
B C 1. Find the perimeter of a square with side 5cm.

Solution
A D
O Method 1 5cm
Perimeter of square
E = 4L , but L = 5cm 5cm
5cm
Perimeters P = 4 × 5cm = 20 cm
The distance around a plane figure is called its 5cm
perimeter. All polygons have perimeters.
P = 5cm + 5cm + 5cm + 5cm = 20cm
Perimeter of a Triangle 2. Calculate the perimeter of a square of side
To calculate the perimeter of a triangle in which 13cm.
the length of the sides are given, add the lengths
of the three (3) sides. Solution
Perimeter , where
Worked Examples Perimeter = 4 ×13cm = 52 cm
17cm 16cm
1. Calculate the perimeter
3. A square has a side of 24cm. Calculate its
of the figure below; 20cm perimeter.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 624
Solution L 3cm
P = 4L = 4 × 24 cm = 96 cm
2. A square has a perimeter of 28cm.Calculate
4. What is the perimeter of a square with side
the length of a side of the square.
11cm?
Solution
Solution
Method 1 L , but P = 28cm and L = ?
P L= 7cm
P = 44cm

3. Find the length of the side of a square whose

Method 2
perimeter is 132cm.
P = 44cm
Solution
5. The length of a side of a square is 9cm.
Calculate the perimeter of the square. L , but L =? and P = 132cm
L 33cm
Solution
Method 1
P , but L 9cm Exercises 22.5
P A. Calculate the perimeter of a square with the
P = 36cm 9cm 9cm following sides.
1) 27cm 2) 16cm 3) 19m
9cm
4) 140cm 5) 21cm 6) 40m
Method 2
P = 36cm
B. Calculate the length of a side of a
square with the following perimeters.
Length of a Square Given the Perimeter 1. 104cm 2. 168cm 3. 72cm
To find the length of a square, given the 4. 148cm 5. 284cm 6. 100cm
perimeter, remember that P = 4L. Making L the
subject, L = , where L is the length of the square C. Calculate the area of a square with
and P is the perimeter of the square. the following perimeters.
1. 40cm 2. 28cm 3. 88cm
Worked Examples 4. 140m 5. 36cm 6. 44cm
1. The perimeter of a square is 12cm. Find the
length of a side of the square. Perimeter of a Rectangle
The perimeter of a rectangle is calculate by
Solution adding up all the sides.
L , butP = 12cm and L = ?
Consider the rectangle below;
By substituting the value of P,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 625
 A=L×B
20cm2= 5cm × B
=B
B = 4cm

P=L+B+L+B P = 2(L + B) where L = 5cm and B = 4cm

P = 2L + 2B P = 2(5 cm + 4cm) = 18cm
P = 2(L + B)
4. What is the perimeter of a rectangle with area
(
of 150cm2 and breadth 10cm.

Worked Examples Solution

1. Calculate the perimeter of a rectangle whose Area of a rectangle, A = L × B, where
length and breadth are 9cm and 3cm respectively. B = 10cm, L = ? and A =150 cm2
A=L×B
Solution 150cm2 = L × 10cm
Method 1
L= = 15cm
Perimeter of a rectangle, P (
But L = 9cm and B = 3cm
P = 2(9cm + 3cm) P = 2(L + B) where L = 15cm and B = 10cm
P = 2 (12cm) 9cm P = 2(15 cm + 10cm) = 50cm
P = 24 cm
3cm 3cm 5. The perimeter of a rectangle is 26cm.If its
9cm length is 9cm. Calculate:
Method 2 a. the breadth of the rectangle,
P = 9cm + 3cm + 9cm + 3cm = 24 cm b. the area of the rectangle.

2. A rectangle has a length of 13cm and breadth Solution

7cm. Calculate its perimeter. P ( but L = 9cm, B =? and
P = 26cm
Solution 26cm 2 (9cm + B)
P = 2(L + B), but L = 13cm and B = 7cm (
P = 2(13cm + 7cm) = 40 cm
13cm
13cm – cm B
3. The area a of a rectangle is 20cm2. If its length 4cm or B
is 5cm, find its perimeter.
Area of a rectangle, A = L × B
Solution where L = 9cm and B = 4cm
Area of a rectangle, A = L × B A = 9cm × 4cm = 36cm2
But A = 20cm2, L = 5cm and B = ?
Baffour – Ba Series, Core Maths for Schools and Colleges Page 626
Exercises
1. The length of a rectangle is 5cm more than its 2.
12cm
width and the area is 50 cm2. Find the length, 10cm
width and the perimeter. Ans: 5, 10, 30
13cm
13cm
2. The length of a rectangle is greater than its 14cm
breadth by 3cm. If its area is 10cm2, find the
Solution
perimeter. Ans 14cm
P = 12 + 13 + 14 + 13 = 62cm

3. The perimeter of a rectangle is 116 cm. If the

Exercises 22.6
length is 22 cm more than the breadth, form an
Calculate the perimeter of the figures:
equation and use it to find the length and breadth
1. 12cm
of the rectangle.

10cm
4. A table is three times as long as it is wide. If it 15cm
was 3m shorter and 3m wider, it should be a
2.
square; how long and how wide is it? 2cm
9cm

5. The sides of a rectangle with area 135cm2 are

13cm
3x cm and (x + 4) cm long. Form an equation in x, 5cm
and solve it. Write down the length and breadth
7cm
of the rectangle.
3. 23cm

Perimeter of Other Polygons 16cm 13cm

To find the perimeter of other polygons‟
I. Identify all the dimensions of the polygon II. 20cm
Add up all the dimensions to obtain the perimeter
4.
of the polygon 8cm
20cm
Worked Examples
6cm
Calculate the perimeter of the figures;
18cm
1.
24cm
Perimeter of a Circle (Circumference)
18cm 20cm The perimeter of a circle is called its
Circumference. The circumference, C, of a circle
26cm
is calculated by the formula.
Solution
P = 18cm + 26 cm + 20cm + 24cm = 88 cm 1

Baffour – Ba Series, Core Maths for Schools and Colleges Page 627
Where d is the diameter of the circle. 3. A circle has a radius of 28cm. Find its
perimeter (
But
Substitute d = 2r in eqn (1)
 Solution
C = 2𝛑r but r = 28cm
C=2× × 28 =176 cm
2
Solved Past Question
Where r is the radius of the circle. Thus, any of 1. A circle with centre O, has radius 12 cm. A
the formulas: C = πd and C = 2πr can be used to
calculate the circumference of a circle. The chord PQ of the circle is 10cm long. Calculate:
implication is simple: i. the distance of the chord from the centre O.
I. when the diameter is given, use C = πd ii. angle POQ
II. when the radius is given, use C = 2πr
Solution O
Worked Examples i. Sin θ = 12
xx
1. Calculate the circumference of a circle with θ= ( ) P 5 5 Q
diameter 14cm ( ) θ = 250

Solution ii. < POQ = 2θ = 2 × 250 = 500

Method 1
Given the diameter, d=14cm, we use Exercises 22.7
= 44cm Calculate the circumference of the circles
1. 2. d = 14cm 3. d = 56cm
4. r = 42cm 5. d = 21cm 6. r = 10 cm
Method 2

Area of a Triangle
= 7cm The area of a triangle is calculated with reference
C=2× × 7 cm = 44 cm to the nature (dimensions) of the triangle.
1. Given three sides, a, b and c of the triangle, the
2. What is the circumference of a circle with area is given by ;
radius 21cm? ( A=√ ( ( ( a c
Where p is half of the perimeter,
Solution Or p =
b
Given the radius, we use the formula;
C = 2πr 2. Given two sides a. and b, and an included
C=2× × 21 cm =132 cm angle, θ, the area of the triangle is calculated by
the formula:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 628
A = ab sin θ Solution
a b = 6cm and h = 3cm, substitute in
A = bh
θ
b A = × 6cm × 3cm = 18cm2
3. Given the length of the base and the height of
the triangle, the area is calculated by the formula;
Exercises 22.7B
A = bh. Find the area of each triangle.
h 1. 2.
800
10cm 13 cm
b 8cm
0
Here, when given two sides, you can find the 45
height by the use of Pythagoras theorem. 18 cm
6cm
3. 4.
Worked Examples 7 cm 17cm
Find the area of the 9cm
4cm
triangle below; 6 cm 12 cm
15 cm
7cm Challenge Problems
Solution 1. Two chords and a diameter form a triangle
Let a = 9cm, b = 4cm and c = 7cm inside a circle. The radius is 5cm and one chord is
p= = cm = 10 cm 2cm longer than the other. Find the perimeter and
area of the triangle. Ans 24, 24cm2
a = 9cm, b = 4cm , c = 7cm and p = 10cm
A=√ ( ( ( 2. The length and width of a rectangular garden
A=√ ( ( ( are 150m and 120m. A foot path of regular width
A=√ ( ( ( =√ = 13cm2 is added to the boundary of the garden and the
total area of the garden becomes 2,800m2 more
2. Calculate the area 9cm than its original area. Find the width of the foot
of the triangle. path. Ans 5
0
30
7cm 3. An isosceles triangle is inscribed in a circle in
Solution
a = 9cm, b = 7cm, θ = 300, substitute in such a way that its longest side, which goes
A = ab sin θ through the centre is √ cm. Find the area of the
triangle. Ans 12.5cm
A = (9)(7) sin 300= 15.75cm2
Area of Quadrilaterals
3. What is the area of the 3cm 1.Area of a Trapezium
triangle? The formula for calculating the area of a
6cm trapezium is; ( ;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 629
where A is the area of the trapezium, Solution
a is the length of the upper base, (
b is the length od the lower base,
But A= 268cm2, b =16, h =11 and a= ?
h is the height.
Makea the subject of (
(

11cm

16cm
(
Put in b = 16cm, h = 11cm and A = 268cm2
Worked Examples
a = 32.73cm (2dp)
1. Calculate the area of the trapezium below;
21cm
Note: In some cases, it is first necessary to
6cm calculate the height, h, by the use of Pythagoras
theorem; and then substitute in the trapezium
18cm formula ; (
Solution where a and b are the opposite sides of the
( trapezium

( 12 cm + 18 cm) × 6cm cm2 More Worked Examples

Determine the area of the following figures;
2. Calculate the area of the figure ABCD. 1. 30cm

A 12cm B 15cm

6cm
10cm
b
C 18cm D 49cm Not drawn to scale
Not drawn to scale
Solution Solution
Considering the right – angled triangle at the
Area = (a + b) h
extreme left,
a = 12cm, b = 18cm, h = 6cm.
30cm
A = (12 cm + 18 cm) × 6 cm = 90cm2
15cm
3. The lower base of a trapezium is 16cm and its
height is 11cm. If it has an area of 268cm2,
9cm 30cm 10cm
find its upper base.
By Pythagoras theorem

Baffour – Ba Series, Core Maths for Schools and Colleges Page 630
152 2
+ 92 Area of the rectangle = L × B
152 – 92 = h2 But L = 60cm and B = h = 24cm
2 15cm Area = 60cm × 24cm
225 –
m A = 1,440cm2
144 = h2
2 9cm
√
12cm Total Area of the figure;
= Area of ∆ + Area of the rectangle,
= 384cm2 + 1440cm2
Substitute , in (
= 1,824cm2
Where a = 30cm, b = 49cm and h = 12cm
( Method 2
A= = 474 cm2 By Pythagoras theorem, h = 24cm.
Substitute h = 24cm in the formula for the area of
2. a trapezium = (a + b)h
60cm
But a = 60cm, b = 92cm and h = 24cm
40cm A = (60 + 92) × 24
A = 1,824cm2
92cm
Not drawn to scale
Exercises 22.8
Solution
A. Calculate the area of trapeziums;
Method 1
1. 42cm
Consider the triangle below;.
402 = h2 + 322 39cm
402– 322 = h2 40cm
2
15cm 42cm
1600 – 1024
h2 = 576 2.
32cm 20cm
h=√ = 24cm

Area of the triangle; 9cm

A= bh 30cm
Substitute b = 32cm and h = 24 cm
Challenge Problems
Area of ∆ = × 32cm × 24 cm = 384cm2 Calculate the area of the following figures;
1. 12cm
Consider the rectangle below; 4cm 9cm
60cm
6cm
h = 24cm
18cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 631
 3. The area of a rectangle is 15cm2. If the length
6cm
2. of the rectangle is 5cm. Calculate :
2cm i. the breadth of the rectangle.
5cm ii. the perimeter of the rectangle
2cm
Solution
6cm i.
2 5cm
3. 8cm

4cm 2 15cm
2
4cm 10cm =

8cm

Some Solved Past Questions

1. The area of a rectangle is 18cm2. One of its
Area of a Rectangle sides is 9cm long. Find its perimeter.
The area of a rectangle is calculated by
multiplying the length by the breadth. Solution
A = L ×B, but A = 18cm2, L = 9cm
B 18cm2 = 9cm × B

L B= = 2cm
Area = Length × Breadt
A = L× B or A = LB P = 2(L + B) = 2( 9cm + 2cm) = 22cm

Worked Examples 2. An equilateral triangle has side 16cm. A square

1. The length and breadth of a rectangle is 15cm has the same perimeter as the equilateral triangle.
and 10cm respectively. Find its area.
What is the area of the square?
Solution
Solution
A=L×B
Perimeter of the triangle;
P = 16cm + 16cm +16cm = 48cm.
2

Perimeter of the square = L + L + L + L = 4L.

2. A rectangle has a length of 12cm and breadth But perimeter of square = perimeter of ∆.
of 6cm. Find its area. ⇒4L = 48cm
L= = 12cm
Solution 12cm
But Area of a square;
A 6cm A = L2 = L × L
A 2
A = 12cm × 12cm = 144cm2

Baffour – Ba Series, Core Maths for Schools and Colleges Page 632
3. A square of side 6cm has the same area as ⇒60cm2 = 5 (x cm + 4cm)
rectangle of length 9cm. Find the breadth of the 60 = 5x + 20
rectangle. 60 − 20 = 5x + 20
40 = 5x,
Solution x= = 8cm
Let, B, be the breadth of the rectangle.
Area of the square = Area of the rectangle
Exercises 22.9
⇒L2 = L × B
1. The length of a rectangle is 20 cm and the
But length of the square = 6cm and the length of
breadth 7cm, calculate the area.
the rectangle = 9cm
2. The area of a rectangle 117cm2.. If its length is
By Substitution,
13 cm, what is the breadth?
62 = 9 × B
3. Find the length of a rectangle whose area is
6 cm× 6cm = 9 cm × B
135cm2 and breadth 9 cm.
B= = 4cm 4. Find the breadth of a rectangle in which the
area is 810 cm2 and the length 45cm.
4. The perimeter of a rectangle is 24cm. If the 5. A rectangle has an area of 36 cm2 and width of
breadth of the rectangle is 4cm. Find the area of 3cm. Find its perimeter .
the rectangle
Area of a Square
Solution A square is a rectangle with all sides equal.The
P = 2 (L + B). But P = 24cm, B = 4cm area of a square is found by multiplying the
24 cm = 2 (L + 4cm) length by itself. Since a square has all sides equal,
24 = 2L + 8cm it implies that if one side is L, then all other sides
24cm – 8cm = 2L is L. That is; A= L × L = L2
L= = 8cm L

But Area of a rectangle A = L × B. L L

L = 8cm and B = 4cm
A = 8cm × 4cm = 32cm2 L
Worked Examples
5. The area of the figure below is 60cm2. Find the 1. A square has a length of 11cm. What is its
area?
value of x
xcm 4cm Solution 11cm
5cm A = 2,
2
Not drawn to scale ⇒ ( 11cm
Not drawn to scale
A = 11 × 11 = 121 cm2
Solution
A=L×B 2. A square has an area of 169cm2. Find its
A = (x + 4) × 5 (But A = 60cm2) length.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 633
Solution Consider the parallelogram below:
2
,
169cm2
2
But ,
2
⇒ = L2 =?
2 h
√ = √ 2,
L = 13 cm
The height „h‟ is the perpendicular distance
3. The area of a square is 49cm2. Find the between any two parallel sides and the base is
perimeter of the square. one of these two parallel sides connected by the
perpendicular.
Solution
A = L2 = L × L Worked Examples
But A = 49cm2 1. Calculate the area of the parallelogram
49 = L2
2 2
√ cm = √ , 10cm Not drawn to scale
L = 7cm.
25cm
Perimeter of a square,
P = 4L Solution
But L = 7cm ( (
⇒P = 4 × 7 = 28cm. where l = 25cm and h = 10cm
A = 25cm × 10 cm = 250 cm2
Exercises 22.10
A. Calculate the area of a square with the 2. The length and height of a parallelogram are
following lengths; 27cm and 30cm respectively. What is its area?
1) 5cm 2) 12cm 3) 16cm
4) 25cm 5) 23cm 6) 41cm Solution

B. Find the length of a square with the But

following areas; = 810 cm2
1) 81cm2 2) 49cm2 3) 225cm2
4)169cm2 5) 400cm2 6) 121cm2 Exercises 22.11
A. 1. A parallelogram has a base of length 10cm
Area of a Parallelogram and a perpendicular height of 4cm. Find its area.
The area of a parallelogram is calculated by the 2. Calculate the area of a parallelogram whose
formula: , length is 23cm and perpendicular height of 17cm.
A is the area; 3. What is the perpendicular height of
L is the length of the base; parallelogram of area 810m2 which has length of
H is the height od the parallelogram; 30m?

Baffour – Ba Series, Core Maths for Schools and Colleges Page 634
B. Find the areas of the figures: Area of rectangle, A1 =
2
1. A1 =
4cm 5cm
Area of ∆ A2 = bh
7cm
A2 = = 9cm2
2.
25cm
40cm Total area of the shape, A
A = Area of rectangle (A1) + Area of ∆(A2)
3 0cm A = A1 + A2
3. A = 48cm2 + 9cm2 = 57cm2.
10cm
24cm
2. Find the total area and the perimeter of the
garden below;
13cm E
4. 48.5m 50m
20m
22cm A B
44.8m 45.2m
14cm 10cm 70m
Area of Complex/Combined Shapes D 90m C
Complex shapes are formed when two or more
shapes are combined or put together to form a Solution
single shape. The areas of complex shapes are Split the figure to obtain the following shapes.
calculated by finding the sum of the areas of the E
separate solids or finding the difference of the A B
areas of the separate solids. 48.5m
70m m 20m 50m
Worked Examples C 90m D
Calculate the area of the figures below. A 44.8m 45.2m B
1. 8cm Area of rectangle ABCD,
A1
3cm A1 = 6300 m2
6cm
Area of triangle ABE (A2);
Not drawn to scale
Solution A2 = bh
Split the shape into a rectangle and triangle as A2 = ( = 900 m2
shown below and find the area of each.
8cm Total area = Area of ABCD + Area of ABE
6cm 3cm = A1 + A2
6cm
= 6300m2 + 900m2= 7200m2

Baffour – Ba Series, Core Maths for Schools and Colleges Page 635
Perimeter = /AD/ + /DC/ + /CB/+ /BE/ + /EA/ and /YZ/ = 28cm, find
= 90 + 70 + 50 + 48.5 + 70 = 328.5 cm the area of the remaining X
part of the circle (π = )
3. Find the area of the shaded region in the
figure below: O 
Solution Y
Area of the circle, A = π r2
4cm But r = = = 17.5, and π = Z
9cm 6cm

A= × 17.5 × 17.5 = 962.5 cm2

14cm
Not drawn to scale
For ∆XYZ,
Solution /XZ/ = 35 cm /YZ/ = 28 cm, /XY / = ?
Let the area of the large rectangle be A1 and By Pythagoras theorem,
the area of the small rectangle be A2 /XZ/2 = /YZ/2 + /XY/2
2
A1 = 352 = 282 + /XY/2
2
A2 /XY/2 = 352 – 282
Area of shaded region
/XY/ = √ = 21 cm
A = A1 – A2
A=( ) cm2 = 48cm2
Area of ∆ XYZ ;
4. In the figure below, find the area of the shaded = /XY/ /YZ/ sin
region ( = × 21 × 28 sin 900= 294 cm2

7cm Area of the remaining part;

11cm = Area of circle – Area of triangle
= 962.5 cm2 – 294 cm2 = 668.5 cm2
16cm
Solution 6. The diagram below shows a trapezium MNOT,
Let the area of the rectangle be A1 and the area of in which MN // TO, /MN/= 14 cm, < MTO = 600
the circle be A2. and /MT/ = /NO/ = 12 cm. If the semi – circle
A1 = L × B = 11cm × 16 cm = 176cm2 MPN is removed from the trapezium, calculate
A2 = πr2 = × 7 × 7 = 154 cm2 correct to the nearest cm2, the area of the
remaining portion (π = ).
Area of the shaded region;
M 14 cm N
A = Area of the rectangle – Area of the circle
A = A1 – A2
A = 176 cm2 – 154 cm2 = 22cm2 12 cm
h cm P h cm
0
60
5. In the diagram below, XYZ is cut off from the T S R O
circle, center O. If /XZ/ = 35 cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 636
Solution is shaded with some portions of the square,
2
Area of semi – circle = π r calculate the area of the shaded portions.(π = )
But r = = = 7 cm
Solution
Area of semi – circle = × × 72 = 77cm2
Area of total shaded portion;
=Area of square – Area of unshaded sector
Area of trapezium = (a + b) h
= L2 – × πr2
= (/MN/ + /TO/) h
Consider ∆MST,
/MT/ = /NO/ = 12 cm, h = ? /TS/ = ? But r = 7cm , L = 2r =2(7cm) = 14cm, and
θ = 3600 – 800 = 2800 , substitute in
sin 600 =
L2 – × πr2
⇒h = 12 sin 600 = 10.3923 cm
⇒ 142 – × × 72
By Pythagoras theorem, = 196 – 119.7778 = 76.22cm3 (2 d.p)
/TS/2 + h2 = MT2
/TS/2 + (10.3923)2 = 122 Exercises 22.12
/TS/2 = 122 – (10.3923)2 A. Determine the area of the following;
/TS/2 = 36 1. All diagrams are not drawn to scale
/TS/ = √ = 6cm

⇒But TS = RO = 6cm (∆MST ∆NRO) 14m

/TO/ = TS +SR + RO
25m
/ TO/= 6 + 14 + 6 = 26 cm 2.
6m
Area of trapezium; 28m

= (/MN/ + /TO/) h 32m

3.
= × (14 + 26) × 10.3923 12cm

= × 30 × 10.3923 155.8845 cm2 12cm

10cm
Area of shaded region
= Area of trapezium – Area of semi – circle
= 155.8845 cm2 – 77cm2
= 78.8845 cm2 = 79 cm2 B. Determine the area of the shadedregions in
following diagrams;
7. The figure shows a 7 cm 1.
O
circle of radius 7cm 800
7 cm 12cm
inscribed in a square.
Ifa portion of the circle
12cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 637
 C
2. 5cm
7 cm 7 cm

A B
12cm

20cm
i. Calculate the angle CAB correct to the nearest
3. degree.
ii. Find the perimeter of the shaded region.
4cm
9cm 10cm 8cm iii. Find the area of the shaded region.

6cm
9cm
8. In the diagram below, O is the centre of the
circle and AT is a tangent.
4. T
7cm
3cm
9cm  O A

12cm
C. 1. If the radius of the circle is 21cm and the Calculate the area of the shaded region
area of the rectangle is 800cm2, find the area of
the shaded region in the diagram below; Word Problems
When problems involving area of complex shapes
are stated in words, it is advisable to make a
sketch representing the problem before
attempting to solve.

Worked Examples
2. In the diagram below, what is the area of the 1. Mr. Martin had a plot of land of size
shaded region? On this piece of land, he cultivated
plantain that covered an area of 31m , and
23cm cassava that covered an area of
7cm
Find the area of the land that was not cultivated.

32cm
Solution A1
Plantain A2 21m
Challenge Problems
1. The diagram below shows two circles of radius 31m
140m
7cm with centres A and B. The distance AB is A
40m
12cm and the point C lies on both circles. The 35m
Cassava A3
common region to both circles is shaded.
160m Not drawn to scale

Baffour – Ba Series, Core Maths for Schools and Colleges Page 638
Let the area of the piece of land be A1, the area of Total area of land cultivated;
land cultivated with plantain be A2,the area of the = A2 + A3 = 1,296m2 + 3,850m2 = 5,546m2
land planted with cassavabe A3and the area of
land that was not cultivated be A Area of land that is not cultivated;
A1 = L × B = 140m × 160m = 22,400m2 A = A1 – (A2 + A3)
A2 2
A = 7,700 m2 – ( 1,296m2 + 5,146m2) = 2,554 m2
2
A3
3. The length and width of a rectangular garden
Total Area of land cultivated; are 150m and 120m. A foot path of regular width
Area of land planted with plantain + Area of is added to the boundary of the garden and the
land planted with cassava. total area of the garden becomes 2800m2 more
than its original area. Find the width of the foot
Total Area = A2 + A3 path. Ans : 5
= 651m2 + 1,400m2 = 2,051 m2
Solution
Area of land that is not cultivated (A); x x
A = A1 – (A2 + A3) x x
120
A = 22,400m2 – 2,051m2 = 20,349 m2 150
x x x x
2. Mrs. Doe has a plot of land that measures100m
× 77m. If she cultivates maize that covers an area Area of first rectangle
of 36m × 36mand also construct a circular fish = 150 × 120 = 18,000 m2
pond of radius 35m,calculate the area of her land
that is not cultivated. ( )
Let the width of the foot path be x
Area of the second rectangle;
Solution A1 (2x + 150) (2x + 120) = 18000 + 2800
4x2 + 240x + 300x + 18000 = 20800
Fish pond A2 4x2 + 540x + 18000 – 20800 = 0
77 m A3 36m maize
35m 4x2 + 540x – 2800 = 0
A 36m x2 + 135x – 700 = 0
100m Not drawn to scale
(x – 5) ( x + 140) = 0
m x = 5 or x = - 140
Let A1 be the area of the piece of land, A2 be the x=5
area of the land planted with maize, A3 be the The width of the foot path is 5m
area of land used for fish pond and A be the area
of land that is not cultivated. Exercises 22.13
A1 = L × B = 77 m × 100 m = 7,700m2 1. The area of a rectangular plot is 70m × 90m.
A2 = L × B = 36 m × 36 m = 1,296m2 If it is cultivated with groundnut that covers an
A3= πr2 area of 25 m × 40m, find the size of the plot of
A= × 35m × 35m = 3,850m2 land thatis not cultivated.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 639
2. Mr. Brown rented a room of size 250 cm × fraction of the room is not covered with carpet.
300 cm. His bed occupied a space of 130 cm ×
145 cm and his refrigerator occupied 80cm × 2. The area of a square exceeds twice that of
60cm space. Calculate the area of the empty space another by 56cm2. If the difference of the
in the room. perimeter between the two is 24cm, find the area
of the smaller square. Ans 100 or 4
3. Mr. Henry purchased a plot of land that
measured 120m × 120m. He put up a storey 3. A rectangular floor measures 10m by 6m. A
building that occupied a space of 40m × 60m. He piece of rectangular carpet measuring 6m by 4m
then constructed a circular swimming pool of is placed in the middle of it. Calculate the area of
radius on the same piece of land. Calculate the uncovered floor.
the area of land that was left bare.
4. A rectangular room holds a rectangular carpet
4. A circular plot has a radius of 63m. On this in the centre measuring 8m by 16m. The width
piece of land, shadrack wants to cultivate from the edge of the carpet to each wall is the
pineapple that will cover a space of 43m × 32m, same. The area not covered by the carpet is
Meshach needs 50m × 50m for 112m. What is the width of the uncovered area
groundnutproduction and Abednego requires a around the carpet?
circular piece of land of radius 14m for tomato
production.Calculate the area of land left for 5. A carpet is laid around a rectangular dance
Daniel‟s cocoa production. floor measuring 10m by 8m. If the width of the
carpet is 2m, find its area.
5. A square plot of land 96m wide is used to
cultivate 625m2 of cassava and 40m × 30m beans 6. A rectangular piece of carpet is placed on the
production. Calculate the area of plot that is not floor of a rectangular room leaving a margin of
cultivated. 1m around it. If the room measures 8m by 6m,
find the area of thr room nit covered by the
Challenge Problems carpet.
1. A piece of rectangular paper measures 8x by
12x. If a circle of radius 3x is cut out of the 7. Carpet cost Ghȼ16.00 a square foot. A
middle of the paper, what will be the area of the rectangular floor is 16 m long by 14m wide. How
remaining piece of paper in terms of π. much will it cost to carpet the floor?

B. A floor measures 15m by 8m is to be laid with 8. A rectangular bathroom measures 9.5m by

tiles measuring 50cm by 25cm. 12m. The floor is covered by rectangular tiles
i. Find the number of tiles required. measuring 1.5cm by 2cm. How many tiles are
ii. If the carpet is laid on the floor so that a space there on the bathroom floor?
of 1m exist between its edges of the floor. What

Baffour – Ba Series, Core Maths for Schools and Colleges Page 640
23 PLANE GEOMETRY II Baffour – Ba Series

Tangent to a Circle x + y + 300 = 1800, but x = 900

A tangent to a circle is: y + 900 + 300 = 1800
i. A straight line which meets the circumference y = 1800 – 900 – 300 = 600
at one point only.
ii. Perpendicular or 900 to the diameter or radius 2. The figure below shows a circle, center Z with
drawn to the point of contact. radius 5cm long and a tangent XY 12cm long.
How far is Y from the center of the circle?
Consider the figure below;
X
A 12
5
Z∙ Y
O ∙
Solution
X M Y < ZXY = 900. Therefore triangle XYZ is a right –
AM is the diameter which is the axis of angled triangle.
symmetry, and O is the center of the circle. XY is By Pythagoras theorem,
the line parallel to the chords, at the end of the + =
diameter. YZ = √
Since all the chords are perpendicular to AM, XY =√ =√ = 13cm
is also perpendicular to AM and is called a
tangent to the circle. M is called the point of 3. In the diagram, ̅̅̅̅ is the diameter of the circle
contact O.
P
Worked Examples
1. In the figure below, CP is a tangent to the O 720 Q
circle, center O. Find x and y
R
S

oθ∙ 0 If ̅̅̅̅ is a tangent to the circle at P and < POR =

y
720, calculate the value of < PQR.
X0 300
C P
Solution
Solution < POR + < SOR = 1800
CP is perpendicular to OC, 720 + < SOR = 1800
Therefore, <OCP = x = 900 < SOR = 1800 – 720 = 1080

Baffour – Ba Series, Core Maths for Schools and Colleges Page 641
∆SOR is an isosceles; < BTD + < ODC + < DBT = 1800
⇒< OSR = < ORS = x ⇒900 + 33.50 + < DBT = 1800
2x + 1080 = 1800 < DBT = 1800 – 900 – 33.50 = 56.50
2x = 1800 – 1080
2x = 720 Exercises 23.1
x = 360 1. In the figure below, PR and QS are tangents at
opposite ends of the diameter PQ. Find a, b and c.
< PQR + < QSP + < SPQ = 1800 Q S
< PQR + 360 + 900 = 1800 0 0
bb c
0

< PQR = 1800 – 360 – 900 = 540

O∙
4. In the figure below, O is the center of the 0
aa
circle, TA is a tangent and < OAT = 230. 0 40
0
4
Calculate < DBT P 0 0
R

D
2. In the figure below, OA, OB and OC are radii,
and PCQ is a tangent to the circle.
O
C
A
O B
1000 1200
T B A

Solution P C Q
< OTA + < OAT + < TOA = 1800
900 + 230 + < TOA = 1800 Copy the diagram, mark two right angles and fill
< TOA = 1800 – 900 – 230 = 670 in the sizes of all the other angles.

< TOA + < COD = 1800 3. In the figure below, GE is a diameter, and DEF
670 + < COD = 1800 is a tangent to the circle. DE = 8cm, GE = 6cm
< COD = 1800 – 670 = < 1130 and GF = 6.5cm. Calculate the lengths of DG and
EF.
G
But ∆ OCD is an isosceles triangle
⇒< ODC = < OCD
< COD + < ODC + < OCD = 1800 6 6.5
1130 + 2 < ODC = 1800
2 < ODC = 1800 – 1130
2 < ODC = 670 D 8 E F
< ODC = = 33.50 4. In the diagram below, AB is a straightline
touching the circle center O at A. If /AB/ = 12cm,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 642
/BC/ = 8cm and angle OAB = 900, calculate the 3. AB……..
radius of the circle.
8cm A
5cm
C

O
C
8cm B
B
4. CB
A 12 cm B

5. Two chords and a diameter form a triangle

C
inside a circle. The radius is 5cm and one chord is 
2cm longer than the other. Find the perimeter and
area of the triangle. Ans 24, 24cm2
B A
10√
5. FA….
6. Determine the value of x
i.
C
8 
 C
x
16 A
D B 4cm
ii. 2cm

77
 C F A
55 O

6 8cm
x
B
B. Solve for the missing information
1. CB…….
The Tangent – Kite

P
D C B
o
12cm O∙o T

Q
A
2. CB……….
B
In the figure above, tangents from T touch the
C circle, center O, at P and Q.
0
45 A
D OPTQ is called a tangent – kite, and has axis of
symmetry OT. Two tangents to a circle from a

Baffour – Ba Series, Core Maths for Schools and Colleges Page 643
point outside the circle are equal in length. The
C
line PT = QT. Thus, < OPT = < OQT 700
0
20
200
O∙ M A
Worked Examples 200
200

1. AB and AC are tangents to the circle with 700

center O in the figure below. If < OBM = 200, fill B
in the sizes of all the angles in the figure.
2. In the diagram below, O is the center of the
C
circle, YT is a tangent, < TYZ = 200, and XO// YZ
i. Find, < XOY
O∙ M A ii. Calculate the
0
20
value of
the reflex O
B
angle XOZ
X Z
Solution
OCM = 200
0
20
< OCA = 900 (tangent to a circle) y T
⇒ < OCM + < MCA = < OCA Solution
200 + < MCA = 900 i. /OY/ is perpendicular to /YT/
Therefore, < OYT = 900 and ZYT = 200
< MCA = 900 – 200 < OYZ + < ZYT = < OYT
<MCA= 700 < OYZ + 200 = 900
< OYZ = 900 – 200 = 700
< AMC + < MCA + CAM = 1800
But, AMC = 900 and < MCA = 700 < XOY and < OYZ are alternate angles
⇒ 900 + 700 + < CAM = 1800 But < OYZ = 700
< CAM = 1800 – 900 – 700 = 200 Therefore, < XOY = 700
< OCM + < OMC + < COM = 1800
But OCM = 200 and OMC = 900 ii. < OYZ = < YOZ = 700 (Base angles of an
⇒ 200 + 900 + < COM = 1800 isosceles triangle)
< COM = 1800 – 200 – 900 = 700 < XOZ + < XOY + < YOZ = 3600
But < XOY = 700 and < YOZ = 700
OA is the axis of symmetry. It implies that ∆OCA < XOZ + 700 + 700 = 3600
= ∆OBA < XOZ = 3600 – 700 – 700
⇒< OCM = < OBM = 200 < XOZ = 2200
< COM = < BOM = 700
< OMC = < OMB = 900 3. In the figure below, TA and TB are tangents to
< MCA = < MBA = 700 the circle O. Given that < ATB = 200, find:
< CAM = < BAM = 200 i. < AOT ii. < PAT

Baffour – Ba Series, Core Maths for Schools and Colleges Page 644
 A iv. a pair of right – angles;
v. if OR = 3cm, and OT = 8cm, calculate the
length of the tangents RT and ST to 1 decimal
P
O T place.

B 2. Figure ABCD below is a tangent- kite and <

BCD = 400. Name two isosceles triangles and
Solution calculate the size of each. B
i. < ATB = 200
< OTA = < OTB = 100
< OAT + < OTA + < AOT = 1800 C ∙A
900 + 100 + < AOT = 1800
< AOT = 1800 – 900 – 100 = 800 D
3. In the figure below, OPQR is a tangent – kite.
ii. ∆OAP is Isosceles P
< AOP = < AOT = 800
< OAP = < OPA O∙ S Q
< OAP + < OPA + < AOT = 1800
2 < OAP + 800 = 1800
2 < OAP = 1800 – 800 R
i. Name the axis of symmetry and explain why
2 < OAP = 1000 PR is perpendicular to OQ
< OAP = = 500 ii. If < SPQ = 700, calculate the size of each angle
in the diagram
But < OAP + < PAT = 900 iii. If OQ = 7.2cm and OP = 2.7cm, calculate the
500 + < PAT = 900 length of the tangents to two significant figures
< PAT = 900 – 500 = 400
4. AB is a tangent touching a circle, center O, at
Exercises 23.2 B. AO = 17cm and OB = 8cm
1. In the figure below, ORTS is a tangent – kite. i. Calculate the length of AB.
ii. If AB is produced 12cm to C, calculate how far
R C is from the center O to 1 d.p.
iii. Use your tables to calculate the angles of
S
triangle AOB.
O∙ T

5. In the figure below, R

S PQRT is a circle.
Name:
i. one pair of congruent triangles; /PQ/ = /PR/, O Q
ii. two pair of equal lines; < PQR = 400 T 600

iii. three pair of equal angles; and < POT = 600, 400
P

Baffour – Ba Series, Core Maths for Schools and Colleges Page 645
Find: i. < TOR iii. < PSQ iii. If ∆ ABC has BC = 6cm and AB = AC = 8cm,
Calculate the radii of the three touching circles
Challenge Problems
1. In the figure below, tangents BC, CA and AB 3. In the figure below, a circle touches the sides
to the circle touch it at X, Y and Z. of ∆ ABC at D, E and F
i. Write down three pair of equal lines in the
A
figure.
z ii. If BF = 6cm, and CE = 4cm, calculate the
y length of BC.
iii. If the perimeter of the triangle is 30cm,
calculate the length of AF.
B x C B
i. Name three pair of lines of equal length in the F
figure.
6 E
ii. If BZ = 4cm, XC = 5cm and AY = 3cm,
4
Calculate the lengths of AB, BC and CA.
iii. If the perimeter of ∆ ABC is 42cm, BX = 6cm A D C
and CX = 7cm, calculate the lengths of AB and 4. A point P is 22cm from the centre O of a circle.
AC The angle between the tangents from P is 660.
iv. If the perimeter of ∆ ABC is 50cm, AZ = 10cm Calculate the radius of the circle.
and CX = 12cm, calculate the length of BC.
v. If the perimeter of ∆ ABC = 48cm and AZ = Properties of Angles in a Circle
6cm, calculate the length of BC. I. Angles at the centre of a circle is double any
angle at the circumference subtended by(or
2. Three circular discs, with centres A, B and C standing on) the same arc.
and radii 5cm, 4cm and 3cm respectively, are in C
contact as shown below; 0
q
C
C 0
O
∙ ∙
2q
0
2q
q

A
A∙ ∙B
A
B B

Worked Examples
i. If the first two circles touch at X, why are A, X 1. In the figure below, find the value of the angle
and B collinear? (Hint: Think of a common marked x. C
tangent at X) 0
62
ii. Calculate the lengths of BC, CA and AB, and o
hence make an accurate drawing of the A ∙
x
arrangement
B

Baffour – Ba Series, Core Maths for Schools and Colleges Page 646
Solution A
Let q = 620, then x = 2q
x = 2 × 620 = 1240 C 700 y T
o x
2. Find the value of x in the diagram below:

B
A B 2 × 700 = x,
x = 1400
0
102
∙o y + x + 900 + 900 = 3600 (angles in a quadrilateral),

x
But x = 1400
Solution y + 1400 + 900 + 900 = 3600
Let x = q, then 2q = 1020 y = 3600 –1400 + 900 + 900 = 400
⇒ 2x = 1020
x= = 510 Exercises 23.3
1. In the figure below, name two angles at the
circumference which stand on (or are subtended
3. In the figure below, calculate the sizes of all
by): E
the angles in terms of x and y. Show that < AOB
i. the minor arc BC,
= 2 < ACB
ii. the minor arc DE, D
C
iii. the arc AED. A
0
X 0
A y
∙
O B C
2. In each of the figures below, find the sizes of
P the angle marked with letter y:
B

Solution
< AOB = < AOP + < POB 0
70
= 2x0 + 2y0 = 2(x + y)0 = 2 < ACB ∙y
4. TA and TB are tangents to a circle centre O. If
< ACB = 700, calculate angle ATB
3. In the figure below, calculate < AOB, if
A < ACB = 380
C
C 700 T
o

∙O
B A
Solution
B
Let < AOB = x and < ATB = y

Baffour – Ba Series, Core Maths for Schools and Colleges Page 647
4. Calculate < ACB in the figure below, if 8. In the figure below, TA and TB are tangents,
< AOB = 520 and O is the center of the circle.
A
C
0
∙ T
C
0
50 O ∙
A
B
B
5. In the diagram below, O is the center of the
circle. The points A, B, C are points on the i. If < ACB = 500, calculate the size of <ATB
circumference of the circle. Angles CAO and ii. If, ACB = x0, calculate the size of < ATB (in
AOB are 320 and 1400 respectively. terms of x)
Calculate: C
i. angle OBC Challenge Problems
ii. angle COB 1. ∆XYZ is inscribed in a circle with center O, and
0 o
32
140
0 OY and OZ are joined. If <YXZ = 600, calculate
A B the sizes of the angles of ∆ YOZ

6. In the figure below, calculate the reflex 2. ∆PQR is inscribed in a circle with center O,
< AOB, if < ACB = 1150 and OP, OQ and OR are joined. If < PQR = 1100
C and < QOR = 1300, calculate the sizes of the
angles of ∆PQR
B
O
A
∙ 3. Below is a figure with
O C
center O and < ACB = 300
300

A B
7. In the figure below, O is the center of the circle
and OM = MN = ON. Show that AB is equal in length to a radius of the
K circle

II. Every angle in a semi – circle is a right angle

O
∙ C

M N
A ∙O B
Calculate the sizes of the angles at the center
and the circumference subtended by: i. < AOB = 1800
i. the minor arc MN ii. < ACB = 900
ii. the major arc MKN

Baffour – Ba Series, Core Maths for Schools and Colleges Page 648
Worked Examples B. 1. AB is a diameter of a circle and C is a point
1. In the figure below, find the value of the angle on the circumference. < BAC = 450.
marked x. i. What is the value of < ACB?
C ii. Calculate < ABC.

0
x - 13 2. In the diagram below, C is the center of the
A
32
∙O B
circle. PQRS is a parallelogram inscribed in the
circle. STQ and PCQ are straight lines and RS is
parallel to QP. Angle SPC = 650.

Solution S R
<ACB = 900 (angle subtended at the
T
circumference by a diameter). 650
P Q
x - 13 + < ACB + 320 = 1800 C
x –130 + 900 + 320 = 1800
x = 1800 – 900 – 320 + 130 = 610
Show that triangles CQT and RST are similar and
Exercises 23.4 hence, find : < CRQ
A. In the figures below, O is the center of the
circle. Mark in the sizes of all the angles in the 3. In the diagram below, O is the center of the
circle. circle. /SN/ = /NM/ M
1. R
and < LMS = 440,
x
Find the value of x
200
P 650
O
∙ y Q L
S O
x
N

S
Challenge Problem
2. A
AB is a diameter of a circle of radius 10cm. BC is
x
a chord of this circle of length 16cm.
350
i. Calculate the length of AC.
∙O D ii. If D is the image of B under reflection in AC,
B z y
calculate the area of ∆ ABD.
iii. If E is the image of C under reflection in AB,
C calculate the area of quadrilateral ACBE.
3.
F
III. Angles in the same segment of a circle are
O
∙
z equal
x y A segment of a circle is a region of the circle
E G
bounded by a chord and an arc. Every chord
divides a circle into two segments.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 649
Angles at the circumference of a circle standing x = y (angles in the same segment of a circle), but
on the same arc are often described as angles in y = 420. Therefore, x = 420
the same segment and they are equal. < BEC + y + 580 = 1800, but y = 420
< BEC + 420 + 580 = 1800
Q
< BEC = 1800 – 420 – 58 = 800

P O < BEC = z (vertically opp. angles)

R
But < BEC = 800
Therefore z = 800
A B
∆ADE is similar to ∆BEC
In the figure above, < APB = < AQB = < ARB
2. Find the values of angles m and n in the figure
Similarly, in the figure below, below;
D
B m
E
C O
n
C x 1200
x
D
A A
B
< ABC = < CED, < BAC = < CDE and
Solution
< ACB = < ECD
2m = 1200 (property I)
∆ABC is similar to ∆DCE
m= = 600
Worked Examples m = n (angles in the same segment of a circle),
1. In the figure below, find the values of angles x, but m = 600. Therefore, n = 600
y and z. Which two triangles are similar?
3. Determine the value of the angles marked p
D and q in the figure below:
x
C B
z E y E
p
580 q
0
330 470 62
A C 500
B
D
A
Solution Solution
330 + 470 + 580 + y0 = 1800 620 = < DCE (Vertically opp. angles)
y = 1800 – 330 – 470 – 580 = 420 < DCE + 500 + q = 1800, (But < DCE = 620)
620 + 500 + q = 1800

Baffour – Ba Series, Core Maths for Schools and Colleges Page 650
q = 1800 – 620 – 500 = 680
F E
But p = q
⇒p = 680

4.i. Prove that triangles ABC and EDC in the C D

figure below are similar.
ii. If AB = 8cm, BC = 6cm, AC = 5cm and DE = 2. In the figure below, O is the center of the
12cm, calculate the lengths of CD and CE. circle Q
D

B P O
6 R
12
8
C
5
A B
A E

Solution i. If < AOB = 800, what are the sizes of angles

i. < = < (angles subtended at the APB, AQB and ARB
circumference by the same arc AE) ii. If < AOB = 720, what are the sizes of angles
< = < (angles subtended at the APB, AQB and ARB
circumference by the same arc BD) iii. If < AOB = 2x, what are the sizes of angles
< =< (vertically opposite angles) APB, AQB and ARB
iv. What can you say about the sizes of all angles
ii. /DE/ / / like APB, AQB and ARB, which are subtended by
/DE/ = k /BA/ the arc AB?
= k (Scale factor)
3. B
k= = 1.5
500
/CD/ /AC/ A
/CD/= k /AC/, but k = 1.5 and /AC/ = 5cm C
300
/CD/ = 1. 5 × 5cm = 7.5cm
D
/BC/ /CE/
In thediagram above, A, B, C and D are points on
/BC/ = k /CE/, but k = 1.5 and /CE/ = 6cm
a circle. /AB/ = /AC/, < DBC = 500 and < ADB =
/BC/ = 1. 5 × 6cm = 9cm
300. Calculate the value of < CAD

Exercises 23.5
4. In the diagram below, WXYZ is a circle with
1. In the figure below, mark four pairs of equal
center O. XZ and WY intersect at V. <XOY = 1100
angles at the circumference
and angle < YVZ = 1000. Calculate:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 651
i. angle XZY ii. < WXZ
X Y
Z
W
Z

0
V 100 35
0

A B
 0
110
Challenge Problems
Y 1. i. Explain why the triangles AEB and DEC
X
in the figure below are similar.
ii. If AB = 9cm, AE = 7.5cm, BE = 6cm and CE =
5. In the figure below, find the values of x, y, p 4cm, calculate the lengths of DE and DC
and q.
A
D
600 7.5
200
p 9 E
q 5
6
C
40 0 B
y x

2. i. Prove that triangles PQR and PST in the

6. In the figure below, fill in the sizes of all the figure below are similar.
angles ii. If TQ = 6cm, QP = 4cm, and PR = 8cm,
N calculate the lengths of PS
J 30
0

0
T
70
H
0 Q
40 M R
K
S P

7. Fill in the sizes of all the angles in the 3. Find the values of the angles named with
figurebelow. Why are the triangles similar? letters in the figure below;
S R
Z
60
0
c
Y
d g b
e m 0 T
50
0 0 25
h 0n 100
a f 40
Q W
P V

8. In the figure below, XY is parallel to AB. 4. In the figure below, < BPC = 500, < BRC = 600
Calculate the sizes of as many angles as possible and < BCR= 700. Calculate the angles of ∆ ABC

Baffour – Ba Series, Core Maths for Schools and Colleges Page 652
 B Worked Examples
1. In the figure below, O is the center of the
P circle. Calculate:
500
0 60
0
i. Reflex < BOD ii. x iii. y iv. x + y
70 R
C A A
Q
x
5. i. In the figure below, < MKN = < GKH. Prove
that triangles MKN and GKH are similar. O
0
ii. Name another pair of similar triangles in the 140 D
B
figure, and prove that they are similar.
y
K
Solution C
x x
i.< BOD + 1400 = 3600(sum of angles in a
circle)
M N
H < BOD = 3600 – 1400 = 2200

ii. 2x = 1400 (Property I)

G
Cyclic Quadrilaterals x= = 700
A quadrilateral whose vertices lie on the
circumference of a circle is called a cyclic iii. x + y = 1800(opposite angles of a cyclic quadrilateral),
quadrilateral. The vertices are said to be But x = 700
concyclic points. 70 + y = 180
y = 1800 – 700 = 1100
Properties
I. The opposite angles of a cyclic quadrilateral iv. x = 700 and y = 1100
are supplementary (sum up to 1800) x + y = 700 + 1100 = 1800 (supplementary)

x 2. In the diagram below, ABCD is a cyclic

a 0
x + y = 1800 quadrilateral. DAE, CBE and DCF are straight
0
a + b = 1800 lines. If h + m = 960, find the value of x.
E
y b
0 0
m

Similarly, in the figure below, A

x
B
B A 0 0 0 x
x i. x + y = 180
0 0
D ii. <CBA + <CDA = 180 h
o ∙ D
C F
0
y
C

Baffour – Ba Series, Core Maths for Schools and Colleges Page 653
Solution If < DQC = 760, < BPC = 520 and < BCP = x,
From ∆ABE, calculate the value of x
< BAD = x + m (exterior angle theorem)
Solution
From ∆CFB, From ∆DCQ,
< DCB = h + x (exterior angle theorem) < ADC = x + 760 (exterior angle theorem)
< BAD + < DCB = 180
(x + m) + (h + x) = 1800 From ∆BCP,
2x + m + h = 1800 (But m + h = 960) < ABC = x + 520(exterior angle theorem)
2x + 960 = 1800 < ADC + < ABC = 1800 ( *
2x = 1800 – 960
(x + 760) + (x + 520) = 1800
2x = 840
x + 760 + x + 520 = 1800
x= = 420 2x + 1280 = 1800
2x = 1800 – 1280
3. Find the values of angles x and y in the figure 2x = 520
below; x= = 260
C
D y B
x 0
110 2. The diagram below shows a circle PQRS with
Solution
center O, quadrilateral OPSR is a rhombus.
x + 1100 = 1800
< QPO = < ORQ = x and < PQR = q.
0 x = 1800 – 1100
50 Q Find: i. q
x = 700
ii. x
A q
y + 500 = 1800 iii. < QRS
y = 1800 – 500= 1300
O
Solved Past Question x
x R
In the diagram below, ABCD is a cyclic
quadrilateral. AB and DC are produced to meet at P
P, AD and BC are produced to meet at Q. S
Q Solution P

i. PORS is a rhombus, opposite angles are equal

0
76 < POR = < PSR = 2q
D Q

q
A C
x
O
2q x
x y
B 0 y
R
52
P 2q

P S
P
Baffour – Ba Series, Core Maths for Schools and Colleges Page 654
q + 2q = 1800 (Opp <s of a cyclic quad) Solution
3q = 1800 i. Consider □PRST;
q = 600 x + 1180 = 1800
x = 1800 – 1180 = 620
2q = 2 × 600 = 1200
Consider □QRST;
Consider OPQR, y + 1180 = 1800
< POR = 3600 – 1200 = 2400 y = 1800 – 1180 = 620
x + < POR + x + q = 3600 ( *

x + 2400 + x + 600 = 3600 Consider ∆ OQR (isosceles);

2x + 3000 = 360 < ROQ + y + y = 1800
2x = 3600 – 3000 < ROQ + 620 + 620= 1800
2x = 600 < ROQ = 1800 – 620 – 620
x = 300 < ROQ = 560

iii. < QRS = x + y + y z + < ROQ = 1800 (< s on a staright line)

Consider ∆POR z + 560 = 1800
y + y + 2q = 1800 z = 1800 – 560 = 1240
2y + 1200 = 1800
2y = 1800 – 1200 Now x = 620, y = 620 and z = 1240
2y = 600 620 + 620 = 1240
y = 300 x+y=z
S
< QRS = x + y + y ii. 0
118
< QRS = 300 + 300 + 300 n n R
T
x m m
< QRS = 900
Z
O
3. In the diagram below, O is the center of a x y
circle, /TS/ = /SR/, angle TPR = x, < TQR = y, P Q
<TOR = z and < TSR = 1180. < STP = x + m + n
S
0 From ∆ TSR;
118
T
R 2n + 1180 = 1800
2n = 1800 – 1180
Z 2n = 620
O n = 310
x y
P Q
From ∆ TOR;
i. Find the relationship between x, y, z 2m + 1240 = 1800
ii. Find angle STP. 2m = 1800 – 1240

Baffour – Ba Series, Core Maths for Schools and Colleges Page 655
2m = 560
S
m = 280
x

< STP = 620 + 280 + 310 O

< STP = 1210
0
y R
P 40
0
Exercises 23.6 100
1. In the figure below, O is the center of the Q
circle. 5. In the figure below, R
calculate the sizes of c
D S 115
0
Calculate: the angles
b Q
y C
i. Reflex < BOD marked a, and b
O 80
0 a
A x ii. x
P
iii. y
B iv. x + y
6. The diagram shows a circle PQRS with center
2. In the figure below, O is the center of the O. The reflex angle at O is 2040, angle QRS = 540
circle. Find the sizes of the angles marked m and and angle OPS = x, find the value of x.
n. H Q

m P
R
O x 0
0 24
E 100
0
0
240

n
F G
S K
3. Find the size of angle y in the diagram below
7. In the figure below,
O is the center of the
O
y
circle. ∙
100
0
x N
FO M
0 J
0
32
72 Calculate the sizes of:
i. < MKN (standing on the minor arc MJN)
ii. < MJN (standing on the major arc MKN
4. P, Q, R, S are four points on a circle center O
below; < PQR = 1000 and < OPQ = 400, find the Challenge Problems
value of y 1. In the diagram below, O is the centre of the
circle, < OQR = 320< MPQ = 150. Calculate:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 656
 i. < QPR
P Opposite Interior angle
15
0 ii. < MQO
M m

O

Exterior angle
0 n
32
Q
From the diagram above, m = n
R
2. The points A, B and C are all on the Worked Examples
circumference of a circle, andn O is the center. In the figure below, calculate the sizes of the
A angles named with letters.
T
C x O D a S
220 b
B
0 0
Work out the size of angle x 77 96
Q
R U
3. In the figure below, calculate the size of angle
Solution
< ABC in two different ways.
a = 960 ( *
D
C b + 770 = 1800 ( *
0
50
b = 1800 – 770 = 1030

0 2. Find the values of u v

45
A B angles u, v, w and x
in the figures
1140 x w 1100
below;
4. Calculate the sizes of all the angles in the
figure below in which ABC and DEF are double
Solution
chords
A v = 1140and u = 1100 ( *
0 B
100 C ⇒ u + w = 1800 ( *

75
0 But u = 1100
D F 1100 + w = 180
E
w = 1800 – 1100 = 700

II. An exterior angle of a cyclic quadrilateral is

x + v = 1800, but v = 1140
equal to the interior opposite angle
x + 1140 = 1800

Baffour – Ba Series, Core Maths for Schools and Colleges Page 657
x = 1800 – 1140 = 660 Solution
From the diagram,
Exercises 23.7 b = 480 and a = 330
1. In the figure below, calculate < BCD.
a + c + 480 = 1800,
A 1100 B0 E 330 + c + 480 = 1800 ( But a = 330)
70
c = 1800 – 330 – 480 = 990

D C 2. In the figure below, calculate the values of

angles x, y, z and a.
2. In the figure below, O is the center of the
circle. Calculate the angles of quadrilateral
PQRS. a
y
P Q T x
0
80
0
0 57
O 65 z
1500

R
Solution
S 650 + 570 + z = 1800 (< s on a straight line)
z = 1800 – 650 – 570 = 580
III. An angle between a tangent and a chord
through the point of contact is equal to any An angle between a tangent and a chord through
angle in the opposite segment . the point of contact is equal to any angle in the
opposite segment. Therefore; z = y, but z = 580
m
⇒y = 580
m
n
n Also, a = 650
m n m
x = 650 + 570 = 1220

n 3. In the diagram, WXYZ are points on the

Worked Examples circumference of a circle center O. <XOY = 600,
Find the values of the angles a, b and c in the <YWZ = 450 and < XYW = 800. Calculate < ZYW.
figure below; W
450
a Z
480
0
60
0
b c 330 80
X Y

Baffour – Ba Series, Core Maths for Schools and Colleges Page 658
Solution circle. If angle CBE = 350 and angle ADB = 630,
2 × < XWY = 600 calculate the following;
< XWY = = 300 D C i. ACB
ii. ACD
0
63 iii. ABC
XWZY is a cyclic quadrilateral and that opposite
iv. BAD
angles are supplementary (1800 ) 0
35 E
< XWY+ < YWZ+ < XYW + < ZYW = 1800
A B
0 0 0
But < XWY = 30 , < YWZ = 45 , < XYW = 80 and
< ZYW = ? 2. In the figures below, find the values of the
300 + 450 + 800 + < ZYW = 1800 angles marked with letters
< ZYW = 1800 – 300– 450 – 800 = 250 E
r
4. In the figure below, AB = AC and the tangent q D
0
92
BC touches the circumcircle of triangle ABC at B. F
Given that ̂ C = 700, find: i. ̂ T ii. ̂ B p 0
A 63
C T B
A C

0
3. In the diagram below, PQR is a tangent to the
70
circle at Q. < YQR = 540 and < XYQ = 410. Find
B the size of < XQY.
Solution X
i. < ABC = < BCA = 700 (base < s of an Isosceles ∆ )
Y
< ABC + < BCA + < CAB = 1800 410
700 + 700 + < CAB = 1800
< CAB = 1800 – 700 – 700 = 400
540
P
< CBT = < CAB = 400 Q R
< BCT = < TAB + < ABT 4. In the figure below, TB touches the circle at B
< BCT = 400 + 700 = 1100 and BD is the diameter. Angle TBA = 310 and
angle BAC = 690. Calculate:
ii. < ABT + < TAB + < ATB = 1800
T
700 + 400 + 400 + < ATB = 1800 A
< ATB =1800 – 700 – 400 – 400 = 300 D 69
0 i. < ADC
0
31 ii. < ABC
C iii. < CAD
Exercises 23.8 B
1. In the figure below, BE is a tangent to the
Ans i. 1000 ii. 800 iii. 210
circle and BD is a line through the center of the Not drawn to scale

Baffour – Ba Series, Core Maths for Schools and Colleges Page 659
Review Exercises 23.9 Given that < OCB = 500 and < OAB = 280,
In each of the following diagrams, find the calculate < BXC and < XBC.
value of the angles represented by letters.
1. 8. In the diagram below, AB is a diameter of the
circle and the tangent at C meets AB produced at
T. Given that < CAB = 310, calculate;
C
200 y x 300
A B
A 310 T
2. 3. B
N
i. < CBA
O O y
  ii. < CTA
136
0 L
J 420
x L M
9. Calaculate the angles of x and y in the
K diagrams below
4. M 5. D
K C

550 y
N 10. Calaculate the angles of x and y in the
b
diagramsO below O

N 350
25
y a c0
38 x A 80 C
250 700
P L 0 M B
A B
O

6. In the figure below, O is the center of the

circle. Find x and y x
0
O O 140
0
42
y
O 5x

x
11. Calaculate the angles of x and y in the
7. In the diagram below, O is the center of the diagrams below
circle. AOX and ABC are straight lines.
C
X 220
O O
o B
2x + 28

3x – 70 x
280 0
50 670
B
D
A B C
A

Baffour – Ba Series, Core Maths for Schools and Colleges Page 660
24 TRIGONOMETRY I Baffour – Ba Series

Trigonometry 1. Sine or sin

It is a branch of a mathematics that deals with 2. Cosine or cos
measurement of lengths and angles of right- 3. Tangent or tan
angled triangles.
B1
B1
Labeling a Right - angled Triangle Hypotenuse B1
A right- angled triangle is labelled as follows:

Opposite
B
I. The side that faces the right angle, also known
as the longest side is called the Hypotenuse (H),
II. The side that faces the acute angle, , is called O
the Opposite (O) A A1 A1 A1
III. The third side is called the Adjacent (A) Fig. I
Adjacent

From fig. I, the ratios are defined as,

θ
sin θ = =
Adjacent

= = =

Opposite cos θ = =

Exercises 24.1 = = =
A. For each of the following right – angled
tan θ = =
triangle, indicate the side that represents the
hypotenuse, opposite and adjacent.
a = = =
1. 2.

b c
These ratios can easily be remembered by the
θ θ acronyms: SOH, CAH, TOA; where
SOH = sin θ = ,
B. In triangle XYZ, x is a right angle
1. Which side is the hypotenuse? CAH = cos θ =
2. Which side is opposite to angle y? TOA = can θ =
3. Which side is adjacent to angle y?
4. Which side is opposite angle Z? Trigonometric Ratios of 300 and 600
5. Which side is adjacent to angle Z? The trigonometric ratio of 300 and 600 are derived
from an equilateral triangle of dimensions 2 –
Sine, Cosine and Tangent of Angles units as shown below;
The three basic trigonometric ratios are:
Baffour – Ba Series, Core Maths for Schools and Colleges Page 661
By Pythagoras theorem,
H C
22 = 2
+ 12
2 450
= 22 – 12
2
0
30 30
0 √
=3 1
2 2
=√
450 Fig. III
0 0
60 60 Fig. II A 1 B
1 A 1
The dimensions of fig. II can be shown below: From this fig. III, the trigonometric ratios of 450
H are obtained as follows;
√
cos 450 = =
√
0 0
30 30 √
2
sin 450 = =
2 √
√
tan 450 =
0 0
60 60 Fig. II
1 A 1 The trigonometric ratios of 900 are obtained as
0
From this figure, the trigonometric ratios of 30 follows
and 600 are obtained as follows: cos 900= = 0,
0 0 √
sin 30 = , cos 60 = √
sin 900 = = 1,
0 √ 0 √ √
sin 60 = , cos 30 = √
0 0
tan 900 = = undefined
tan 30 = , tan 60 = √
√

These ratios can be used instead of a calculator,

Trigonometric Ratios of 450 and 900 especially when answers are required in the form
C of surds.
450
Summary Table:
1
Ratios of 300, 450, 600 and 900
450
A 1 B 300 450 600 900
By Pythagoras theorem, sin √ √
2
= 12 + 12 = 2
cos √ √ 0
=√
The dimension of the figure is shown below; tan √ 1 √ -

Baffour – Ba Series, Core Maths for Schools and Colleges Page 662
Trigonometry Ratios for Any Angle Memory Aid: (Christian name, Surname)
The Unit Circle = (cosθ , sin θ ) = ( x, y)
The unit circle has its center at the origin(0, 0)
and the length of the radius is 1. Note: Using Pythagoras theorem:
Take any point p(x, y), on the circle, making an cos2θ + sin2θ = 1
angle of Ɵ, from the center.
y Values of Sin, Cos and Tan for 00, 900, 1800,
Cos θ = = x, 2700 and 3600

·
P ( x, y)
sinθ = = y, Both diagrams below represent the unit circle but
1 y θ using two different notations to describe any
θ tanθ = , tanθ =
x θ point p on the circle.
x

··
0
( cos 900, sin 90 )
P ( cos Ɵ, sin Ɵ)

· ·
1 Sin Ɵ
0 0
( cos180 , sin 180 ) ( cos 00, sin 00)
This very important results indicates that the θ
Cos Ɵ ( cos 3600, sin 3600)
coordinates of any point on the unit circle can be

·
represented by p(cosθ, sin θ), where θ is any
angle ( cos 2700, sin 2700)
y

··
(0, 1)

·
P ( cos Ɵ, sin Ɵ)
P (x, y)
1

·
1 Sin Ɵ y
(-1, 0) Ɵ (1, 0)
Ɵ
Cos Ɵ
x x

·
(0, -1)

As the point p rotates, θ changes. These By comparing corresponding points on both unit
definitions of cosθ and sin θ in terms of the circles, the values of sin, cos, and tan for 00, 900,
coordinates of a point rotating around the unit 1800, 2700 and 3600, can be read directly without
circle apply for all values of the angle θ0, the use of a calculator.

(Cos 00, sin 00) = (cos3600, sin 3600) = (1, 0) (Cos 900, sin 900) = (0, 1)
Cos 00 = Cos 3600 = 1 Cos 900 = 0
sin 00 = sin 3600 = 0 sin 900 = 1
tan 00 = tan 3600 = = 0 tan 900 = ( undefined)
(Cos 1800, sin 1800) = (-1, 0) (Cos 2700, sin 2700) = (0, -1)
Cos 1800 = Cos 3600 = -1 Cos 2700 = 0
sin 1800 = 0 sin 2700 = -1
tan 1800 = = 0 tan 2700 = tan 3600 = (undefined)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 663
The x and y axes divide the plane into four cos θ = x, sin θ = y and tan θ =
quadrants. Consider the unit circle below,
y cos (-θ) = x, sin (-θ) = - y, tan (-θ) = –
Thus;
2 1 cos (-θ) = cos θ
x- x+ sin (-θ) = - sin θ
y+ y+ tan (-θ) = - tan θ
x
3 4
x- x+ cos θ = x and sin θ = y Worked Examples
y- y- Ɵ Find without using tables or calculators
tan θ = =
1. sin (– 600) 2. cos (–300) 3. tan (–2250)
By examining the signs of the four quadrants, the
sign of sin θ, cos θ, tan θ for any value of θ can Solutions
be found y √
1. sin (– 600) = – sin 600= –
√
Summary of Signs S A 2. cos (–300) = cos 300 =
x 3. tan (–2250) = – tan 2250 = – tan 450 = – 1
T C
x
x Exercises 24.2
1st quadrant: sin cos and tan are ALL +
positive 1. tan (–1500) 2. tan (–2100) 3. cos (–1200)
2nd quadrant: sin is positive, cos and tan are 4. cos (–1350) 5. sin (-2700) 6. sin (-1350)
negative. y
-
3rd quadrant: tan is positive, sin and cos are Methods for Finding the Trigonometric Ratio
negative. - for Any Angle between 00 and 3600
4th quadrant: cos is positive, sin and tan are Method 1
negative. y
I. Draw a rough diagram for the angle in the x – y
-
plane, taking measurement in the anticlockwise
A very useful memory aid, CAST, in the diagram direction from the positive x - axis
above shows the ratios that are positive for the II. Determine in which quadrant the angle lies
angles between 00and 3600. III. Find its related acute angle to the nearest
horizontal line (x - axis)
Negative Angles IV. Use the trigonometric ratios of the related
Consider the unit circle showing angles θ and -θ angle
V. Use the CAST diagram below to find the signs
( cos Ɵ, sinƟ)
of the ratios
x y y
Ɵ
-Ɵ x S A
-y
T C
( cos (-Ɵ), sin(-Ɵ)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 664
Note:sin2 A = ( etc e.g. Sin 3000 = sin (3600 – 3000)
√
= – sin 600 = –
Method 2
b. cos ( 3600 - θ) = cos θ
If θ is the general angle, then
E.g. cos 3000 = cos (3600 – 3000)
1. In the first quadrant, (00 < θ < 900), all the
= cos 600 =
trigonometry ratios are positive
2. In the second quadrant, (900 < θ < 1800), only
c. tan ( 3600– θ) = – tan θ
sin is positive. That is:
E.g. tan 3000 = tan (3600 – 3000)
a. sin (180 – θ) = sin θ
= – tan 600
eg. sin 1200 = sin (1800 – 1200)
=–√
= sin 600
Summary
b. cos (1800 – θ) = – cos θ
eg.cos 1200 = cos (1800 – 1200 ) 900
= – cos 600
Sin 1800 - x All
0 x
c. tan (180 – θ) = – tan θ
eg.tan 1200 = tan (1800 – 1200 ) 1800 00 3600
= – tan 600 = √ x - 1800 3600 - x
Tan
Cos
3. In the third quadrant, (1800< θ < 2700), only
2700
tan is positive. That is
a. sin ( θ– 1800) = – sin θ Worked Examples
e.g. sin 2100 = sin (2100 – 1800) 1. Find cos 2100, leaving your answer in surd
= – sin 300 = – form:

Solution
b. cos ( θ – 1800) = – cos θ
Method 1
e.g.cos 2100 = cos (2100 – 1800)
√ I. The diagram for the angle is shown below;
= – cos 300 = –

c. tan (θ–1800) = tan θ

e.g.tan 2100 = tan ( 2100 – 1800 ) 300

= – tan 300 = –
√
Related angle Start
0 0
4. In the fourth quadrant, (270 < θ < 270 ), II. 2100 is in the 3rd quadrant
subtract the given angle from 3600 to obtain cos is negative in the 3rd quadrant
the related angle. That is, Related angle = (3600 – III. Related angle is 300
θ). Only cos is positive. That is explained below: √
IV.  cos 2100 = - cos 300 = –
a. sin ( 3600 - θ) = - sin θ

Baffour – Ba Series, Core Maths for Schools and Colleges Page 665
Method 2
Let θ = 2100 300
2100 is in the 3rd quadrant where only tan is 60
0

positive
Related angle = (θ – 1800)
= (2100– 1800) = 300
Related angle
⇒ 2100 is related to 300
Sin 2100 = – cos 300 = –
√ tan is negative in the 2nd quadrant
tan 1200 = - tan 60 = - √
2. Find sin 3150, leaving your answer in surd
Method 2
form:
Let θ = 1200
1200is in the 2nd quadrant, where only sin is
Solution
positive
Method 1
Related angle = (1800–θ)
3150 lies in the 3rd quadrant as shown in the
= (1800– 1200) = 600
diagram below;
⇒ 1200is related to 600
√
0
sin1200 = sin 600 = ,
45 Related angle
0
45
Simplifying and Evaluating Trigonometric
Expressions
Replace each angle with it‟s ratio and perform the
sin is negative in the fourth quadrant
included operation. If answer is required as a
Sin 3150 = – sin 450 = – surd, do so by rationalizing the denorminator
√

Worked Examples
Method 2
1. Evaluate sin 300 + 2 tan 450 without using
Let θ = 3150 calculators or tables
3150is in the 4th quadrant, where only cos is
positive Solution
Related angle = (3600–θ) Sin 300 + 2 tan 450
= (3600– 3150) = 450
= + 2 (1) = + 2 = 2 =
⇒ 3150 is related to 450
sin3150 = - sin 450 = –
√ 2. without using tables or calculator evaluate cos
450 + sin 300.
3. Find without using tables, tan 1200,
Solution
Solution cos 450 + sin 300
Method 1 √ √
+ =
1200 lies in the 2nd quadrant

Baffour – Ba Series, Core Maths for Schools and Colleges Page 666
3. Without using tables or calculators, simplify ⇒ 2 sin 2100 = 2 (- sin 300)
sin 1350 – sin 3150 = 2(– ) = -1
⇒3 cos 315 – 2 sin 2100 0
Solution √
sin 1350 – sin 3150 – (-1 ) = + 1 =
√ √
√ √ √ √ √
–(– )= + = =√
Method 2
2 0
4. cos 45 + sin 30 0 3 cos 3150 – 2 sin 2100
√ √ √
√ = ( ) – 2( )= +1 =
=( ) + = + =1

Solved Past Question 2. Simplify: , leaving your answer

1. Without using mathematical table or calculator, in surd form.
evaluate 3 cos 3150 – 2 sin 2100, leaving your
answer in surd form.
Solution
Solution
Method I √
3150 lies in the 3rd quadrant as shown in the =
√
diagram below; √ √
= Rationalise the denorminator
√ √
(√ (√
450 Related angle =
0 (√ (√
45
√ √
= = + =2+√
cos is positive in the fourth quadrant
√
cos 3150 = cos 450 = = 3. Evaluate , leaving your answer
√
in surd form.
⇒ 3 cos 3150 = 3 ( ) =
√ √

Solution
0
The diagram for angle 210 is shown below; ,
√
√
= √
(solving the numerator)

300 √ √
= √

Related angle Start √ √ √

=
2100 is in the 3rd quadrant and the related angle is =
√ √
×
√
300, but sin is negative in the 3rd quadrant
( √ √
 sin 2100 = - sin 300 = – =–
√

Baffour – Ba Series, Core Maths for Schools and Colleges Page 667
 ( √ √
=– 5. Without using table, evaluate .
√
Rationalise the denorminator
( √ √ √ 6. Express in surd form.
=– ×
√ √
√ ( √ √
=– (√ )(√
C. 1. Find + , if x = 3 cos 600 and y = 2 sin 450.
√
=– 2. Given that x = sin 450 and y = cos 450, find the
√
=– value of , without using tables or
calculators.
4. If = a + b√ , find the values of a and b.
3. If x = cos 600 and y = sin 600, evaluate ,
Solution without the use of tables or calculators

√
4. If p = sin 300 and q = cos 300, evaluate ,
= = √
=2 = 2× =
√ ⁄ √ √ without tables or calculators.
√
= × Rationalise the denorminator
√ √ 5. If x = tan 300 and y = tan 450, find the value of
( √
= .
( √ ( √
√
=
Inverse of Trigonometric Ratios
= 8 + 4√ It is the reverse process of determining the angle
⇒8 + 4√ = a + b√ given the value of one of the trigonometric ratios.
Therefore a = 8 and b = 4 The inverse of a trigonometric ratio is written as (
- 1
). Thus, is pronounced “inverse sine x”
Exercises 24.3
or “sine inverse x” or “arc sine x”
A. Evaluate, leaving your answer in surd form
where necessary Consider the triangle below;
1. cos 1200 + cos 2250 2. tan 2400– tan 3300
3. sin2 600 + cos2 450 4. tan2 300– sin2 600
2
√
B. 1. Without using tables or calculators, find the θ
value of sin 600cos 450 + cos 600 sin 450 1
2. Without using tables or calculators, find the Ratio Angle
√ √
value 1 + cos2 30. sin θ = θ=
3. Express as a surd. cos θ = θ=
4. Express as a surd. tan θ = √ θ= √

Baffour – Ba Series, Core Maths for Schools and Colleges Page 668
However, a problem arises with the sin-1x, cos-1x tan y = √ = 1.732
and tan-1x notation. ⇒y = tan -1(1.732) = 600, 2400

Consider the following equations; Exercises 24.4

If sin θ= , then θ = = …,-330 , 0
Find the values of θ, where 00 θ 900
-2100, 300, 1500, 3900, 5100,… 1. cos θ = – 2. cos θ = 0.6
If tan θ = , then θ = =…,-3150, -1350, 3. sin θ = -0.7660 4. cos θ = 0.8
450, 3150, 4050, 6750, …
Hence, if there is no restrictions on the value of θ, Trigonometric Equations
then the equation sin θ = and tan θ = 1 have an Given a trigonometric equation, the intention is to
infinite number of solutions. find the value of the variable (in degrees)
This problem is overcomed by restricting the that satisfies the equation. The satisfactory value
value of θ to the range - 900 θ 900. is called the truth set or solution set of the
The angles within this range are often called equation.
“principal values”. I. Identify the left hand side (L.H.S.) and the right
hand side (R. H. S.)of the given equation.
Using these restrictions for θ, we often obtain a II. Identify the location of the trigonometric ratio
√ and divide both sides of the equation by its co-
single value for our answer. Thus, = 600, efficient, if any.
not also 1200, as 1200 is not in the range -900 to III. If the trigonometric ratio is at the L.H.S, find
900 its inverse at the R. H. S and vice – versa
to obtain an angle.
Worked Examples IV. Expand brackets if any and group like terms
1. Find the principal values of each of the to work out the value of the variable with
following; restrictions to the given range.
i. ii. iii. ( )
√ √
Worked Examples
1. Given that tan (x + 250) = 5.145, where 00 x
Solution
900, find to one decimal place the value of x.
i. = ( = 300
Solution
ii. = ( = 450
√ tan (x + 250) = 5.145
iii. ( )=- ( ) = - 300 x + 250 = (5.145)
√ √ 0 0
x + 25 = 79
√ x = 790 – 250 = 540
2. Given that sin y = - and tan y = √ . Find y.
2. Find θ if cos (θ + 600) = 0.0872, where 00 x
Solution 900
The only quadrant in which sin is – ve and tan is
+ ve is the 3rd quadrant.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 669
Solution Solution
cos (θ + 600) = 0.0872 5 sinx = 4.33
θ + 600 = (0.0872) sin x =
0 0
θ + 60 = 85
θ = 850 – 600 = 250 x= ( ) = 600

2. If 8 sin x + 2 = 5, find x correct to the nearest

3. Find the truth set of 1.414 sin p = 1, where 00
degree.
p 900
Solution
Solution 8 sin x + 2 = 5
1.414 sin p = 1 8 sin x = 5 – 2
8 sin x = 3
sinp =
sin x =
p= ( ) = 450
x= ( ) = 220

4. If 1.5 cos x = 0.75, find x if 00 x 900 Exercises 24.5

A. Solve the following for 00 θ 900
Solution
1. sin ( = 0.8 5. sin (2 12) =
1.5 cos x = 0.75
cosx = 2. tan ( θ = 6. cos θ =
√
x= ( ) = 600 3. 8 cos θ = 3 7. sin (θ =
4. 4 tan θ = 3 8. 3 cos (θ – 100) = 1
5. If x cos 600 = 1.5 and y sin 300 = 2, evaluate;
√ B. 1. If 10 sin (x – 45)0 = 5, where 00< x < 900,
find the value of x.
Solution 2. Given that 2 cos x = √ , where 00 < x < 900,
x cos 600 = 1.5 find the value of x.
x= =3 3. Solve the equation 2 sin2 θ = sin θ, for the
values of θ from 00 to 1800.
y sin 300 = 2 4. Solve the equation 2 cos2 θ = cos θ, for the
values of θ from 00 to 1800.
y= =4

Equality of sine and Cosine of Complementary

Substitute x = 3 and y = 4 in √ Angles
⇒√ =√ =5 Complementary angles are two angles that sum
up to 900
Some Solved Past Questions
1. Given that 5 sin x = 4.33, where 00 x 900, Consider the triangle below;
find x correct to the nearest degree.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 670
 R
2. Given that cos (2x – 23)0 = sin 470. Find the
value of x where 00 < x < 900
p
q
Solution
cos (2x – 23)0 = sin 470
P r Q
cos (2x – 23)0 = cos (90 – 470)
sin R = and cos Q = ⇒2x – 230 = 430
2x = 230 + 430
Hence sin R = cos Q = cos (900 – R)
2x = 660
x = 330
Similarly, cos R = sin (900 – R). Therefore, the
sine of an angle is equal to the cosine of its
3. Given that sin (5x – 28)0 = cos (3x – 50)0, find
complementary angle and vice – versa. For
the value of x where 00< x < 900
example:
sin 750= cos (900 – 750)
Solution
sin 750= cos 150
sin (5x – 28)0 = cos (3x – 50)0
cos 300 = sin (900 – 300)
But sin (5x – 28)0 = cos 900 – (5x – 28)0
cos 300= sin 600
= cos 90 – 5x + 280
Worked Examples = cos (900 + 28o– 5x)
1. If sin (x + 300) = cos 400, find x where 00< x < 900 = cos (118 – 5x)

Solution ⇒cos (118 – 5x) = cos (3x – 50)0

Method 1 1180 – 5x = 3x – 500
Expressing cos 400 as a sin 1180 + 50 = 3x + 5x
sin (x + 300) = cos 400 1680 = 8x
sin (x + 300) = sin (900 – 400) x = 210
sin (x + 300) = sin 500
x + 300 = 500 3. Without, using tables or calculators, find the
x = 500 – 300 = 200 value of – .

Method 2 Solution
Expressing sin (x + 300) as a cos
–
sin (x + 300) = cos 400
cos [900 – (x + 300)] = cos 400
cos (900 – x – 300) = cos 400 But cos 500 = sin (900 – 500) = sin 400
cos (60 – x)0 = cos 400 cos 220 = sin (900 – 220) = sin 680
600 – x = 400 ⇒ – =2–1 =1
x = 600 – 400
x = 200 Exercises 24.6
1. If sin 5x = cos 200, find the value of x.

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2. If sin 250 = cos(y + 50)0, find the value of y Solution
3. If cos 3y = sin 2y and 0 0
y 0
90 , find the cos 1500 = cos (90 + 60)0
But cos (x + y)0 = cosxcosy – sin x sin y
value of y. ⇒ cos (90 + 60)0
4. If sin(x – 10 )0 = cos(x + 10)0 and 00 x 900, = cos900cos 600 – sin 900 sin 600
√ √ √
calculate the value of x. =( ( ) –( ( )=0– =–

Expressing a Given Ratio as a Sum or 3. Express cos 150as adifference of two

Difference of Two Trigonometric Ratios trigonometric ratios and hence find the value of
A given angle θ, can be expressed as a sum of cos 150, if cos (α – β) = cosα cosβ + sin α sin β.
two special angles, α and β. ⇒θ = (α + β).
Likewise, θ, can be expressed as a difference of Solution
two special angles, α and β. ⇒θ = (α – β). Hence, cos (α – β) = cosα cosβ + sin α sin β
note the following identities: cos 150 = cos (45– 30)0
1. sin (α + β) = sin α cos β + cos α sin β cos (45 – 30)0 = cos 450 cos 300 + sin 450 sin 300
2. sin (α – β) = sin α cos β – cos α sin β √ √
= ( ) ( )+ ( ) ( )
√

3. cos (α + β) = cos α cos β – sin α sin β √ √

4. cos (α – β) = cosα cosβ + sin α sin β = +
√ √
5. tan (α – β) = =

6.tan (α + β) = Exercises 24.7

Express the following as a sum or difference of
Worked Examples two trigonometric ratios and simplify leaving
1. Express 1350 as a sum of two trigonometric the answer in surd form:
ratios and find its value if sin (a + b) = sin a cos b 1. cos 750 2. sin 1650 3. sin 300
+ cos a sin b. 4.cos 1650 5. cos 150 6. sin 1050

Solution Applications to Right – angled Triangles

sin 1350 = sin (90 + 45)0 Finding the Value of the Interior Angles of a
But sin (a + b) = sin acosb + cosa sin b Right Triangle from Two Given Sides
⇒ sin (90 + 45)0 I. Name an angle as θ, if not given in the diagram
= sin 900 cos 450 + cos 900 sin 450 II. Apply the appropriate trigonometric ratio that
√ √
= (1) ( ) + (0) ( ) =
√ links the two given sides.
III. Find the inverse of the ratio applied to
2. Express cos 1500 as a sum of two trigonometric determine the value of the angle named θ.
ratios and hence find the value of cos 1500, if cos IV. Make use of the fact that the sum of interior
(x + y)0 = cosxcosy – sin x sin y. angles of a triangle is 1800.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 672
Worked Examples ii. Find the height above the ground at which the
1. Find sin θ and the value of θ in the right – upper end of the ladder touches the wall.
angled triangle below;
Solution
Solution 8cm i. Let the angle the ladder makes with the wall be θ
From the diagram, 5cm sin θ = =
O = 5cm and H = 8cm
θ θ= ( )
sin θ = = Ɵ
0 10m
0 θ = 27 Wall
θ= ( ) = 39 (to the nearest degree)

2. Find cos θ and the value of θ in the figure

below; 4.5m

12cm ii.The height above the ground at which the upper

Solution
end of the ladder touches the wall be y
From the diagram,
A = 6cm and H = 12cm θ cos 270 =
6cm
cos θ = = cos 270 =
10m 270
θ= ( )= y = 10 cos 27 0 Wall
y
y = 8.9 (2 s. f.)
3. In the diagram below, find sin and the value
of θ 4.5
m
Solution 7cm 2. A ladder is leaning against a vertical wall. The
Let the unknown side bea θ ladder is 6m long and the foot of the ladder is 2m
5cm
+ from the base of the wall. Find the angle the
49 + 25 ladder makes with the wall.
= 74
a = √ = 8.60cm (2 d.p) Solution 6m θ
Let the angle the ladder makes
⇒O = 7cm and H = 8.60cm
with the wallbe
sin θ = = 2m
sin θ =
θ = sin-1( ) = (to the nearest degree)
θ= ( )

Word Problems θ=
1. A ladder 10 m long rests against a vertical wall
so that the distance between the foot of the ladder Exercises 24.8
and the wall is 4.5m. In the diagrams below, find the ratio for angle
θ and the value of angle θ
i. Find the angle the ladder makes with the wall.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 673
1. 2. III. Make the unknown side the subject of the
6cm equation and workout for its value.
15cm θ
9cm
8cm Worked Examples K
θ
1. In the diagram below, 62
0

3. 4. KGM is a right triangleand

angle GKM = 620. Find the
8cm 18cm
length of KM and KG. M
G 9cm
θ
θ 20cm 15cm Solution
8c
m against
sin =
B. 1. A ladder which is 8m long is placed
a vertical wall. If the ladder reaches 6.3m up the = = 10.2cm (3 s. f)
wall, calculate the angle the ladder makes with
the wall. By Pythagoras theorem,
= +
2. A ladder is12m tall. If the ladder is placed But = 10.2
against a vertical wall 7.5m high, calculate the = +
angle the ladder makes with the wall. – =
= 23.04
3. A ladder is placed against a vertical wall of
=√ = 4.8cm
height 13m. Calculate the angle the ladder makes
(sin, tan and cos can be used to find the side KG)
with the horizontal, if the ladder is 25m high.

2. A ladder leans against a wall. The end of the

4. Linda is flying a kite. She pins the end of the
ladder touches the wall 12m from the ground.
kite to the ground so that the string is pulled
The foot of the ladder is 9m away from the foot
straight as the wind pushes on the kite. If the
of the wall.
length of the kite string is 7.8m and the kite has a
i. What is the length of the ladder?
vertical height of 5.9 above the ground, what
ii. Calculate the angle the ladder makes with the
angle does the string of the kite makes with the
ground. Wall
horizontal ground?

Solution Ladder x
The Length of the Sides of a Right – angled 12cm
i. By Pythagoras theorem,
Triangle Given the Value one Acute Angle and
x2 = 92 + 122
a side θ
I. Identify the given side, the unknown side and x=√ 9cm
the given acute angle. x=√ = 15m
II. Find the ratio of the angle that relates the other
ii. Let θ be the angle the ladder makes with the
two sides (unknown side included).
ground

Baffour – Ba Series, Core Maths for Schools and Colleges Page 674
O = 12cm and A = 9cm, TOA is applicable height of the kite above the ground when the
tan θ = string is taut and its inclination to the horizontal
is 650.
θ = tan-1( ) = 530 (Nearest degree)
Solution
3. A ladder 8.5meters long leans against a vertical Let k represent the kite,KG represents the height,
wall. The top (T) of the ladder makes an angle of h, of the kite above the ground and HG be the
580 with the wall. How far, correct to one decimal ground level; K
place, is the foot (F) of the ladder from the wall? sin 650 =

Solution sin 650 =

58m hm
Let the distance between the ladder and the foot h = 58 sin 650 = 52.5
of the wall be x; T Hence the kite is
53m high (2 s. f) 650
580 H G
From the diagram
= 8.5m 6. In the figure below ABCD is a rectangle in
x = 8.5 sin 580 which AD = 43.7cm DE = 100cm, and E ̂ C =
x = 7.2m 270. Calculate:
F x i. AB ii. EC iii. E ̂ B
4. In the diagram below, O is the center of the
P Solution
circle, <QPR = 500
i. From ∆CDE;
and /QR/ = 10cm
cos 270 =
Calculate to one
decimal place,
the radius of the Q 10cm
∙
O
500

DC = 100 cos 270 = 89.1cm

100cm
E

circle. But DC = AB 0
27
R (length of a rectangle) D C
Solution AB = 89.1cm 47.3cm
<PQR = 900 (angle in a semi – circle)
Let PR = x ii. From ∆CDE; A
0
B
sin 50 = sin 270=
x= = 13.05cm /EC/ = 100 sin 270 = 45.4cm

iii. From ∆ABE,

But the radius, r = x
Let < EAB = θ
= × 13.05 = 6.5cm
tan θ =
But BE = EC + CB = 45.4 + 47.3 = 92.7cm
5. A kite in the air has its string tied to the
AB = 89.1cm
ground. If the length of the string is 58m, find the

Baffour – Ba Series, Core Maths for Schools and Colleges Page 675
tan θ = From ∆OAB
sin 850 =
θ= = 460
But AP = 7.1 cm
6. A protractor , in the shape of a semi circle, sin 850 =
center O and radius 5cm, is held in vertical plane /CP/ = 7.1 sin 850 =7.1cm
with one corner on a table as shown below;
P
Note:
Always watch the degree of accuracy permitted in
5cm
answers to problems involving measured
m
quantities. In the above example,
O
0
40 (no 5), the given measures 58 and 65 have only
two significant digits. This as a matter of fact
The straight edge makes an angle of 400 with the
limits our answer to two significant digits
horizontal. Calculate:
i. the height of O above the table,
ii. the height of P above the table. Some Solved Past Questions
1. In the diagram below, XYZ is a triangle, /XZ/
Solution = 3cm, /YZ/ = 5 cm and <YXZ = < XNZ = 900.
i. Rename the figure as shown below; Find /XN/
P X

0 3cm
45
5cm
m Y N Z
0 5cm O 5cm
45
400
A C B Solution
/OP/ = /OA/ = 5cm Applying Pythagoras theorem on ∆ XYZ,
∆OAP is isosceles, ⇒ base angles are equal = +
<A= 450 and <P = 450 But = 5cm, = 3cm,
Let the height of O above the table be /OB/ = +
From ∆OAB; – =
X
=
sin 400 =
= √ = 4cm
/OB/ = 5 sin 450 = 3.5cm 4cm 3cm

ii. From ∆OAB, From ∆ XYZ,

Ɵ
sin θ = Z
/AP/ 2 = 52 + 52 Y N
/AP/2= 25 + 25 5cm
AP = √ = 7.1cm θ= ( )
θ = 370 (2. s. f)
Let the height of P above the table be CP

Baffour – Ba Series, Core Maths for Schools and Colleges Page 676
From ∆ XYN, 2. A ladder, 8m long leans against a wall and
X
0
sin 37 = makes an angle of 600 with the ground;
530 37
0 a. how high up the wall does the ladder reach?
/XN/ = 4 sin 370 3cm
4cm b. how far from the wall is the foot of the ladder?
= 2.4 cm
370 530
Y Z 3. A ladder 12m long, leans against a wall and
N
makes an angle of 530 with the ground;
5cm i. how high up the wall does the ladder reach?
ii. how far from the wall is the foot of the ladder?
2. In the following diagram, PR is perpendicular
to QS, angle QPR = 600, angle PSR = 300 and /
B. Find the angles and sides indicated by
PQ / = 8cm, find the length of QS
letters in the following:
P 1. 2.
13
600
n
0
8 16 h 7
300 2
Q R S 520
Solution
Let /QR/ = xand /RS/ = y 3.
15 18
⇒ /QS/ = x + y p
0
sin 600 = 47
m0
4.
sin 600 =
x = 8 sin 600 = 7 cm

/PR/2 + /RQ/2 = /QP/ 2 r

/PR/2 = /QP/ 2 – /RQ/ 2 u0 550
/PR/2 = 82 – 72
10 12
/PR/2 = 15
/PR/ = 4 5.
0
tan 30 =
y= = 7 cm q

0 0
25 40
But /QS/ = x + y = 7 cm + 7cm = 14 cm
t
50 m
Exercises 24.9
A. 1. In a right – angled triangle DEF, the Determing Other Ratios from a Given One
hypotenuse /DE/ = 35cm long, and angle D is Draw a right – angled triangle and name one
400. Find /EF/ and /DF/. acute angle as θ.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 677
 (
i. sin θ = = .
(

Ɵ O = 3 and H = 5, Adjacent (A) = ?

Then label the dimensions of the triangle by 5
By Pythagoras theorem,
observing the following; 3
I. Given sin θ= ,ais the side that facesθ Ɵ

(opposite) and b is the side that faces the right – = 16

angle (hypotenuse) A=√ = 4cm

a b Now O = 3cm and H = 5cm, A = 4cm

Ɵ (
cos θ = =
(
(
II. Given cosθ = , a is the side containing the tan θ = =
(
right angle and angle θ(adjacent) and b is the side
that faces the right angle (hypotenuse)
ii. Perimeter of the triangle;
P = O + A + H = (3 + 4 + 5)cm = 12cm
b

iii. The interior angles of the triangle are the

θ
a a. right – angle =

IV. Given tan θ = , a is the side that facesθ b. sin θ =

(opposite) and b is the side containing the right ( )= (to the nearest degree)
angle and angle θ (adjacent). c. ( =
 The interior angles are 900, 370 and 530
a
θ 2. Given that sin y = , where 00 y 900, find
b
tan y. Hence, find the truth set of cos x = 1 – tan
V. Apply Pythagoras theorem to get the length of y, giving your answer correct to the
the third side. nearestdegree.
VI. Find the required trigonometric ratio.
Solution 8
(
7
Worked Examples sin y = = .
( y
1. In a right – angled triangle, sin θ = . Find:
O = 7 and H = 8, A = ?
i. cos θ and tan θ
ii. the perimeter of the triangle By Pythagoras theorem,
iii. the interior angles of the triangle

Solution

Baffour – Ba Series, Core Maths for Schools and Colleges Page 678
 – = 15 ⁄
= = ( ⁄ )
A=√ 4cm (to the nearest cm)
⁄ ⁄
= = =
Now O = 7cm and H = 8cm, A = 4cm ⁄
(
tan y = =
(
4. If 4 tan x = 3,where 00 x 900,
cosx = 1 – tan y, i. find the value of x
cosx = 1 – ii. evaluate –
cosx = 0.75
x=- ( = 1390 ( Nearest degree)
Solution
4 tan x = 3
3 h
3. If sin x = , find the value of
tanx =
x

Solution x= ( ) = 370 4
17
Method 1 8

sinx = x ii. Method 1

Let the hypotenuse of the triangle beh and
Let the third side of thetriangle be a
x = 370
By Pythagoras theorem,
172 = a2 + 82 cos 370 = 3 h

a2 = 172 – 82 8
17 370
2 h= = 5cm 4
a = 225
x
a=√ = 15 a = 15 cos370 = , but x = 370

⇒cosx = , substitute in
tanx = and 2tan x = 2 × ( ) =
⁄ ⁄
Substitute the values of tan x and 2 tan x = = = =
⁄ ⁄
⁄ ⁄
in = = = =
⁄

ii.Method 2
Method 2 Let the hypotenuse of the triangle be h
sinx = 17 By Pythagoras theorem,
8
h2 = 42 + 32
x= ( ) = 280 x
h2 = 25 3 h
Let the third side of the triangle be a h = √ = 5cm x

tan 280 = 4
17 cos x = substitute in
a= = 15 8
⁄ ⁄
0 280 = = = =
⇒tan 28 = ⁄ ⁄
a

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6. Given that sin x = , 00 < x < 900 find correct 2. If cos p = , where 00< p < 900, find the value
to 3 decimal places of

Solution 3. Given that tan x = , where 00 < x < 900, what

sinx = , but sin x= is the value of sin x + cosx.
⇒O = 5 and H = 8
By Pythagoras theorem, 5 8 4. sinθ = and θ is obtuse, find the value of

x .
a 5. If sin x = , where 00 x 900, find the value of :
–
– i. 1 – cos2 x, ii. 1 + cos2 x,
5 8 iii. ( , iv. (
a=√ x
6 6. If sin θ = 0.65, find the following for 00 x 900;
a=6
i. θ ii. tan θ iii. iv. ( )
cosx = ⇒cos x =
7. If cosx = 0.8, find the value of the following
tanx = ⇒ tan x =
for 00 x 900:
sinx = ⇒ sin x = i. cos2x – sin2x ii. –

⁄ ⁄ ⁄
= = = × = = 0.075 Application to Other Triangles
( ⁄ ) ⁄
Notation
The diagram below shows the usual notation for a
Exercises 24.10 triangle in trigonometry.
A
A.1. Find without using tables, the values of Angles: A, B, C c b
Length of sides: a, b, c
cos θ and tan θ, if sin θ = B C
2. Find without using tables, the values of cos θ a
The length of the sides are donated by a lower
and tan θ, if sin θ = case letters, and named after the angle they are
3. In a right – angled triangle, tan θ = , opposite. i. e. a is opposite to angle A, b is
opposite to angle B and c is opposite to angle C
i. Find sin θ and cos θ
ii. What is the perimeter of the triangle?
Sine and Cosine Rule, Area of a Triangle
iii. Find the interior angles of the triangle
Sine Rule:
A
= = c b
B.1.If sin A = and cos B = , find the value of
sin A cos B – cos A sin B, B C
a

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Alternatively, triangle, but with right – angled triangles the
basic trigonometric ratios are used.

4. The largest angle of a triangle is opposite the

Cosine Rule (for length of sides):
largest side and the smallest angle is opposite the
= + – 2bc cosA
shortest side. There can be only one obtuse angle
= + – 2ac cosB
in a triangle.
= + – 2ab cosC
Tackling Problems in Trigonometry
Alternatively (for size of angles);
I. If not given, draw a diagram, and put in as
cos A = much information as possible.
II. If two, or more, triangles are linked redraw the
cos B =
triangles separately.
cos C = III. Watch for common sides which link the
triangles (i.e. carry common values from one
Area of ∆ abc triangle to another triangle).
A = ab sin C = ac sin B = bc sin A IV. Use the sine or cosine rule as needed.

Worked Examples
When to Use the Sine Rule
1. In ∆ abc, /ab/ = 3, /ac/ = 5 and /bc/ = 7.
Use the sine rule if you know:
Calculate:
1. Two angles and one side.
i. the measure of the greatest angle of the triangle.
2. Two sides and an angle opposite one of these
ii. the area of ∆ abc, leaving your answer in the
sides.
√
form , where b is a prime.
When to Use the Cosine Rule
Use the cosine rule if you know: Solution
1. Two sides and an included angle. i. The largest angle is opposite the largest side
2. The lengths of the three sides. Using the cosine rule,
= + – 2bc cos A
Note: = 5 + 32 – 2(5)(3) cos A
2

49 = 25 + 9 – 30 cos A b
1.As a general rule, if you cannot use the sine rule
then use the cosine rule. 30 cos A = 25 + 9 – 49
30 cos A = - 15 7
3
2. If two angles are given, workout the third angle cos A = =– A
c a
straight away, as the three angles in a triangle add 5
A= ( ) = 1200
up to 1800.

3. The sine and cosine rules and the area of the ii. Area of ∆ abc = bc sin A
triangle formulas also apply to a right – angled = (5)(3) sin 1200 = (5)(3)( ) =
√ √

Baffour – Ba Series, Core Maths for Schools and Colleges Page 681
2. In the diagram below, /pq/ = 4cm, /pr/ = 5cm,
A
/qr/ = 6cm and / <psr / = 220
p
24cm
4 13cm
5
220 B 17cm D
q r s C
6 Solution
Find /ps/, correct to two places of decimals Let x represent /CD/
A

Solution
24cm A
Work on the two angles triangles separately
13cm
I. Consider ∆ pqr, p
Given three sides, B 17cm D C 24
4 5 x
use the cosine rule; 13

cos R =
R Consider ∆ ABD, D
q 6 r B 17 D
Using the cosine rule,
cos R = = = = + – 2(17)(13) cos D
( (
576 = 289 + 169 – 442 cos D
R= ( ) = 41.410 (2 d. p) 576 = 458 – 442 cos D
442 cos D = 458 – 576
 / <prs / = 1800 – 41.410 442cosD = -118
= 138.590 cos D = A
p
D= ( ) = 1050 15
0

24cm
4 5 13cm
138.590 0 0 0 0
41.41 0 22 tan 15 = 105 75
q r s B
6 17cm D x C
tan 150 =
II. Consider ∆ prs, x = 13 tan 150 = 3.5
Having two angles and one side, use the sine rule Hence /CD/ = 3.5cm
to find /ps/, p
= r Exercises 24.11
Unless otherwise stated, where necessary, give
r= 5
138.590 the lengths of sides and areas correct to 2
220
r = 8.82849883 r s decimal places and give angles correct to one
place of decimals
Thus /ps/ = 8.83cm (2 d. p)
A. 1. In ∆ pqr, p = 7 cm, P = 300, Q = 840. Find
R, q and r
3. In the figure below, /AB/ = 24cm, /BD/ = 2. In ∆ abc,b = 8 cm, c = 10cm, and A = 600. Find a
17cm, AD = 13cm and < ACD = 900. Find /CD/

Baffour – Ba Series, Core Maths for Schools and Colleges Page 682
3. In an acute – angled triangle ABC, the lengths tables or calculator, find the area of the triangle.
of AB, AC are 10.3 cm , 15.7 cm respectively, and
the angle C is 400, calculate to the nearest tenth of 9cm
a degree, the angle A
A
4. In ∆XYZ, XY = 17.2 cm, YZ = 21.3 cm, ZX =
10cm
16.0 cm. Find the three angles and the area of the
triangle 3. The area of the triangle below is 6cm2. Find the
5. In ∆ABC, a = 9.5 cm, b = 8.4 cm, C = 730. value of x.
Calculate; x
i. the area the triangle,
ii. the side c, 300
iii. angles A and B. (x + 2) cm
4. In the diagram below,
B. Without using calculator, find the area; p
1.
3√ 4 5

450 220
q r s
5 6

√
/pq/ = 4cm, /pr/ = 5cm, /qr/ = 6cm and / <psr / =
2. 220. Find /ps/, correct to two places of decimals
600
3. 4
5. In ∆ ABC, A = 51, B = 680, and c = 4.5m.
Calculate; i. side b, ii. the area of ∆ABC.
4
1200
5√ Angles of Elevation and Depression
4. M Suppose you are on the ground and you decide to
see a bird on top of a tower. You have to turn
24
your head through a certain angle before you see
the bird. If you are looking in the horizontal
25
K 0 P L direction before you turn your head up to see the
32 bird, then the angle through which you turn your
C. 1. Calculate the three angles in triangle pqr head is called theangle of elevation.
below, correct to 2 places of decimals Bird
p

2 3

q 4 r

2. In the triangle below, cos A = . Without using

Angle of elevation

Baffour – Ba Series, Core Maths for Schools and Colleges Page 683
On the other hand, the bird will also have to turn tan θ =
its head downwards to see you. If it is also
looking in the horizontal direction before it turns θ= ( ) = 570
to see you from the top of the tower, then the
angle through which the bird turn its head is 2. A bird stands on top of a story building which
called the angle of depression. is 42m tall. A girl standing 7m away from the
story building observes the bird. Calculate the
Angle of depression angle of elevation of the bird from where the girl
Bird stands

Solution
Let the angle of elevation be θ
From the diagram,

In general, the angle of elevation is equal to the TOA is applicable

42m
angle of depression. tan θ =
θ= ( )
θ Angle of depression 7m
θ = 810
θ Angle of elevation
3. An airplane is flying at a height of 4 miles
To solve problems involving elevations and above the ground. The distance along the ground
depressions; from the airplane to the airport is 9 miles. What is
I. Make a sketch of the diagram to represent the the angle of depression from the airplane to the
problem. airport?
II. Indicate the dimensions and angles if given
III. Represent the unknown sides by any preferred Solution
variable and an unknown angle by θ Let the angle of depression be θ
tan θ = AIRPLANE
θ
Worked Examples θ= ( )
1. A surveyor checking his instruments is 0 4
θ = 24 θ
standing 75m away from the bottom of a tower AIRPORT
block of flats which he knows is 114m high. 9
What is the correct angle of elevation of the top
4. A boy on top of a tree views a sea boat which
of the block from where he stands?
is 27m away from where the boy is. If the ground
distance between the boat and the tree is 16m,
Solution 114m
what is the angle of depression of the boy to the
Let the angle of elevation be θ;
boat?
From the diagram,
TOA is applicable 75m

Baffour – Ba Series, Core Maths for Schools and Colleges Page 684
Solution 2. A bird sits on top of a lamppost. The angle of
Let the angle of depression be θ depression from the bird to the feet of an observer
BOY standing away from the lamppost is 340. The
θ
distance from the bird to the observer is 24m.
27m
How tall is the lamppost?
θ
BOAT Solution
16 m
Let the height of the lamppost be h
From the digram, CAH is applicable From the diagram, SOH is applicable
cos θ = sin 340 = 34
0

θ= ( ) = 540 h = 24 sin 340 = 13 cm

24 cm
h
0
Calculating the Height / Distance Given the 34
Angle of Elevation or Angle of Depression
3. From the top of a tower, the angle of
Given the angle of depression or the angle of
depression from a point on the ground 10m away
elevation and the length of one side,
I. Represent the unknown side by any preferred
from the base of the tower is 600.
variable
i. How tall is the tower?
II. Find a trigonometric ratio (SOH, CAH, TOA)
ii. What is the angle of elevation of a point M,
that links or relates the given angle, the
half – way up the ladder?
known side and the unknown side
III. Make the unknown side the subject to obtain
Solution
its value
i. Let the height of the tower be x
0
60
Type 1
Here, elevation or depression is taken from the tan 600 =
x
ground x = 10 tan 600
0
x = 17.3 60
Worked Examples 10m
1. The top of a cliff 5 km away has an angle of x = 17m(2 s. f)
elevation 140. Calculate the height of the cliff in
meters. ii. Let the angle of elevation be θ M

8.5m
Solution tan θ =
Let the height of the cliff be h θ
θ= ( ) 10 m
From the diagram, 0
TOA is applicable θ = 40.3
h
Hence, the angle of elevation is about 400
tan 140 = 140
⇒h = 5 tan 140 = 1.2m 5 km

Baffour – Ba Series, Core Maths for Schools and Colleges Page 685
4. A building is 45m high. At a distance away 3. A boy whose eye is 2m above the ground is
from the building, an observer notices that the standing at a distance of 24m from a tall building
angle of elevation to the top of the building is on a level ground. Find the height of the building
510. How far is the observer from the base of the from the ground, if the angle of elevation from
building? the top of the building is 20o

Solution Solution
Let the distance between the observer and the Letthe height of the building from the ground
base of the building be x =x+2
From the diagram, TOA is applicable Given O = x and A= 24cm,
tan 510 = TOA is applicable;
x
tan 200= ,
x= = 36m 200
45m x = 24 tan 200
E 24 2
x = 8.7m
510 24
Type 2 xm
But the height of the bulding;
Here, elevation or depression is taken from an =x+2
additional height (in the case of a person, the eyes) = 8.7m + 2 m = 10.7m

Worked Examples 4. A pole 10m stands on the same horizontal level

1. A man 1.5m tall stands on top of a vertical cliff with a tree. From the top of the pole, the angle of
which is 30m high and sees a fish boat on the sea elevation of a bird on top of the tree is 360. If the
at an angle of depression of 150. How far is the pole and the tree are 220m apart, calculate the
boat from the cliff? height of the tree.
Solution
Solution
Let the distance between the cliff and the boat be x x
Let the height of the
360
150 tree be (x + 10)m
MAN 1.5m 220m
360 =
10m 10m
30m x = 220 tan 360 Pole Tree
cliff 15 0
boat
x = 160m (3 s.f)
x
Opposite (O) = 31.5m, Adjacent (A) = x But the height of the tree;
From the diagram, TOA is applicable = (x + 10)m = (160 + 10)m = 170m
tan 150=
Exercises 24.12
x= = 118m 1. The angle of elevation of the highest point of a
Therefore, the cliff is 118m high tree viewed by a boy of height 1.5m standing

Baffour – Ba Series, Core Maths for Schools and Colleges Page 686
18m from the foot of the tree is 370. Find the Worked Examples
height of the tree from the ground. Type 1
1. A boat is sailing directly towards a cliff. The
2. Find the angle of elevation of the sun when a angle of elevation of a point on the top of the cliff
vertical pole 2m high cast shadow 1.5m long on and straight ahead of the boat increases from 100
the horizontal ground. to 150 as the boat sails a distance of 50m. What is
the height of the cliff?
3. The angle of elevation of the top of a church
tower is viewed by a boy of height 1.65m Solution
standing 20m from the foot of the tower is 400. Let y be the height of the cliff
Find the height of the tower from the ground.
0
5 750
4. The angle of depression of a ship measured
from the top of a light house 15m above the sea y
level is 120. Find the horizontal distance of the
100 150
ship from the light house.
50 x
m
5. The angle of depression of a roundabout from tan 100 = ……………….(1)
the 8th floor of a block of a flat is 180. Each floor tan 150 = …………………..(2)
is 3m high and the observation is made by a girl
of height 1.5m. Find the distance of the
From eqn (1)
roundabout from the flats.
y = (50 + x) tan 100…………(3)
6. The angle of depression of a roundabout from
From eqn (2)
the 10th floor of a block of flat is 150. Each floor y = x tan 150 ……………….(4)
is 350cm high and an observation is made by a
man of height 125cm. Find the distance of the equating eqn(3) and eqn (4)
roundabout from the block of flats. (50 + x) tan 100 = x tan 150
50 tan 100 + x tan 100 = x tan 150
Double Elevations and Depressions 50 tan 100 = x tan 150 – x tan 100
I. Make a sketch of the diagram showing the two 50 tan 100 = x (tan 150 – tan 100)
angles of elevations or depression or both and the x= = 96m
–
given dimensions.
II. Represent unknown sides by any preferred
variable. Put x = 96 in eqn (4)
III. Form two trigonometric equations involving y = 96tan 150 = 26m
the angles of elevation or depression and the Hence, the height of the cliff is 26m
given dimensions.
IV. Solve the equations simultaneously 2. A surveyor at sea level observed that the angle
of elevation of the top of a

Baffour – Ba Series, Core Maths for Schools and Colleges Page 687
mountain, M, from two pointsN Put y = 12,018m in eqn (3)
M
and P due west of it are160 12,018 tan 180 = x
and 180 respectively as x = 3,905m
shown inthe diagram.  The height of the mountain is 3,905m

Method 2
180
The sine rule can also be used by working
160
N P Q separately on each triangle
1600m

3. From the top of a 100m storey building. A man

If /NP/ = 1600m, and the base of the mountain, observes a car moving towards the building. If
Q, is vertically below M, calculate the height of the angle of depression of the car changes from
the mountain. 220 to 460, during the period of observation, how
far does the car travel?
Solution
Let the height of the mountain be x and /PQ/ = y Solution
as shown in the diagram M Let the distance the car moved be x + d

0
MAN
0
22
46
x 100 m

0 0
CAR 22 46
160 180
N P y Q x d
1600m

tan180 = ……………………...(1) tan 460 = ……….(1)

⇒d = = 97m
tan160 = ( …………..…(2)

From eqn (1) tan 220 = ……….(2)

y tan 180 = x………………….(3)
Put d = 97m in eqn (2)
From eqn (2)
(1600 + y) tan 160 = x………....(4) tan 220 = ……….(2)
⇒ + =
eqn (3) = eqn (4)
y tan 180 = (1,600 + y) tan 160 x=( ) – 97 = 151m
y tan 180 = 1,600 tan 160 + y tan 160
y tan 180 – y tan 160 = 1,600 tan 160 ⇒x + d = 151m + 97 m = 248 m
y(tan 180– tan 160)= 1,600 tan 160
Type 2
y= = 12,018m 1. A man standing 60m away from a tower
–
notices that the angles of elevation of the top and

Baffour – Ba Series, Core Maths for Schools and Colleges Page 688
bottom of the flagstaff on top of the tower are 640 ii. Let the height of the pillar be x
and 620 respectively. Find the height of the tan 420 = ………………..(1)
flagstaff, correct to one decimal place.
tan 490 = …………….(2)
Solution
Let the height of the flagstaff be x and the height From eqn (1)
of the tower be y y tan 420 = x………………(3)

x flagstaff From eqn(2)

y tan 490 = x + 2…………..(4)

20 y Tower
620 Put x = y tan 420 in eqn (4)
60m y tan 490 = y tan 42 + 2
y tan 490 – y tan 420 = 2
tan 620 = ………………..(1)
y(tan 490 – tan 420 ) = 2
y = 60 tan 620
y= = 8m
–

tan 640 = ………………(2)

0
60 tan 64 = x + y Put y = 8 in eqn (1)
x = 8 tan 420 = 7.20m
Put y = 60 tan 620 in 60 tan 640 = x + y Therefore, the height of the pillar is 7.2m
60 tan 640 = x +60 tan 620
x = 60tan 640– 60 tan 620 = 10.2m (1 d. p) iii. By Pythagoras theorem
2. A tank 2m tall stands on top of a concrete = x2 + y2
pillar. From a point (P) on the same horizontal But x = 7.2 and y = 8
ground as the foot of the pillar, the angle of = (7.2)2 + (8)2
elevation of the top (T) and bottom (B) of the = 115.84
tank are 490 and 420 respectively. =√ = 10.7m
i. Draw a diagram to represent this information.
ii. Calculate, correct to one decimal place, the Type 3
height of the pillar. Worked Examples
iii. Calculate, correct to one decimal place, /PB/ 1. From an airplane in the air and at a horizontal
distance of 1050m, the angles of depression of
Solution the top and base of a control tower at an instant
i T are 360 and 410 respectively. Calculate correct to
2m the nearest whole number, the;
i. height of the control tower,
B ii. shortest distance between the aeroplane and the
x 0
70
42 base of the control tower.
y P

Baffour – Ba Series, Core Maths for Schools and Colleges Page 689
Solution i. the shortest distance between the pilot and the
i. Let the height of the tower be y and the distance base of the control tower;
between the aeroplane and the base of the control ii. the height of the control tower.
tower be AB
AEROPLANE (A) Solution
0
36 i. Let the shortest distance between the pilot and
0
41
the base of the control tower PB
x PILOT (P)
0
30
0
0
33
36
1050m x
TOWER

y
0
41 30
0
B
8.5km

TOWER
1050 m
y
0
33
tan 360 = ……….(1) B
8.5km
⇒x = 1050 tan 360 = 763 m 0
cos 33 =
tan 410 = ……….(2) /PB/ = = 10.1 m
The shortest distance between the aeroplane and
1050 tan 410 = x + y the base of the control tower is 10.1m
But x = 763 m
⇒1050 tan 410 = 763 + y ii. Let the height of the control tower be y
y = 1050 tan 410 – 763 = 150m tan 300 = ……….(1)
The height of the control tower is 150 m ⇒x = 8.5 tan 300 = 4.90 km

ii. The shortest distance between the aeroplane tan 330 = ……….(2)
and the base of the control tower = AB 0
8.5 tan 33 = x + y
0
cos 41 = y = 8.5 tan 330 – x
/AB/ = = 1391 m y = 8.5 tan 330 – 4.90 = 0.620km
The shortest distance between the aeroplane and The height of the tower is 0.620m
the base of the control tower is 1391m
Type 4
2. From a horizontal distance of 8.5km, a pilot Worked Examples
observes that the angles of depression of the top 1. A man 2m tall stands on the same level ground
and base of a control tower are 30o and 33o as a vertical pole. He observes that the angle of
respectively. Calculate, correct to three elelvation of the top of the pole is 210 and the
significant figures: angle of depression of the foot of the pole is 60.
Find, correct to the nearest meter;

Baffour – Ba Series, Core Maths for Schools and Colleges Page 690
i. how far the pole is from the man,
ii. the height of the pole. A

Solution 240
B
i. Let the distance between the pole and the man

46.5 m
620

28.5 m
be x and the height of the pole be h+ 2

48m

30m
660 280
h

1.5 m
1.5m
a b

1.5m
POLE

x 210
60 From the diagram,
2m
0
2m
MAN tan 240 =
6
x a= tan 240 = 20.7m

tan 60 = ……….(1) tan 620 =

⇒x = = 19m b = 28.5 tan 620 = 53.6m

ii. From the diagram, TOA is applicable, /AB/ = a + b

/AB/ = 20.7m + 53.6m = 74.3m
tan 210 = ……….(2)
But x = 19m, put in eqn (2) 2. Two points A and C, on opposite sides of a
tan 210 = vertical pole, are on the same level ground as the
h = 19 tan 210 = 7 m foot of the pole, B. The angles of elevation of the
top of the pole D from A and C are 300 and 480
⇒h + 2 = 7m + 2m= 9 m respectively. If the distance between A and C is
The height of the pole = 9 m 50m, find /BD/, the height of the pole.

Some Solved Past Questions Solution

1. Two towers A and B are 48m and 30m high Let /BD/ = h, /CD/ = y and /AD/ = x as shown in
respectively. Tower A lie to the west and B to the diagram;
the east of a man 1.5m tall. From the man‟s eye D
level, the angles of elevation of the top of A and Using sine rule,
450 600
B are 660 and 280 respectively. Calculate to three (
=
y x
significant figures, the distance between A and B h
=
450 300
Solution x= C B A
Let /AB/ = a + b 50m
x = 36.60m(4 s. f)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 691
sin 300 = , but x = 36.6m From ∆ EHG
sin 550 =
sin 300 = ,
/EH/ = 25.4 sin 550 = 20.8cm
h = 36.6 sin 300 = 18.3m
But /BD/ = h = 18.3 m
Therefore, the height of the pole is 18.3m sin 350 =
/GH/ = 25.4 sin 350 = 14.6cm
Alternatively,
Using sine rule, From ∆ HFG
(
= sin 250= , but /GH/ = 14.6

sin 250 =
=
/HF/ = = 34.5cm
y= = 25.88m(4 s. f)

EF = EH + HF
0
sin 45 = , but y = 25.88m EF = 20.8 + 34.5 = 55.3cm
sin 450 = ,
4. A vertical pole AB is erected on a level ground.
0
h = 25.88 sin 45 = 18.30m A man 1.7mtall stands at C, 24 m away from the
But /BD/ = h = 18.3 m foot Bof the pole. The angle of elevation of the
top A from the man is 540. Calculate, correct to
3. In the figure below, angle GEH = 350, angle one decimal place the height of the pole.
EGF = 1200, /EG/ = 25.4cm and GH is
perpendicular to EF. Solution
G
Let the heightof the man to the end of the
1200 vertical pole be x
25.4cm
A

350
E H F x
540
Calculate /EF/, correct to three significant figures 1.7m 1.7m
B C
24 m
Solution
Draw the triangles separately as shown below; From the diagram;
G G The height of the pole = (x + 1.7)m
But x = 24 tan 540 = 33.0m (1 d.p)
650
24.5cm 550
Substitute x = 33.0 in (x + 1.7)m
The height of the pole = (33.0 + 1.7)m = 34.7m
350 250
E H H F

Baffour – Ba Series, Core Maths for Schools and Colleges Page 692
Exercises 24.13 6. The feet A and C, of two vertical poles AP and
1. Two ships, S and T, 40km apart, CR are on the same horizontal plane as the foot B
simultaneously observe by a radar of an of a vertical flag pole, QB = 5m high. A is 10m
aeroplane A at an angle of elevation of 100 and 60 east and C is 10m west of B. The heights of the
respectively.Given that the aeroplane is directly poles are /AP/ = 6m and /CR/ = 5m. Calculate;
above the line joining the ships, calculate; i. /AR/,
i. the distance AS, ii. the angle CP makes with the horizontal,
ii. the height of the areoplane above sea – level. correct to the nearest degreeiii. the angle the
plane ACQ makes with the horizontal, correct to
2. A surveyor wishes to measure the height of a the nearest degree.
round tower. Measuring the angle of elevation, he
finds that the angle increases from 220 to 360 after Trigonometry and Bearings
walking 25 m towards the base of the tower. Revision: Refer to page 264 to 269 (Other
Calculate the height of the tower, correct to the application of bearings).
nearest meter. I. Draw a diagram to represent the problem.
II. If the triangle formed is not a right angled
3. The length of the shadow of a pole on a level triangle, use cosine rule or cosine rule to
ground increased by 60 meters when the angle of determine the value of an unknown side and the
elevation of the sun changes from 540 to 320. values of unknown angles.
Calculate the height of the pole, correct to three III. To find the bearing of say A from B, is to find
significant figures. the total angle turned through in the clockwise
direction from the north pole of B to the direction
4. From a Swiss mountain rescue station, 897m of A.
up a mountain, a manobserves that the angles of
depression of two villages on the valley floor are Worked Examples
530 and 280 respectively. How far apart are the 1. B is 48km south – west of A, and C is 66km
villages? from B on a bearing of 0800. Find, the distance
and bearing of A from C.
5. Two vertical electricity poles, each 18m high
stands at the points A and B along a straight Solution A
0
y
horizontal road. A straight footpath meets 45
theroads at B from a point C on the foot path, 43 48km C
meters from the road. The angles of elevation of 0
45 350 66km
the top of the poles at A and B are respectively
14.60 and 16.40. Calculate, correct to one decimal B
place;
a. the distance ̅̅̅̅ and ̅̅̅̅, Let the distance of A to C be y
b. the distance between the poles. From the diagram, ∆ABC is not a right – angled
triangle. Therefore, use the cosine rule to find the

Baffour – Ba Series, Core Maths for Schools and Colleges Page 693
value of the unknown side and unknown the ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
= R
angles. θ
⇒y2 = 662 + 482 – 2 (66)(48) cos 350 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 0

y 2 = 1469.85 P 53 370
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 22.7768km 48
0

y=√ = 38 km
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 23km
15km 0
42 12
0
From the diagram, < C (Nearest kilometer)
482 = 662 + 382 – 2 (66)(38) cos C Q
482 – 662 + 382 = - 2 (66)(38) cos C
-3496 = - 5016 cos C ii. Let the ships final distance from P be ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
cos C = = R
(
θ
C = cos –1( ) = 460(nearest degree) ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
= 0
P 53 370
0
350 + 450 + C + = 1800 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 48

350 + 450 + 460 + = 1800 23km

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 18.4972km 15km 0
= 1800 – 1260 = 540 42
0 12

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 18km
But + θ = 900
Q
54 + θ = 90
Alternatively;
θ = 90 – 54 = 360
Using the cosine rule;
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ =152 + 22.7768)2 – 2(15)(22.7768) cos 540
Bearing of C from A ;
= 90 + θ = 90 + 360 = 1260 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 743.7826 – 401.6360
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = 342.1466
0
2. A ship sails on a bearing of 138 from a port P ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = √ = 18.4972 = 18km
for a distance of 15km to a port Q and then sails
on a bearing of 0120 to a port R. If the bearing of Revision Exercises
R from P is 0530, find: 1. An aircraft flies 100 km from A to B on a
i. the distance from Q to R; course 0750, then 200 km from B to C on a course
ii. the ships final distance from P. 3430. How far north and how far east are B and C
from A.
Solution
i. Let the distance from Q to R be ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 2. An object Q is 6km from P on a bearing
From the diagram, N200E, and object R is 7.5 km from P on a
(42 + 12)0 + (37 + 48)0 + θ = 1800 bearing N750E. Calculate the distance and
θ = 1800 – 540 – 850 = 410 bearing of Q from R.

Using sine rule, 3. A ship sails due east at a steady speed of

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 24km/h while a fishing boat is towed north – east
=
(

Baffour – Ba Series, Core Maths for Schools and Colleges Page 694
at a steady speed of 10km/h from the same point.
They start at the same time; Ladder before slipping

a. find the bearing of the ship from an observer Wall

Ladder after slipping
on the boat at the end of 2 hours;
b. find the distance between the ship and the boat Ɵ
x
at that time (2 hours after the starting time. “Upwards”
When a ladder or a stake slips backward:
4. Town Q is 20km due north of P and the I. A new angle of elevation, θ, is formed.
bearing of town R from Q is 1400. If R is 8km II. There is ahorizontal adjustment (reduction) in
from Q, calculate; the distance between the foot of the wall and the
a. the bearing of R from P, to the nearest degree, ladder causing an increase in the height of the
b. how far north of P, R is, correct to two wall. This increment is represented by x in the
significant figures. diagram.
x
5. Kaya rides a bicycle to school everyday. From
home, he rides for 3.5km on a bearing of 0370 Ladder before slipping
Wall
and then 1.4km on a bearing of 3350 before
Ladder after slipping
arriving at school. Ɵ
a. What is the total distance Kaya rides from
home to school? To tackle involving problems:
b. How far (i)north (ii) east I. Make a sketch of the figure.
is Kaya‟s school from home? II. Draw the figures separately as “before the
c. find the bearing and distance of Kaya‟s school ladder slipped” and “after the ladder slipped.”
from his home. III. Work for the interior angles and the length of
the sides of each triangle.
A Slipping Ladder or Inclined Plane IV. Put the separate figures together with its
A ladder or any straight object (stake) leaning
dimensions.
against a vertical wall may either slips backward
or upward as shown below;
Worked Example
1. A ladder 5m long leans against a vertical wall
“Backwards” at an angle of 700 to the ground. The ladder slips
When a ladder or an object slips backward:
down the wall 2m. Find correct to two significant
I. A new angle of elevation, θ, is formed
figures.
II. There is an upward adjustment (reduction) in
i. the new angle which the ladder makes with the
the length of the wall causing an increase in the
ground,
distance between the ladder and the wall. This
ii. the distance the ladder has slipped back on the
increment is represented by x in the diagram.
ground from its original position.

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Solution ii. To calculate for a,
i. Let the new angle the ladder 2m tan θ = , butθ = 330 5m
makeswith the ground be θ and 0
2.7m
5m tan 33 =
the distance the ladder has h 330
slipped back be x a= = 4.2m 4.2m
0
Ɵ 70
Alternatively
x b
By Pythagoras theorem,
Draw the figures separately as “before the ladder 52 = a2 + 2.72,
slipped” and “after the ladder slipped”. 52–( 2
= a2 2m
a2 = 17.71
“Before”
a=√ = 4.2m 5m
Let the height of the wall be hand the distance
between the foot of the wall and the ladder be b 2.7m
Below is the
sin 700 = dimension 330 70
0
0
20 x 1.7m
h = 5 sin 700 = 4.7m of the figure;
5m
sin 200 = h From the diagram, 4.2m
b = 5 sin 200 = 1.7m x + 1.7 = 4.2
700
x = 4.2 – 1.7 = 2.5
b
Alternatively Therefore, the ladder slipped back 2.5m
0
By Pythagoras theorem, 20
2 2 2
5 = b + h , but h = 4.7 5m 3. In the diagram below, PS and QT are two
4.7m ladders 10m and 12m long respectively, placed
52 = b2 + ( 2

52 - ( 2
= b2 700 against a vertical wall TR. PS makes an angle of
2
b = 2.91 1.7m 600 with the horizontal and PQ = 0.5m.
T
b=√ = 1.7m T

“After” S
S
Height of wall = (4.7 – 2)m = 2.7m
Let the distance between the 600
ladder and the foot of P 0.5cm Q R
the wall be a and the 0.5cm
CalculateP to twoQ significantRfigures;
new angle the 5m 2.7m
ladder makes with i. the angle which QT makes with the horizontal,
Ɵ
the ground be θ a ii. the height of point T above the horizontal.
sin θ =
θ= ( )= 330

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Solution
T
Let the angle QT makes sin θ =
with thehorizontal be θ, b But θ= 680
/QR/ = x, /SR/ = y, /ST/ = b S ⇒ sin 680 =
12
12 sin 680 = b + 8.7
10 T
y 12 sin 680 – 8.7 = b
b = 2.4m b
600

P 0.5cm Q x R 12
Alternatively 8.7
Draw the triangles separately as shown below and By Pythagoras theorem, 680
calculate all the angles and dimensions of each; 2 2 2
12 = (4.4) + (8.7 + b) , Q 4.4 R
122 – (4.4)2 = (8.7 + b)2,
From ∆ PRS, S 124.64 = (8.7 + b)2,
sin 600 = 300
0
√ = √( T
y = 10 sin 60 10 y
√ = 8.7 + b 220
2.4 cm
y = 8.7m (2 s. f.) 60 0
b=√ – 8.7 = 2.4 S
P x R 300
0.5
cos 600 = 8.7 cm
0
Dimensions
10 cos 60 = 0.5 + x of the figure; 60 0
68 0

10 cos 600 – 0.5 = x P 0.5cm Q 4.4 cm R

x = 4.5m i. The angle QT makes with the horizontal is 680

Alternatively, ii. The height of point T above the horizontal

By Pythagoras theorem, = RS + ST
102 = y2 + (0.5 + x)2, = (8.7 + 2.4)cm = 11cm
But y = 8.7
T
102 = (8.7)2 + (0.5 + x)2 Exercises 24.14
102 – (8.7)2 = (0.5 + x)2 b 1. A pole 25m long is placed against a vertical
24.31 = (0.5 + x)2 12 wall such that its lower end is 7m from the foot of
√ = √( 8.7 the wall on the same horizontal ground. If the
Ɵ upper end of the pole is pushed down by 2 m,
√ = 0.5 + x
Q x = 4.4 R calculate correct to two significant figures:
x=√ – 0.5 a. how much further away from the wall the
T
x = 4.4m lower end will move,
b
b. the angle the pole now makes with the
From ∆QRT,
12 horizontal.
cos θ = 8.7
0 680
= ( ) = 68 2. A ladder ̅̅̅̅ , of length 13m, rests against a
(2. S. f) Q 4.4 R vertical wall with its foot on a horizontal floor at

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a distance of 5m from the wall. When the top vertical tower of height h. The angles of elevation
slips down a distance x meters the foot moves out of the top of the top of the tower from p and q are
x meters. Find x α and , respectively
s
A
x
A1 h

13 q r
α
p

B B1 i. If /α/ = 600 and / / = 300, express /pr/ and /qr/

O 5
x in terms of h
3. A ladder 22m long leans against a vertical wall ii. Find /qp/ in terms of h, if tan <qrp = √
at an angle of 380 to the ground. The ladder slips
down the wall 3.2m. Find correct to two Solution
significant figures; Redraw right – angled triangles prs and
i. the new angle which the ladder makes with the qrsseparately
ground, i. tan 600 =
ii. the distance the ladder has slipped back on the s
ground from its original position. √ =
4. A ladder of length 20m, rests against a vertical √ / pr / = h
wall with its foot on a horizontal floor / pr / = 600
√ p r
at a distance of 12m from the wall. When the top
h
slips down a distance x meters the foot moves out
tan300 =
twice as much as x meters.
= h
i. Find the value of x. √
ii. How far does the foot moves out of the wall? 300 r
/ pr / = √ h q

Three Dimensional Problems

ii. Re - draw triangle qpr separately as
Type 1 √
When solving problems in three dimensions: q r

I. Redraw each triangle separately.

shown below;
II. Apply the sine or cosine rule to these triangles. √
III. Watch for common sides that links the p
triangles.
Using the cosine rule on ∆ qpr
IV. Carry common values from one triangle to
= + -2 . cos < qrp
another.
But cos < qrp is obtained as shown below
Worked Examples
√
p, q and r are points on level ground. [s,r] is a Given tan < qrp = √ = , its triangle is shown
below;

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By Pythagoras theorem, q Using the sin rule,
=3
=
 cos < qrp =
3 x= = 25m
√

p r
Similarly,
1
= A
36m
= |√ | + | | –2 |√ |.( ) ( ) 25m
√ √ y= 0 0
2 55 35
= 3h + – h y = 36m B 44 m C

=
Consider ∆ ABD,
= Using the cosine rule,
h2 = 222 + 252 – 2(22)(25) cos 520 D
=√ =√ h h2 = 484 + 625 – 677
21
h2 = 432 m 22m
2. In the figure below, ∆ABC is a right – angled h=√ m = 21m 0
A
52
25
triangle on a horizontal ground. AD is B
a vertical tower. Consider ∆ACD
D
D
50m 21m 50m
22m

A
A θ
520 36m
350
B 44m C C
Using the cosine rule,
< BAC = 900, < ACB = 350, < ABD = 520 /BD/ = 212 = 362 + 502 – 2(36)(50) cos θ
22m, /DC/ = 50m and / BC/ = 24m. Find: 2(36)(50) cos θ = 362 + 502 – 212
3600 cos θ = 3355
i. the height of the tower, cos θ =
ii. the angle of elevation of the top of the tower
from C. θ= ( ) = 210

Solution Exercises 24.15

Let /AB/ = x, /AC/ = yand /AD/ = h 1. In the figure below, PQR is a right – angled
Consider ∆ ABC, triangle on a horizontal ground. PT is a vertical
A tower. < QPR = 900, /QR/ = 61m, /PR/ = 60m and
y
x /QT/ = 80m.
55
0
35
0 Find: a. the height of the tower b. /TR/
B 44 m C

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 5. X, Y and Z are three points on a level ground.
T The bearing of Y from X is 0300 and the bearing
of Z from Y is 1200. A vertical tower XT stands at
X and the angle of elevation of T from Y is 68.20.
80m P
60m Given that XY = 40 m and YZ = 30m, calculate:
m i. XZ,
ii. the height of the tower.
Q 61m R
m
Type 2
2. The figure below represents a rectangular box. Worked Examples
Given that AB = 6cm, AD = 3cm and AE = 2cm. 1. An aircraft is flying at a height of 120km over
Calculate the length of the diagonal AG. a plain. It is immediately above town A. Town B
B is on a vector ( ) from A. What is the
A distance of the aircraft from town B to the nearest
2cm C kilometer?
D
F
E AIRCRAFT
P
3cm
H 6cm G Solution
120 km
3. ABC is an isosceles triangle in which AB = AC
N
= 10 cm and BC = 4cm. The mid points of AB
A 43km C
and BC are X and Y respectively, and the line W E
25 km
through X perpendicular to AB meets AY at O. S B
Calculate angle BAC and the lengths of AO and
OY From the diagram, the „distance of the aircraft
from town B‟ is the distance PB.
4. A right pyramid has a base ABC which is an Since PA, AC and CB are perpendicular to each
equilateral triangle of side 5 cm. The height of the other;
vertex V above the center O of PB2 = PA2 + AC2 + CB2
the base is 4m. Calculate: PB2 = 1202 + 432 + 252
i. the length /OA/, correct to two decimal places. PB2 = 16,874 km
ii. the angle of inclination of the edge ̅̅̅̅ to the PB = 130 km (nearest kilometer)
base ABC, correct to the nearest degree.
iii. the volume of the pyramid, to three significant
figures.

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25 SEQUENCES AND SERIES Baffour – Ba Series

Sequence Series
A sequence is an ordered list. It is a collection of When the numbers or terms of a sequence are
numbers arranged in a definite order, connected added or considered as a sum, it is called a series.
by a simple rule. For example, perfect squares Thus we have a series of even number as 2 + 4 +
are listed as 1, 4, 9, 16, 25, 36…The dots 6 + 8… and the series of odd numbers as 1 + 3 +
indicates that it is impossible to list all of them 5+7…
since there are indefinite number of them.
A series may end after a definite number of
However, this collection could be represented in terms. Such a series is called a finite series. e.g. 2
several ways, one of which is to write 1, 4, 9, +4+6+8
…n2, n∈N. A list of numbers such as this is On the other hand, a series may not end and it is
generally called a sequence. Other examples of then called an infinite series.
sequences are even numbers: 2, 4, 6,.. odd e.g. 2 + 4 + 6 + 8…
numbers: 1, 3, 5… However, the prime numbers:
2, 3, 5, 7…do not form a sequence because the Types of Sequences
numbers or the terms do not follow any definite There are two main types of sequence namely:
order. 1. Arithmetic Progression (A. P) or Linear
sequence
Worked Examples 2. Geometric Progression (G.P) or Exponential
Write down the next three terms in each of the Sequence
following sequences:
1. 1, 3, 5, 7,… 2.4, 2, 0, -2… The nth Term of an Arithmetic Progression
(A. P) or Linear Sequence
3. , , , … 4. , , , …
An arithmetic progression is a series in which any
term is obtained from the previous term by
Solution adding a certain number called common
1. 1, 3, 5, 7, 9, 11, 13 2.4, 2, 0, -2, -4, -6, -8 difference.

3. , , , , , , 4. , , , , , , Consider the sequence: 1, 3, 5, 7, 9

1. The sequence is said to have five terms,
Exercises 25.1 denoted by n. Thus n = 5
Write down the next three terms in each of the 2. Each term is represented by U. Therefore, 1, 3 ,
following sequences: 5 , 7 , 9= U1, U2 ,U3, U4, Un
1. 1, 2, 4, 8… 2. 1, 2, 6, 24, 120… In this case:
U1 is called the first term of the sequence,
3. , , , … 4. - , - , - …
U2 is called the second term of the sequence,
5. 12, 8, 4,….. 6. -82, -88, -94… U3 is called the third term of the sequence,
U4 is called the fourth term of the sequence,
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Un is called thenth term of thesequence and of Un = a + (n – 1) d
course the last term of the sequence. U20 = 7 + (20 – 1) (– )
3. The digit representing the first term is denoted
= 7 + 19 × (– )
by a. That isU1 = a
4.U2 – U1 is called the common difference =7+– = =–
denoted by d. U2 – U1 = d. The common
Therefore, the 20th term is –
difference d can be positive or negative.
5. The sum of the first n terms of a linear
sequence is denoted by Sn. Finding the General Term of an A.P.
Thus, for U1, U2, U3, U4, … Un , The general term of an AP is the formula
Sn = U1 + U2 + U3 + U4 + … Un representing a given A.P. The formula is found as
follows:
If any term is obtained from the previous term by I. Identify the given sequence.
adding the common difference, d, then it implies II. Identify the values of a and d.
that if: III. Substitute the values of a and d in
U1 = a Un = a + (n – 1) d and simplify if possible.
U2 = a + (2 – 1) d = a + d
Worked Examples
U3 = a + d + d = a + (3 – 1) d = a + 2d
1. Find the general term of the sequence:
U4 = a + d + d + d = a + (4 – 1) d = a + 3d
2, -1, -4…
U5 = a + d + d + d + d = a + 4d
Un = a + (n – 1) d or Un = U1 + (n – 1) d
Solution
Therefore, the nth term, also called the general
From the sequence: 2, -1, -4…
term of a linear sequence, denoted by Un is given
a = 2, d = -1 – 2 = -3
by; Un = a + (n – 1) d
Substitute in Un = a + (n – 1) d
Un = 2 + (n – 1) (-3)
Worked Examples
1. Find the twelfth term of the sequence: = 2 – 3n + 3
19, 14, 9, 4… = 5 – 3n
Therefore, the general term is Un = 5 - 3n
Solution
In the sequence, 19, 14, 9, 4… 2. The first term of a linear sequence is 13 and the
a = 19, d = U2 – U1 = 14 – 19 = -5, n = 12 common difference is 8. Find :
Substitute in Un = a + (n – 1) d i. the general term, ii. the tenth term.
U12 = 19 + (12 – 1) -5
U12 = 19 + 11(-5) = 19 – 55 = -36 Solution
i. If a = 13 and d = 8
2. Find the 20th term of the sequence 7, 6 , 5 … ⇒ Un = 13 + (n – 1) 8
Un= 13 – 8 + 8n
Solution Un= 8n + 5
Let a = 7, d = 6 – 7 = – and n = 20 Therefore, the general term is Un= 8n + 5

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ii. To find the tenth term, Finding the First Term, a, and the Common
a = 13, d = 8 and n = 10 Difference, d of an A.P.
Substitute in Un = 8n + 5 Given the value of a term other than the first term
U10 = 8(10) + 5 = 85 and the common difference, d, the first term is
calculated as follows:
3. Write down the 12th and nth terms of the A.P, I. Write an expression for the given term and the
+ +… common difference
II. Equate the expression to the given value of the
Solution term
III. Solve for the value of a in the equation to
From the sequence: + + …
obtain the first term of the A.P
a= ,d= – = = , n = 12
Substitute in Un = a + (n – 1) d Worked Examples
The fourth term of an A.P is 18, and the common
Un = + (n – 1)
difference is -5. Find the first term.
U12 = + (12 – 1)
Solution
U12 = + (11) = + = =
U4 = 18, d = -5, n = 4
Substitute in Un = a + (n – 1) d
ii. The nth term, Un = a + (n – 1) d U4 = a + (4 – 1) (-5)
But a = , d = U4 = a + 3(-5)
By substitution, U4 = a – 15
Un = + (n – 1)
But U4 = 18
Un = + ( – ) ⇒a – 15 = 18
a = 18 + 15 = 33
Un= – +
Therefore, the first term of the A.P is 33
Un= - +
Exercises 25.3
Exercises 25.2 1. In an A.P, the third term is 12, and the
1. Write down the first five terms of each of the common difference is 7. Find the first term.
following linear sequence;
i. first term 5, common difference 3 2. Find the first term of an A.P if the tenth term of
ii. first term 3, common difference the A.P is 77, and common difference -12
2. Write down a formula for the nth term of each
3. The fourth term of an A.P.is 20, and the
of the following linear sequences:
common difference is 6. Find the first term.
i. 3, 6, 9, 12… ii. 0,5, 10, 15…
ii. 12, 9, 6, 3… iii. 2, 7, 12, 17….
4. What is the first term of an A.P. whose sixth
In each case , write down the 20th term.
term is 42 and common difference, 4.

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Finite Arithmetic Sequence ⇒ 2n = 46
It is an A.P. that comes to an end. In other words, n = 23
it an A.P., that has a given last term (L). Therefore 46 is the 23rd term
Eg. 2, 4, 6, ..50
2. Find the number of terms of the A.P:
For a finite A.P; 407 + 401 + … –133
= L
⇒ a + (n – 1) d = L Solution
By changing the subject to n, the number of terms From the A.P = 407 + 401 + … –133
of an A.P, given the last term can be calculated. a = 407, d = 401 – 407 = -6, L = -133, n =?
Substitute in Un = a + (n – 1) d
Worked Examples Un = 407 + (n – 1) -6
Determine the number of terms in the sequence; Un = 407– 6n + 6
5, 11, 17….77 and hence find the 30 the term. Un = 407 + 6 – 6n
Un = 413 – 6n
Solution
5, 11, 17….77 But Un = L = -133
a = 5, d = 6 = L = 77 ⇒ 413 – 6n = -133
6n = + 133
=5+6(n–1)
6n = 546
=5+6n–6 n = 91
= 6n–1 Therefore the A.P. has 91 terms
⇒ 6 n – 1 = 77
6n = 77 + 1 3. Find the number of terms of the linear
6n = 78 sequence 3, 7, 11…31
n = 13 terms
Solution
= 6 (30) – 1= 179 From the linear sequence 3, 7, 11…31
a = 3, d = 7 – 3 = 4, n = ? and l = 31
2. The first term of an A.P. is 2 and the common
Un = 3 + (n – 1) 4
difference is 2. Find the term that is equal to 46.
Un = 3 + 4n – 4
Un = -1 + 4n
Solution
a = 2, d = 2, L = 46
But Un = L = 31
Substitute in Un = a + (n – 1) d
⇒ 31 = -1 + 4n
Un = 2 + (n – 1) 2
31+ 1 = 4n
Un = 2 + 2n – 2
32 = 4n
Un = 2n
n=8
Therefore, the number of terms is 8
But Un = L = 46

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4. The nth term of the sequence; 5, 8, 11…is 383. term is 21. Find the common difference and the
Find n first term.

Solution Solution
From the sequence; 5, 8, 11… Un = a + (n – 1) d
a = 5 , d = 8 – 5 = 3, Un = 383 U2 = a + (2 – 1) d = 15
Substitute in Un = a + (n – 1) d
Un = 5 + (n – 1) 3 ⇒a + d = 15……….(1)
Un = 5 + 3 (n – 1) U5 = a + (5 – 1) d = 21
Un = 5 + 3n – 3 ⇒a + 4d = 21…………….(2)
Un = 5 – 3 + 3n eqn (2) – eqn (1)
Un = 2 + 3n (a – a) + (4d – d) = (21 – 15)
3d = 6
But Un = 383 d=2
⇒2 + 3n = 383
3n = 383 – 2 Put d = 2 in eqn (1)
3n = 381 a + 2 = 15
n = 127 a = 15 – 2 = 13
First term is 13 ; the common difference is 2
Exercises 25.4
A. Find the number of terms of each of the 2. Find the linear sequence whose eighth term is
following linear sequence: 12 and 12th term is -8.
1. 2, -9, -20, …-141 4. 2 – 9 – … – 130
2. 2.7 + 3.2 + … + 17.7 5. 8, 10, 12,…32 Solution
3. 50 + 47 + 44 + … + 14 6. 39, 33,…, -63 Un = a + (n – 1) d
U8 = a + (8 – 1) d = 12
First Term, a, and Common Difference, d, of an ⇒a + 7d = 12……….(1)
A.P from Two or more given Terms U12 = a + (12 – 1) d = -8
Given the values of two or more terms of an A. P. ⇒a + 11d = -8………(2)
the first term, a, and the common difference, d,
can be calculated as follows: eqn (2) – eqn (1)
I. Identify the values of the given terms. (a – a) + (11d – 7d) = (-8 – 12)
II. Write two (or more) equations, each for the 4d = -20
given terms. d=-5
III. Solve the equations simultaneously to
determine the respective values of a and d. Put d = - 5 in eqn (1)
a + 7(- 5) = 12
Worked Examples a – 35 = 12
1. The second term of an A.P is 15 and the fifth a = 12 + 35 = 47
The first term is 47, common difference is -5

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The sequence is 47, 42, 37, 32, 27, 22, 17, 12, 7, U3 = a + 2d
2, -3, -8
8th term is five times the third term
3. The 8th and the 22nd terms of a linear are 38 ⇒a + 7d = 5(a + 2d)
and 108 respectively. Find: a + 7d = 5a + 10d
i. the first term and the common difference, a – 5a = 10d – 7d
ii. the general term of the sequence. - 4a = 3d……………(1)

Solution U7 = a + 6d
Un = a + (n – 1) d U4 = a + 3d
U8 = a + (8 – 1) d = 38
⇒a + 7d = 38……….(1) The 7th term is 9 greater than the fourth term
U22 = a + (22 – 1) d = 108 ⇒a + 6d + 9 = a + 3d
⇒a + 21d = 108………(2) a – a + 6d – 3d = -9
3d = -9
eqn (2) – eqn (1) d = -3
(a – a) + (21d – 7d) = (108 – 38)
14d = 70 Put d = -3 in eqn (1)
d=5 - 4a = 3(-3)
- 4a = -9
Put d = 5 in eqn (1) a = 2.25
a + 7(5) = 38 ⇒The first five terms of the sequence is:
a + 35 = 38 2.25, - 0.75, - 3.75, - 6.75, -9.75
a = 38 – 35 = 3
First term is 3 and the common difference is 5. 2. If 3, x, y, 18 are in arithmetic progression, find
the values of x and y.
b. Substitute a = 3 and d = 5 in Un = a + (n – 1) d
Un = 3 + (n – 1) 5 Solution
Un = 3 + 5n – 5 From 3, x, y, 18
Un = 5n – 5 + 3 a = 3, U4 = 18, d = ?
Un = 5n – 2 Un = a + (n – 1) d
Therefore, the general term is Un = 5n – 2 U4 = 3 + (4 – 1) d
U4 = 3 + 3d
Some Solved Past Questions
1. The 8th term of an A.P. is five times the third But U4 = 18
term whilst the 7th term is 9 greater than the ⇒ 3 + 3d = 18
fourth term. Write the first five terms of the A.P. 3d = 18 – 3
3d = 15
Solution d=5
Un = a + (n – 1 )d
U8 = a + 7d ⇒x = 3 + 5 = 8
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y = 3 + 5 + 5 = 13 2 + 4 + 6 + 8… + 50 + 52
The sequence is 3, 8, 13, 18 If the sum is represented by S, we have,
S = 2 + 4 + 6 + 8… + 50+ 52
3. 2k + 1, 3k, 5k – 5… are the first three terms of
an arithmetic sequence. Reversing the order of the terms:
i. Calculate the value of k S = 52 + 50 + 48 + 46 + 44… + 2
ii. Find the next two terms of the sequence
Adding the two sums,
Solution S= 2 + 4 + 6 + 8…+ 50 + 52
i. 3k – (2k + 1) = (5k – 5 ) – 3k 52 + 50 + 48 + 46 + 4…+ 2
3k – 2k – 1 = 5k – 5 – 3k 54 + 54 + 54………. 54 + 54
3k – 2k – 5k + 3k = 1 – 5
-k=-4 In this sum, we have 26 terms
k=4 Therefore 2S = 54 × 26
S= = 702
ii. 2(4) + 1, 3(4), 5(4) – 5
9, 12, 15… Now let U1 be the first term, d the common
Common difference = 12 – 9 = 3 difference and Sn the sum of n terms.
9, 12,15, 18, 21 ⇒ Sn = U1 + ( + ( + [(
( ]
Exercises 25.5 Reversing the order of the terms, we have
1. Find the first term, the common difference and Sn = [( ( ]+( +( + U1
an expression for the nth term for each of the A.P. Add the two sums,
below: 2Sn= [ ( ] + [ ( ] +
i. second term, 3 and third term 5, [ ( ]
ii. second term, 4 and fifth term 16.
For n terms,
nd th
2. The 2 and 5 terms of a linear sequence are 2Sn = n [ ( ]
26 and 62 respectively. Find: [ ( ]
Sn =
i. the first term and the common difference,
ii. the general term of the sequence. Sn = [ ( ]

But U1 = a
3. Find the 28th term of a linear sequence whose
⇒Sn= [ ( ]
fourth term is -18 and ninth term is 12.
Sn= [ ( ]
th
4. Find the 10 term of the linear sequence whose But ( =l
2nd term is 28 and 17th term is -2. ⇒Sn = ( )

Sum of a Linear Sequence or A.P. Given the last term Un, the sum of the linear
Suppose we want to find the sum: sequence is given by the relation,
Baffour – Ba Series, Core Maths for Schools and Colleges Page 707
Sn = [ ] OR Sn = [ ( ] 351= ( )
351 = 5a + 255
Worked Examples 351 – 255 = 5a
1. Find the sum of the first 10 terms of the A.P: 96 = 5a
4, 2, 0, -2… A = 19.2

To find the value of d, use the formula;

Solution
an = a + (n – 1)d
A.P = 4, 2, 0, -2…
a10 = 19.2 + (10 – 1)d
From the sequence, U1 = a = 4, d = 2 – 4 = -2 and
a10 = 19.2 + 9d
n = 10
51 = 19.2 + 9d But a10 = 51
Substitute in Sn = [ ( ] 51 – 19.2 = 9d
S10 = [ ( ( ( ] 31.8 = 9d
S10 = [ ( ( ] d = 3.5
S10 = 5 ( )
S10 = 5 ( ) 4. Find the sum of the first 20 terms of the linear
S10 = -50 sequence 3, 5, 7, 9…

Solution
2. Find the sum of the first 20 terms of the
From the linear sequence: 3, 5, 7, 9…
arithmetic sequence whose first term is 2 and
a = 3, d = 5 – 3 = 2 and n = 20
whose common difference is 4.
Substitute in Sn = [ ( ]
Solution S20 = [ ( ( ]
n = 20, a = 2, and d = 4 S20 = 10 [ ( ]
Substitute in Sn = [ ( ] S20 = 10 (44)
S20= [ ( ( ] S20 = 440
S20= 10 [ ( ]
Exercises 25.6 A
S20= 10 (80)
1. Find the sum of the first 20 terms of the
S20= 800
linear sequence 4, 2, 0, -2…
3. The sum of the first ten terms of an A.P. is 351 2. Find the sum of the A.P. 3, 6, 9 … 300.
and the tenth term is 51. Find the first term and
the common difference. 3. How many terms of the A.P;. 15 + 13 + 11 +…
are required to make a total of -36.
Solution
S10 = 351, a = ?, a10 = 51, n = 10, d = ? 4. Given the arithmetic sequence:
Sn = [ ] y – 3, 2y – 4 and 23 – y
351= [ ] i. Determine the value of y.

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ii. Find the sum of the first 26 terms of the preceding term by a fixed/constant number or
sequence. common factor called common ratio, r.

5. The first three terms of an arithmetic sequence Consider the sequence: 1, 3, 9, 27, 81…,
are 2k – 7, k + 8 and 2k – 1. It is seen that each term is obtained by a
i. Calculate the value of the 15th term of the multiplication of the preceding digit by 3;
sequence. 1. The sequence is said to have five terms,
ii. Calculate the sum of the first 30 even terms of denoted by n. Thus n = 5
the sequence
2. Each term is represented by U. Therefore, 1, 3,
Exercises 25.6B 9 , 27, …n = U1 , U2 ,U3 , U4,… Un
1. The sixth and eleventh terms of a linear In this case:
sequence are respectively are 23 and 48. U1 is called the first term of the sequence,
Calculate the sum of the first 20 terms of the U2 is called the second term of the sequence
sequence. U3 is called the third term of the sequence U4 is
2. The fifth term of an A.P. is 17 and the third called the fourth term of the sequence
term is 11, Find the sum of the first seven terms. Un is called the nth term of the sequence and of
course the last term of the sequence
3. The first and thirtieth terms of an A.P. are 71
and -16 respectively. Find the sum of the first 50 3. The digit representing the first term is denoted
terms of the sequence. by a. That is: U1 = a

4. The sum of the first three terms of an A.P. is 3 4. = is called the common ratio denoted by r.
and the sum of the first five terms is 20. Find the
first five terms of the progression. i.e. = = = r. The common ratio, r, can be
positive or negative
5. The fourth term of an A.P. is 15 and the sum of
the first five terms is 55. Find the first term and 5. The sum of the first n terms of a geometric
the common difference, and write down the first sequence is denoted by Sn
five terms. Thus, for U1, U2, U3, U4, … Un ,
Sn = U1 + U2 + U3 + U4 + … Un
6. The sum of a number of consecutive terms of (Note Un = L, last term)
an A.P. is , the first term is and the
common difference is -3. Find the number of If any term is obtained from the previous term by
terms. multiplying the common ratio, r, then it implies
that if:
The nth Term of a Geometric Progression U1 = a
(G.P) or Exponential sequence U2 = U1r = = ar
It is the type of sequence in which each term after U3 = U2r = =
the first is obtained by a multiplying the U4 = U3r = =

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U5 = U4r = = ratio is . Find the sequence.
Un = =
Un = = Solution
Therefore, the nth term of a geometric sequence Generally, a G.P. is of the form a, ar, ar2,
denoted by Un is given by; Un= ar3…where a is the first term and r is the
common ratio
Worked Examples ⇒a = 20
1. Find the 8th term of the geometric sequence;
a2 = 20 × = 25
12, 24, 48…
a3 = 20 × ( ) =
Solution
From the sequence:12, 24, 48… a4 = 20 × ( ) =
n = 8, a = 12 and r = =2 Therefore, the sequence is; 20, 25, ,

Substitute in Un = Exercises 25.7

U8 = ( = 12 × = 1,536 A. For each geometric sequence, find the
constant ratio and the indicated term
2. Find the ninth term of the G.P: , , … 1.4, 12, 36…(U12) 3. 16, 64,256…(U8)
2. , , ……(U20) 4. 3, - , ….(U10)
Solution
From the sequence: , , …
B. 1. Find a formula for the nth term of the
n = 9, a = , andr = ÷ = × = geometric sequence with r = 2 and a1 = 3
Substitute in Un =
2. Find the nth term of the geometric sequence,
U9 = ( ) = ( ) = ( )( )=
a = , n = 10, r = 2
3. Find a formula for the nth term of the
3. Find the 8th term of the geometric sequence:
geometric sequence: 192, – 48, 12, – 3
1, , , ,…
4. For the geometric sequence: 5, -20, -80, find U9
Solution
From the geometric sequence 1, , , ,… 5. In the following G.P, find the value of y : 3, 12,
48, 5y + 7
a = 1, r = ÷ = × = and n = 8
Substitute in Un =
A Finite Geometric Progression
U8 = ( ) =( ) = A finite geometric progression is the one that has
a given last term. For all finite G.P:
4. The first term of a G.P. is 20 and the constant Un = arn – 1 = L

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Worked Examples 2. In the geometric sequence; , ….
1. Find the number of terms of the G.P. 2 + 4 +
, determine the number of terms.
8…+ 512

Solution 3. Determine the number of terms of the

From the G.P., 2 + 4 + 8…+ 512, geometric sequence, - 4 , 2, -1… .
a = 2, r = = 2, L = 512 = Un
4. The first term of a geometric sequenece is 2 ,
Un =
and the common ratio is 3, find the fifth term.
512 = 2(
=( 5. The first term of a geometric sequenece is 16 ,
256 = ( and the common ratio is , find the fouth term.
28 =
8=n–1 Unknown First Term of a G.P, Given a Term
n=8+1=9 and the Common Ratio,
The number of terms is 9 Given the value of a term other than the first term
and the common ratio, r,
2. Find the number of terms of the G.P. I. Write an expression for the given term and the
, , ….. common ratio.
II. Equate the expression to the given value of the
Solution term.
III. Solve for the value of a in the equation to
G.P.= , , …..
obtain the first term of the G.P.
a= ,r= = ,L= ,n=?
n–1 Worked Examples
Substitute in ar =L
1. The third term of a G.P is 5 and the common
( )n – 1 =
ratio is . Find the first term of the G. P.
n–1
× ( ) = ×-4
Solution
( )n – 1 = Un =
n = 3, r = and U3 = 5
( )n – 1 = ( )
U3 = ( ) = a( ) = a( )
n–1=5
⇒ a( ) = 5 (But U3 = 5)
n=5+1=6
=5
Exercises 25.7B a = 4 × 5 = 20

1. Find the number of terms of the geometric

Exercises 25.8
seqence: 0.03, 0.06, 0.12 … 1.92.
1.Find the first term of the G.P whose fourth term

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is 9 and common ratio, . ⇒ 16 = ………………..(1)

2. The third term of a G.P is -8 and the common U9 =

ratio is 4. Find the first term. U9 =
But U9 = 256
3. A geometric sequence has a fifth term of 6 and ⇒ 256 = ………………..(2)
a common ratio – . What is the first term of the
eqn (2) ÷ eqn (1)
sequence?
=
4. Determine the value of the first term of a G.P. 16 =
whose sixth term is -10 and constant ratio, -3.
r= √
5. A geometric sequence has a common ratio – . r=2
If the fourth term of the sequence is , find the
first term of the sequence. Put r = 2 in eqn (1)
16 = (
Finding the First Term, a, and the Common 16 = 16a
Ratio, r, of a G.P., Given the Values of Two or a=1
More Terms The first term is 1 and the common ratio is 2
Given the values of two or more terms of a G. P.
the first term, a, and the common ratio, r, can be b. The seventh term,
calculated as follows: U7 =
I. Identify the values of the given terms U7 =
II. Write two (or more) equations, each for the
given terms But a = 1 and r = 2
III. Solve the equations simultaneously to U7 = ( ( = =
determine the respective values of a and r
2. The second and fourth terms of an exponential
Worked Examples sequence of positive terms are 9 and 4
1. The fifth and ninth terms of an exponential respectively. Find:
sequence are 16 and 256 respectively. Find: i. the common ratio,
i. the first term and the common ratio , ii. the first term.
ii. the seventh term.
Solution
Solution i. Un =
i.U5 = U2 =
U5 = U2 = ar

But U5 = 16 ButU2 = 9

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⇒ 9 = ar ……….(1) ÷ =
U4 =
U4 = ar3 =

( ) =
ButU4 = 4
⇒r =
⇒4= ……….(2)
eqn (1) ÷ eqn (2)
Put r = into eqn (1)
=
= = a( )
= a
r=√ =
4a = 4
a=1
ii. Put r = into eqn (1) The first term is 1 and the common ratio is

9= (a)
ii. The eighth term U8 =
9 × 3 = 2a
But a = 1 and r =
a= = 13.5
Therefore the first term is 13.5 U8 = ( ) =( ) =

3. The third and sixth terms of an exponential Exercises 25.9

sequence are and respectively. Find: 1. The fifth and ninth terms of an exponential
sequence are 16 and 256 respectively. Find:
i. the first term and the common ratio,
i. the first term and the common ratio,
ii. the 8th term of the sequence.
ii. the sixth term.

Solution
2. The first and sixth terms of an exponential
Un =
sequence are 13 and . Find the commom ratio.
U3 =
U3 = ar2
ButU3 = 3. The sum of the first two terms of a G.P.is 3,
and the sum of the second and third terms is
⇒ = ar2……….(1) – 6. Find the first term and the commom ratio.

U6 = Sum of a Geometric Progression or G.P

U6 = ar5 Consider the geometric sequence with first term,
ButU6 = and common ratio, r :
, , , , ,
⇒ = ar5……….(2)
Suppose we want to find the sum of the GP:
Sn = + + + ….+ …(1)
eqn (2) ÷ eqn (1) Multiply both sides of the equation by r

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rSn = + + ….+ ……(2) iii. the sum of the five term.

eqn (1) – eqn (2) Solution

Sn - rSn = - i. G. P = ,
Factorize both sides of the equation
U1 = U2 = U3 = 2
Sn (1 – r) = (1 - )
( ⁄
Sn = r= = =4
– ⁄

On the other hand, eqn (2) – eqn (1) ii. a = , r = 4

rSn – Sn = arn – a
U5 =
U5 = = = × 256 = 32
Factorize both sides of the eqnto obtain
Sn (r – 1) = a(rn – 1)
Make Sn the subject to obtain iii. a = , r = 4, and n = 5
( –
Sn =
(
Sn = , for r> 1
Therefore, the sum of the first n terms of a G.P. ( (
with first term, a, and common ratio, r is given S5 = = = 127.88
by:
( 3. Find the sum of the first eight terms of
1. Sn = , for r < 1
–
the G.P. : 5 + 15 + …
(
2. Sn = , for r > 1
Solution
Worked Examples a = 5, r = = 3, n = 8
1. Find the sum of the first ten terms of the G.P (
Sn = , for r> 1
whose first term is and common ratio, 2
( ( (
S8 = = = = 16,400
Solution
a = , r = 2, n = 10 Some Solved Past Questions
(
The sum of the first and third terms of a G.P is 40
Sn = , for r> 1 while the fourth and the sixth terms are
( ( in the ratio 1: 4. Find the :
S10 = = = ( = 511.5
i. common ratio, ii. the fifth term.

2. An exponential sequence is given by Solution

i. First term, U1 = a
, . Find:
Third term, U3 = ar2
i. the common ratio, Sum of first and third terms,
ii. the fifth term,

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a + ar2 = 40………………..(1) II. Identify whether the problem is a series or a
sequence or both.
Fourth term, U4 = ar3 III. Form a mathematical statement for the
Six term, U6 = ar5 problem using series and sequence notations.
IV. Workout the problem and provide the
Fourth term: six term required answer.
ar3 : ar5 = 1 : 4
= Worked Examples
1. A worker started saving Gh¢2,500.00 the first
=
year, Gh¢5,000.00 the second year,
=4 Gh¢10,000.00 the third year and continued
r= √ = 2 or - 2 doubling his savings each year. How much will
be saved in 11th year?
ii. a + ar2 = 40………………..(1)
a( 1 + r2 ) = 40 Solution
But r2 = 4 Gh¢2,500, Gh¢5,000, Gh¢10,000…..
a( 1 + 4 ) = 40 A a = 2,500, r = = 2, n = 11
5a = 40
Un =
a=8
U11 = (
Therefore, the first term is 8
U11 = (
U11 = (
Exercises 25.10
1. Find the sum of the given geometric series n = U11 = 2500 × 1024
20, r = -2 and a1 = 1 U11 = 2,560,000

2. Calculate the sum of the six terms of the 2. Mr. Jimmy starts a job with an annual salary
geometric sequence: 2 + 12 + … + 972 of Gh¢2,500.00 which increases by Gh¢500.00
every year. After working for 6 years, Mr. Jimmy
3. Find the sum of the first ten terms of the is promoted to a new position associated with a
geometric series: 18 + 6 + 2 + …. new annual salary of Gh¢4,200.00 which
increases by Gh¢220.00 every year. Calculate;
4. The third term of a G.P. is 10 and the sixth i. Mr. Jimmy‟s annual salary in the tenth year of
term is 80. Find the common ratio, the first term service.
and the sum of the first six terms. ii. his total earnings at the end of the tenth year of
service.
Word Problems
For word problems involving series and Solution
sequence, i.a = 2,500, d = 500, n = 6
I. Carefully read the problem. Un = a + (n – 1) d
U6 = 2,500 + (6 – 1) 500

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U6 = 2,500 + 500(5) material in the sample at the beginning of the 7th
day.
U6 = 5,000
Annual salary at the end of the sixth year is 3. A culture of bacteria doubles every 2 hours. If
Gh¢5,000.00 there are 500 bacteria at the beginning, how many
bacteria will there be after 24 hours?
For the next four years
a = 4,200, d = 220, n = 4 4. Tim earned Gh¢105.00 on the first day. If he
U4 = a + (n – 1) d earned two times the amount of money earned the
U4 = 4,200 + (4 – 1) 220 day before each day, how much did he earn in the
U4 = 4,200 + ( 3) 220 first 7 days?
U4 = 4,860
Mr. Jimmy‟s annual salary in the tenth year of 5. A man starts savings on first April. He saves
service is Gh¢4,860.00 Gh¢1.00 the first day, Gh¢2.00 the second day,
Gh¢4.00 the third day, and so on doubling the
ii. His total earnings at the end of the tenth year amount every day. If he managed to keep on
of service = S6 + S4 saving under this system until the 15th of that
Sn = [ ( ] month, how much would he have saved?

S6 = [ ( ( ] 6. Doris was left a legacy of Gh¢600,000.00 by

S6 = 3 × 7500 = 22,500 his father. She decided to let her children have a
share of it but they have to earn it by catching
S4 = [ ( ( ] flies. For the first fly they caught, she gave them
S4 = 2 × 9060 = 18,120 Gh¢1.00, for the second fly, she gave them
Gh¢2.00, for the third fly, Gh¢4.00, for the fourth
But total earnings = S6 + S4 Gh¢8.00 and so on. By the end of the day, her
= 22,500 + 18,120 children had caught20 flies. How much of the
= Gh¢40,620.00 Gh¢600,000.00 was left for herself?

7. If a lab technician has a salary of

Exercises 25.11
Gh¢22,000.00 her first year and is due to get a
1. The sum of the interior angles of a triangle is
Gh¢500.00 raise each year;
180º, of a quadrilateral is 360º and of a pentagon
i. what will her salary be in the seventh year?
is 540º. Assuming this pattern continues, find the
ii. what is the total salary for seven years of work
sum of the interior angles of a dodecagon (12
of the technician?
sides).
8. On the first day of October, a teacher suggests
2. A mine worker discovers an ore sample
to his students that they read five pages of a novel
containing 500 mg of radioactive material. It is
and everyday thereafter, increases their daily
discovered that the radioactive material has a
reading by two pages. How many pages will they
half-life of 1 day. Find the amount of radioactive

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read during October if the students follow the Jimmy is promoted to a new position associated
instruction?
with a new annual salary of Gh¢7,500.00 which
9. If an air conditioning system is not completed increases by Gh¢360.00 every year. Calculate;
by the agreed upon date, the contractor pays a i. Mr. Jimmy‟s annual salary in the tenth year of
penalty of Gh¢500.00 for the first day that it is service
over due,Gh¢600.00 for the second day, ii. His total earnings at the end of the tenth year
Gh¢700.00 for the third day and so on. If the of service
system is completed 10 days late, then what is the
total amount of penalties that the contractor must Challenge Problem
pay? Consider yourself, your parents, your grand
parents, your great – grand parents, your great –
great grand parents, and so on, back to your grand
10. Mr. Jimmy starts a job with an annual salary parents with the word “great” used in front 40
of Gh¢3,000.00 which increases by Gh¢400.00 times. What is the total number of people you are
every year. After working for 10 years, Mr. considering?

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26 GEOMETRICAL CONSTRUCTION Baffour – Ba Series
Introduction
A B
A mathematical set which contains the required
6cm
instruments is needed for geometrical
constructions. Most at times students are 2. with the pin – point of your compass at A and a
expected to use a pair of compass and a ruler distance more than half of the line AB draw an
only. arc above the line and another arc below the line.

Construction of a Line Segment 3. Maintaining the same distance, set the compass
A line segment is a straight line that has two point at B and construct an arc
endpoints. That is to say, it starts from one point, above and below the line to cut the first pair of
say ( and ends at another point say ( . arcs drawn.
To construct a line segment of length
i. With a ruler and a pencil, draw a straight line
which is presumably more than 4cm

A 6cm B
ii. With a ruler and a pencil, cut the line towards
the left end and label that point A

4. Join the points of intersection of the arcs with a

A
straight line to get the bisector of line AB.
iii. Measure a distance of 4cm on the ruler and
with the compass point at A draw an arc to cut
the line at the right end and label that point B.
A B
4cm A 6cm B

Thus = 4cm above is the required line

segment
Exercises 26.2
Note: Do not rub the extensions. Bisect the following line segments.
1. = 8.5cm 4. = 7.5cm
Bisecting a Line 2. = 8cm 5. = 7cm
To bisect a line means to divide the line into two
equal halves. Construction of a Perpendicular at a Given
To bisect a line = 6cm. Point on a Straight Line (AB)
1. Draw = 6cm I. Draw the required straight line AB.

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II. Mark the point says M on the line.
III. With the compass point at M and at any
convenient radius, construct two arcs to cut the
line at the left (c) and right (d ).
IV. Using any suitable radius, step on the left arc
C D
(c) and right arc ( d ) respectively and draw, two
A B
arcs above the line to intersect (meet) with each
other at E. Fig. 2
V. Draw a line from E to the point M to
get the required perpendicular line. Exercises 26.4
Construct a perpendicular at the end of the
E
following lines.
1. = 6cm at A 2. = 6cm at B.
3. = 7.5cm at Q 4. = 7.5cm at P
5. = 7cm at M 6. = 7cm at N

Construction of a Perpendicular from a Given

A c m d B
Fig. 1 Point (K) to a Given Line AB
1. Choose any point say, K above the line AB.
Exercises 26.3 2. With the compass pin at K, draw an arc to cut
Construct a perpendicular at the given line AB at C and D.
points on the following lines. 3. With C as the centre and any convenient
1. = 7cm, m = 2cm from B. radius draw an arc below line AB, with the same
2. = 8cm, n = 3cm from P. radius and D as the centre , construct another arc
3. = 8cm, n = 3cm from Q. below line AB to meet the first arc at point E.
4. = 6.5cm, r = 2.5cm from M. 4. Join K and the arcs at E with a straight line to
obtain the perpendicular from K to AB.
Construction of a Perpendicular at the End of
a Given Line (AB) K
1. Draw the given line AB
2. To draw the perpendicular at the end A, extend
the line to the left of A.
3. With the compass point at A and at any
convenient radius, construct a semi- circle below A C D B
the line to cut the line at C and D respectively.
4. Using any suitable radius step at C and D Fig. 3
respectively and construct two arcs above the line E
CD to meet at E. Constructing a Line Parallel to a Given Line AB
5. Join the arcs and the point A with a straight line 1. Construct the line AB and mark a point K
to get the required perpendicular at the end point A. outside (on top of) AB.

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2. Construct a perpendicular from K to line AB as 5. Draw a straight line from B through P. BP is
in figure 3. the angel bisector of ABC.
3. Extend the line KE through K and step at K A
with a convenient radius , mark two points
above ( L ) and below ( M ) the line KM. S P
4. Bisect the line segment LM and identify the B
meeting points of the arcs as C and D. T
5. Draw a line through C and D. Line CD is C
therefore the line parallel to AB. Exercises 25.6
Bisect the following angles;
L 1. A 2. B

K B
C A
D
C
C
M
C D 2. 3.
B
A A C
B

E Fig. 4
B A C
Exercises 26.5
Construct a line parallel to the following lines. Construction of Angles (900, 450, 1200, 600, 300,
1. = 7.5cm 4. = 9cm 150, 750, 1050) at a Given Point Using a Ruler
2. = 8.5cm 5. = 6.5cm and a Pair of Compasses Only
3. = 8cm 6. = 7cm 1. Construction of 900 and 450 at the point A on
line AB
Angles i. Draw the line segment AB with extensions.
Bisecting an Angle ii. With A as centre and at any convenient radius,
To bisect an angel means to divide the angel into draw a semi–circle on top of the horizontal line to
two equal halves. cut it at C and D.
iii. With the compass point at C and the same
Steps: radius, draw an arc to cut the semi – circle at E.
1. Draw an angle ABC. iv. With the compass point at D and the same
2. With the compass point at B draw an arc to radius (as in ii & iii) draw an arc to cut the semi
cross AB and BC at S and T respectively. – circle at F.
3. With the compass point at S and a suitable v. With the pin at E and F respectively and at
radius construct an arc inside the angel ABC. any convenient radius, construct two arcs above
4. With the same radius, place the point at T and the semi – circle using the same radius, so that
construct another arc to meet the first one at P. the arcs intersect with each other at.

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vi. Draw a straight line through the arcs at G to d. Maintaining the same radius, step at D and
the point A to get an angel of 900at right. draw an arc to cut the semi – circle at F.
e. Draw a straight line from A to pass through E
G
to obtain 1200 at right.
Figure V
E F
H
E F 1200
C A D B
C D Figure VII.
A B

To construct angle 450, construct 900 and bisect f. To get 600, draw a line from A to pass through
the 900 by placing the compass point at D and H F.

G
E F

H 600
E F C A D B
Figure VIII
450
C D 3. Construction of 300 and 150 at a point A on
A B
line AB
i. Draw line AB and construct 600 at A.
Exercises 26.7 ii. Construct a bisector of angle 600 by placing the
Given that = 5cm, construct the compass point at D and F.
following, all on separate diagrams iii. A line from A through the arcs gives an angle
1. 900 at B 2. 900 at A of 300.
3. 450 at B 4. 450 at A,

2. Construction of 1200and 600 at a Point on F

Line AB
a. Draw the line AB with extensions. 30
A 0 D B
b. With A as centre and at any convenient radius
Figure IX
draw a semi – circle on top of the horizontal line
to cut it at C and D. To construct 150, go through the process of
c. With the compass point at C and the same constructing 300 and bisect the 300 to get 150
radius, construct an arc to cut the semi- circle at E. as shown below.

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 Exercises 26.9
Using a ruler and a compass only, construct the
following angles at the given points on separate
diagrams.
1. 750 at A and B
2. 750 at M and N
150 3. 1050at P and Q
A D B
Figure X
The Idea of Locus
Exercises 26.8 1. The locus of a point which is equidistance from
Construct the angles at the given point. one point means, draw a circle. Thus the locus of
1. 1200 at A and B. point, L1 which is 6cm from A, means draw a
2. , 600 at P and Q.. circle with radius of 6cm from A, with A as
3. 300 at A and B. centre.
L
4. 150at P and Q.
1
A
4. Construction of 750 at a point A on line AB
a. Draw the line AB and construct 900
at A as shown below. 2. To construct locus of points L 2 which is
b. Bisect the angle equidistance from two points A and B means
between H and F to draw the bisector or the mediatorof AB as shown
get 750 at the right. below.
H
E F
750
C A D B A B
0
5. Construction of 105 at a point A on a given
line AB
L2
i. Draw the line AB and construct 900 at A as
shown in figure below. 3. To construct the locus of points L3 which is
ii. Bisect the angle between E and equidistance from lines AB and AC means, place
H to get 1050 at the compass point at the common point A and
the right. bisect the angle between lines AB and AC.
B

E H
P
0 A L
105
3

A B C

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4. The locus of a point L4 equidistance from three III. Maintaining the same radius, replace the point
points A, B and C means bisect AB, BC and AC of the compass at B and draw another arc to cut
and use the point of intersection of the bisectorsto the first at C .
draw a circle through the three points A, B and C. IV. Join A, B and C to form the required triangle.

C
C

O 3cm 3cm
A

A 3cm
B
B 2. Isosceles triangles
In this case only two sides are equal. For
example to draw a triangle ABC with sides 6cm,
5. The locus of points at a fixed distance d, from 6cm and 4cm.
a line line , L is a pair of parallel lines, d distance I. Draw line AB.
from L and on either side of L. II. Open the compass to 4cm and mark the points
A and B on the line so that /AB/ = 4cm
d III. Set the compass to 6cm.
L IV. Place the compass at A and draw one arc.
d
V. Maintaining the same distance, place the
compass at B and draw another arc which cuts or
6. The locus of points equidistance from two meets the first at C.
parallel lines , L1 and L2 is a line parallel to both VI. Draw lines AC and BC.
L1 and L2 and mid way between them. C
Not drawn to scale
L1
d
6cm 6cm
d
L2

Constructing Triangles A 4cm B

1. Equilateral triangles
Set your compass to the length of the side of the 3. Scalene triangle
required equilateral triangle. For example to draw A scalene triangle has three unequal sides. For
an equilateral triangle of sides 3cm example to construct triangle ABC with sides
I. Draw a line and mark a point A. Measure 3cm. 6cm, 5cm and 4cm.
II. Place the point of the compass at A and draw a I. Draw a line and mark the point A near to one
generous arc, 3cm from A, which cuts the line at B. end of it.

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II. With your compass set at 6cm, mark the point Exercises 26.10
B on the line exactly 6cm from A. Using a pair of compass and ruler only,
III. Adjust your compass to 5cm and place the construct the triangles with the following sides
compass point at A and draw an arc above the and name the type of triangle drawn in each
line. case.
IV. Adjust your compass to 4cm, place the 1. Triangle PQR with each side 7cm.
compass point at B and draw another arc to cut or 2. /AB/ = 8cm, /BC/= 7cm and /AC/ = 7cm
meet the first one at C. 3./AB/ = 7cm, /BC/= 8cm and /AC/ = 8cm
V. Join C to A and B to obtain the diagram shown 4./AB/ = 9cm, /BC/= 7cm and /AC/ = 8cm
below.
C The Circum-circle
If the perpendicular bisectors of all the three sides
of a triangle are constructed, they meet at a point
5cm 4cm
O. With O as centre, a circle can be drawn to
touch the three vertices of the triangle. The circle
is said to be circum-scribed and it is called a
A 6cm B “circum-circle”. It‟scentre is the circum-centre.

4. Right – Angled Triangles

C
It has one angle as 900. For example to construct
a right-angled triangle of ,
O
and the right angle at C. A
I. Draw a straight line AB
II. Using your compass mark point B, 8cm from A .
III. Construct a right angle at B.
B
IV. Open the compass to measure 8cm
V. Place the compass point at A and draw an arc
to cut across the perpendicular line at C.
VI. Join C and A. C Worked examples
1. Using a ruler and a pair of compasses only,
construct triangle PQR such that ,
0
and angle PQR = 90
8cm
ii. Construct the perpendicular bisectors of PQ
and QR and name the intersection O.
iii. Draw a circle, with O as center and OQ is
7cm radius.
A B
iv. Measure: (a) (b) angle QPR

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Solution

O
O 4cm

P 6cm Q

Not drawnto Worked examples

scale 1. a. Using a ruler and a pair of compasses only,
, Angle QPR = 340 construct triangle ABC in which /AB/ = 6.5cm,
/BC/ = 7.5cm and /AC/ = 8cm
2. a. i. Using a ruler and a pair of compasses b. Construct angle bisectors of angles ABC, BCA
only, construct PQR such that angle and CAB to meet at O
0 0
RPQ = 90 and angle PQR = 30 , measure . c. With O as centre, draw a circle to touch the
ii. Construct the perpendicular bisector sides of the triangle.
(mediator) of RQ. Let it meet RQ at O
b. With O as centre and radius OP, draw a Solution
circle. Measure
C
Solution
R Not drawn to scale

8cm 7.5cm
O
O

30
0 A 6.5cm B
P 8cm Q
2. a. Using a ruler and a pair of compasses only,
/RQ/= 9.4cm, and /OP/ = 4.4cm construct triangle ABC such that ,
0
and angle ABC = 60 . What type of
The Inscribed Circle triangle is triangle ABC?
When the angle bisectors of a triangle are b. Construct the bisector of angle BAC to meet
constructed, the three bisectors meet at a point O. BC at D. Measure
With O as centre, a circle can be drawn to touch c. Construct the perpendicular bisector of BA
all the three sides of the triangle. This is called to meet AD at O.
the “inscribed circle”.

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d. Using O as centre and radius OD, draw a circle To construct a rectangle ABCD of length 8 cm
to touch the three sides of the triangle. and breadth 6 cm;
I. Draw a line segment AB = 8 cm
Solution II. Construct perpendiculars at A and B
C
III. With A as centre and radius 8 cm, draw an arc
to cut the perpendicular at A at the point D
IV. With B as centre and radius 8 cm, draw an arc
to cut the perpendicular at B at the point C
8cm V. Join C to D with a ruler to obtain square
D ABCD.
O
D C

600
A 8cm
B
Not drawn to scale 8 cm

a. Triangle ABC is an equilateral triangle and

A 10 cm B

Exercises 26.11
Drawing Quadrilaterals 1. Costruct a square ABCD of side 6.5 cm.
1. A square and a Rectangle Measure the length of the diagonal
Set your compass to the length of the side of the
required square. For example, to construct 2. Construc a rectangle with length 10 cm and
asquare ABCD of sides 4.5cm breadth 7.5 cm. Measure the length of its
I. Construct a line segment AB = 4.5 cm diagonal
II. Construct perpendiculars at A and B
III. With A as centre and radius 4.5 cm, draw an 2. Constructing Parallelograms
arc to cut the perpendicular at A at the point D a. Two Sides and the Length of the Diagonal.
IV. With B as centre and radius 4.5 cm, draw an For example, to draw parallelogram ABCD in
arc to cut the perpendicular at B at the point C which AB = 6 cm, BC = 4 cm and diagonal 7 cm
V. Join C to D with a ruler to obtain square
ABCD. I. Draw AB = 6cm
II. With A as center and radius 7 cm, draw an arc
D C III. With B as center and radius 4 cm, draw
another arc cutting the previus arc at C
IV. Join BC and AC
4.5 cm
V. With A as center and radius 4 cm, draw an arc
VI. With Cas center and radius 6 cm, draw
A 4.5 cm B another arc cutting the previously drawn arc at D

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VII. Join DA and DC to form parallelogram iii. With center B and radius 7, cut BQ at C
ABCD iv. Construct < 750 (supplementary angle of < 1050)
at A
D C v. With A as center and radius 7 cm, cut AP at D
vi. Join C to D to obtain parallelogram ABCD
7 cm P Q
4 cm

D C

A 6 cm B

b. One side and the length of Two diagonals

Note that the diagonals bisect each other. For 0
0
75 105
example to construct parallelogram ABCD whose
A 10 cm B
side is 5 cm and diagonals 6 cm and 6.6 cm
i. Draw AB = 5 cm Exercises 26.12
ii. With A as centre and radius 3.3 cm, draw an 1. Using a ruler and a pair od compasses only,
arc construt a parallelogram PQRS with PQ = 9 cm,
iii. With B as center and radius 3 cm, draw QR = 6 cm and < PQR = 1200. Measure the
another arc cutting the previous arc at O length of QR
iv. Join OA and OB
v. Produce AO to C such that OC = AO and 2. Using a ruler and a pair of compasses only,
produce BO to D such OD = OB construct parallelogram ABCD in which AB = 7
vi. Join AD, BC and CD to form parallelogram cm, BC = 5 cm and diagonal 8.5 cm
ABCD
3. Using a ruler and a pair of compasses only, to
D C construct parallelogram PQRSin whichPQ = 7.5
6.6 cm cm and diagonals 11 cm and 8.4 cm
O
Constructing a Kite
6 cm
Worked Examples
A 5 cm B a. Using a ruler and a pair of compasses only,
construct triangle ABC with = 10cm, angle
0 0
c. Two sides and One Angle ABC = 30 and angle CAB = 60 .
Make use of the fact that consecutive angles are b. Construct a perpendicular from the point C to
supplementary. Therefore, draw the given angle meet the line AB at P.
as well as its supplementary. For example to draw c. i. Extend the line CP to meet point D such that
parallelogram ABCD with AB = 10 cm, BC = 7 = .
cm and < ABC1050 ii. Join A to D and B to D.
i. Construc AB = 10 cm d. What type of quadrilateral is ADBC?
ii. Construct < 1050 at B

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Solution Solution
C a. Q O

10cm
0 0
60 30
P 10cm
A B
450

M 10cm N
Not to drawn to scale
b. i. /MO/ = 7.3cm ii. /NQ/ = 18cm
D

d. Quadrilateral ADBC is a kite. Drawing Other Figures

Sometimes students are required to construct
Constructing a Rhombus some figures other than a circum-circle and an
Worked Examples inscribed circle. Follow the question carefully
1. a. Using a ruler and a pair of compasses only, and do exactly what it requires from you to obtain
construct triangle ABC with = 8cm, = such figures.
8cm and = 7cm.
b. Bisect angle ABC and let the bisector meet AC Worked examples
at D. Produce to P such that . 1. a. Using a ruler and a pair of compasses only,
Join AP and CP construct triangle XYZ in which Z ,
0
c. Measure i. angle ADB ii. angle XYZ = 60 and XZ . Measure X
d. What kind of quadrilateral is ABCP? b. Construct the mediator of YZ.
c. Draw a circle with centre X and radius 5cm.
Solution d. Measure where A is the point of
P C intersection of the mediator and the circle in
the triangle region XYZ.

D 8cm Solution
7cm

8cm X
Not drawn to scale A B
i. Angle ADB = 900
ii. Quadrilateral ABCP is a Rhombus. 9cm
A
2. a. Using a ruler and a pair of compasses only,
construct a rhombus MNOQ of length 10cm and 60
< MNO = 450 6 cm 0
Z
Y
b. Measure; i. /MO/ ii. /NQ/

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i. ii. 4. Using a ruler and a pair of compasses only,
construct triangle XYZ with = 8cm, XYZ =
0
2. a. using a ruler and a pair of compasses only, 60 and .
draw Measure: i. angle YZX ii.
b. Construct a perpendicular to PQ at Q.
c. Construct angle QPS = 600 at the point P onPQ Solution
X
such that .
d. Construct a line parallel to PQ through Q and
the parallel line through S to meet at R. 9cm
e. Measure .

600
Solution 8cm
Y Z
S R Not drawn to scale

Angle YZX = 650 and

6.5cm
4. a. Using a ruler and a pair of compasses only,
0
60 construct triangle PQR in which
angle QPR = 450 and angle PQR = 900. Measure
P 9cm Q
b. Construct the mediator of PQ to meet PR at the
3. a. Using ruler and a pair of compasses only, point S.
construct triangle ABC such that c. With S as the centre and radius 3cm, construct
angle ABC = 900 and angle BAC = 300. Measure a circle.
the length of BC.
b. Bisect the angle ACB to meet BA at D. Solution
c. What type of triangle is CDA? R

Solution

C
S

450
P 8cm Q
Not drawn to scale

900 0
30 5. i. Using a pair of compasses and a ruler only,
D A
B 10cm construct triangle ABC such that
Triangle CDA is a scalene triangle. and angle ABC = 300 and Measure
angle ACB.

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ii. Construct a perpendicular from C to meet line c. With O as centre and radius OA, draw a circle
AB at D. to touch the vertices of the triangle.
d. Measure angles ABC and ACB.
Solution
2. a. Using a ruler and a pair of compasses only,
C construct triangle XYZ such that ,
8cm /ZY/ = 8.5cm and ZX = 7.5cm.
b. Construct the mediator of line YZ and the
mediator of line XZ.
0
30 c. Locate O, the point of intersection of the
A D 10cm
B mediators of lines YZ and XZ.
d. With O as centre and radius OY, draw a circle.
Ans: 1d. ABC = 780, ACB = 470
Angle ACB = 960 and = 4cm.
e. Measure the radius of the circle drawn in (d)
and calculate the circumference of the circle
6. a. Using a ruler and a pair of compasses only,
( .
construct triangle PQR such that the length of PQ
= 10cm, angle QPR = 900 and angle PQR = 300.
3. Using a ruler and a pair of compasses only,
b. Bisect the angle QRP to meet PQ at M c.
i. Construct ∆ ABC where / AB/ = 7cm, /AC/ =
With M as centre, and radius MP, draw a circle.
8cm and < A = 1050.
d. Measure the radius of the circle.
ii.the locus of points 6cm, from C;
Solution iii. Y, the locus of points equidistant from ⃗⃗⃗⃗⃗
and ⃗⃗⃗⃗⃗ to cut X in P and R.
R Not drawnto scale b. Measure i. |⃗⃗⃗⃗⃗ | ii. |⃗⃗⃗⃗⃗ |

4. a. Using a ruler and a pair of compasses only,

construct triangle ABC with sides /AB/ = 7cm,
/BC/ = 8cm and /AC/ = 9cm
b. Construct the locus L1,of the point equidistance
P M 10cm from A and B, and the locus L2, of the point
Q
equidistance from B and C to meet at O.
c. With centre O and radius OA, draw a circle to
Radius, MP = 3.3cm. pass through the vertices of the triangle.
d. Measure and write down the radius of the
Exercises 26.13 circle you have drawn.
1. a. Using a ruler and a pair of compassesonly,
construct triangle ABC such that =7.5cm, 5. Using a ruler and a pair of compasses only,
BC = 8.5cm and =10cm. i. Construct a ∆ ABC, such that ,
b. Construct perpendicular bisectors of AB and /AC/ = 8cm and angle BAC = 300.
BC to meet at O.

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ii. Construct the bisector of the angles ACB to 8. i. Using a ruler and a pair of compasses only,
meet line AB at D. construct triangle ABC with /AB/ = 10cm, angle
iii. Measure and . ABC = 300 and angle CAB = 600.
iv.Write down the ratio Ans:2:1 ii. Construct the locus of the point which is
Ans.(iii) /AD/ = 4cm, /BD/ = 2cm equidistance from line AC and CB to meet AB at P.
iii. Construct the locus of the point which is
6. a. Using a ruler and a pair of compasses only, 2.5cm from P.
construct triangle ABC such that = 7cm,
0
/AC/ = 7cm and angle CAB = 60 9. Using a ruler and a pair of compasses only;
b. Construct a perpendicular from C to meet AB i. Construct triangle XYZ where /XY/ = 8.5cm,
c. Construct /CD/ = 7cm, parallel to AB through C. angle ZXY = 450 and angle XYZ = 450.
d. Join B to D with a ruler. ii. Construct a perpendicular from Z to pass
e. Name the figure formed. (Parallelogram) through XY to T such that /ZT/ = 9cm.
iii. Join X to T and Y to T to form figure XTYZ.
7.a. Using a ruler and a pair of compasses only, iv. Name figure XTYZ. Ans: Rhombus
construct triangle RST in which = 6.5cm,
0
angle TRS = 45 and angle RST= 750. 10. i. Using a ruler and a pair of compasses only,
Measure . Ans: /RT/ 7.7cm construct triangle ABC in which = 7.5cm,
0 0
b. Construct the locus, L1, which is equidistance angle BAC = 75 and angle ABC = 45
from the points R and S and the locus, L2, which ii. Bisect angles ACB, CBA and BAC. Let the
is equidistance from the points S and T. Locate bisectors meet at O.
the meeting point of L1 and L2 as O. iii. With centre O, draw a circle to touch the
c. With O as centre, draw a circle to touch the sides of the triangle.
vertices of the triangle.

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27 MENSURATION II Baffour –Ba Series

Net of Common Solids Surface and Surface Area

1. Net of a cylinder By surface, we mean an object that has only two
dimensions – length and breadth but no
thickness.

2. Net of a cone Study the examples below;

1. C D

3. Net of a cuboids A B

This parallelogram has a flat surface so it can be

called a plane.

2.

4. Net of a prism In these examples, the surfaces are not flat but
curved and are therefore not planes. All solids
are bounded by a surface or surfaces. Surface
area means the measure of surface.

5. Net of a pyramid Surface Area of a Rectangular Solid

All rectangular solids have six surfaces. The
surface area is found by calculating the area of all
surfaces and then adding them up.

6. Net of a cube b

Total surface area of a rectangular solid,

A = 2(ab + ac + bc )

Baffour – Ba Series, Core Maths for Schools and Colleges Page 732
Worked Examples 3. A rectangular box has length 6cm, breadth
1. Calculate the total surface area of a rectangular 7cm and height 8cm. Calculate the total surface
solid, which has a length, width and height of area of the box.
18cm , 4cm and 6cm.
Solution
Solution
7cm
4cm
Top
8cm
End Front End
s s 6cm 6cm

Area of face = 42cm2

18cm
(Base) Area of top = 48cm2
Area of ends = 8cm × 7cm = 56cm2
Let a = 18, b = 6 and c = 4
Area of front (ab) = 108cm2
Total surface area,
Area of top (ac) = 72cm2
A = 2( 42 + 48 + 56)cm2
Area od end (bc) = 6cm × 4cm = 24cm2
A = 2 × 146 = 292cm2

Total surface area, A

Exercises 27.1
A = 2(108 + 72 + 24)cm2 A.1. Calculate the total surface area of a
A = 2 × 204cm2 = 2
rectangular box that is 18cm long, 12cm wide
and 9cm high.
2. How many square centimeters are in a total
surface area of a rectangular solid of length 2. Calculate the total surface area of a rectangular
15cm, width 3cm and height 2cm. solid, which has length, width andheight of
13cm, 9cm and 7cm respectively.
Solution
15cm
3. How many square centimeters are in a total
2cm surface area of a rectangular solid of length
12cm, width 7cm and height 5cm.
3cm

4. A rectangular box has length 18cm, breadth

Area of face = = 30cm2 21cm and height 24cm. Calculate the total
Area of top = = 45cm2 surface area.
Area of ends = 2cm × 3cm = 6cm2
B. Calculate the total surface are of the
Total surface area, rectangular solid with the dimensions
A = 2( 30 + 45 + 6)cm2 = 2 × 81 = 162cm2
Length Width Height

Baffour – Ba Series, Core Maths for Schools and Colleges Page 733
 1. 8m 3m 2m But l = 25 cm
2. 12cm 2cm 3cm Total surface area = 6 × ( = 3,750 cm2
3. 5 cm 1 cm 2 cm
4.36 cm 5cm 3cm 3. The volume of a cube is 125cm3. Find
the area of one of its faces.

Surface Area of a Cube Solution

Since the cube is a rectangular solid, the method V 3

for calculating the surface area of a rectangular Substitute 3

solid can be used to calculate the surface area of By substitution,

a cube. In a cube, all the edges are equal. 125cm3 3

√ OR
To find the total surface area of a cube, find 3 3
125cm
the area of one face and multiply it (answer)
53 = L3
by 6 (because it has 6 faces).
Equating exponents and bases,
2
Area, A .
Mathematically,
2 Substitute
Area (A) , where
A= ( = 5 cm × 5 cm = 25 cm2
A is the total surface area of the cube
is the length of an edge of the cube.
Exercises 27.2
Calculate the total surface area of a cube
Worked Examples
with the following sides;
1. Calculate the total surface area of a cubewith 1) 13cm 2) 17cm 3) 66m 4) 10cm
sides 4cm long.

4cm
Surface Area of a Cone
Solution The total surface area of a cone is calculated
2 D
Area ,
by the formula; A where
but m 4cm5
2
4cm
2 A represent the total surface area
Area ( = 96 cm cm
p is the perimeter of the base
2. A box is made of equal dimension of S is the slant height and
25cm.Calculate the surface area of one of the b is the area of the base
face and the total surface area of the box.
Slant height
Solution
2
i. Area of one face A
But l = 25 cm base
A= 25 cm × 25 cm = 625 cm2
But the perimeter of the base p = the
2
ii. Total surface area of the box , A = 6l , circumference of the circle.Thus, p = C = 2πr

Baffour – Ba Series, Core Maths for Schools and Colleges Page 734
The area of the base, b = the area of the 3. 16 cm 35cm
circle.Thus, b = πr2 4. 15 cm 14 cm
By substitution, total surface area of the cone:
A 2 B. Find the curved surface area of a cone with
2 the following measurements;
A
1. Slant height = 3cm, diameter = 14cm.
2. Slant height = 24cm, diameter = 14cm.
Curved Surface Area of a Cone 3. Slant height = 12cm, diameter = 9cm.
The curved surface area of circular cone of
4. Find the radius of a cone with a curved surface
base radius, r, and slant height, s, is
area 44m2 and a slant height 4m
calculated by the formula, A = π r S
5. What is the height of a cone with a curved
surface area of 28πm2 and a radius of 4m
Work Examples
1. Calculate the total surface area of a solid with a
Surface Area of a Pyramid
slant height of 15cm and a base radius of 7cm.
The formula for calculating the total surface area
of a pyramid is; A = ps + b , where
Solution
A = rh + r2 A is the total surface area
p is the perimeter of the base.
Substitute π = , r = 7cm , h = 15 cm,
s is the slant height and
A= × 7 × 15 + ×7×7 b is the area of the base.
A = 330 + 154 = 484cm2
Slant edge
2. Find the total surface area of a cone of slant
height 5cm and base radius 7cm. Slant height

Solution
A = rh + r2 A
Base
Substitute π = , r = 7cm , h = 5 cm,
Method 2
A= ×7×5+ ×7×7
I. Draw the net of the pyramid (rectangular) to
A = 110 + 154 obtain four triangles surrounding the base.
A = 264cm2

Exercises 27.3
A. Calculate the total surface area of a cone
with the following dimensions
II. Find the area of each triangle using the
Slant Height Radius of Diameter of
(h) the base, r the base, d formula , A = bh, where h is the slant height of
1. 5 cm 3.5cm the pyramid.
2. 25 cm 21cm
Baffour – Ba Series, Core Maths for Schools and Colleges Page 735
III. Find the area of the base using the formula, Worked Examples
A=L×B 1. Calculate the total surface area of a solid
IV. Total surface area = Area of the base + area pyramid with a square base of sides 10cm, and a
of each triangle. slant height of 12cm.

Surface Area of a Pyramid Solution

2 12 cm
The formula for calculating the total surface area Area of the base (b)
of a pyramid is; A = ps + b , where b = 10 cm × 10 cm
b = 100 cm2
A is the total surface area
p is the perimeter of the base. 10cm
s is the slant height and 10cm
Perimeter of the base
b is the area of the base.
P = 10 cm + 10cm + 10 cm + 10 cm = 40

Slant edge Substitute p = 40 cm, h = 12 cm and b = 100

Slant height A= ps+b
2
A
A
Base Method 2
Area of the base 12
A = 10 × 10 = 100 cm2
Method 2
10
I. Draw the net of the pyramid (rectangular) to
Area of each triangle 10
obtain four triangles surrounding the base.
A = (10)(12) = 60 cm2

Area of 4 triangles
= 4 (60)
= 240cm2

Total surface area;

II. Find the area of each triangle using the = Area of the base + area of each triangle.
= 100 cm2 + 240 cm2
formula , A = bh, where h is the slant height of
= 340 cm2
the pyramid.
III. Find the area of the base using the formula,
2. The height of a pyramid is 24cm. If it has a
A=L×B
square base of side 14cm, find its total surface
IV. Total surface area = Area of the base + area
area.
of each triangle.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 736
Solution Area of trianlges of the net;
By pythagoras theorem, = 2 × (8) (12) + 2 × (6) (13)
S=√ = 96 cm2 + 78cm2 = 174 cm2
S=√
24
25 S = 25 cm Area of the base of the pyramid;
A = 8 × 6 = 48 cm2
7
14
14
Total surface area;
Perimeter of base; = Area of the base + area of each triangle.
P = 14 + 14 + 14 + 14 = 56cm = 48 cm2 + 174cm2 = 222 cm2

Area of base; 4. The base of a pyramid, 16cm high, is a square

b = 14 × 14 = 196 cm2 whose each side is of length 24cm. Find the slant
height and the length of the slant edge of the
Total surface area; pyramid.
A = p s + b = (56) (25) + 196 = 896 cm2
Solution
3. The height of a pyramid is 12cm. If it has a By pythagoras theorem,
rectangular base of length 8cm and breadth 6cm, S=√
find the total surface area of the pyramid. S=√
16 s S = 20 cm
Solution
12 24cm
24cm
12 S By pythagoras theorem,
S 12
SE = √
4 6 20 SE
3 6 SE = √
8
8 SE = 23 cm
Pyramid 1 Pyramid 2 12 12
Exercises 27.4
From pyramid 1, Calculate the total surface area of a pyramid
S=√ 12 with the following dimensions;
S= √ = 12cm
13 13 Slant height Length of side of
6
From pyramid 2, 8 (h) Square base (b)
S=√ 12 15cm 8cm
29cm 25cm
S=√ = 13cm
30cm 15cm
150cm 60cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 737
B. 1. The base of a rectangular pyramid is 10 For an opened cylinder, the total surface area, A
cm by 18 cm. If the height of the pyramid is = 2 rh (lateral surface area)
15cm, find its total surface area.
For a cylinder closed or opened at one end, the
2. The height of a pyramid is 12 cm. Find the total surface area , A = r2+ 2 rh
total surface area of the pyramid if it has a square
base of sides 10cm Worked Examples
1. Calculate the total surface area of a solid
3. A pyramid has a height 5cm. It has a cylinder with a height 10cm and radius 7cm.
rectangular base that measures 6cm by 8cm.
Calculate its total surface area. Solution
A = 2πr2 + 2πrh
4. The pyramid below has arectangular base and Substitute r = , r = 7cm and h = 10 cm
faces that are isoscleles triangles. Finf the total
surface area to the nearest whole number. A=2× ×7×7+2× × 7 × 10
2
A = 308 + 440 = 616cm

8 cm 2. What is the total surface area of a cylinder of

height 10cm and diameter 7cm.
8 cm
4 cm Solution
A = 2πr2 + 2πrh
10cm
Surface Area of a Cylinder Substitute π = ,
The formula for calculating the total surface area
2 r = = cm and h = 10 cm
of a solid cylinder is: ,
7cm
where π is a constant value taken as , r is the
A=2× × × +2× × × 10
radius of the cylinder and h is the height of the
cylinder. A = 77 + 220 = 297cm2
Top

3. The base radius of a closed cylinder is 4m.

A
The height of the cylinder is 7m. Calculate its
Lateral surface area
total surface area ( π = ).
Base
NB: the top and base of the cylinder are circles. Solution
Total surface area of a closed cylinder,
Total surface area, A, A= 2π r (r + h),
= Area of top and base + lateral surface area Substitute r = 4m and h =7m

2
A=2× × 4 (4 + 7) = 276.57m2
+

Baffour – Ba Series, Core Maths for Schools and Colleges Page 738
4. A cylinder closed at one end has radius 7cm ii. V = πr2h
and height 20cm. (π = ). V= × (7) 2 × 10 = 1,540cm3
i. Find the total surface area.
ii. If the cylinder is filled with water to a depth Exercises 27.5
of 5cm, calculate the volume of water in it. A. Calculate the total surface area of a solid
cylinder with the given dimensions;
Solution 1. r = 10.5 cm, h = 22cm 3. r = 35 cm, h = 26cm
i. Total surface of a cylinder, 2. r = 21 cm , h = 23cm 4. r = 7 cm , h = 17cm
A = π r2 + 2 rh
Substitute r = 7cm and h = 20cm B. Find the curved surface area of a cylinder
A= ×7×7 +2× × 7 × 20 with the following measurements
A = 154 + 880 = 1,034 cm2 1. Height = 14cm, radius = 6cm
2. Diameter = 70cm, height = 9cm
ii. Volume of cylinder, v = π r2h 3. Length = 12mm, radius = 2mm.
But r = 7cm and h = 5cm
C. Find the radius of a cylinder with the
V= × (7)2 × 5 = 770cm3
following measurements;
1. Curved surface area = 44cm2 and height =
5. A rectangular sheet of metal has length 44cm 3.5cm
and breadth 10cm. If it is folded to form a 2. Curved surface area = 220m2, length = 7m
cylinder with the breadth becoming the height, 3. A sheet of paper, 20cm by 15cm, is rolled into
calculate: a cylinder, so that its longer edges meet. Find the
i. the radius of the cylinder formed; diameter of the cylinder so formed.
ii. the volume of the cylinder.
Surface Area of a Sphere
Solution There formula for calculating the total surface
Length of metal sheet = 44cm, area of a sphere is; 2
where;
Breadth = 10cm and π = A represents the total surface area,
π is a constant
10cm r is the radius of the sphere.
44cm
C = 44cm = 2πr 2
r
10 cm
1
i. 2πr = 44cm 6
π
=
π π
= = 7cm Worked Examples
1. Calculate the total surface area of a sphere of
radius 14cm.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 739
Solution 2. The general formula for the total surface area
π 2, of a right prism is A = ph + 2B, where p
Substitute r = 14cm andπ represents the perimeter of the base, h represents
2
the height of the prism and B the area of the base.

3. There is no easy way to calculate the surface

2. The diameter of a sphere is 21cm.Calculate its area of an oblique prism in general. The best way
total surface area. is to find the areas of the bases and the lateral
faces separately and add them.
Solution
A π 2. Worked Examples
But cm and π = 1. Find the lateral surface area of triangular
2 prism with base edges 3cm, 4cm and 5 cm and
altutide 8cm.

Exercises 27.6 Solution

Calculate the total surface area of spheres with p = 3 + 4 + 5 = 12cm
the following radii and diameters;
1. r = 7cm 2. r = 5cm 3 . r = 9cm Lateral surface area;
4. d = 35cm 5. d = 21cm 6. d = 28cm A = ph
A = 12(8) = 96cm2
Surface Area of a Prism
The lateral surface area of a prism is the sum of 2. Find the total surface area of a trapezoidal
the areas of its lateral faces. prism with parallel edges of base 6cm, and 12cm,
the legs of the base 5cm each, the altitude of the
The total surface area of a prism is the sum of the base 4cm and a heihght of the prism 10cm.
areas of its lateral faces and its two bases.
Solution
Perimeter of the base 6cm
is the sum of the lengths 4cm
of the sides of the base. 5cm
P = 6 + 5 + 12 + 5 = 28cm 10cm
12cm

Usually, if right or oblique is not mentioned, it is The base is an isosceles trapezoid. The area of
assumed that it is a right prism. base;
B = (a + b)h
1. The general formula for the lateral surface area
B = (6 + 12) (4) = 36cm2
is A = ph, where p represents the perimeter of the
base and h represents the height of the prism.
Total surface area:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 740
A = ph + 2B Practical Problems
A =28(10) + 2(36) = 280 + 72 = 352cm2 For all practical problems,
I. Separate the complex shape into individual
3. Calculate the surface area of the triangular shapes.
prism below: II. Find the appropriate areas.
III. Find the total surface area.
IV. Given the cost per m2, multiply the cost
6cm 7cm 7cm by the total surface area.
12cm
4cm Worked Examples
Solution 1. Calculate the cost of painting the four walls
Perimeter of the base; and ceiling of a room 30m long, 22m wide
p = 7 + 7 + 4 = 18cm and10m high at Gh¢15.00 per 9m2.

Area of the base; Solution

2 2
B = (4)(6) = 12cm Area of 2 long walls (
Area of 2 short walls 2(22 10 440m2
Total surface area: Area of ceiling 30 22 660m2
A= ph + 2B
A =18(12) + 2(12) = 216 + 24 = 240cm2 Total surface area;
600 440 660
Exercises = 1,700 m2
1. A triangular prism has a triangular end with a 1700
base of 5cm and a height of 4cm. The length of Number of square meter 9
each side is 8cm and the width of each side is
6cm. What is the surface area of the prism? A = Cost of painting of Gh¢15.00 per m2
1700
156cm2 = 9 15
= Gh¢28 33
2. A triangular prism has a triangular end with a
base of 8cm and a height of 6cm. The length of 2.
each side is 10cm and the width of each side is
6cm. What is the surface area of the prism? A = 7cm
2
428cm D
25cm Semi
Semi
3. A triangular prism has a triangular end with a
spherical spherical
base of 12cm and a heightof 9cm. The length of
2
each side is 14cm and the width of each side is Area of two semi- spherical ends π
2
9cm. What is the surface area of the prism? A = A
486cm2

Baffour – Ba Series, Core Maths for Schools and Colleges Page 741
Lateral surface area of a cylinder π , the diagram belowwith floor tile each at a cost of
= 550cm2 Gh¢2.00 per square meter.

Total surface area,

2
3m 2m
3.5m
3. Find the cost of covering with cocoa seedlings, 21m
a piece of land with dimensions as shown below 12m
15m 2x –
if the cost of a seedling is Gh¢4.00 per m2
10

25m
8m Ans R
24m

2. The dimensions of a field track is shown in the

18m
diagram below. Calculate the cost of covering
the track with tartan surface, if the cost of
Solution covering 1 square meter is Gh¢12.00.

8m 6m 7m

28m
18m

120m
Area of rectangle (A1) = L × B.
Where 18m and 8m 3. What would be the cost of carpeting the floor
2
A1 18 8 144m of a house with the dimensions shown below, if
the cost of is Gh¢14.00.
1
Area of triangle (A2) ;
2
35m
Where 8m and 6m
1
15m
A2 8m 6m = 24m2 30m2
2

Area of the total plot of the land; 40m

A1 + A2 = 144m + 24 m2 = 168 m2 30m

Cost of seedling; Meaning of Volume

= 168 × Ghȼ4 = Ghȼ672.00 When we want to know how much a container
will hold, we need a measurement of space.
Exercises 27.7 Space involves three dimensions – length,
1. What would be the cost of covering the breadth and height. This measurement of space is
surface area represented by theshaded portions of called volume.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 742
The measurement of volume is based on unit The area is found by finding out how many cubic
cubes. units are contained in the solid as shown below;

Study the figures below;

The volume is found by counting the entire unit

cubes that completely fill the solid.

Volume of a Cuboid /Rectangular Solid

The volume of a rectangular solid is calculated
by the formula; V = LWH, whereV is the volume,
L is the length, W is the width and H is the
height of the solid

Worked Examples
All the above are 3- dimensional solids because
1. Calculate the volume of the rectangular solid
they have length, width and Height.
with length 10cm, width 5cm and height 3cm
All solids contain volume measured in cubic
Solution
meters (m3) or cubic centimetres (cm3).
Consider the solid below: 3cm
5cm
V = LWH 10cm
But L = 10cm, W = 5 and H = 3
3

2. A rectangular box has length 20cm, width 6cm

The solid is 1unit long, 1unit wide and 1unit high. and height 4cm. Find the volume of the box.
It therefore has a volume of 1cubic unit.
Solution
6cm
In the diagram below: 4cm
20cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 743
 3
6. A rectangular box has a length 20cm width
6cm and height 4cm. Find how many cubes of
3. The volume of a rectangular pond is 420cm3. size 2cm that will fit into the box.
If the pond is 12cm long and 7cm wide, calculate
its height. Solution
V = LBH = 20 × 6 × 4 = 450cm3
420cm3
Solution
7cm Volume of 1 cube = 2 × 2 × 2 = 8cm3
12cm
Number of cubes = = 60 cm3
V = LWH
But V = 420 cm3, L= 12cm, W = 7cm and H = ?
420 cm3 = 12 cm × 7 cm × H
Exercises 27.8
420 cm3 = 84 cm2 × H A. 1. Calculate the volume of a rectangular solid
of which is 18cm long, 2cm wide and 3cm high.
H= = 5 cm
2. The length, width and height of a rectangular
tank is 16cm, 9cm and 11cm respectively.
4. A rectangular notebook has a volume of Calculate the volume of the tank.
96cm3.If it is 4cm wide and 3cm high, how long 3. A pit is 25m deep, 17m wide and 8m long.
is the book? What is the volume of the pit?
4. Find the volume of a rectangular box that is
Solution 10cm long, 8cm wide and 2cm high.
V = LWH 5. The dimensions of a water tank in the form
But V = 96 cm3, L = ? W = 4cm and H = 3cm of a cuboid are 60cm by 15cm by 20cm. Find
96 cm3 = L × 4cm × 3cm the capacity of the tank
420 cm3 = L × 12 cm2
L= B. 1. The volume of a rectangular tank is
3cm
L = 8 cm 96cm 3
4096cm3. If its width and height is 16cm and 8cm
4cm respectively, calculate its length.
L=?

5. The volume of water in cuboids is 9m3, the 2. A rectangular tower has a volume of 968cm3.
length of the cuboids is 3m and breadth is 2m. If it has a length of 16cm and width 11cm,
Calculate the depth of the water in the cuboids. calculate its height.

Solution 3. A door has a length of 30cm and height 90cm.

V = L B H, Calculate the width of the door if its volume is
But V = 9 m3,L = 3m , B = 2 m and H = ?. 8,100cm3.
9=3×2×H
9=6H 4. A storey building is 15m high and 5m wide. If
the storey- building occupies a space of 1,425m3,
H= = 1.5m
calculate its length.
.
Baffour – Ba Series, Core Maths for Schools and Colleges Page 744
Volume of a Cube Let x be the side of the cube.
A cube is a rectangular solid in which the length, 27 cm3 = x3
width and height are the same. =
Therefore the formula can be written 3 cm = x or x = 3cm
as where is the length of the
side of the cube. Area of one face = x × x = x 2 = 32
A = 3cm × 3cm = 9cm2
In the diagram below;
V=L×L×L 4. The volume of water in a rectangular tank is
V = L3 30cm3. The length of the tank is 5cm and its
breadth is 2cm. Calculate the depth of water in
the tank.

Worked Examples
Solution
1. Calculate the volume of a cube of side 6cm.
V=L×B×H
But V = 30cm3, L = 5cm, B = 2cm
Solution
30cm3 = 5cm × 2 cm × H
V=L×L×L
30cm3 = 10cm2 × H
V = L3
But L= 6cm H= = 3cm.
3

Exercises 27.9
2. Find the length of the side of a cube whose A. Calculate the volume of a cube with the
volume is 1728cm3. following side;
1) 11.5cm 2) 12cm 3) 20cm
Solution B. Find the length of the side of a cube with the
Method 1 following volumes:
V = L3, 1) 343cm3 2) 2,744cm3 3) 1,331cm3
But V = 1,728 cm3 = L3
12 × 12 × 12 = L × L × L C.1. The volume of a cube is 64cm3. Find the
length of one of its sides.
⇒ L = 12 cm
2. The volume of a cube is 27cm3. Find the area
of one face.
Method 2 3. Find the volume of a cube whose length is
3 3
14cm
3
√
Volume of a Cylinder
3
3. The volume of a cube is 27cm . Find the area The formula for finding the volume of a cylinder
of one of its faces. is 𝛑 2 , where;
V is the volume of the cylinder,
Solution 𝛑( is a constant with value or 3.142,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 745
R is the radius of the cylinder , 4. The volume of a cylinder is 220cm3 the radius
h is the height of the cylinder. of the cross-section is 2.5cm. Find the height of
the cylinder ( )

2
Solution
2
Volume of a cylinder,
But V = 220cm3, r = 2.5cm, ,h =?

Worked Examples 220 = × (2.5)2 × h

1. If a cylindrical tank is 21cm in diameter and 7 × 220 = 7 × × (2.5)2 × h
40cm in height, what is its volume? 1,540 = 137.5h
h = 11.2cm
Solution
2
5. A cylinder has diameter 4cm and height of
Where =
14cm. (π = ). Find:
i. the circumference of the base
= 1,3860cm3 ii. the area of the base
iii. the volume of the cylinder.
2. Calculate the volume of the figure below
( ) Solution
i. c = 2πr,
d = 4 and = r = = 2cm
C=2× × 2 = 12.57cm
Solution
2
ii. Area of the base, A = π r2
But A = × (2) 2 = 12.57cm2
2
( = 616 cm3
2
iii.
3. A timber log of height 35m has a base radius of V= × (2)2 × 14 = 176cm3
4m. Calculate its volume.
6. A rectangular sheet of metal has length 44cm
Solution
and breadth of 10cm. If it is folded to form a
cylinder with the breadth becoming the height,
calculate:
V = r2h i. the radius of the cylinder formed,
But r = 4cm, h = 35cm and = ii. the volume of the cylinder ( ).
2 3
( ) = 1,760cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 746
Solution 8. A water tank in the form of cuboids of height
i. Length of metal sheet = 44cm 22cm and a rectangular base of length 7cm and
Breadth = 10cm, width 5cm is filled with water. The
water is then poured into a cylindrical container
44cm
of diameter 14cm. Calculate the height of the
10cm water in the cylindrical container

Solution
C = 2πr = 44cm
Volume of water in the cuboids = L B H
2 π r = 44cm
V = 7 × 5 × 22 = 770cm3
= ,
Volume of water in the container = Volume of
r= = 7cm water in the cylindrical container
⇒770 cm3 = πr2h
2
ii. But r = = = 7cm
2 3
V= × (7) × 10 = 1,540cm 770 = × 7 × 7× h
7 × 770 = 154 h
7. A rectangular tank of length 22cm, width 9cm h= = 5cm.
filled with water. The water is poured into a
cylindrical container of radius 6cm, calculate the:
Exercises 27.10
i. volume of the rectangular tank
ii. the depth of water in the cylindrical container For each question, take
1. Find the volume of a cylindrical tank of radius
Solution is 42cm and height 45cm.
1. Volume of water in the rectangular tank,
2. The volume of water in a cylindrical container
3
V = L B H = 22cm × 9cm × 16cm = 3,168cm is 308cm3. If the height of the wateris 8cm. Find
the radius of the base of the cylinder.
ii. Volume of cylindrical container = π r2 h. 3. The volume of a cylindrical is 1,100cm3. If the
But volume of water in the rectangle tank diameter of its base is 10cm, find its height.
= volume of water in the cylindrical container
⇒3,168cm3 = π r2 h, Volume of a Pyramid
But r = 6. The formula for the volume of a pyramid is
3168 = × (6)2 × h where;
7 × 3168 = ×6×6×h V is the volume of the pyramid,
b is the area of the base of the pyramid,
h= = 28cm
h is the height of the pyramid.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 747
 3. The diagram below shows a right pyramid with
a rectangular base WXYZ and vertex O. If /WX/
= 8cm, /ZW/= 6 cm and /OX/ =13 cm, Calculate:
a. the height of the pyramid;
h O
b. the value of <OXZ,
correct to the nearest degree;
c. volume of the pyramid.

The height of the pyramid is the distance from the Solution Z

Y
centre of the base to the apex. a. By Pythagoras theorem; M
/xz/2 = 62 + 82 6cm
Worked Examples /xz/2 = 100 W 8cm X
1. Calculate the volume of a pyramid with a /xz/ = √ = 10cm
square base of sides 3m and height 4m.
But /MX/ = /XZ/
Solution = (10) = 5cm O

To calculate the volume, By Pythagoras theorem; 13cm

first find b 4m /OM/2 + /MX/2 = 132
Z
(the area of the base) /OM/2 = 132 – 52 Y
6cm M
Area of base ( 2
. /OM/2 = 144
But 3m /OM/ = √ W 8cm X
2 OM = 12cm
Area of base (b) = O

Volume of pyramid, V bh b.To find the value

2 3 of < OXZ, consider the 12
Z 13
diagram below:
2. The height of a pyramid is 15cm. The side of cos θ = M
θ
the square base is 5cm. Find the volume. 5
θ= ( )
X
0
θ = 67
Solution
Volume of pyramid, V bh
V = bh
But area of square base, b,
15cm But area of base, b = L B
b = 5cm × 5 cm = 25 cm2
A = 6 cm × 8cm = 48cm2
2
V
V = 125 cm3 5cm V bh × 48 × 12 = 192cm3

Baffour – Ba Series, Core Maths for Schools and Colleges Page 748
Exercises 27.11 V= × × 62 × 14
1. The length of the side of a square base of a V= 3

pyramid is 5cm. If the height of the pyramid is

15cm. Calculate its volume. 2. Calculate the volume of a cone which has a
base radius of 7cm and a height of 12cm.
2. A pyramid has a square base of side 8cm long.
Its height is 10cm. Find its volume.
Solution
3. A pyramid has a rectangular base 15cm by V = πr2h
20cm and its height is 9cm. Find: But r = 7 cm, h = 12 cm
a. the area of the base
b. the volume of the pyramid V= × × 72 × 12 = 3

4. Calculate the volume of the figures below: 3. If the volume of a cone is 1,056cm3 and its
a. b. height is 63cm, find its radius?( )
6 20
Solution
3 5 2
,
10 But h = 63cm, r = ?, V = 1,056cm3 and =
1,056cm3 2
The Volume of a Cone
1,056cm3 2
The formula for the volume of a cone is :
2
, where; r2 =
V is the volume of the cone r2
r is the radius of the cone r √
𝛑 (pi) is a constant with value
4. The volume of a cone of height 7cm is 66cm3.
h is the height of the cone.
Find the radius of the cone.
2
Solution
2
Worked Examples ,
3
1. Find the volume of a cone which has a base
radius of 6cm and a height of14cm.( ) 66cm3 2

198cm3 = 22 × 2

Solution
2
r2 = =9
V = πr h
√
But r = 6 cm, h = 14 cm
r = 3 cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 749
5. The volume of a cone with base 5cm is Recasting a Solid
1,650cm3. Calculate the height of the cone. Recasting is the act of giving a solid or metal
object a different form by melting it down and
( )
reshaping it. The amount of substance contained
in the new solid is equal to the amount of
Solution substance in the old solid formed from it. Thus,
2 though the shape of the object is changed, the
,
volume remains the same.
3
and
Worked Examples
1,650cm3 ( 2
1. A solid metalcube of side 6cm is recast into
1,650cm3 asolid sphere. Find the radius of the sphere.
1,650cm3 2
×h
Solution
Volume of cube = Volume of Sphere
L3 = πr3
6. A solid cone has base radius of 7cm. If the
volume of the cone is 1,078cm3, find the vertical 63 = πr3
height of the cone (π = ) 216 = πr3
3 × 216 = πr3 × 3
Solution
2 648 = 4πr3
,
3 =
2 162 = π r3
× × h,
r3 =
h
r3 =
3 × 1,078 = 154h
h= = 21cm r=√ = 3.72 cm

Exercises 27.12 2. A cone is 8cm high and radius 2 cm. The cone
1. Calculate the height of a cone with volume is melt and recast into a sphere. Find the
1,848cm3 and radius 7cm. diameter of the sphere.
2. What is the height of a cone whose volume is
88cm3 and radius 22cm? Solution
3. A cone has a radius of 14cm and height 28cm. Volume of cone = Volume of sphere
Find its volume. πr2h = πr3
4. A cone has a height of 5.6cm and a radius of
5cm. Find the volume of the cone correct to π (2)2(8) = πr3
1decimal place (Ans 146.7cm3) × 32π = πr3
32 = 4r3

Baffour – Ba Series, Core Maths for Schools and Colleges Page 750
r3 = Number of smaller spheres made from larger
sphere
r= √ = 2 cm
= = = 27 spheres
⁄
But diameter of the sphere = 2r = 2 × 2 = 4cm
4. A solid rectangular block with dimension
3. A solid metal cylinder of radius 6cm and 15cm, 12cm and 8cm respectively is melted
height 18 cm is melted down and recast as a cone down completely and recast into a solid right
of radius 6cm. What is the height of the cone? circular cone of base radius 12cm. Calculate:
i. correct to two significant figures, the height of
Solution the cone (π = 3.142) .
Volume of cylinder = π r2h ii. Use your results in (i) to calculate the slant
Volume of cone = height and hence the curved surface area of the
Volume of cylinder = Volume of cone cone, correct to two significant figures.
But radius of cylinder = 6 cm,
height of cylinder = 18 cm, Solution
radius of cone = 6 cm,
12cm
height of cone = h
V = LBW 8 cm
π r2 h = 15cm
V= 15cm × 12cm × 8cm = 1,440cm3
π (6)2h =
Volume of rectangular solid = Volume of cone
= 18
2
⇒LBW h
h = 3 × 18cm = 54cm
2
1,440 = ( ( h
4. A solid metal sphere 12 cm in diameter is
h=( (
= 9.5 (2s.f.)
melted and recast into small solid metal spheres
4cm in diameter. How many of these smaller
solid metal spheres will there be? ii. Let the slant height of the cone be h
By Pythagoras theorem,
Solution h2 = 9.5 2 + 122
Volume of sphere = πr3 h2 = 234.25
Large spheres h=√
h = 15cm (2 s.f) 12
r= = = 6cm
V = × π(6)3 = 288π The curved surface area of the cone,
A = rl
Small spheres A = (3.142) × 12cm × 15.4 cm = 580cm2
r= = = 2cm
5. From a sphere of radius 10cm, a right circular
V = × π(2)3 = π cylinder whose base diameter is 12 cm is carved

Baffour – Ba Series, Core Maths for Schools and Colleges Page 751
out. Calculate the volume of the right circular 6. A metallic box of length 16cm, breadth 8cm
cylinder correct to two decimal places. and width 4 cm is recast into a solid cylinder of
height 8cm. Calculate the surface area of the
Solution cylinder, to two decimal places.
From the figure,
10 cm C
OB = OC = 10 cm O Challenge Problem
(radius of sphere) 6 cm The surface area of a solid metallic sphere is 1,256
A B
In ∆ABC, cm2. It is melted and recast into a solid right
OB2 = OA2 + AB2 circular cone of radius 2.5 cm and height 8cm.
OA = √ =√ =√ = 8 cm Calculate;
i. the radius of the solid sphere
Height of cylinder = 2 × OA = 2 × 8cm = 16 cm ii. the number of cones recast(π = 3.14)

Volume of right circular cylinder = πr2h 2. A 4cm × 4cm × 2 cm slab of cholcholate is

= × 62 × 16cm melted then made into twenty thin cylindrical
sticks of diameter 4mm. If 10% of the chocolate is
= × 6cm × 6 cm × 16cm wasted during the process, what is the length of
= 1,810.29cm3 one chocolate stick?

Exercises 27.12B Frustum of a Right Circular Cone

1. A solid metal ball of radius 2cm is melted and A frustum of a cone or truncated coneis the
the metal obtained recast to form a right circular results of cutting a cone by a plane parallel to the
cone of radius 5 cm. Find the height of the cone. base and removing the part parallel to the apex as
shown in the figure below:
2. Twenty – seven ball bearings of diameter 1 cm r = radius of upper base
are melted down and recast to form a single R = radius of lower base
sphere. Find its diameter. s = slant height
3. A circular cone of height 12 cm and radius 9 cm 1. s = √(
is recast into a solid sphere. Calculate the surface
area of the sphere. 2. Volume = πh(

4. A cone is 8.4cm high and the radius of its base 3. Lateral area/ curved surface area;
is 2.1cm. It is melted and recast into a sphere. = π (R + r)s r
Find the radius of the cone.
= π(R + r) √(
h s
5. A metallic sphere of radius 10.5cm is melted
4. Total surface area, A R
and recast into small cones, each of radius 3.5cm
A=π[ ( ]
and height 3 cm. Find the number of cones thus
obtained.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 752
Worked Examples V = πh(
1. Calculate the lateral area, surface area and
V= × × 12( (
volume of a truncated cone of radii 2 cm and
6 cm and a height of 10cm. V= × ×1 (
V= × × 1,116 = 1,169 cm3
Solution
R = 6 cm, r = 2 cmand h = 10 cm and
Method 2
Lateral area = π (R + r) 2 Complete the cone as shown
But = √( From the completed cone,
s
= √( 10 similar figures are formed
S = 10.77 cm 6 ⇒ = x
4(x + 12) = 7x
Substitute s = 10.77, 4
4x + 48 = 7x
R = 6 cm, r = 2 cm , in 48 = 7x – 4x 12
Lateral area = π (R + r) 48 = 3x 7
⇒ (2 + 6) 10.77 = 270.79 cm2 x = 16 cm

Total surface area, A Volume of frustum;

=π[ ( ] = Vol of the complete cone – vol of the smaller cone
= [ ( ] = × × 7 2× ( – × × 4 2× 16
= [ ]= 396.50 cm2 = 1, 437.33 – 268.19 = 1,169 cm3

3. The figure below shows a cone whose upper

V = πh(
part has been cut off. The base radius is 6 cm and
V= × × 10( ( the upper radius is 3 cm.
V= × × 10(
3 cm
3
V= × × 520 = 544.76 cm 4 cm

6 cm
2. In the figure below, find the volume of the
frustum, if the radii of the bases are 4 cm and 7
cm and the height is 12 cm( ) If the height of the remaining portion is 4 cm,
4 calculate correct to the nearest whole number, the
volume of the original cone.
12
7 Solution
Solution
From the complete cone: similar right triangles
Method 1 are formed
R = 7, r = 4, h = 12 and substitute in

Baffour – Ba Series, Core Maths for Schools and Colleges Page 753
⇒ = Solution
3
6x = 4 × 3
6x =12 x
But But r = 3cm and =
x=2 3 cm h
3
= 113.14cm3
4 cm
But h = 4 + x
6 cm
⇒h = 4 + 2 = 6 cm 2. Calculate volume of a gold ring of radius
21cm.
V = πr2h = × × 62 × 6 = 226 cm3
Solution
3 21cm
Exercises 27.13 m
1. Find the volume of the frustum of a cone with But r = 21cm and =
bases of radii 5 cm and 9 cm, given the height of
3
the frustum as 6cm. ( = 38,808cm3

2. Calculate the lateral area, surface area and 3. A circle has a radius of 14cm. Find
volume of a truncated cone of radii 10 cm and 12 i. the surface area
cm and a slant height of 15 cm. ii. the volume of the circle.

3. A frustum of height 5 cm is formed from a Solution

2
right circular cone whose height is 10 cm and i. Surface area, A
whose base radius is 4 cm. Calculate the volume 2 2
(
of the frustum.
3
ii.
4. In the figure below, calculate:
8
i. the lateral area of the figure, (14 cm)3
15
ii. the surface area of the figure,
12 V = 11,498.67cm3
iii. the volume of the figure.

Volume of a Sphere Exercises 27.14

The formula for finding the volume of a sphere A. Calculate volume of a sphere with the
3 following radii;
is: where;
v is the volume, 𝛑is a constant and ris the radius 1) 4cm 2) 5cm 3) 6cm
4) 7cm 5) 8cm 6) 10cm
Worked Examples
1. The radius of a sphere is 3cm. Calculate its B. Calculate the radius of a sphere with the
following volumes.
volume.
3cm 1) 36cm3 2) 105cm3 3) 400cm3
4) 88cm3 5) 2150cm3 6) 616cm3

Baffour – Ba Series, Core Maths for Schools and Colleges Page 754
The Hemisphere ii. total surface area of the reservoir ( π = ).
A hemisphere is half of a full sphere.
1. Volume = π Solution
r b. Volume of the reservoir;
2. Curved Surface Area; A = 2 π = Vol of the cone + Vol of the Hemisphere
= π h+ π
3. Total surface area, A = 3π
= π 6x + π

Worked Examples
A hemisphere has a radius of 3 cm. Calculate its: But Volume of the reservoir = 333 π
i. volume, ⇒ π 6x + π = 333 π
ii. curved surface area,
π + π = π
iii. total surface area ( π = ).
π + 2 π = 1000π
+ 2 = 1000
Solution
⇒ 8 = 1000 6x
r = 3cm and π =
= x
i. V = π
= 125
3
V= × × = 56.57 cm x=√
x=5
ii. Curved Surface Area, Radius of the cone/hemisphere, r = 5m
A=2π
A=2× × 32 = 56.57 cm2 Height of the cone, h = 6x = 6 × 5 = 30 m
i. Volume of the hemisphere , V = π
iii. Total surface area, V= ×
A = 3π V = 262 m3 (nearest whole number)
A= 3× × = 84.86 cm2
ii. Total surface area of the reservoir;
2. A water reservoir in the form of a cone = Curved surface area of the cone + curved
mounted on a hemisphere is built such that the surface area of the hemisphere
plane face of the hemisphere fits exactly to the = πrl + 2π
base of the cone and the heiht of the cone is 6
times the radius of the its base. But slant height of the cone ,
a. Illustrate this information in a diagram. l=? l
30 m
b. If the volume of the reservoir is 333 π m , 3 By Pythagoras theorem,
l2 = 302 + 52
calculate, correct to the nearest whole number,
l2 = 925 5m
the
l = 30.4138 m
i. volume of the hemisphere,

Baffour – Ba Series, Core Maths for Schools and Colleges Page 755
Substitute r = 5, π = and l = 30.4138 in cylinder is not a prism because it has curved
sides. The sides of a prism are parallelograms.
A = πrl + 2π (Total surface area of reservoir)
A= × 5 × 30.4138 + 2 × ×
Cross
A = 477.9311 + 157.1429 Prism Shape section
A = 635 m2 (nearest whole number)

Exercises 27.15 Triangular

1. A toy in the form of a cone mounted on a
hemisphere of same radius 7cm. If the total
height of the toy is 31 cm, find the total surface Square
area.

2. A solid wooden toy in the shape of a rght

circular cone mounted on a hemisphere. If the Cubical
radius of the hemisphere is 4.2 cm and the total
height of the toy is 10.2 cm, find the volume of
the wooden toy?
The volume of a prism is the product of the area
of one end and the length of the prism. This is
3. A vessel is in the form of a hollow hemisphere
expressed as :
mounted by a hollow cylinder. The diameter of
V= Area of cross section × length
the hemisphere is 14 cm and the total height of
the vessel is 13 cm. Find the inner surface area of
Worked Examples
the vessel.
A prism has a uniform cross section in the shape
of a right – angle triangle whose two shorter sides
4. A medicine capsule in the shape of a cylinder
are of lengths 2cm and 1.5cm. The length of the
with two hemispheres stuck to each of its ends.
prism is 6cm. Write down:
The length of the entire capsule is 14 cm and the
i. the area of the cross section;
diameter of the capsule is 5 cm. Find its surface
ii. the volume of the prism.
area.
Solution
Volume of a Prism i. Area of the cross section;
A prism is a solid object with identical ends, flat A= b h = × 1.5 × 2 = 1.5cm2
sides and the same cross section all along its 6cm
length. 2cm

A cross section is the shape made by cutting Alternatively 1.5cm

straight across an object. A prism is said to be a A = b h sin 900

polyhedron, meaning all sides are flat. Hence, a
A = (1.5) (2) sin 900 = 1.5cm2

Baffour – Ba Series, Core Maths for Schools and Colleges Page 756
ii. The volume of the prism, V= Al Solution 4cm
But V = 1.5and l = 6
V= 1.5cm2 × 6cm= 9cm3
7cm
2cm 2cm
Exercises 27.16 m
1. The figure below shows a prism of length 30 5cm

cm whose cross section is an equilateral triangle

4cm
of side 4cm.
Total volume of solid;
Volume of cylinder volume of cuboid

Calculate:
2
i. the area of the triangular end; = 352cm3
ii. the total surface area of the prism;
iii.the volume of the prism.
Volume of cuboids
cm,
2. What is the volume of a prism whose ends 3
have an area of 25m2 and which is 12m long
3 3 3
=
3. In the triangular prism below,
8cm
2. In the figure below, calculate the total volume
6cm

4cm
32cm
a. What is the area of the cross section of the 10.5cm
35cm
prism?
Solution
b. What is the volume of the prism?
Total volume;
4. A rectangular prism is 13 cm long. If the
= Volume of cylinder +Volume of cone
volume of the prism is 936cm3, find the area of
one end of the prism 2

Volume of Combined Solids But

Two or more solids may be combined to form a
complex solid. In such solids, the volume is Volume of cylinder;
calculated by finding the sum of the volumes of ( cm)2 3

the separate solids.

2
Volume of cone, V ,
Worked Examples
1. Calculate the volume of the solid below. But h = 10.5cm

Baffour – Ba Series, Core Maths for Schools and Colleges Page 757
V = 2,816cm3 = 24 + 96 + 144 + 120 = 384 cm2

Exercises 27.17
Total Volume;
3 3 Calculate the total volume of the solids below
= 30,976 cm3
𝛑
3. The diagram shows a hut used in storing grains 1. Diagrams are not drawn to
12cm
which is in the shape of a triangular prism scale

mounted on a cuboid. If the dimensions are as

32cm
shown in the diagram,

R Q 36cm

3m 5m 2.
S 3cm
T P
6m
M
8m
N 12m O 9cm
6cm
Calculate;
12cm
a. the volume of the hut.
b. total external surface area of the hut. 3. A model house is made by sticking a
triangular prism on top of a rectangular block as
Solution shown below:
Volume of Prism, R Q
V = Area of cross – section × length
5m 9m
S
Area of cross – section, A, T P
= Area of triangle + area of rectangle 7m
M
10m
A = bh + L × B N 14m O
2 Calculate the volume of the model house.
A = × 8 × 3 + 8 × 6 = 12 + 48 = 60m

4. A cone is contained in a cylinder so that their

But length, L = 12m
bases and heights are the same as shown below:
V=A×L
V = 60m2 × 12m = 720m3

ii. Total surface area = area of the net

10 km
2(SRT) + 2(NMST) + 2(NOPT) + 2(QRTP)
2( BH) + 2(LB) + 2(LB) + 2(LB) 3 cm
= 2( × 3 × 8) + 2(8× 6) + 2(12 × 6) + 2(5 × 12)

Baffour – Ba Series, Core Maths for Schools and Colleges Page 758
Calculate the volume of the space lying in – circumscribing about a prism with a square base.
between the cylinder and the cone (π = )
1 cm
6. A tent in the shape of a cylinder surmounted by
a conical top. If the height and the diameter of the 3 cm
cylindrical part are 2.1 m and 4 m respectively,
and the slant height of the top is 2.8 m, find:
i.the area of the canvas used for making the tent.
ii. the cost of the canvas of the tent at a rate of
If the side of the square base of the prism is 1 cm
Gh¢500.00 per m2
and its height is 3 cm, find the volume and total
surface area of the cylinder
5. The figure below shows a cylinder

Baffour – Ba Series, Core Maths for Schools and Colleges Page 759
28 LOGIC AND REASONING Baffour– Ba Series

Statements Notation of Statements

A statement is a sentence which is either true or In mathematics, statements are denoted by letters
false, but not both. such as P, Q and R. For example, P: 5 + 6 = 13,
simply means P is the statement 5 + 6 = 13. P is
Consider the following sentences: obviously a false statement because 5 + 6 13.
1. Please sit down.
2. How old are you? Negation of a Statement
3. Mr. Osei is a hard working teacher. Consider the statement P: Linda is a girl. The
4. Kumasi is the capital town of Ghana statement “Linda is not a girl” is called the
5. A triangle has four sides. negation of P and it is denoted by P, where the
6. The baby is either a boy or a girl. symbol means “not” and P is read as “not P”.
Hence P:“Linda is not a girl”. Obviously, when
Sentences 1 and 2 are not statements because they P is true, the negation of P is false and vice –
are neither true nor false. versa.
Sentences 3, 4 and 5 are statements because they
can either be true or false, depending on the Worked Examples
context. The sixth sentence, is obviously a true Write down the negation of the statements
statement. 1. P : 5 ∈ {primes} 2. 7 ∉ {even integers}.
3. P : x < 2 4. P : 3 < x < 4
Exercises 27.1
A. Which of the sentences are statements? Solution
1. Ghana is in Africa. 1. P : 5 ∉ {primes} 2. P : 7∈ {even integers}
2. All primes are odd numbers.
3. P : x 2, or P: x 2 4. P : x 3, or x 4
3. Will you go to school tomorrow?
4. 5 = 3 mod 3.
Exercises 27.2
5. 3 is an even number.
Write down the negation of each of the
6. 10 is a multiple of 5.
following statements and state whether the
negation is true or false;
B. Which of the following sentences are
1. P : 3∈ {primes} 2. P : 4 ∈ {factors of 10}
statements? If they are statements, state
3. P : 5 < 3 4. P : x > 3
whether they are true or false
5. P : 4 = 2 6. P : - 4 x 5
1. Do your mathematics exercises.
2. There are 52 weeks in a year.
Statements and Venn Diagrams
3. Drive with care.
Similar to sets, statements can be represented in
4. sin 300 = sin 600.
Venn diagrams. Identify the following from the
5. All prime numbers can only be divided by 1
given statements:
and itself.
1. Universal set.
6. 22 is a composite number.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 760
2. Subsets. 2. P : All my friends are intelligent.
3. Complement. Q : Intelligent people are quite people.
R : Dufie is a quite person but not intelligent.
Worked Examples
Represent each of the following statements on a 3. P : A vegetarian does not eat meat.
Venn diagram. Q : All my friends are vegetarians.
i. P : Kofi is a good boy. R : Jonathan is not a vegetarian but does not eat
ii. Q : 4 is a factor of 100 but 3 is not. meat.
S : Rosemond eats meat.
Solution T : Brown is a vegetarian but he is not my friend.
i. U = {boys} A = {Good boys} k = Kofi
k∈ {Good boys} 4. P : Rich men live in big houses.
U
Q : Mr. Owusu is a rich man.
A R : Mr. Owusu lives in a big house.
kk
5. P : All boys play football.
Q : Danny plays football.
R : Danny is a boy.
ii. U = {Integers} A = {factors of 100}
4 ∈ {factors of 100} A1 = {3} Compound Statements
U
A compound statement is a combination of two or
A more statements in a single statement. They are

4 usually formed by the conjunction “and” and
“or”. For example, P : x > – 4, Q : x < 3. The
3
compound statement P and Q is the statement “x
Exercises 27.3 > – 4 ” and “x < 3”written as P and Q : - 4 < x < 3
Represent each of the following statements on
a Venn diagram. In general, P and Q is denoted by P˄Q, where “
1. P : A square is a quadrilateral. ˄” means “and”
2. Q : Rich people are hardworking people.
3.R : Every science student studies mathematics If P is a statement about a subset A and Q is a
but not history. statement about subset B, then P˄Q,is a
4. S : All my friends are intelligent people. statement about the subset A B represented in
5. T : Every multiple of 10 is also a multiple of 5. the diagram below:
B U
A
B. Illustrate on a Venn diagram.
1. P : All football players play volley ball.
Q : A volley ball player does not play hockey
Blay plays volleyball but not football.
A∩B

Baffour – Ba Series, Core Maths for Schools and Colleges Page 761
The compound statement “P and Q” is true only 3. P : 2 ∈ {odd numbers}, Q : 2 ∈ {factors of 15}
if both P and Q are true and false only if either P
or Q is false. Solution
A. 1. P˄Q : A square has four sides, and a
Similarly, the compound statement P or Q is the triangle has three sides. P˄Q is true, because P
statement “x > - 4 ” or “x < 3” written as: and Q are both true.
P and Q : - 4 < x < 3
2. P˄Q : 6 + 7 = 13 and 5 + 4 > 8. P˄Q is true,
In general, P or Q is denoted by P˅Q, where “ ˅ because both P and Q are true.
” means “or”
If P is a statement about a subset A and Q is a 3. P˄Q: 4 ∈ {primes} {even numbers }
statement about subset B, then P˅Q, is a P ˄Q is false since P is false.
statement about the set A ∪ B.
B. 1. P˅Q: 4 ∈ {primes}∪{even numbers}
A B U ⇒ 4 is either a prime or an even number
P ˅Q is true, since P is true

2. P˅Q : 3 is a factors of 27 or 5 is a multiples of

10.
AUB
P ˅Q is true since Q is true
The compound statements “P or Q” is true only if
either P or Q is true and false if P is false and Q 3. P˅Q: 2 ∈ {odd numbers} ∪ {factors of 15}
is also false. ⇒ 2 is either an odd number or 2 is a factor of 15
P ˅ Q is false since P and Q are false.
Worked Examples
A. From the following pair of statements, form Exercises 27.4
a compound statement using “and” and A. From the following pair of statements, form
determine whether the compound statement is a compound statement using “and” and
true or false. determine whether the compound statement is
1. P : A square has four sides, Q : A triangle has true or false.
three sides 1. P: 2 ∈ {primes}, Q : 2 ∈ { even number}
2. P : 6 + 7 = 13 , Q : 5 + 4 > 8
2. P: 3∈ {factors of 6}, Q :3∈ { factors of 9}
3. P : 4 ∈ {primes}, Q : 4∈{even numbers }
3. P : 2 + 5, Q : –12 + 16 > 3
B. From the following pair of statements, form a 4. P : 6∉{primes}, Q : 6∈ { composite}
compound statement using “or” and determine
5. P : 10∈ {even}, Q : 5∈ { primes}
whether the compound statement is true or false.
1. P : 4 ∈ {primes}, Q : 4 ∈ {even numbers}
B. From the following pair of statements, form
2. P : 3 ∈ {factors of 27}, Q : 5 ∈ {multiples of
a compound statement using “or” and
10}

Baffour – Ba Series, Core Maths for Schools and Colleges Page 762
determine whether the compound statement is Worked Examples
true or false Which implications are true or false;
1. P : ∈ {integers}, Q : 2 ∈ { prime numbers} 1. x2 > 4, then x > 2
2. If a number is a perfect square, then it is
2. P : 5 ∈ {factors of 14}, Q : 5 ∈ {multiples of
positive
10}.
3. P : 3 , Q: 2 3
Solution
4. P : 15 ∈ {primes}, Q : 5∈ { prime} 1. False, (-3)2 > 4, (But -3 < 2)
5. P : 7 ∈ {perfect squares}, Q : 8∈ { perfect 2. True
cubes}
Converse of an Implication
Implication For all implications, the arrow “⇒” always points
Consider the following statements; at the “then clause” as shown in the following
1. If it is a dog, then it is an animal with four legs. implications:
2. If two triangles are congruent then the two 1. It is a dog⇒It is an animal with 4 legs.
triangles are similar. 2. Two triangles are congruent ⇒ Two triangles
3. If a = b, then a2 = b2. are similar.
4. If he has passed the examination, then he is 3. a = b, ⇒a2 = b2
promoted.
If the arrow is reversed like this “⇐”, the
Each of these sentences consist of two clauses – implications appear as follows:
the “if clause” and the “then clause”. Each of the a. It is a dog ⇐ It is an animal with four legs.
two clauses is a statement in itself. b. Two triangles are congruent ⇐ Two triangles
Let us, consider the “if clause” as statement P and are similar.
the “then clause” as statement Q. It is observed c. a = b⇐a2 = b2
that in each case, if statement P is true, then it Implications a, b and c are the respective
implies that statement Q is also true. This is converses of the implications 1, 2 and 3
expressed as P implies Q, symbolically P ⇒
Sentences of the type P ⇒ Q are referred to as For converse (a);
implication. a. It is a dog ⇐ Ithas four legs
This means that “if it has four legs, then it is a
Statements 1, 2 and 3 may appear as follows: dog”
1. It is a dog ⇒ It is an animal with 4 legs
2. Two triangles are congruent ⇒ Two triangles The implication is false, because not only dogs
are similar have four legs. Since this implication is not true,
3. a = b, ⇒a2 = b2 it is written as:
It is a dog ⇐ It has four legs
Note that the arrow “⇒” always points at
the“then clause” Equivalent Statements
Consider the statement below:

Baffour – Ba Series, Core Maths for Schools and Colleges Page 763
„If he passed the examination, then he is Since P ⇒ Q and Q ⇒P are true, P ⇔Q is also
promoted‟ true and hence P and Q are equivalent.
1. The implication is:
„He has passed the examination⇒ He is 2. Two statements P and Q are defined by
promoted‟. This is a true statement. P : x ∈{even numbers}
Q : 4x ∈ {even numbers }
2. The converse is ; Determine whether P and Q are equivalent
He has passed the examination⇐ He is promoted.
This is also a true statement. Solution
This implies that the implication as well as its P ⇒Q : If x is even , then 4x is even
converse is true. The implication and its converse Q ⇒P : If 4x is even , then x is even
can be simplified by the use of the arrow “⇔”. P ⇒Q is true, but Q ⇒ P is false since 4 × 1
Thus, the implication appears as: ∈{even numbers} but 1 ∉ {even numbers}
„He has passed the examination⇔ He is Hence P ⇔ Q is false. Therefore P and Q are not
promoted‟ equivalent

Generally, when an implication P ⇒ Q and its Exercises 27.5

converse Q ⇒ P are both true, the statements P A. For the following pair of statements P and
and Q are said to be equivalent, expressed as Q, determine whether P and Q are equivalent
P⇔Q 1. P : The sides of a triangle are equal
Q : The sum of two angles of a triangle is 1200
Note: 2. P : ∆ ABC is isosceles triangle
P ⇒ Q means, P implies Q. Q : ∆ ABC is an equilateral triangle
P ⇐ Q means, P is implied by Q. 3. P : A number is divisible by 10
P ⇔ Q means, P implies and is implied by B. Q : The factors of a number include 2 and 5
4. P : x2 – 3x + 2 = 0, Q : x = 1 or x = 2
Worked Examples 5. P : x + 2 > 3, Q : x + 1 > 3
1. Two statements P and Q are defined by;
P : an angle is 900 B. State whether the following implications are
Q : an angle is a right angle true or false
Write down the following implication in full and 1. x = 4 ⇒x2 = 16 2. x2 = 16 ⇒x = 4
determine whether P and Q are equivalent 3. x < 0 ⇒x2 > 0 4. x > 0 ⇒x2> 0
a. P ⇒Q b. Q ⇒ P c. P ⇔ Q 5. x2 > 0 ⇒x > 0
6. A polygon has n sides ⇔It has n angles
Solution 7. Each angle of a triangle is 600⇔It is an
a. P ⇒Q : If an angle is 900, then it is a right equilateral triangle
angle 8. is the mediator of ⇐ is perpendicular
b. Q ⇒P : If an angle is a right angle, then it is 900 to
c. P ⇔Q : If an angle is 900, then it is a right
angle and vice – versa.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 764
B. Fill in the blank space with ⇒, ⇐ or ⇔
C U
where appropriate, if none is applicable, use B
⇔ A

n(A) = n(B) A and B are

equivalent sets
(a + b)2 = 16 (ab)2 = 16 It is observed that A ⊂ B ⊂ C ⇒ A ⊂ C
x+y>3 x > 3 or y > 3 Thus, if a number is greater 7, then it is also
Area of a square The perimeter of
greater than 3.
is 25cm2 the square is 20cm
A= {x : 1< x < 6, A = {2, 3, 4, 5}
x is an integer} Generally, if P ⇒ Q and Q ⇒ R, the conclusively,
It is a It is a square P ⇒ R. This is called the chain rule of
quadrilateral implication, written as P ⇒ Q ⇒ R.
with 4 right
angles The chain rule can be extended to more than three
statements .
C. Fill in the blanks of the following:
1. If he is a taxi driver, then he drives well
Worked Examples
Paa is a taxi driver, therefore …..
The following implications are true in a certain
college;
2. If a triangle has two equal angles, then it is an
* If a student studies science subjects, he studies
isosceles triangle
mathematics
In ∆ABC, < A = < B. Therefore …
* If a student studies mathematics, he does not
study history.
3. If a quadrilateral is a parallelogram, then its
i. Represent the implications symbolically and
diagonal bisects each other. A rhombus is a
use the chain rule of implication to deduce a valid
parallelogram. Therefore ….
implication.
ii. Illustrate in a diagram.
The Chain Rule of Implication
Consider the statements that P, Q and R are
Solution
defined by :
i. P : a student studies science subject
P : x > 7, Q : x > 5, R : x > 3 and the implications
Q : a student studies mathematics
P ⇒Q : x > 7 ⇒x > 5 R : a student does not study history
Q ⇒R : x > 5 ⇒ x > 3 The implications are:
P ⇒ Q; Q⇒R
This can be represented in a Venn diagram as The implication P ⇒ R is valid, because P ⇒R : If
shown below:
a student studies science subjects, he does not
∪ = {numbers} , A = {numbers > 7} study history
B = {numbers > 5}, C = {numbers > 3}

Baffour – Ba Series, Core Maths for Schools and Colleges Page 765
ii. U = {students in the college} Truth Tables
A = {Science students} A truth table is a table which gives all the truth
B = {mathematics students} values of a compound statement. The table is
C = {non – history students} filled in by considering all possible combinations
of true and false for P and Q and then filling in
C U
B the results for the various connectors mentioned
A above.

The truth table for Negation Statement

Exercises 27.6 P P
1. Consider the following statements: T F
P : x ∈{natural numbers } F T
Q : x ∈{rational numbers } If P is true, then its negation P is false and vice
R : x ∈{ integers } versa.
i. Form valid implications from the statements P,
Q and R The “and” Truth Table
ii. Form a chain of implication. Below is the truth table for the conjunction of two
simple statements p and q.
2. The following statements are true of
inhabitants of Toase; P Q P˄Q
T T T
S1 : only persons over 21 years pay basic rate.
T F F
S2 only persons who pay basic rate can register as F T F
voters. F F F
S3 : only persons who register as voters can vote
in an election. Worked Examples
i. Express the statements S1, S2 and S3 as Use truth table to show whether each of the
implications. following statement is true or false.
ii. Construct a chain of implications. i. 10 is divisible by 5 and 10 is a multiple of 2.
iii. Which of the following implications are valid ii. 2 is a factor of 15 and 10 is a multiple of 2.
deductions from S1, S2and S3? iii. 10 is a multiple of 2and 2 is a factor of 15
a. P1 : every voter in an election is over 21 years iv. 10 is not a multiple of 2 and 2 is a factor of 15
old.
b. P2 : every person who pays basic rate can vote Solution
in an election. i. 10 is divisible by 5 and 10 is a multiple of 2.
c. P3 : every person who registers as a voter pays P : 10 is divisible by 5 = T
basic rate. Q : 10 is a multiple of 2 = T
d. P4 : every person over 21 years of age can The statement is true.
register as a voter .

Baffour – Ba Series, Core Maths for Schools and Colleges Page 766
 P Q P˄Q The “or” Table
T T T Below is the truth table for the disjunction of two
simple statements p or q.
ii. 2 is a factor of 15 and 10 is a multiple of 2.
P : 2 is a factor of 15 = F P Q P˅Q
Q : 10 is a multiple of 2 = T T T T
The statement is false. T F T
F T T
P Q P˄Q F F F
F T F
Worked Examples
iii. 10 is a multiple of 2 and 2 is a factor of 15. Use truth table to show whether each of the
P : 10 is a multiple of 2 = T following statement is true or false.
Q : 2 is a factor of 15 = F 1. 2 is either an odd number or a factor of 15
The statement is false.
Solution
P Q P˄Q 2 is either an odd number or a factor of 15
T F F P : 2 is an odd number = F
Q : 2 is a factor of 15 = F
iv. 10 is not a multiple of 2 and 2 is a factor of 15.
P Q P˅Q
P : 10 is not a multiple of 2 = F F F F
Q : 2 is a factor of 15 = F
The statement is false. The statement is false

P Q P˄Q 2. A triangle has three sides, or a parallelogram

F F F has three sides.

Exercises 27.7 Solution

Use truth table to show whether each of the P : A triangle has three sides = T
following statement is true or false. Q : A paralellogram has three sides = F
1. 4 is a perfect square and four is the square root
of 16. P Q P˅Q
2. x is odd, and x2 is odd. T F T
3. x is a prime number greater than 2 and x is
The statement is true
odd.
4. x is a prime factor of 30 and x is a prime factor
Exercises 27.8
of 40.
Use truth table to show whether each of the
5. x is a prime factor of 18 and x is a prime factor
following statement is true or false.
of 24.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 767
1. An acute angle is 900 or a right angle is
between 900 P Q P ⇒Q P ⇐Q ⇔
2. The sum of interior angles of a triangle is less T T T T T
than 1000, or an octagon has seven vertices
3. x is even or x2 is odd. The statement is true.
4. 6 is a perfect square or eight is the square root
of 16 2. If 2 is an odd number then 2 is not a prime
5. The product of two numbers is even, or the number.
sum of both numbers is even.
6. 83 is an odd number or 38 is a composite Solution
number. P : 2 is an odd number F
7. 169 is a perfect square or 255 is a perfect Q : 2 is not a prime number F
square. The statement is true
3. If the number 21 is divisible by 10, then 10 is
If, Only if, and If and Only if Table divisible by 21

P Q P ⇒Q P ⇐Q ⇔ Solution
T T T T T P : 21 is divisible by 10 F
T F F T F Q : 10 is divisible by 21 F
F T T F F The statement is true
F F T T T
4. If the number 121 is divisible by 11, then 5 is
Note that the P ⇒ Q column can be obtained divisible by 3
from (P⇒Q) ˄ (P ⇐ Q).
Solution
Worked Examples P : 121 is divisible by 11 T
Use truth table to show whether the statement is Q : 5 is divisible by 3 F
true or false. The statement is false
“The product of two numbers is even if and only
if both numbers are even” 5. If a triangle is a polygon, then a polygon is a
triangle
Solution
P : the product of two numbers is even Solution
Q : both numbers are even , P : A triangle is a polygon T
Q : A polygon is a triangle F
P ⇒Q : If the product of two numbers is even The statement is false
then both numbers are even. T
Exercises 27.9 A
P ⇐Q : If two numbers are even, then their Use true or false for each
product is even. T

Baffour – Ba Series, Core Maths for Schools and Colleges Page 768
1. A whole number is divisible by ten if and only B. Negate the following statements:
if its decimal numeral ends with zero. 1. P ˄ Q
2. Two lines are parallel if and only if they do not 2. (P ˅ Q) ˄ (R˄ S)
meet, however far they are extended. 2. (P ˄ ¬ Q) ˅ (¬R ˅ S)
3. Two figures have the same area if they are
congruent. Valid Arguments
4. The product of two numbers is even if and only A valid argument is a series of statements in
if both numbers are even. which each statement follows logically from the
preceding ones. Thus, an argument begins with a
B.1. If two acute angles are congruent and given statement and ends with a final statement
complementary then their measure i 450. T (conclusion).
2. If 8 is divisible by 5, then 8 is a factor of 5. T
3. If a rectangle is a parallelogram, then a square The student‟s duty is to determine whether the
is parallelogram argument is valid/logical or invalid/ illogical by
4. If 6 is a factor of 36, then 6 is a perfect square analyzing the given statement to see if it matches
5. If 14 is a prime number, then 97 is a an odd with the conclusion. This is easily determined by
number representing the statement in a Venn diagram.
6. If a kite has congruent angles, then a rectangle
has congruent angles In order to avoid making false deductions or
conclusions, mathematicians assumes that all
Negation and Contrapositive Table the premises (generalization, definitions..) are
true
P Q ¬P ¬Q P⇒ Q ¬Q⇒ ¬P
T T F F T T Worked Examples
T F F T F F 1. Determine whether the following statement is
F T T F T T valid or not:
F F T T T T If he is a good tennis player, he can run well.
Nana Amoah is a good tennis player. Therefore,
Exercises 27.9B Nana Amoah can run well.
A. In the following exercises, P, Q, R, and S will
represent truth statements. Solution
Construct the truth tables for the following Rewrite the statement as follows:
statements, and give their converse and P : All good tennis players can run well.
contrapositive: Q : Nana Amoah is a good tennis player.
1. ¬P ⇒Q Therefore, Nana Amoah can run well.
2. P ⇒ ¬ Q
3. (P ˅ Q) ⇒ R The statement P can be represented in the
following Venn diagram:
4. (P ˄ Q) ⇒ (R˅ S)
∪ = {people}A = {good tennis players}
5. (P ⇒ Q) ⇐ (R⇒
B = {goodrunners}

Baffour – Ba Series, Core Maths for Schools and Colleges Page 769
Nana Amoah = k H = {honest people} U
k ∈ A, U b = Tom, b∈F, H
B R
A F⊂ R⊂ H F
k k

Exercises 27.10
A ⊂ B (because all good tennis players are good A. Study the arguments and state whether they
runners) are valid by using Venn diagrams to illustrate
your answers;
From the diagram, since k ∈ A, we conclude that 1. All odd numbers are whole numbers.
k ∈ B. This confirms that Nana Amoah is a good 3 is an odd number.
runner.  3 is a whole number.

2. Obtain a valid conclusion from the following 2. All boys over 15 years acan wear long pants to
statements: school.
P : Every science student study Chemistry. Ablo is 16 years old.
Q : Brown does not study chemistry.  Ablo can wear long pants to school

Solution 3. All flowers are beautiful to look at.

U = {students}, C = {chemistry students} Roses are flowers.
S = {science students}, Brown = k, k ∉C,  Roses are beautiful to look at
C U
S 4. All locally manufactured goods are marked
“Made in Ghana”
k k The portable radio is marked, “Made in China”
 the portable radio is not manufactured in Ghana
From the Venn diagram, we conclude that
„Brown is not a science student‟ 5. All squares are rhombuses
ABCD is not a rhombus
3. i. Illustrate the statements P Q and R on a Venn  ABCD is not a square
diagram
P : All my friends are religious 6. All triangles having at least two equal sides are
Q : Religious people are honest people isosceles triangles.
R : Tom is an honest person since he is my friend An equilateral triangle has three equal sides.
ii. Determine whether R is a valid conclusion  an equilateral triangle is an isosceles triangle.
from P and Q
B. In each case,draw a Venn diagram to
Solution determine whether the statement, Q is a valid
i. U = {people} conclusion from the statement P.
F = {my friends} R = {religious people} 1. P : Lazy students fail their examination.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 770
 Q : Aku is lazy since she failed her C. Each case, determine by drawing Venn
examination. diagrams whether the statement R is a valid
2. P : In Tanokrom, every old man uses a walking conclusion from the statements P and Q
stick. 1. P : Rich men live in big houses
Q : Atiah is an oldman since he uses a walking Q : Mr. Owusu lives in a big house
stick. R: Mr. Owusu is a rich man
3. P : My friend are all clever people
Q :Amuzu is a clever person since he is my 2. P : Allboys like football
friend. Q : Dzifa likes football
R : Dzifa is a boy
4. P : People who lives in glass house donot
throw stones. 3. P : Basketball people are tall people.
Q : David lives in a glass since he does not throw Q : Neos is a basketball player.
stones. R : Neos is tall.

5. P : When it rains , there is no school 4. P : Vegetarians do not eat meat.

Q : Yesterday it rained since there was no school. Q : Hindus are vegetarians.
R : Hindus do not eat meat.

Baffour – Ba Series, Core Maths for Schools and Colleges Page 771
29 TRIGONOMETRY II Baffour – Ba Series

Graphs of Trigonometric Functions Worked Examples

Here is the graph of y = sin : 1. Copy and complete the table of values for y =
1 – 4 cos x, to one decimal place for 00 x 3000

x 00 300 600 900 1200 1500

y -3.0 1.0
x 1800 2100 2400 2700 3000
y 4.5 -1.0

c.1 Use the graph

i. to solve the equation 1 – 4 cos x = 0
The graph of y = cos
ii. find the value of y when x = 1050
iii. find x when y = 1.5

Solution
Iny = 1 – 4 cos x
When x = 300, y = 1 – 4 cos 300
y = - 2.5
When x = 600, y = 1 – 4 cos 600
y = - 1.0
When x = 1200, y = 1 – 4 cos 1200
The graph of y = cos x is the graph of y = sin
y = 3.0
shifted, or translated
When x = 1500, y = 1 – 4 cos 1500
y = 4.5
The graph ofy = tan When x = 1800, y = 1 – 4 cos 300
y = 5.0
When x = 2400, y = 1 – 4 cos 2400
y = 3.0
When x = 2700, y = 1 – 4 cos 2700
y = - 1.0

x 00 300 600 900 1200 1500

y -3.0 -2.5 -1.0 1.0 3.0 4.5
x 1800 2100 2400 2700 3000
y 5.0 4.5 3.0 1.0 -1.0

Baffour – Ba Series, Core Maths for Schools and Colleges Page 772
b. d. Find, correct to one decimal place, the value of
6 y for which x = 720

5
Solution
4 a. When x = 00, y = 9 cos 00 + 5 sin 00
y= 9.0
3
When x = 600, y = 9 cos 600 + 5 sin 600
2 y = 8.8
1
When x = 900, y = 9 cos 900 + 5 sin 900
y = 5.0
0 When x = 00, y = 9 cos 1500 + 5 sin 1500
0 30 60 90 120 150 180 210 240 270 300
-1 y = -5.3

-2
x 00 300 600 900 1200 1500
-3 y 9.0 10.3 8.8 5.0 -0.2 -5.3
-4
b.
Comparing 1 – 4 cos x = 0 with y = 1 – 4 cos x, y
12
=0
From the graph, when y = 0, x = 750 and x = 2850 10

8
ii. From the graph, when x = 1050, y = 1.9
6
iii. From the graph, when y = 1.5, x = 970 and x =
2640 4

2. a. Copyand completethe table of values for y = 0

0 30 60 90 120 150 180
9 cosx + 5 sin xto one decimal place
-2

x 00 300 600 900 1200 1500 -4

y 10.3 -0.2
-6

b. Using a scale of 2 cm to 300 on the x – axis and c. i. comparing 9 cos x + 5 sin x = 0 with y = 9
a scale of 2 cm to 2 units on the y – axis, draw the cos x + 5 sin x , y = 0
graph of y = 9 cosx + 5 sin x for 00 x 1500
From the graph, when y = 0, x = 1190
c. Use your graph to solve the equations
i. 9 cosx + 5 sin x = 0 ii. comparing 9 cos x + 5 sin x = 3.5 with y = 9
ii. 9 cosx + 5 sin x = 3.5 cos x + 5 sin x , y = 3.5

Baffour – Ba Series, Core Maths for Schools and Colleges Page 773
From the graph, when y = 3.5, x = 990 + 2 cos x, correct to 1 decimal place

d. From the graph, when x = 720, y = 7.6

x 00 300 600 900
Exercises 29 y 2.2
1. Construct a table of values for y = 3 sin x + 2
cos x for 00 x 3600 x 1200 1500 1800 2100 2400
y -1.2 -2.0
b. Using a scale of 2 cm to 500 on the x –axis and
2 cm t 1 unit on the y – axis, draw the graph of b. Using a scale of 2cm to 300 on the x – axis and
for y = 3 sin x + 2 cos x for 00 x 3600 2 cm to 0.5 units on the y axis, draw the graph of
y = sin x + 2 cos x for 00 x 2400
c. Use your graph to solve the equations; c. Use your graph to solve the equations;
i. 3 sin x + 2 cos x = -1 i. sin x + 2 cos x = 0
ii. 3 sin x + 2 cos x< 1.5 ii. sin x = 2.1– 2 cos x
d. Find y when x = 1710.
2. a. Copy and complete the table for y = sin x

Baffour – Ba Series, Core Maths for Schools and Colleges Page 774
 ANSWERS TO EXERCISES
1. Sets and Properties of Sets

Ex 1.8 Ex 2.23
A. 1. {5, 10, 15.., 30} 2. {1, 2, 3…8} 3. {2, 6, 8, 10, 12} 4. {1, 3, 4, A. 1. 1.9 2. 0.15 3. 3.49 4. 11.3 5. 31.0571
5, 6, 7, 8} B. 1. 1.447 2. 90.093 3. 6.075 4. 1.145 5. 2.108
Ex 2.24
B. 1. E ∪ F= {5, 7, 9} 2. E ∪ G = {1, 7, 9, 13} 1. 0.46 2. 14.52 3. 3.762 4. 7.866 5. 3.1815 6. 0.0288
3. F ∪ G = {1, 5, 7, 9, 13} 4. E ∪ (F ∪ G) = {1, 5, 7, 9, 13}
Ex 2.25
Ex 1.10 1. 2 2. 60 3. 20 4. 1.32 5. 0.9 6. 0.002
i. A1 = {11, 12} ii. B1 = {1, 2, 4, 5, 7, 8, 10, 11}
iii. C1 = {3, 5, 6, 7, 9, 10, 11, 12} Ex 2.26
A. 1. 8.3 × 2. 1.4 × 3. 9.8 × 4. 3.5607 × 5. 3.915
Ex 1.11 × 6. 3.425 × 7. 4.68702 × 8. 3.493 × 9. 4.3 ×
6.i. = {1, 2, 3…18}, A = {2, 4, 6, …18}B = {3, 6, 9…18}A∩B =
{6,12}, A∪B = {2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 15, 16, 18}(A∪ B. 1. 3.0 × 2. 3 × 3. 6.0 × 4. 4.0 ×
B)1 = {1, 5, 7, 11, 13, 17} 7. i. A = {5, 10, 15..95} B ={7, 14, 21, 28,
35, 42, 49, 56, 63,70, 77, 84, 91, 98}A ∩ B = {35, 70}ii. Multiples of Ex 2.27
35 from 35 to 70iii. n(A) = 19, n(B) = 14 n(A B) = 2 , n(AUB) = 1. 34250 2. 0.1173 3. 0.0053 4. 7180300 5. 402000 6. 900
n(A) + n(B) – n(A∩B)⇒31 = 19 + 14 – 2
Ex 2.28
Ex 1.12 A. 1. 5.3 × 2. 3.95 × 3. 87.525 × 4. 1.0 ×
1. U = {50}, E = {30} G = {17} i. 4 ii. 26 iii. 13 iv. 39 B. 1. 7.3050 × 2. 4.405 × 3. 9.9 × 4. 5.0 ×
2. i. ii. 17 iii. 13 3. i. ii. 15% iii. 85% 4. x = 10 5. i. 39 ii.
676. 160 7. i. ii. 9 boys 8. i. 6 ii. 28 iii. 66 9. x = 13 10. x = 12 Ex 2.29
1. 1.2 × 2. 1.8 × 3. 2.52 × 4. 6.24 ×
Ex 1.13
1. n(A B C) = 2 2…. 3. a. 20 b. 50 c. 10 4. 30 5. 20 6. 12 Ex 2.30
7. 43 8… 9. x = 26 10. 4 11. 24 12. n = 12 13. ii. 24 iii. 21 A. 1. 8 × 2. 1.2 × 3. 3 × B. 1. 4 × 2. 1.4 ×
14. i. 270 ii. 125 iii. 0 iv. 205 v. 250 vi. 0 vii. 25 viii. 395 3. 6 × C. 1.2 × 2. 2.8 ×

2. Real Numbers
Ex 2.35
Ex 2.13 1. -18 2. 6 3. 16 4. 18 5. 89 6. -27
1. 2. 3. 4. 5. 6. 7. 8. 9.
Ex 2.36
Ex 2.14 1. 2. - 3. - 4. 5. 6.
1. 2. 3. 4. 5. 6. 6
Ex 2.39
Ex 2.15 1.0. ̇ 2. 0. ̇ ̇ 3. 1. ̇
1. 2. 4 3. 4. 5.
Ex 2.40
1. 2. 3. 4. 5. 6.
Ex 2.16
A. 1. 2. 3. 4. 5. B. 1. Gh1,500.00 2. 3. 4.
Ex 2.41
5. 1.i. a = 2, b = 5 ii. Associative

Ex 2.17 Ex 2.42
1. i. ii. 14 2. i. ii. 16 3. i. ii. T = Gh2,100.00 , G = 1.a. 23 b. 0 2. a. 32 b. 2 3. i. 39 ii. 55 iii. 5811 iv. P = 1 4. i.
Gh1,800.00 P = Gh900.004. i. ii. Gh231.00 5. Gh60.00 ii. iii. 5.a. b. c. 80
6. i. a.
Ex 2.18
1. 0.6 2. 0.2 3. 0.58 4. 1.25 5. 0.011 6. 0.06 7. 0.03 8. 0.009 Ex 2.48
9. 0.009 i. x2y ii. xy + xz iii.Distributive2. i. {3, 5} ii. {3, 5} iii. is
distributive over∪
Ex 2.19
1. 1 2. 3 3. 4 4. 2 5. 3 Ex 2.49
Ex 2.20 1. -2 2. 0 3. 2.25 4. i. 0 ii. commutative 5. i. ii. yes iii. b = 8
1. 0.2 2. 0.064 3. 0.05 4. 0.11

Baffour – Ba Series, Core Maths for Schools and Colleges Page 775
3. Algebraic Expressions Ex 3.20
Ex 3.4 A.1. 39 2. 6 3. -28 4. 50 5. 97 B. 1. 14 2. -12 3. 27
A. 1. 15 2. 80 3. 20 4. 12 p 5. 5 6.
6 b B. 1. 2. 3. 5t 4. - 5. 6 6. 7. 4. Surds
Ex 4.2
1. 5√ 2. 5√ 3. 3√ 4. 22 5. 4√ 6. 4√ 7. 10√ 8. 7√
9. 10√ 10. 3√ 11. 10√ 12. 6√ 13. 7√ 14. 11√
Ex 3.7 15. 5√
1. x = 0 or x = 3 2. x = -2 or x = -5 3. x = 2 or x = 5 4. x = 2 5. x = -
1 or x = 4 6. x = 2 or x = 1 Ex 4.7
A. 1. -2 + √ 2. 2 + 5 √ 3. 84 – 60 √ 4. 24 - 18 √ 5. 10
Ex 3.8
1. x = 9 or x = -9 2. x = 4 or x = -4 3. x = 0 or x = -5 4. x = 1 5. x √ -5√ 6. 24 – 18√ B. 1. 11 2. 23 3. -9 √ + 15 √ 4.
23 5. 24 √ + 21 √ 6. 23
= - 6. x =2 or x = 5
Ex 4.8
Ex 3.9 1. 8 – 2√ 2. 9 – 6√ 3. 59 – 24√
Ex 3.10
A. 1. m2 – 14m + 40 2. 10a2 – 4a + 6 3. 12ab + 3ad + 4bc + cd 4.
Ex 4.9
Pq – 36 5. 3y2 + 13y + 12
1. 3 + 2√ 2. 5 + 2√ 3. 50
B. 14y2 – 44y + 35 2. 14m2 + 2mn - 7n2 3. -3a2 + ab – 5b2
4. -6pq + 4q2 5. 7qp Ex 4.10
1. 5√ 2. 6√ 3. 4 4. 45 5. 2√ 6. 6 7. 6√ 8. 40
Ex 3.11
A.1. 2ny(1 + 3y) 2. 2y(2xy – 9) 3. m (51 + 15m) 4. 12q(p + 3q) 5. Ex 4.11
5u(2u + 7t) 6. st(t – 28) 1. 25.452 2. 0.2598
Ex 4.12
√ √ √ √ √ √
B. 1. 4yt(3t – 4y) 2. 7pq(3q – 7p2) 3. 6pr(1 + 7p) 4. 8rs(2 + 8s2) 5. A. 1. 2. 3. 4. 5. 6. 7.
9m(3k2 – 1) 6. 5cd(3d + 5c) √ √ √ √ √
8. √ 9. B. 1. 2. 3. 4.
C. 1. 2xy(2x – 3y + 4t) 2. 3c(ab + 33c – 5a2c) 3. 6km(2k + 3n + m2) C. 1. 0.707 2. 0.577 3. 0.466 4. 3.46 5. 4.476. 1.15 7. 1.34 8.
4. 4pq (4pq – 5r – 2) 5. px ( ) 14.1 9. 0.671 10. 0.566D. 1. 19.48 2. 0.354 3. 0.3534 4. 3.732

Ex 4.13
Ex 3.12
1. (a – b)(2p + q) 2. (x + b)(x – a) 3. (x + 2q)(y – 3c) A. 1. √ 2. √ 3. √ 4. 3 + √ B. 1. 1 + √ 2. -1 +
4. (p – q)(2r + 3s) 5. (p – 2q)(r + 3s) 6. (4y – 8d)(x – 2y) √ 3. 24 + 12 √ 4. 7 – 4 √ 5. √ + √ 6. √ – √ 7. 3 √ + 3√
8. √ – √ 9. -1 + √ 10. 2 + √ 11. -2 + √ 12. -3√ – 3√ C.
Ex 3.14 √ √
1. -1 + √ 2. - 3. 5 + 2√ 4. + 5. -1 + √ 6. -1 –
A. 1. (x + 3) (3x + 2) 2. (3x + 5) (2x – 1) 3. (3x + 3) (3x + 1 ) 4. (5x
– 2) (x – 2 ) 5. (x + 1) (7x + 2 ) 6. (2x + 1) (5x + 3 ) √ 7. 9 + 4 √ 8. 4 – 2 √

B. 1. (x – 4) (4x – 5 ) 2. (3x – 1) (4x – 1) 3. (x – 6) (3x – 2 ) 4. (x – 3) 5. Number Bases

(5x – 2 ) 5. (2x – 1) (x – 3 )6. (6x + 1) (x + 2 )
Ex 5.4
Ex 3.15 A. 1. 3203 2. 11333 3. 21205 4. 22110 5. 1414
1. (2x + 3y) (x + y ) 2. (6x – 2y) (x – y ) 3. (2x - 3y) (7x – y ) 4. (3x - B. 1. 222 2. 43
4y ) (x – y ) 5. (5x + 5y) (x - 6y )
Ex 5.5
Ex 3.16 A.1. 1002 2. 3140 3. 12 4. 2215 5. 1024 6. 3001
A. 1. (x + 5)2 2. (x + 2)2 3. (x + 9)2 4. (x - 16)2 5. (x - 13)2 6. (x - B. 13 2. 3401 3. 223 4. 323
4)2B. 1. (3x - 2)2 2. (2x - 5)2 3. (2x - 9)2 4. (2x - 1)2
Ex 5.6
Ex 3.18A 1. 1011 2. 110000 3. 1111 4. 111000 5. 1010100
1. (x + y – 2) (x + y + 2) 2. (a – 6 – 4b)(a – 6 + 4b
3. (2x + 9 – 5b) (2x + 9 + 5b) 4. (2x – 5 – a)(2x- 5 + a) 5. (-2b – 3)(- Ex 5.7
2b + 3) 1. 1 2. 10110 3. 110 4. 101 5. 100 6. 111

Ex 3.18B (Accept alternate answers) Ex 5.8

1. 7772 - 2232 2. 6742 - 3262 3. 6442 - 3562 A. 1. 20202 2. 4023 3. 204400 4. 22203 5. 1227 6. 6122
4. 23562 - 22562 5. 7922 - 6922 6. 18672 – 17672 B. a. 10100 b. 1101100 c. 111100C. 1. 11001 2. 11110 3. 1111
4. 1001110
Ex 3.19
1. k = -3 2. m = -1 Ex 5.9

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A. 1. 97 2. 543 3. 485 4. 110 5. 147 6. 1012B. 1. 11 2. 14 3. 2 22 101 130 204
5 4. 27 5. 12 6. 37C. 1.17 2. 67.641 3. 135 4. 8.519 5. 72 6. 3 33 124 220 311
448 7. x = 35, y = 45 z = 104 w = 202 8. a = 7, b = 5 c = 4D. 1. 4 44 202 310 413
1021 2. 2200 3. 11 4. 21 5. 20111
E. 1. 150 2. 405 3. 43 4. 701 5. 234 6. 2176F. 1. 4 2. 5 3. 5 3. a. b. i. 8 ii. 13 c. i. 6 ii. 3
4. 7 5. 5 6. 7 G. 1. 7.76 2. 154.192 3. 192.719
4. 22.5625 6. Relations and Functions

Ex 5.10 Ex 6.6
A. 1. 103 2. 1122 3. 304 4. 314 5. 402 6. 1132 7. 1434 8.
1. y = - 3x + 15 2. y = x + 3. y = 3x + 17
110B. 1. 111011 2. 1001101 3. 1100110 4. 1000010 C. 1. 11102
2. 11021 3. 11100 4. 452 5. 1300 4. y = 5. y = 6. y = 7. y = x + 8. y = - 2x + 3
9. y = -2x – 4
Ex 5.11
A. 1. 123 2. 23 3. 101001 4. 2010 5. 111110
Ex 6.7
B. 1. 11011 2. 111001 3. 100011 4. 101110
C. 1. 24 2. 142 3. 41D. 1. 304 2. 212 3. 38 4. 2010 1. y = 2. y = ( ) 3. y = 4. y = 2( 5. y = ( )

Ex 5.12 Ex 6.8
A. 1. 63 2. 108 3. 106 4. 1714 5. 5014 B. 1. 23 2. 84 3. 130 1. y = x2 + 2x 2. y = 2x2 + 5x – 3 3. y = 3x2 – 5x + 2 4. y = - 5x2 –
4. 6e4 5. 20teC. 1. 1219 2. 1t653 3. 3t 4. 2266 3x – 5 5. y = x2– 6

Ex 5.13 Ex 6.9
b. 14 c. i. n = 4 ii. n = 5 1. y = , k = 20 2.y = 3x2 – x + 4, 3044 3.y = 5(3)x - 1 , 32805
2. i. b. i. 31 ii.14 c. i. m = 3 ii. m = 4
4. y = ,
Ex 5.14
A. 1. 11102 , 2213 , 30114 2. 136, 405 , 101102 3. 2324 , 2305 , 22023 ,
3326 B. 1. 2245 , 20023, 556 , 1410 2. 110112 , 1013 , 115 3. 4036, 22013, Ex 6.11
118 A. 1. 126 2. 3. – 7 4. 5. 15 6.
Ex 5.15 B. 1 . a. 2, , - 8 b. 2 2. a. – 10 b. 0 c. 7 d. 3. a. i. 2 ii. 5 b. 1
A.1. 7 2. 5 3. 5 4. 3 5. 2 6. 6 B. 1. 2 2. 3 3. 4 4. 2 5. 1 c. 10 4. i. 22 ii. – 3 iii. – 2 iv. – 3 5. i. 24 ii. 3 6. 60 7. -
6. 1C. 1. 6. 2. 2 3. 5 4. 4 D. 1. 9 2. 8 3. 4 4. 3
8. 9. 15, 1 10. 15, 6 11. 8 or 9 12. -2 <x< 3 13. i. -2, , - 1.27
Ex 5.16 ii. 2 14. i. Q = {2, -1, -2, -1, 2}
1.
+ 0 1 2 C. 1. {-7, - 4,-1, 2} 2. {- 8, - 5, 0, 7, 32} 3. 20, 24, 27, 30, 32}
0 0 1 2 Ex 6.12
1 1 1 10 1. a = 5, b = - 8 2. b = 3 c = 5, x = -2 or x = 3. a = -3, b = 2
2 2 10 11
4. b = 5. b = -2
× 0 1 2
0 0 0 0 Ex 6.13
1 0 1 2 1. 0 2. - 3. 3 4. 3 5. 5 or 2 6. 4 or 3
2 0 2 11
Ex 6. 14
2. 100000two cm ; 100111two 3. 4 4. x = 3, y = 7 5. 0, 2, 3, 4, 5, 6, 7, 1. 2. 2 or -5 3. 3 4. 3 or 4 5. 6 6. – 15
8, 9
C.P: Ex 6.17
2. A(9,6), B (3,8 ) C-3,8) D(-1, 4) E(4, -10), F(-5, -8),G(-3,-5), H(0, -
Base 12 10 8 5 3 2 6) I(-2, -4 ) J(8, -2)
23 27 33 102 1000 11011
2e 35 43 120 1022 100011 Ex 6.21
25 29 35 104 1002 11101 1. a. b. i. (2, - 4) ii. – 4 iii. x = 2
2. a. b. i. (a, b) = (1,5) ii. (0, 3) iii min : 3 iv. No zeroes v. 3 f(x)
3.110111; 55 4. 2020three mm, 22100three mm2 19 vi. x = 0 3. a. b. i. P(2, 6) and P(4, 4) ii. (2.5, 6.25) iii. 6.25
iv. x = 0 or x = 5 v. -6 f(x) 6.5 vi. x = 2.5
Ex 5.17 4. a. b. (-3, 0) and (1, 4)
1. i. 44 ii. 30
2. a. b. i. 204 ii. 413 iii. 23 c. i. 4 ii. 3 Ex 6.23
1. (3, 2) 2. (-7, - 6) 3. (3, 0) 4. (0, 3.5) 5. – 2, -1 ) 6. (9.2, - 4.5)
* 1 2 3 4
1 11 23 40 102 Ex 6.24

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1. (-8, 3) 2. (7, 6) 3. (8, 5) 4. (- 11, 12) Ex 7.12
1. 23cm 2. 60m 3. 10 inches 4. 7cm 5. 5 cm 6. A = 72cm2,
Ex 6.25 P = 33.96cm 7. 0.8cm
A. 1. 5 2. 4.47 3. 3.61 4. 13.34B. 1. 9.23 2. 4 3. 17 C.P. 1. 8√ 2. H = √ , A = √ 3. i. 5 ii. 60cm2
C.p: 2. i. (-1, -1), (0, 0), (1, 1) etc ii. y = c iii. (2, 2) 3. t = 2 or t = -2
4. k = 14 or k = -2 Ex 7.13
1. 100cm 2. 26.8cm 3. 10.58cm 4. 8.05 cm
Ex 6.26
1. PQ = PR = √ , QR = √ , M = (2.5, 4.5) 2. AB = AC = 5, Ex 7.14
BC = 7 3. AB = AC = √ , BC = √ 4. AB = AC = 13, BC = 14 5. 1. 26cm 2. 8 cm 3. 12.12cm 4. 7.6cm 5. h = 50cm, LM = 50.35 cm
AB = AC = 5, BC = 3 6. 40cm 7. a. 4 b. 8

Ex 6.34 Ex 7.15
1. 4y + 3x – 27 = 0 2. 4y + 5x – 7 = 0 3. 11y + 3x – 35 = 0 1. x = 5cm, a = 30cm2 2. 4:1 3. 17cm 4. 3cm 5. 1.6cm and 2.6 cm
4. 5y – x + 19 = 0 5. 5y + 12x – 1 = 0 6. 3.95 cm and 6.95 cm 7. 6cm, 8cm, 10cm

Ex 6.35 Ex 7.16
1. y – 3x – 11 = 0 2. 3y + x – 13 = 0 3. y – 2x + 3 = 0 4. 3y – x + 1. 22 ft 2. 4.2m 3. 9 miles 4. 93ft 5. 5 cm 6. 5m 7. 6.38m 8. i.
17 = 0 5. y + 5x + 22 = 0C.P. 1. i. 3y – x – 1 = 0 ii. y + 2x – 12 = 0 6.40km ii. a. 6.40km b. 12.8km 9. 25ft 10. i. /AB/ = 22.5m ii.
iii. Q(5, 2) and R(1, - 3 ) 2.i. 3y = - 4x ii. OA = OB = 5, A = 10 sq. /BD/ = 33.1m 11. AB = 28.7m, BD = 44.6m 12. 4ft
units iii. 20 sq. units
Ex 7.17
Ex 6.36 A. 1. 600 2. 600 3. x = 650, y = 1150 4. x =700, y = 1100
1. y = 3x + 2 2. y = x + 4 3. y = 4x + 6 4. y = x – 4 5. y = -3x B. 1. 220 2. 850, 580, 950 3. x = 980, 490, 1470, 65.30

–2 6. y = x + 7 Ex 7.19
A. 1. 21600 2. 36000 3. 72000 4. 59400 5. 27000 6. 36400
Ex 6.37 B. 1. 1200 2. 880 3. 1590 4. 480
A. 1. 4y – 3x – 1 = 0 2. y – 7x + 28 = 0 3. 5y + 2x + 14 = 0 4. 2y –
5x – 16 = 0 5. 4y – 3x – 1 = 0 Ex 7.20
B. 1. i. 2y – 5x + 19 ii. 5y + 2x + 4 = 0 2. i. 3y – 2x + 14 = 0 ii. 2y 1. 8 2. 10 3. 18 4. 22 5. 19 6. 29
+ 3x – 8 = 0 3. 3y – 4x + 13 = 0 4. x2 + y2 = 25
Ex 7.21
Ex 6.38 A. (f, e) (g, h), (l, k) and (m, n)B. x = 1150, y = 100z = 1700
1. Not 2. Yes 3. Yes 4. NotC. P: 1. i. 6.32 ii. (3, 1) iii. m = 3; y
– 3x + 8 = 0 iv. r = -14 2. p = 3, q =1 Ex 7.22
1. 152.300 2. 154.280 3. 158.820 4. 161.050 5. 162.850
Ex 6.39
1. (7, -7) 2. (1, 5) 3. (- 0.5, - 3.5) 4. (-1.5, – 5.5) 5. (4, -7) C.P. 1. Ex 7.23
i. (2, - 1) ii. 2y – 5x– 8 = 0 iii. yes 3. (-1, -1) 4. (5, 4) 1. 350 2. x = 37.50, y = 67.50 3. 240 4. 46.70

7. Plane Geometry 1 Ex 7.24

Ex 7.5 1. 12 2. 6 3. i. Ext < = 300, Int < = 1500 ii.18000
1. … 2. 850
8. Equations and Inequalities
Ex 7.6 Ex 8.15
A. 1. z = 580 , y = 580, x = 1180 2. e = 820 3. y = 1180, x = 620 4. a = A. 1. 4 < x < 7 or (4, 7) 2. -3 ≤ x < 2 or [-3, 2) 3. -3 ≤ x < 2 or (-3, 2)
400, b = 1200, c = 200B. 1. <PRQ = <PQR = 800 2. x = 580, y = 220, z 4. -1 < x < 3 or (-1, 3) 5. -4 < x ≤ 2 or (-4, 2] 6. 2 ≤ x ≤ 3 or [2, 3] B.
= 880 3. x= 800, y = 600 1. -20 ≤ x < -4 or [-20, -4) 2. 0 < x < 1 or (0, 1) 3. -5.5 < x < 1.5 or [-
5.5, 1.5) 4. -1 < x < 5 or (-1, 5) 5. - ≤ m ≤ 2 or (- , 3]
Ex 7.7
A. 1. r = 500, t =700 2. a = 700, b = 700, c = 400 3. y = 600, 4. 500 5. 9. Vector and Bearings
k = 560, m = 540 6. b = 1080, c = 550, a= 430 7. d = 1020 8. x = 200 Ex 9.1
9. a = 1300 10. b = 1100 11. d = 300 , e = 900 12. k = 300B. 1. y = 1. A = 3550 , B = 0170 , C = 0570, D = 1520, E = 0900 , F = 2850
2800 2. a = 730 3. x = 750 , y = 300 4. x =300 , y = 200 5. d = 650 , e
= 480 , f = 670 Ex 9.2
A. 1. 2130 2. 3350 3. 3000 4. 150 5. 1070B. S 500 W 2. N 380 E
Ex 7.8 (every answer is in cm) 3. S 250 E 4. S 750 E 5. N 570 W 6. N 50 W
1. 12 2. 24 3. 36 4. 36 5. 6.7 6. 7 7. x = 20 Ex 9.3
B. 1. 5.2 2. 23.05 B. 1. (7km, 1050) 2. (6km, 3150) 3. (11km, 0650) 4. (3km, 2500)
Ex 7.11 Ex 9.4
1. 13m 2. 17.89m 3. 7.42m 4. 6.92ft 5. 15.49ft A1. 2250 2. 1450 3. 1200 4. 3050B. 1. 2970 2. 0400 3. 1350 4.
C.P. 1. 5m 2. i. 8m ii. 4m 0270C. i. 290 ii. 042 iii. 0350 iv. 1470

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Ex 9.5 10. Statistics I
1. (20km, 3420) 2. (46km, 2630) 3. i. 63km ii. (63km, 2340) 4. i. 23 Ex 10.2
km ii. (23km, 2660) 5.i. 146 km ii. (1146km, 2870) 6. i. 116km ii. 1. -4 2. 6 3. 48 4. i. 3 ii. 3 iii . 3.33 5. 15
(120km, 2600) b. 113km 7.i. 19km ii. 21km, 0870 iii 4km 8. a. i. 47m
ii. 17m b. i. 108m ii. 34m c. 1280 d. 113.6m Ex 10.3
1. 3 2. 38 3. 23yrs 3 months 4. -4 5. 408 6. Gh¢4800 7. 12
Ex 9.6
1. ( ) 2. ( ) 3. ( ) 4. ( ) Ex 10.6
1. 2. a. b. 3. a. b.
Ex 9.7
1. (- 4, 1) 2. (2, 5) 3. (- 6, 7) 4. (2, 5) 5. (- 3, 5) 6. ( -4 , 1) Ex 10.8
1. i. 1200 ii. 45mins 2. i. 400 ii. Gh¢1260.00 iii. 3. i. 1200 ii 30
Ex 9.8
A. i. c and f ii. a = (2, -3), (18, - 27), (8, - 12) … b = (20, 14), (30, iii. 90 iv. 11% 4. i. 50 ii…. 5. i. 36 0 ii. Gh¢900.00 iii.
12), (40, 28)…c = (- 1 , - 3), ( - 8, - 24 ), ( - 5 , - 15) Gh¢60.00
B. i. (20, 0) ii. ( - 8, - 10) 2.i. a. b – a b. 2b c. 2a d. 2(a – b) ii. ⃗⃗⃗⃗⃗ =
Ex 10.12
-2 ⃗⃗⃗⃗⃗ 3. ii. N is the mid point of AC
1. 27.4 2. i. 21 iii.
Ex. 9.9
1. 6 2. 7.8 3. 8.1 4. 5 5. 8.5 6. 10 Ex 10.13 assumed mean
C.P. /u/ + /v/ = 2, /u + v/ = 1 1. 68.07 2. 73.7 3. 30.25 4. 33.25

Ex 9.15 11. Rigid Motion

1. i. ( ) ii. ( ) iii. 6.4 units 2. ( ) 3. i. ( ) ii. ( ) iii. ( ) Ex 11.1
1. P1(-3, 12), Q1(-6, - 4), R1(2, 9) 2. A1(8, 11) 3. (4, - 5) 4. R1(-5, 8),
iv. √ 4. i. a = ( ), b = ( ) ii. ( ) iii. 5 5. a . B = (1, 4), C = S1(2, 9), T1(4, 0) 5. (-1, 4)
(8, 4) b. (0.5, 3) 6. a. ⃗⃗⃗⃗⃗ = ( ), ⃗⃗⃗⃗⃗ = ( ), D= (6, -1) b. / ⃗⃗⃗⃗⃗⃗
Ex 11.2
/ = 4 7. i. Q = ( ) S = ( ) ii. 13.6 units 8. a. i. D = ( ), B =
1. A1(-4, 8), B1(-7, 11), C1(-9, -2) 2. (4, - 2) 3. (-3, - 3) 4. (6, - 4)
( ) ii. ⃗⃗⃗⃗⃗ = ( ), ⃗⃗⃗⃗⃗ = ( ) b. equal vectors 9.i. ⃗⃗⃗⃗⃗ = ( ), ⃗⃗⃗⃗⃗ = 5. (5, 3)
( ) , ⃗⃗⃗⃗⃗ = ( ) ii. /PQ/ = /QR/ = 10 10. Q = (10, 2)
C.P. 1. h = 2, k = - 1 2. S(4, 2) or S(2, -6) 3. C(0, 4) D((-5, 3) Ex 11.3
1. (-8, - 3) 2. (6, - 1) 3. (-3, - 10) 4. A (1, 4), B (7, 9), C (1, 9)
Ex 9.16
1. (7, 7) 2. V = 18, U = 8 3. a = 9, b = - 2 4. X = 3, y = - 1 5. (1, Ex 11.4
2) 6. i. 13.5 units ii. ( ) 1. ( -6 , - 14) 2. ( , ) 3. A1 (15, 12) , B1 (3, 9) , C1 (- 6, 6), D1
(3, 6)
Ex 9.17
1. (p, q) = (2, 3) 2. i. (m, n) = (6, -1) ii. 6.4 units 3. i. ( ) ii. m = Ex 11.5
3.5, n = - 1 .5 4. a. m = 3 , n = - 1 b. 8.5 units 1. (40, -28) 2. A1 (0, - 10), B1 (8, -14), C1 (4, 6) 3. R1 ( , 1)
S1 ( , ), T1 (2, ) 4. (6, - 8)
Ex 9.18
A. 1. ( ) 2. ( ) 3. ( ) 4. ( ) Ex 11.6
1. A1 (1, - 1), B1 (-4, 9) , C1 (-9, 14) 2. A1 (1, 1) , B1 (3, 0), 3. P1 (0,
B. 1. ( ) 2. ( ) 3. ( ) 4. ( ) 5. ( ) 6. ( ) 2) . Q1 (4, 2), R1 (4, -6) , S1 (0, - 2) 4. i. (3, 2) ii. (12, 11)

Ex 9.19 Ex 11.7
1. 540 2. 300 3. 590 4. 130 5. 690 6. 300 1. (5, 2) 2. (3, 1) 3. P(- 3, 1) , Q(0, ), R(6, 8)
Ex 9.20
Ex 11.8
A. 1. ( ) 2. ( ) 3. ( ) 4. ( ) 5. ( )
1. (-3, 0) 2. U(- 6, 1) V(-3, 3) W(- 4, 7)
B. 1. ( ) 2. ( ) 3. ( ) 4. ( ) 5. ( ) Ex 11.9
1. A( - 11, 1) B(- 5, -4 ) C(- 8, - 4) D(- 13, - 2) 2. i. P1 (- 3, 8) , R1
Ex 9.21 (0, 12), T1 (3, 8) ii. O2 (4, 0) , P2 (7, 8), R2 (4, 12), T2 (1, 8) 3. A1(1,
A. 1. (14 units, 0600) 2. (9 units, 1980) 3. (10 units, 3070) 4. (12 -3), B1(4, -5), C1(9, -5) , D1(6, -3), A2(-5, -3), B2(- 8, - 5) , C2(-13, -5),
units, 1800) 5. (9 units, 320)6. (11 units, 2430) B. 1. (10 units, 1910) D2(-10, -3) 4. A1(2, - 6), B1(- 8, - 8) , C1( - 5, - 11), A2(6, - 8) , B2(11,
2. (18 units, 0380) 3. (15 units, 1270) 4. (22 units, 1170) 5. (10 -8 ) , C2( - 3, 9)
units, 450)6. (19 units, 2980)
Ex 11.10
Ex 9.22 1. A1(4, -3), B1( -2, -5), C1(-3, 9), 2. A2(- 3, - 4), B2(5, -2), C2(9, 3),
1. a. 20.52km b. 0630 2. a. 1252km b. 1380 3. a. 22.67km A3(3, 4), B3(5, - 2), C3(- 3, 9), 4. i. P1(3, 4 ), Q1(2, -1), R1(5, -3), ii.
b. 2430 P2(4, -3), ii. P2(4, -3), Q2(1, -2), R2(-3, -5),
Ex 11.11

Baffour – Ba Series, Core Maths for Schools and Colleges Page 779
1. Rotation through 900 about the origin 2. 1. Rotation through - 900 A. 1. Gh¢ 7,611.94 2. Gh¢2,537.31 3. Gh¢ 15,223.88 4.
about the origin 3. 1. Rotation through 1800 about the origin 4. 1. Gh¢101,492.54B. 1. Gh¢7,515.79 2. Gh¢25,052.63 3. Gh¢3,364.21
Rotation through 900 about the origin 4. Gh¢16,821.05

12. Ratio and Rates Ex 12.17

i. 18 mins ii 8 : 20 am iii. 1 hr iv. 8 : 02 am v. 0.375 km/h
Ex 12.3
A. 1. Proportion 2. Not a proportion 3. Proportion 4. ProportionB. Ex 12.18
1. 2. 3. C. 1. , 2. 3. 1. 20 2. 8 3. 113,520 4. 109,020

13. Percentages 1
Ex 12.4
A. 1. 20 2. 5 3. 6 B. 1. 9 2. 12 3. 3 4. 30 5. 6 6. 3C. 1. 8 2. 4
Ex 13.44
3. 8 4. 15 5. D. 1. -1 2. 1 3. 4 4. – 4 5. 16 1. 2% 2. 50% 3. 5.3% 4. 7% 5. 20% 6. 6.3%

Ex 12.5 Ex 13.5
1. 120 2. 32.5 3. 30 4. 80 5. 504 A. 1. 60 2. 1,300 3. 1,206 4. 18,000 5. 10, 000
B. 1.Gh¢6,100 2. Gh¢4,787.33 3. Gh¢6,000 4. Gh¢250
Ex 12.6
1. 45 2. 36 mins 3. 24hrs 4. 4 days 5. i. 14 ii. 50 mins Ex 13.6
1. 40% 2. 75% 3. 75% 4. 7.5% 5. 40% 6. 100
Ex 12.7
A.1. i. Gh¢400, Gh¢200 ii. Gh¢360, Gh¢240 iii. Gh¢480, Gh¢120 Ex 13.7
iv. Gh¢350, Gh¢250 v. Gh¢375, Gh¢225 2. i. Gh¢1,960, Gh¢280 ii. 1. Gh¢22.22 2. Gh¢91 3. Gh¢160 4. i. 20c. ii. 16 cm
Gh¢1, Gh¢540, Gh¢700 iii. Gh¢1,400 , Gh¢560, 280 iv. Gh¢1,120, 5. Gh¢13,110 6. Gh¢2,944 7. 23% 8. 3,600 votes
Gh¢840, Gh¢280
B. 1. Shamo = Gh¢8,000.00, Bako = Gh¢20,000.00, Aspa = Gh¢12, Ex 13.8
000.00 ii. Gh¢12, 000.002. Kate = Gh¢7,470.00, Comfort = Gh¢9, A. 1. Gh¢7,000 2. Gh¢1,700 3. Gh¢1,800 4. Gh¢20,900
960.00 3. Gh¢20,000.00 4. Gh¢20,000.00 5. Gh¢18,700.00, B. 1. Gh¢1,650 2. Gh¢25,900 3. Gh¢18,300 4. Gh¢387,000
Gh¢22,100 6. 8.75 7. i. Okra = Gh¢55,000.00, Harry = C. 1. Gh¢1,402.50 2. i. Gh¢1,640 ii. Gh¢2,772 iii. Gh¢4,412 3.
Gh¢77,500.00 , Regina = Gh¢31,000 ii. Gh¢22,500.00 8. i. Naa = Gh¢3,840 ii. Gh¢6,8160
Gh¢24, 000.00, Ayeley = Gh¢40,000.00 ii. Gh¢16,000.00 9. i. Gh¢
8,000.00, Gh¢8,500.00, Gh¢9,500.00 ii. Gh¢,600.00, Gh¢5,950.00, Ex 13.9
Gh¢6,650.00 10. i. Tom = Gh¢100,000.00, Ben = Gh¢30,000.00, Pak 1. 87.5
= Gh¢50,000.00 ii. Gh¢100,000.00 iii. Gh¢50,000.00
Ex 13.10
Ex 12.8 1. Gh¢ 3,176.47 2. Gh¢50,000
1. 2102. 233. Gh¢400,000.00 4. 1125. 13,155 C. P. i.
Ghȼ39,000.00ii. An = Ghȼ 7,800, Nt = Ghȼ 5,200, As = Ghȼ 26,000 Ex 13.11
iii. Ghȼ 18,200.00 A. 1. Gh¢1,414.40 2. Gh¢91.65 3. Gh¢18.36 4. Gh¢704
5. Gh¢ 20.90B. 1. Gh¢12,500 2. Gh¢1,083.90 3. Gh¢990
Ex 12.9 4. Gh¢4,743 5. Gh¢1,665.6
1. Ghȼ 300.00 2. Ghȼ 100.00 3. i. Ghȼ 32.00 ii. Ghȼ 64.00 4.
Ghȼ 267.00 5. Ghȼ70.006. Ghȼ2.07 7. 22words8. 23 deaths 9. 6 Ex 13.12
mins C.P. 1. 4,270.00 2. Gh¢47.00 A. Gh¢40,000 2. Gh¢240,000 3. Gh¢350,000 4. Gh¢23,000

Ex 12.10 Ex 13.13
1. i. 350 km/h ii. 97.2 m/s 2. i. 3.33m/s ii.0.2 km/min 3. 1. 2% 2. 30% 3. 12.5% 4. 15% C.P 1 . i. Gh¢42 ii. Gh¢1250 2.
1110km/h 4. i.125km/h ii. 34,7m/s 5. i.48km/h ii. 40km/h iii. Gh¢41.25
38km/h Ex 13.14
1. 3.5 2. 59.34 3. 1711 4. 7639 5. 230
Ex 12.11
1. 1.8 hr 2. 3 hr 3. 300 mins 4. 21,600s Ex 13.15
1. 84% 2. 35.5% 3. 54.5% 4. 25% 5. i. 16 ii. 50% 6. 25% 7.
Ex 12.12 9% 8. 27.8% 9. 30%, 23.1% C.P: 1. 12%
1. 720km 2. 337km 3. 4800 m 4. 360 km 930km
Ex 13.16
Ex 12.13 A. 1. Ghȼ 4,800.00 2. Ghȼ270.00 3. a. Ghȼ 3.22 b. Ghȼ.38
1. i. 8,000km ii. 3,500km iii. 10,300km iv. 2,800km2. a. 1 : 50 b. 4. Ghȼ1,800.005. Ghȼ440.80B. 1. Ghȼ 200.00 2. Ghȼ 50,090.91 3.
i. 3 cm ii. 5.6cm 3. 1,200km 4. L = 7200, B = 4500, A = Ghȼ 15,000.00 4. Ghȼ32, 579.19
32,400,0005. i. 1 : 2 ii. 25km 6. i. 220km ii. 7. 8.
Ex 13.17
Ex 12.14 1. Ghȼ13,750.00 2. Ghȼ1,100.00 3. Ghȼ27,000.00
1. 1 : 20,000 2. 1 : 30,000 3. 1 : 50,000 4. 1 : 85 5. 1 : 4,000,000 4. Ghȼ330,000.00 5. a. Ghȼ59,888.00 b. Ghȼ1,247.67
Ex 12.15

Baffour – Ba Series, Core Maths for Schools and Colleges Page 780
Ex 13.18
1. Ghȼ3,000.00 2. Ghȼ23,809.52 3. Ghȼ5,000.00 4. Ghȼ8,500.00 Ex 15.9
1. 1 + 2. 3. 4. 2 + 5. 2 + 3
Ex 13.19
1. 8% 2. 60% 3. 30% 4. 8% Ex 15.10
A. 1. 7 2. 0 3. 3 4. 5 5. 6. 7. 8. 2 9. 5B. 1. 2 2. 1 3.
Ex 13 .20
1. 4yrs 2. 10yrs 3. 2yrs 4. 40 yrs 5. Gh¢3,369.50 4. 4 5. 6. -2 7. 1 8. -1 9. C. 1. 0.67 2. 0.85 3. 0.51 4.
1.5
Ex 13.21
1. Gh¢3,200.00 2. a. Gh¢400x b. 1.12x c. 12% Ex 15.11
1. 2. 2 3. 2.712 4. 1 5. 0.3010
Ex 13.22
1. i. Gh¢ 23,000.00 ii. Gh¢ 3,000.00 iii. 68.58% 2. 6.15% 3. i. Ex 15.12
Gh¢17,200.00 ii. Gh¢3,200.00 iii. 156% 4. 37.5% 5. i.
Gh¢1,243.55 ii. Gh¢10, 878.30 iii. 23.62% 6. i. Gh¢1,475.00 ii. 1. 2 2. -2 3. 4. 1 5. 6. 3
Gh¢27,700.00 iii. 10.8% 7. 27.27% 8. 34.66% 9. i. Gh¢8093.25 ii.
Gh¢83,875.50 iii. 6.88% 10. i. Gh¢22,800.00 ii. 14% iii. 32% Ex 15.13
11. 25.85% A. 1. 1.465 2. 0.6309 3. 0.1981 4. 1.161 5. -2.585 6. 4.419
B. 1. 2.161 2. 7.122 3. 4.052 4. 4.954 5. 3.878 6. 4.358 7.
14. Modulo 1.757 8. 1.635

Ex 14.2B Ex 15.14
1. a. Feb b. Apr 2. a. June b. Nov 3. a. Sat b. Wed c. Wed d. A.1. 1.079 2. 0.5927 3. 1.255 4. 0.1761 5. 0.6532 6. 1.6019 7.
Mon 4. A. Mon b. Thu c. Sat d. Sun 5. a. 116 b. 562 2.68 8. -2.372B. 1. 1.113 2. 2.046 3. 1.293 4. 1.544 5. 0.341 6.
-0.251 7. -0.862 8. 2.862 9. 0.251
Ex 14.3
1. 1 pm 2. i. 9 days ii. 36 days iii. 27 days 3. i. wed ii. 7 wks (35 Ex 15.15
days) 4. i. mon, fri, tue, sat, wed, sun, thu ii. sat iii. sun iv. sat v. A. 1. 2 2. 4 3. -2 4. 1 B. 1. 2. 3. 4.
28 days5. sat 6. Wed
5. 6.
Ex 14.5
1. 2 2. 3 3. 0 4. 27 5. 8 6. 2 7. 31 8. 5 Ex 15.16
A. 1. 0 2. 3 3. 2.5 4. 5. 4 6. 8 7. 7 or 9 8. 6.5 or 2.4 9. 10 or -
Ex 14.7
A. 1. 1 2. 6 3. 10 4. 8 5. 8 6. 4B. 1. 5 2. 4 3. 13 4. 22 5. 1 B. 1. 1 2. 3 3. 13 4. 5. 0.24 or -4 25 C. 1. 10 2. 2
13 6. 1C. 1. 3 2. 24 3. 8 4. 48 5. 19 6. 9 or -2.5 3. 3 4. - 5. 3.69 or 0.81
Ex 14.8 Ex 15.19
1. -1.3923 2. 1.3923 3. 3.3923 4. -5.6021 5. 0.3586 6. 0.8597
1. 5, 10 2. 7, 21 3. 4, 44 4. 30 5. 11, 33 6. 45

Ex 14.9 Ex 15.20
1. 1, 3 2. 0 3. 2, 4 4. 2, 3 5. 3 6. 5 7. 2 8. 1, 3, 5, 7 1. 4.914 2. 1.251 3. 3167 4. 0.0001596 5. 0.02721 6. 0.7902

Ex 15.21
15. Indices and Logarithms
1. 3.31 2. 1.127 3. 0.002825
Ex15.5
Ex 15.22
A. 1. 3 2. 3. -3 4. 5. 1 6. 7. 8. 3B. 1. 4 2. 2 3. 4 4. 2 A. 1. 0.3865 2. 1.374 3. 25.41 4. 29016 5. 0.2802 6. 0.0107B. 1.
5. 6. 7. -1 8. 7 9. 10. C. 1. 2 2. -1 3. -4 4. -3 5. 3 6. -3 20.21 2. 3.075 3. 0.6559 4. 0.2724 5. 0.0357
7. 0 8. -3 C. 1. 1.447 2. 1961 3. 8.932 4. 932.8 5. 0.04152

Ex 15.6 16. Simultaneous Equations

Ex 16.7
A. 1. = 3 2. = 3. 4. =2 1. (15, 11) 2. (19, 11) 3. 30 and 12 4. ( -5, - 4) 5. P = 5p and R =
5. = 6. = -3B. 1 = 2. 2 = 3. 625 = 4. 15p 6. 92 people 7. 21, 50p and 33, 20p 8. 11. i. 400 ii. 200 9.
320 10. p = 6, q = 4 C.P.
= 5. 3 = 6. 0.1 × =
Ex 16.8
Ex 15.7 1. 35 2. 53 3. 86 4. 45
1. 2. 3. 4. 5.
17. Percentages 2
Ex 15.8 Ex 17.1
1. 2 + 3 2. 2 + 3. 2 + 3 + 4. 1 + A. 1. Gh¢1295.77 2. Gh¢2174.70 3. Gh¢26147.93
+ 5. + 6. + log 2 + B. 1. Gh¢112,394.24 2. I = Gh¢4,210.89, A = Gh¢6,710.89 3. Gh¢
821.35 4. Gh¢ 80,767

Baffour – Ba Series, Core Maths for Schools and Colleges Page 781
Ex 17.2 B. 1. i. 64 ii. 2. 6.75, y = 2x3 3. 31.5 4. y = 3x, 12 5. a. A =
A. 1. Gh¢ 443.94 2. Gh¢ 883.22 3. Gh¢ 18,316.80 2
44.1cm b. r = 3 6. x = 4, y = 2 7. i. 1.5 ii. 16 8. 55 9. 128 10.
B. 1. Gh¢ 40320.00 2. Gh¢ 228,131.25 3. Gh¢ 164,115.72 4.
Gh¢343,637.24 5.Gh¢595,444.90 7 11. 4 12 . s = √ 13. 216 14. 23.35liters 15. i. 15 16. 2.60

Ex 17.3
1. Gh¢ 29,864 2. Gh¢ 3762.00 3. Gh¢ Gh¢1,613.66 4. i. Ex 18.2
Gh¢1,504.03 ii. Gh¢1,575.97 5. Gh¢5,200.93 B. 1. i. 1.5 ii. 2. y = 3. 4. 5. i. y = ii. 6. 7.
7.5 8. 9. 3.6 10. 0.04
Ex 17.4
1. Gh¢1,734.00 2. Gh¢145,800.00 3. Gh¢ 851.84 4. Gh¢547.00
Ex 18.3
5. i. Gh¢1,500.00 ii. Gh¢5,742.19 6. Gh¢19,000, Gh¢17,100.00,
√
Gh¢14,022.00, Gh¢9,815.40 7. i. Gh¢12,609.92 ii. Gh¢954.6 iii. A. 1. i. Z = kx2 y ii. V = khr2 iii. R = B. 1. Z = , 2.
√
Gh¢13,564.52
= ii. 3. 913 4. 189 5. C = , 6. y = , 48 7. i.
Ex 17.5 1 ii. 4.9 8. 6, 3
1. i. D = ȼ31,500, G = ȼ10,500 ii. D = 75%, G = 25% 2. i. B =
Ghȼ10.5 million, W = Ghȼ4.5 million ii. 300% 3. S = ȼ3,876.92, D Ex 18.4
= Gh¢4,123.07 4. Gh¢2,660.00 5. i. 16,000.00 ii. Gh¢1,190.00 B. 1. i. T = 15 + 10r ii. a. 6 b. 20 2. 12 3. y = 4x - 4. a. k = 1
iii. 25% 6. Gh¢14,700.00 7. Gh¢3,300.008. 45 9. 8 months 10.
Gh¢7,500 and c = -1, y = 1 – x b. 1 5. i. c = 6500 + d ii. 7,000 6. Gh1,
090,000 7. S = 2000 + 15n 8. i. C = 6 + ii. 7.3 9. Gh195
Ex 17.6
10. i. C = 10000 + 100nw ii. Gh130,000 11. i. C = 8d 3 + ii.
1. Gh¢1,500.00 2. a. Gh¢ 10,975,610 b. 12.9%
GH512,000 iii., 42 miles 12. i. C = 830 + ii. Gh93013. i. V = t
Ex 17.7A 2
- t ii. – 21
1. Gh¢9,000 2. i. Gh¢385.00 ii. Gh¢2,585.003. i. Gh¢7,320.00 ii.
Gh¢43,920.004.i. Gh¢225.00 ii. Gh¢2,100.00
20. Probability
Ex 17.7B Ex 20.10
1. i. Gh¢355.50 ii. Gh¢44.50 2. i. Gh¢700.00 ii. Gh¢100 3. 1.i. ii. 2. I, Not ii. Ind 3. i. ii. iii. 4. 5. 6.
Gh¢435.55 4. Gh¢3,888.005. Gh¢9,900
7. i. ii. iii. 8. i. 0.6 ii. 0.05 iii. 53 9. 0.57 10. i. ii.
Ex 17.8
A. 1. Gh¢200.00 2. Gh¢115.00 3. Gh¢45.00 4. Gh¢25.00 5. Ex 20.11
Gh¢117.50 B. 1. Gh¢422.00 2. Gh¢2,321.00 3. Gh¢1,181.60 4. 1. 2. 3. i. ii . iii.
Gh¢2,532.00
Ex 20.12
Ex 17.9
1. i. Gh¢4,117.07 ii. Gh¢102.93 2. Gh¢720.00 3.Gh¢1333.76 4. i. 1. 2. 3. i. a. b. ii. a. b. 4. i. ii. 5 . i. ii.
Gh¢27,203.80 ii. Gh¢1,496.20 6. i. ii. iii. 7. i. ii. 8. 9. i. ii 10. i. ii.

Ex 17.10
1. Gh¢918.00 2. Gh¢2,283.00 3. Gh¢1,440.00 4. Gh¢945.00 21. Quadratic equation
5.Gh¢1,200.00 6. Gh¢1,300.00 7. 5% 8. 9% Ex 21.7
1. 2 or -3 2. 17 3. 7 or -12 4. b = 10 5. 13 or – 6 6. 2 or 9 7.
Ex 17.11 16yrs C.P 1. 46 or 57 2. 60km/h 3. 5
A.Gh¢75.50, Gh¢150.00, Gh¢1504.00, Gh¢5940.00B. 1. Gh¢
6,000.002. i. Gh¢ 1,200.00 ii. Gh¢1,020.003.i. Gh¢4,700.00 ii.
Gh¢658.00 iii. Gh¢4,342.004. i. Gh¢540.00 ii. Gh¢1,188.00iii. Ex 21.9
Gh¢3,312.005. 15% 6. 6. i. 15,700.00 ii. 1,000.00iii. 17,000.00 7. a. 1.c. i. x = 1 ii. x = - 0.7 or x = 2.7 iii. - 0.7 <x< 2.7 2. b. i. x = -1 ii. 2
9000.00 b. 122.92 c. 12.3% 8. a. 6752.00 b. 30148.00 iii. -5 x< -1 iv. -1 x< 3 3. b. i. x = -1 ii. y = - 4 iii. -5 x< -1
iv. -1 x< 3 4.b. ii. 1.2<x 5 iii. - 4 <x 1.2iv. -4 < x< 1.2 v. - 4
Ex 17.12 >x> 1.2 5. c. i. x = -1 or x = 3 ii. y = 4 iii. x = 1 iv. 1 <x 5
1. 13,750.00 2. 4,000.00 3. 9,000.00 4. 10,080.00
22. Mensuration 1
Ex 17.13 Ex 22.1
1. Gh¢ 71,000.00 2. i. Gh¢8082.50 ii. Gh¢710.00, iii. 900 units 3.i. 1. i. 2.10 ii. 3.14 iii. 8.10 2. i. 18.84 ii. 9.42 iii. 17.42 3. i.
Gh¢236.00 ii. 560 units 4. i. 34,400.00 ii. 54,800 liters C.P. 265.48 23.9cm ii. 33.9cm2 4. i. 32.97cm ii. 230.79cm2 iii. 60.97cm 5.
35.5
18. Variations
Ex 18.1 Ex 22.2
1. 110 2. 940 3. 1580

Baffour – Ba Series, Core Maths for Schools and Colleges Page 782
Ex 22.3 1. i. < BAC, < BEC, < BDC ii. < DBE, < DCE iii. < ABD, < ACD
1. 2. 2200 3. 760 4. 260 5. i. 380 ii. 1070 7. i. 600, 300 ii. 3000, 1500
Time 1:00 2:30 7:00 8:45 10:30 11:20 8. i. 80 ii. (180 – 2x)0 C.P. 1. <YOZ = 1200, < OYZ = < OZY = 300
Angle 300 1050 2100 262.50 3150 3400 2. < P = 650, < Q = 600, < R = 550 3. ∆ AOB is equilateral

2. 8.18cm2, 3.27cm Ex 23.4

A. 1. x = 700, y = 250 2. x = 450, y = 450, z = 550 3. x = 300, y =300, z =
Ex 22.4 1200 B. 1. i. <ACB = 900, < ABC = 450 2. 250 3. 800 C. P. i. 12cm
1. 0.03 2. 220.87 3. i. 14 ii. 29.3 iii. 44.3 4. i. 7.3 ii. 21.3 iii. ii. 192 cm2 iii. 192 cm2
4.45 5. 14.1
Ex 23.5
Ex 22.5 2. i. all 400 ii. all 360 iii. all x0 iv. all equal 3. 1200 4. i. 550 ii. 250
A. 1. 108 2. 64 3. 76 4. 560 5. 84 6. 160 B. 1. 26 2. 42 3. 5. x = 400, y = 600, p = q = 200 8. <XYZ = <YXZ = < ZAB = 350, <
18 4. 37 5. 71 6. 25C. 1. 100 2. 49 3. 484 4. 1225 5. 81 6. XZY = < AZB = 1100 C.P. 1. i. they are equiangular ii. DE = 5 cm,
121 DC = 6 cm 2. 5 cm 3 …. 4. <A = 400, < B = 600, < C = 800

Ex 22.6 Ex 23.6
1. 37 2. 36 3. 72 4. 52 1. i. 2800 ii. 400 iii. 1400 iv. 1800 2. m = 500, n = 1300 3. i. 450 ii.
22.50 iii. 112.50 4. x = 800, y = 600 5. a = 1150, b = 650 6. x = 3607. i.
Ex 22.7 500 ii. 1300 C.P 1. 580 ii. 410 2. i. x = y = z ii. 1210 3. 8504. < ABE
1. 351.68 2. 44 3. 176 4. 262 5. 66 6. 62.86 = < EFC=1050, <DEB= < BCF = 800, < EBC = 750, < BEF = 1000

Ex 22.7B Ex 23.7
1. 24cm2 2. 96cm2 3. 70cm2 4. 60cm2 1. 700 2. < P = 750, < R = 1050, < PQR = 1000, < PSR = 800
Ex 22.8
1. 1930.5 2. 225 3. 1287 4. 704 CP; 1. 123 2. 22 3. 140 Ex 23.8
i. 630 ii. 270 iii. 820 iv. 900 2. r = 630, q = 880, p = 290 3. 850
Ex 22.9
1. 140 2. 9 3. 15 4. 18 5. 30 Review ex
1. 1300, 500 2. x = 440 3. y = 480 4. y = 250 5. a = 250, b = 550 and c
Ex 22.10 = 620 6. x = 25.70 y = 115.70 7. 400, 390
A. 1. 25 2. 144 3. 256 4. 625 5. 529 6. 1681
B. 1. 9 2. 7 3. 15 4. 13 5. 20 6. 11 24. Trigonometry 1
Ex 24.2
Ex 22.11 √ √ √ √
1. 2. - 3. 4. 5. 6. -
A.1. 40 2. 391 3. 27B. 1. 28 2. 1200 3. 240 4. 418
Ex 24.3
Ex 22.12 – √ √ √
√
A. 1. 504 2. 1288 3. 220 4. 289.71 B. 1. 30.86 2. 150 3. 506 4. A. 1. 2. 3. 4. - B. 1. 2. 3. 4 + 2√ 4. 20 – 10
99C. 1. 586 2. 214 √
√ 5. 2 - √ 6. 2 + √ C. 1. 2. 3 3. 2 4. 5. 1 + √
Ex 22.13
1. 5300 2. 51350 3. 11414 4. 7982 5. 7391 Ex 24.4
1. 1200 2. 53.130 3. – 490 4. 36.870
23. Plane Geometry I
Ex 23.1 Ex 24.5
1. a = 500, b =900 c = 400 2. <OCP= <OCQ= 900, <AOB=140, A. 1. 120 2. –240 3. 680 4. 370 5. 210 6. 410 7. 900 8. 810
<OAB = <OBA = 20, <OAC = <OCA = 40, <OCB = <OBC = 30, B. 1. 750 2. 300 3. 00,1800, 300 1500 4. 600, 900
<ACP = 50, <BCQ = 60 3. DG = 10, EF = 2.5 4. 5cm
Ex 24.6
Ex 23.2 1. 140 2. 150 3. 180 4. 450
1. i. ORT, OST ii. TR = TS, OR = OS iii. < ORT = < OST , < ROT =
< SOT, < RTO = < STO iv. < ORT , <OST v. 7.4 2. CBD, ADB,; < Ex 24.7
CBD = < CDB = 700, < ABD = < ADB = 200, < BAD = 1400 3. i. √ √ √ √ √ √ √ √ √ √
1. 2. 3. 4. 5. 6.
OQ: P↔R ii. < SRQ = < OPS = < ORS = 200; four right angles at S
iii. 6.7cm 4. i. 15 ii. 28
Ex 24.8
CP: 1. i. AY = AZ, BZ = BX, CX = CY ii. AB = 7cm, BC = 9cm, A. 1. 370 2. 530 3. 240 4. 340B. 1. 380, 2. 510 3. 310 4. 4905.a.
CA = 8cm iii. AB = 14cm, AC = 15cm iv. 15cm v. 18cm 2. ii. BC 450b. 6cm c. 1
= 7cm, CA = 8cm, AB = 9cm iii. 5cm, 3cm, 3cm 3. i. AF = AE, BD =
BF, CE = CD ii. 10cm iii. 5 cm 4. i. < PTQ = 540, < POQ = 1260 Ex 24.9
ii. < PTQ = (180 – 2x)0, <POQ = 2x 5. 12cm A. 1. 22cm, 27 cm 2. a. 7m b. 4m 3. i. 9.6m ii. 7.2 m
B. 1. 13 2. 670 3. P = 11, m = 380 4. r = 17, u = 600 5. t = 22m, q =
Ex 23.3 23m

Baffour – Ba Series, Core Maths for Schools and Colleges Page 783
Ex 24.10 6. 75,712 7. i. Gh25,000 ii. Gh164,500 8. 1085 9. Gh7,000
A. 1. cos θ = , tan θ = 2. cos θ = , tan θ = 3. i. sin θ = , 10. 8. i. Gh¢10,740.00 ii. Gh¢91,20011. b
cos θ = ii. 36 iii. 900, 600, 300 B. 1. 0.51 2. 3. 4. 5.i. 0.85 27. Mensuration 2
0 Ex 27.1
ii.1.15 iii 1.92 iv. 11.56 6. i. 41 ii. iii. iv. 0.75 7. 0.28
A. 1. 972 2. 542 2 3. 358 4. 2628 B.1 920 2. 132 3. 34
8. 7.5
4. 606
Ex 24.11
A. 1. R = 660, q = 13.92 r = 12.79 2. 9.17cm3. 61.5 04. 79.80, 52.60, Ex 27.2
47.60, 135cm25.i. 38.2cm2 ii. 10.7cm iii. 58.30, 48.80B. 1. 7.5 sq. 1. 54 2. 294 3. 216 4. 600 5. 150 6. 384
units 2. 3 sq. units 3. 15 sq. units 4. 10 sq. unitsC.1. 1. 28.960,
46.570, 104.480 2. 27 sq. units 3. 4 4. 8.8cm 5. i. 4.77m ii. 8.34m2 Ex 27.3
A. 93.5 2. 1171.5 3. 1842.5 4. 1276 5. 96.7 6. 487.08
Ex 24.12 B. 1. 66 2. 528 3. 113 4. 3.5 5. 7
1 . 15.1m 2. 530 3. 18.43m 4. 71m 5. 69m6. 12,222 cm
Ex 27.4
1. 304 2. 63 3. 160 4. 9600 5. 925 6. 2075
Ex 24.13
1. i. 15km, ii. 3 km 2. 23m 3. 68.7m,50m 4. 1011m
5. a. 71m, 64m b. 26m 6. i. 21m, ii. 170 iii. 270 Ex 27.5
A. 1. 2145 2. 209 3. 5808 4. 13, 420 5. 1, 012
Revision Ex
1. B : 25.9km, 96.6 km, C : 217.1km, 38.2km B. 1. 528 2. 1980 3. 150.72 C. 1. 2 2. 5 3. 4.8
2. 6.4km3. a. 1130 b. 36.69 km4. a. 0210b. 14km 5. a. 4.9 km b. i.
4.1km iii. 1.5km c. (4.4km, 0200) Ex 27.6
1. 616 2. 314.3 3. 1018.3 4. 3850 5. 1386 6. 2464
Ex 24.14
Ex 27.8
1. a. 4.9m 2. 7m b. 620 3. i. 290ii. 2.2m 4. i.4m ii. 8m.
A. 1. 108cm3 2. 1584cm3 3. 840m3 4. 160cm3
5. 18000cm3B. 1. 32cm 2. 5.5cm 3. 3cm 4. 19cm
Ex 24.15
1.a. 79m, b. 4740m 2. 7cm 3. 230 5. i. 50m ii. 100m
Ex 27.9
A. 1. 1520.88 2. 1728 3. 8000 B. 1. 7 2. 14 3. 11
25. Series and Sequence C. 1. 4 2. 3 3. 2744
Ex 25.1
Ex 27.10
Ex 25.5 1) 49896cm3 2) 249480cm3 3) 15400cm34) 3.5cm 5) 14cm
1. i. a = 1, d = 2, 1 + 2(n – 1) ii. a = 0 , d = 4, 4(n – 1)
2. i. a = 14, d = 12 ii. Un = 14 + 12(n – 1) 3. 99 4. 126 5. a = 30, d Ex 27.11
= -2, U10 = 12 1. 125cm3 2. 213.33cm3 3. a. 300cm2b. 900cm3 4. a. 15 b. 633

Ex 25.6 Ex 27. 12
1. 2550 2. -300 3. 15150 4. 18 5. 910 6. 98 7. 125 8. 6, 1, - 4, 1. 36cm 2. 21cm 3. 468.91cm3 4. 366.33
- 9, - 14 9. a = 3, d = 4; 3, 7, 11, 15, 19 10. 8 11. 3.5, 2 12. i. a =
9, d = 3 ii. 351 Ex 27. 12B

Ex 25.7 1. 3.84cm 2. 3cm 3. 489.54cm3 4. 2.1cm 5. 126 6. 353.57cm2

A. 1. r = 3, 708,588 2. r = , 1,108 3. r = 4, 65, 536 4. - , - 5.
Ex 27.13
7B.1. Un = 3(2)n-1 2. 128 3. Un = 192( )n – 1 4. 327,680
1. 949 2. 1037, 1804, 5338.7 3. 146.67 cm34. i. 1006 ii. 1659 iii.
4777
Ex 25.7B
1. 7. 2. 8 3. 13 Ex 27.14
A. 1. 268cm3 2. 524cm3 3. 905cm3 4. 1437cm3 5. 2146cm3 6.
Ex 25.8 4190cm3B. 1. 2.04cm 2. 2.93cm 3. 4.57cm 4. 2.77cm 5. 8cm
1. 243 2. - 3. 4. 5. - 6. 5.3cm

Ex 25.9 Ex 27.15
1. 858 cm22. 266 cm3 3. 572 cm24. 220 cm2
1. i. a = 1, r = 2 ii. 32 2. 3. -3, -2
Ex 27.16
Ex 25.10 1. i. 6.9 ii. 373.86 iii. 207 2. 300 3. a. 12 b. 96 4. 72
1. – 349,525 2. 18,662 3. 4. a = , r = 2, S = 157.5 Ex 27.17
1. 32,153.60cm32. 761.14cm3 3. 1330 cm3 4. 188.57 5. V = 1.713
Ex 25.11 cm3 A = 26.47 cm26. i. 44m2 ii. Gh¢22,000.00
1. 2160 2. 7.8 3. 1,024,000 4. Gh 3,360 5. Gh16,384

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Baffour – Ba Series, Core Maths for Schools and Colleges Page 785
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Baffour – Ba Series, Core Maths for Schools and Colleges Page 787
Baffour – Ba Series, Core Maths for Schools and Colleges Page 788
REFERENCES
George Humphrey, New Concise Mathematics 4, Mathematics Composition Setters Ltd. Salisbury , Wiltshire,
2003

V. W. Ferris and J. N. Busbridge, Modern Mathematics for Secondary Schools, Book 3, 4 & 5, Evans Brothers
Limited, Ibadan, 1978

Scottish Mathematics Group, Mordern Mathematic for Schools, Second Edition, Butler and Tanner Limited,
1974

New Mathematics for Secondary Schools, West African Editon 4, Afram Publications, Accra, 1981

P. J. F Horril, Pure Mathematic 1 & 2, Longman Group Limited, Essex, 2008

GRE Premier, Kaplan Publishing, New York, 2015

Teacher Education Division, Further Algebra, Ministry of Education Youth and Sports, Accra, 2007

Joint School Project Mathematics, Metric Edition 4S and 5S, Longman Group Limited, 1980

Ghana Senior Secondary School Mathematics, Books 1, 2 and 3, Ministry of Education, Accra,1993

Mathematical Association of Ghana, Mathematics for Senior Secondary Schools Books 1, 2 and 3, Addison
Wesley Longman Limited, 1997

Revise for GSCE Mathematics, Intermediate Tier, University Press, Cambridge 1997

James K. Morton, Richard J. Swinney, GCSE Mathematics, Intermediate Level Key Stage 4

Baffour – Ba Series, Core Maths for Schools and Colleges Page 789