CHAPTER ONE

1.1 Types of Numbers

Natural numbers (N): This is also referred to as the set of counting numbers. That is
N = 1, 2 , 3 , . . .  .

Whole numbers: The set W = 0,1, 2 , . . .  is called the set of whole numbers. This is just the
natural numbers including zero.

Integers (Z): Any number that is positive or negative without a fraction.

Z = (. . . , − 3 , − 2 , −1, 0 ,1, 2 ,3 , . . . ) . The set of positive integers is Z + = ( 0 ,1, 2 ,3 , . . . ) .
The set of positive integers is Z − = (. . . , − 3 , − 2 , −1 ) .

a
Rational numbers Q: Numbers of the form , b  0 . The set of rational numbers includes the
b
integers. Examples of rational numbers are −3 , 54 , 6 , 5 , − 7 etc.

Any fraction whose decimal representation terminates or follows a regular pattern is also called
rational number examples (1) 0.3333 … (2) 0.242424… (3) 0.25

If the decimal representation does not terminate and no do not follow any regular pattern, we call
it an Irrational Number ( Q ) .
Examples: 3 , 3 5 , 6 7 etc.

Real Numbers (R): The set of rational numbers together with the irrational numbers is called
real numbers.

Even Numbers: Numbers that are divisible by 2.

i.e. −6 , − 4 , − 2 , 0 , 2 , 4 , 6 , 8 , . . .  .

Odd Numbers: Numbers that are not divisible by 2 are called odd numbers

−7 , − 5 , − 3 , −1, 1, 3 , 5 , 7 , 9 , ...

Numbers with only two factors are called prime numbers. The only factors are 1 and the number
itself.
Examples 2 , 3 , 5 , 7 , 11 , 13 etc.

1
1.2 Some Basic Concepts in Algebra
Least Common Multiple (L.C.M) and Highest Common Factor (H.C.F)

When we say multiples of 2, we just mean 2 times.

Multiples of 2: 2 , 4 , 6 , 8 , 10 , . . .

Multiples of 5: 5 , 10 , 15 , 20 , 25 , . . .

The Common Multiples of 2 and 5 are

10, 20, 30, 40, …

The smallest of this is 10. We say that the L. C. M between 2 and 5 is 10.

Exercise 1.1.
Find the L.C.M of the following.
(a) 3 and 5 (b) 4 , 5 and 6 (c) 7 and 8 (d) 7 , 8 and 10

1.2.1 Factors
When we say factors of K, we mean all the numbers that can divide K without a remainder.

Example
Find the factors of
(i) 20 (ii) 15 (iii) 12 (iv) 60

Solution
(i) 20: Find any two numbers whose product will give you 20
1  20
2  10
45
Hence factors of 20 = 1, 2 , 4 , 5 , 10 , 20
(ii) Factors of 15 = 1, 3 , 5 , 15
(iii) Factors of 12 = 1, 2 , 3 , 4 , 6 , 12

Now HCF of 20 and 15 = 5

HCF of 12 and 60 = 12

A technique that can be employed to find L.C.M between two numbers is as follows:
L.C.M between a and b is given by
ab
L.C.M =
HCF ( a , b )

2
 12  60
Examples L.C.M of 12 and 60 =
HCF (12, 60 )
12  60
=
12
= 60

20  15
L.C.M of 20 and 15 = = 60
5

Addition of two Numbers

(a) To add two numbers with the same sign, find the sum of the two numbers and give the
answer the same sign.

Example 1
(a) + 7 + (+6) = +13 = 13 (b) + 5 + (+13) = +18 = 18
(c) -5 + (−8) = −13 (d) − 4 + (−6) = −10
Also a − (−b) = a + b

Example 2
(a) 3 − (−5) = 3 + 5 = 8 (b) 27 − (−15) = 27 + 15 = 42

(b) To add two numbers with different signs, find the difference between the numbers and
then give the answer the sign of the larger number.

Example 3
(a) 12 + (−5) = 12 − 5 = 7 (b) − 12 + 5 = −7
(c) − 13 + 25 = 12 (d) − 25 + 13 = −12
(e) − 16 + 9 = −7 (f) 17 + (−8) = 9

We can combine (a) and (b) , so let’s take some examples.

Example 4
(a) − 15 − (−9) + 12 = −15 + 9 + 12 = −15 + 21 = 6
(b) -12- ( -6 ) + (−8) = −12 + 6 − 8 = −20 + 6 = −14
(c) − 14 − (−2) + (−10) = −14 + 2 − 10 = −24 + 2 = −22
(d) − 17 − 6 + (−9) = −32

3
 Multiplication and Division of Numbers
(a) When two directed numbers with the same sign are multiplied or divided, the answer is
positive.

Example 5
(a) 7  5 = 35 (b) − 5  −6 = 30
(c) 24  4 = 6 (d) − 20  −5 = 4

(b) When two numbers with different signs are multiplied or divided together, the answer is
negative.

Example 6
(a) − 8  (+5) = −40 (b) 7  (−4) = −28
(c) − 45  (+9) = −5 ( d) 20  ( − 5) = −4

1.3.0 Types of Fractions.

Proper fraction: This is a fraction whose numerator is less than the denominator
2 3 5
E.g. (i) (ii) (iii) etc.
5 7 7
Improper fraction: This is a fraction whose numerator is greater than the denominator.
7 3 8
Example: (i) , , etc.
5 2 5
Mixed fractions: They are numbers that are a composition of a whole number part and a
fractional part. e.g. 2 53 , 6 14 , 5 72

1.3.1 Addition and Subtraction of Fractional Numbers

In this we may have to find the least common multiple before performing the addition.

Examples 1
Work these out.
a c ad + bc
(1) + =
b d bd

3 2 3 5 + 4  2 15 + 8 23
(2) + = = =
4 5 20 20 20

3 4 21 + 20 41 6
(3) + = = = 1
5 7 35 35 35

4
Exercise 1.3.2 Addition and Subtraction

Find the following:

2 1 3 3 1 5 2 16 1
(a) + − (b) − + (c) + +
5 4 7 4 2 8 7 5 4

4 7 1 17 1 6 1 1 4 5
(d) + − (e) + − + (f) − +
3 5 2 5 4 5 3 8 3 7

5 1 7 2 6 7 4 2 15
(g) + − (h) − − (i) + −
3 4 16 3 5 10 7 3 8

4  −5  3
(j) −  −
3  2  7

1.3.3 Mixed Fractions.

For a mixed fraction, we may have to convert it to improper fraction before we carry out the
workings.

Example 1

Convert the following to improper fractions

(i) 4 18 (ii) 3 34 (iii) 3 53 (iv) 6 56

Solution

4  8 + 1 32 + 1 33 3  4 + 3 12 + 3 15
(i) 4 18 = = = (ii) 3 34 = = =
8 8 8 4 4 4

3  5 + 3 15 + 3 18 6  6 + 5 36 + 5 41
(iii) 3 53 = = = (iv) 6 56 = = =
5 5 5 6 6 6

Example 2

Simplify the following sums:

(i) 4 23 + 6 14 − 1 32 (ii) 4 18 + 3 43 + 5 12

(iii) 6 12 − 3 23 + 6 34 (iv) 4 12 − 3 13 + 8 23

5
Solution

(1) We shall use two methods:

Methods I

2 1 2
4 + 6 −1
3 4 3

Add the whole number first: 4 + 6 – 1 = 9

Now add the fractions:

8+3−8 1
= .
12 4

1 1
Hence the answer is 9 + =9
4 4

Method II

Convert each to improper fraction before finding the LCM.

14 25 5
4 23 + 6 14 − 1 23 = + −
3 4 3

56 + 75 − 20 111 1
= = =9
12 12 4

For the rest of the examples we shall use the first method.

(ii) 4 18 + 3 43 + 5 12

4 + 3 + 5 = 12

Fraction:

1 3 1
+ +
8 4 2

1 + 6 + 4 11 3 3 3
= =1 Hence answer is 12 + 1 = 13
8 8 8 8 8

(iii) 6 12 − 3 23 + 6 34 Whole number part: 6 − 3 + 6 = 9

6
Fraction

1 2 3
− +
2 3 4

6−8+9 7 7 7
= ,Answer is 9 + =9
12 2 12 12

3 −2 +4 5
(iv) 4 12 − 3 13 + 8 23 , Solution , 4−7 + 8 = 9 , =
6 6

5 5
Ans: 9 + =9
6 6

Exercise 1. 3. 4

Simplify the following fractions:

1 6 2 1 1 1 8 1 3
(a) 2 −5 +8 (b) 17 − 16 + 8 (c) 17 − 15 + 19
4 7 3 2 3 4 7 2 4

1 3 3 3 4 1 2 3 1
(d) 2 −8 +6 (e) 6 + − 18 (f) 16 − 17 + 17
4 7 5 8 5 7 3 4 5

2 1 6 1 1  7 1  1 1
(g) 4 − 17 + 8 (h) 6 − 3 −  −3  (i) 6 −  −2  + 3
3 2 7 4 4  8 5  5 8

1 1 1 1  1  1 1 1 1
(j) 15 − 16 + 18 (k) 6 −  −3  −  7  (l) 3 +6 −8
2 3 4 5  4  4 4 4 2

1 2 3 1 3 6 3  1 3
(m) 18 + 3 + 16 (n) 1 −8 +3 (o) 15 −  −7  + 6
4 5 4 2 4 7 4  3 5

1 1 3
(p) 2 + 3 + 17
5 2 16

7
1.3.5 Multiplication and Division of Fractions

a c ac ac
 = =
b d b  d bd

Find
7
3 5 3  5 15 7 21 49
(i)  = = (ii)  =
4 7 4  7 28 39 13 5 65

Division

a c a d ad
 =  =
b d b c bc

Example

Find:

4 2 5 6 1 1
(1) 4  10 (2) 6 5 (3) 13  18
7 3 6 7 3 2

1 3 3 1 7 89
(4) 5 3 (5) 5  15 (6) 7 1
15 5 4 3 15 135

1 1 1 2 3 1
(7) 1 1 (8) 2  (9) 4 9
2 4 9 9 5 5

2 2
(10) 7 4
5 15

Solution

We shall solve few and the rest will be given as exercise.

4 2 32 32 32 3 3
(1) 4  10 =  =  =
7 3 7 3 7 32 7

5 6 41 41 41 7 7 1
(2) 6 5 =  =  = =1
6 7 6 7 6 41 6 6

1 1 40 37 40 2 80
(3) 13  18 =  =  =
3 2 3 2 3 37 111

8
1.3.6 Further Exercises on fractions

Simplify the following:

3 2 2 2 2  1 1  3
(1) 4 − 7 + 11 − 6 (2) 5 −  −6  + 7 −  −12 
4 3 5 3 3  4 2  4

−5 1 3 2  1 3  1 3
(3) + 8 − 6 + 16 (4)  4  3  + 5  3 
4 5 7 15  8 4   15 5

1 1 5 6 1 1
5 + 11 6 5 5 −2
(5) 3 6 (6) 6 7 (7) 3 2
1 1 2 1 5
13  18 3  12
3 2 3 2 12

1  1 1 2 1 5 1  1 1
(8) 7 −  −12  + 11 (9) 4 + 11 − 2 (10) 3  3  2 
2  3 2 9 3 6 4  8 2

2 1 1 1  1 2 1 2
4 +6 −3 4 +  −3  − 3 1 
3 2 4 2  4 3 3 3
(11) (12) (13)
1 3 1 2 1 1
8 −6 7 +5 1
2 4 2 3 4 2

3  3 5
3 + 1 
4  8 6
(14)
1 1
2 −1
8 2

9
1.4.0 Approximations

In approximations, we shall consider

(1) decimal approximations (2) Significant figures and (3) whole number approximations.

1.4.1 Whole number approximations.

The numbers 0, 1, 2, 3 and 4 are nearer to 0 than 10 while 5, 6, 7, 8 and 9 are nearer to 10 than 0.
So if we want to write numbers as only multiples of 10, then we say to the nearest 10.

Example

Approximate these to the nearest 10

(a) 11 = 10 (b) 35 = 40 (c) 47 = 50 (d) 409 = 410

(e) 234 = 230 (f) 196 = 200

Note that the final answer should be divisible by 10.

Exercise 1. 4. 2.

Approximate the following to the nearest:

(a) 10 (b) 100 (c) 1000

(1) 140 (2) 1573 (3) 1484 (4) 178367

(5) 15 (6) 1473

1.4.3 Decimal Approximation

Rule

If we have zero after the decimal point and is not followed by a non zero number, then the zero
in not counted

Example

(i) 21.1000 (1dp) (ii) 22.10100 (3pd)

(iii) 0.00005 (5dp) (iv) 15.1236 (4dp)

To approximate a figure to about 2 decimal places (say), we consider the third decimal number.
If it is less than 5, we don’t add any number to the 2nd decimal number.
10
Example 1

Approximate the following to 2 d.p.

(i) 0.2436 (ii) 24.8134 (iii) 0.67453

Solution

(i) 0.2436 = 0.24 (2d.p) (ii) 24.8134 = 24.81 (2d.p)

(iii) 0.67453 = 0.67 (2d.p)

On the other hand if the digit is 5 or more, then we add one to the number before it.

Example 2

Approximate the following to 2 d.p.

(i) 0.2456 (ii) 24.8164 (iii) 0.67584

Solution

(i) 0.2456 = 0.25 (2d.p.) (ii) 24.8164 = 24.82 (2d.p.)

(iii) 0.67584 = 0.68 (2d.p.)

1.4.4 Significant Figures

In significant figures, we ignore decimal points

Example

21.53 (4s.f.)

178.45 (5s.f.)

The only digit that is a trouble shooting point is zero.

Zeros are significant if and only if there is a non zero number before and after it.

11
Example

(1) 21530 – zero is not significant but we can’t remove it.

(2) 0213 – zero is not significant and can be removed. Because when we remove it we still
have 213.

(3) 2018 – zero is significant.

To approximate to 3 significant figures (say) we just consider the fourth significant figure if it is
5 or more, then add one to the number just before it. If it is less than 5, we do not add any
number to the preceding digit.

Example

(1) Approximate 21598 to (a) 1sf (b) 2sf (c) 3sf

(a) For one significant figure, the second digit is 1 which is less than 5. Hence we just
replace the 1 and all other digits after it by 0.

(a) 21598 = 20000 (1sf) (b) 21598 = 22000 (2sf)

(c) 21598 = 21600(3sf)

Example

(i) (a) 1 7843 = 10000 (1sf) (b) 17843 = 18000 (2sf)

(c) 17843 = 17800 (3sf) (d) 17843 = 17840 (4sf)

Exercise 1.4.5

Round each of the following to (a) 2 decimal places (b) 3 decimal places (c) 2 significant figures
(d) 3 significant figures

(1) 15.1473 (2) 184.217843 (3) 16.14154 (4) 17.1734

(5) 16.98988 (6) 0.0098398 (7) 15.9999 (8) 17.19998

(9) 124.9897 (10) 17.696969 (11) 154.348 (12) 0.183  9.1734

153.142
(13) (14) 16.348 15.379
3.143

12
1.4.6 Tenth, Hundredth and Thousandth

10th means one decimal place

100th means two decimal places

1000th means three decimal places

Example 1

Round 0.6394 to the nearest

(a) 10th (b) 100th (c) 1000th

Solution

(a) 0.6394 = 0.6 (1d.p. or nearest 10th ) since the second decimal number (3) is less than 5

(b) 0.6394 = 0.64 (2d.p. or nearest 100th )

(c) 0.6394 = 0.639 (3d.p. or 1000th)

Example 2

Convert 0.6989 to nearest

(a) 10th (b) 100th (c) 1000th

Solution

(a) 0.6989 = 0.7 (10th) (b) 0.6989 = 0.70 (100th)

When we add one to the first 9 it becomes 10 and we have to add one to the 6 and obtain the 7

Example 3

Convert 19.9898 to nearest (a) 10th (b) 100th (c) 1000th

(a) 19.9898 = 20.0 (1d.p. or 10th) (b) 19.9898 = 19.99 (2d.p. or 100th)

(c) 19.9898 = 19.990 (3d.p. or 1000th)

13
Exercise 1.4.7

Round the following numbers to the nearest (a) 10th (b) 100th (c) 1000th

(1) 29.9999 (2) 0.00998 (3) 25.98999 (4) 9.99989

(5) 127.9888 (6) 0.2989 (7) 189.6678 (8) 19.8899

(9) 143.0059 (10) 0.01098

1.4.8 Standard Form

There are two types of numbers in standard form. For the first type the power is positive and the
second one, the power is negative.

A number is said to be in standard form if it has one whole number followed by the decimal
point and unspecified number of decimal places and then times 10 n where n can be positive or
negative integer.

Example 1

The following numbers are all in standard form.

(a) 2.5  10 3 (b) 1.378  10 −5 (c) 2.84  10 6

(d) 1.0  10 7 (e) 5.143  10 3 (f) 1.485  101

The following numbers are not in standard form

(i) 15.23  10 4 (ii) 123.1  10 3 (iii) 0.69  10 5

Reasons

(1) the first one has two whole numbers;

(2) the second one has three whole numbers;

(3) the 3rd one has no whole number.

14
1.4.9 Expressing numbers in standard form

Write 1586.2 in standard from.

Solution

The first thing to remember is that we need one whole number first. So, for step one, we have
1586.2 = 1.5862  10 n

Step II. Determine the value of n.

The decimal point has moved to 3 place to the left and hence n = 3

Whenever we move the decimal places to the left, the power (n ) is positive

:. 1586.2 = 1.5862  10 3

Example 1

Write the following numbers in standard form

(1) 15634 (2) 15.34 (3) 1000,000 (4) 1862.34

(5) 167.2

Solution

(1) 15634 = 1.5634  10 4

Reason

15634 is a whole number and hence the decimal point is after the 4.

ie. 15634 = 15634.0 move it four places to left.

(2) 15.34 = 1.534  101 (3) 1000000 = 1.0 106 (4) 1862.34 = 1.86234  10 3

(5) 167.2 = 1.672  10 2

Steps:

(1) Identify where the decimal point is located;

15
 (2) Remember you need only one whole number;

(3) Decide whether you need to move the decimal point to the left or right;

(4) Count the number of places that the decimal point has moved.

Example 2

Write the following numbers in standard form

(1) 1700 (2) 184.3 (3) 25.3 (4) 1500.8

(5) 600,000

Solution (1) 1700 = 1.7 103 (2) 184.3 = 1.843 10 2 (3) 25.3 = 2.53  101

(4) 1500.8 = 1.5008  10 3 (5) 600,000 = 6.0  10 5

Exercise

Write the following numbers in standard form

(1) 1,5000,000 (2) 1764 (3) 348.24 (4) 16.38 (5) 25,000,000

(6) 13458 (7)16343.2 (8) 3,000,000 (9) 17.783 (10) 163984

1.5.0 Negative Exponents

We obtain a negative exponent if we move the decimal point to the right

Example1

(1) 0.00625 (2) 0.014 (3) 0.00007 (4) 0.00005

(5) 0.05

Solution

(1) 0.00625 = 6.25  10 −3 (2) 0.014 = 1.4  10 −2

16
(3) 0.00007 = 7.0 10 −5 (4) 0.00005 = 5.0  10 −5 5) 0.05 = 5.0  10−2

Example 2

Write the following in standard form

(1) 0.125 (2) 0.000893 (3) 0.015 (4) 0.0000843

Solution

(1) 0.125 = 1.25  10 −1 (2) 0.000893 = 8.93  10 −4 (3) 0.015 = 1.5  10 −2

(4) 0.0000843 = 8.43  10 −5

Exercise 1.5.1 Standard Form

Write the following in standard form

(1) 0.0034 (2) 0.00000078 (3) 0.0000345 (4) 0.008104

(5) 0.000078434 (6) 0.000134 (7) 0.0000505 (8) 0.0008

(9) 0.0314 (10) 0..00125 (11) 0.0000847 (12) 0.000000078

(13) 0.00125 (14) 0.000184 (15) 0.0000115

Exercise 1.2.1
Convert the following to the number of decimal places indicated.
(1) 21.17874 (3dp) (2) 0.9999 (1dp)
(3) 15.0998 (2dp) (4) 0.2994  15.174 (2dp)
2.1947 21.74  0.0087
(5) (2dp) (6) (4dp)
15.1478 14.83
22.83
(7) (3dp) (8) 12.14 134.007 (2dp)
15.2 16.7
14.21 22.84 10.98 15.74
(9) (2dp) (10) (1dp)
7.84  3.213 0.007
17
1.5.0 ALGEBRAIC EXPRESSIONS

1.5.1 Addition and Subtraction

Additions and subtractions are possible only if the variables are the same.

Example 1

(1) 3x + 4x = 7 x

We just add the numbers 3 and 4. For example if we add 3 pens and 4 pens together, we obtain
seven pens. Hence the x is just representing any physical quantity.

What is not possible is when the powers of the variables are not the same.

(2) 3x 2 + 4 x = 3x 2 + 4 x

We cannot add because the powers are not the same.

In addition, we group ‘like’ terms, i.e. those with the same power.

Example 2

Simplify the following expressions

(1) 3x 2 + 5 x + 8 x 2 + 7 x + 4 (2) 6 x + 8 y + 15 x + 21y + 3

(3) 15x 2 + 8xy + 6 y 2 + 21x 2 + 31y + 3 (4) 7 x 2 + 8 y 2 + 4x 2 − 3 y 2

(5) x 2 + 4 x 2 + 5 y + 7 y + 13

Solution

(1) 3x 2 + 5 x + 8 x 2 + 7 x + 4

= 3 x 2 + 8 x 2 + 5 x + 7 x + 4 (Grouping like terms)

= 11x 2 + 12 x + 4

18
(2) 6 x + 8 y + 15 x + 21y + 3

= 6 x + 15 x + 8 y + 21y + 3

= 21x + 29 y + 3

(3) 15x 2 + 8xy + 6 y 2 + 21x 2 + 31y + 3

= 15x 2 + 21x 2 + 6 y 2 + 8xy + 31y + 3

= 36 x 2 + 6 y 2 + 8xy + 31 + 3

(4) 7 x 2 + 8 y 2 + 4x 2 − 3 y 2

= 7 x 2 + 4 y 2 + 8x 2 − 3 y 2

= 11x 2 + 5 y 2

(5) x 2 + 4 x 2 + 5 y + 7 y + 13

5 x 2 + 12 y + 3

Exercise 1.5.1 Simplification

Simplify the following:

(i) 8x + 16 + 18x − 6 (ii) 3x − 68x − 4 (iii) x 2 + 2 x + 8 x + 16

(iv) 2 x 2 + 16 x − 6 x 2 − 6 x (v) − x 2 + x − 3x 2 + 6 x

(vi) 9x + x + 8x + 4 + 3x (vii) 5 x 2 − 6 x − 12 x 2 − 16 x

(viii) − 4x + 10 + 12x − 20 (ix) 25 x 2 − 16 x + 8 − 32 x 2 + 26 x − 60

(x) − 2 x 2 + 6 x + 12 x 2 + 8 x + 15

19
1.5.2 Expansion

ax m  bx n

(1) We first multiply the coefficients i.e. a  b then x m  x n = x m + n

We write one x and add the powers.

Example 1

(1) 3x  4 x 2 = 12 x1+ 2 = 12 x 3 (2) 6 x 2  4 x 5 = 24 x 7

(3) 4x  5 = 20x (4) x3  4x 6 = 4x9

Expansion of Algebraic Expressions in one bracket

Example 2

Expand a(b + c)

Solution a(b + c) = a  b + a  c

The one outside the bracket will multiply all the others inside.

Example 3

Expand simplify:

(1) 8(x + 2) = 8x + 16

(2) 3x 2 (2 x + 4) = 3x 2  2 x + 3x 2  4 = 6 x 3 + 12 x 2

(3) 3(x − 2) − 4(2x + 1) = 3x − 6 − 8x − 4 = − 5x − 10

(4) 3x(2 x − 4) + 3(x + 5) = 6 x 2 − 12 x + 3x + 15 = 6 x 2 − 9 x + 15

(5) 5(x + 3) + 4(x − 1) = 5x + 15 + 4x − 4 = 9x + 11

(6) 3x(2 x + 7) − 5x(x − 4) = 6 x 2 + 21x − 5x 2 + 20 x = x 2 + 41x

20
Exercise 1.5.2 Expansion

(a) Expand the following:

(i) 8(x + 5) (ii) 2(x − 5) (iii) − 3x(x − 2) (iv) (3x − 5)4

(v) (
7 3 − x2 )
(b) Expand and simplify:

(i) 2x(3x + 4) − 3(x − 5) (ii) 6(3x − 7) + 3(x + 17)

(iii) 3x(7 x − 4) − 8(3x − 5) (iv) − x(x − 1) − 3x(x − 2)

(v) 5(x − 3) − 4(x − 5) (vi) 5x(x − 4) + 7(2x − 3)

(vii) 6 y(2x + y ) − 3x( y + x) (viii) 3a(2a + b) + 4a(2a + b)

(ix) 2x(x + 2) − 3(x + 2) (x) 3x(x + 4) − 2(x + 4)

1.5.3. Expansion of Algebraic Expressions of two term brackets

Expand (a + b) (c + d )

(a + b) (c + d ) = a(c + d ) + b(c + d )
= ac + ad + bc + bd

We take the letter a and use it to multiply the second bracket (c + d ) and then take b(c + d )

Example

Expand and simplify the following.

(1) ( 2x + 5)(3x − 4)
= 2x(3x − 4) + 5(3x − 4)

= 6 x 2 − 8 x + 15 x − 20

= 6 x 2 + 7 x − 20

21
(2) ( 2x + 3 y )( 2x − 3 y )
= 2x(2x − 3 y ) + 3 y(2x − 3 y )

= 4 x 2 − 6 xy + 6 xy − 9 y 2

= 4x 2 − 9 y 2

(3) (3a + 5)(3a + 5) = 3a(3a + 5) + 5(3a + 5)

= 9a 2 + 15a + 15a + 25

= 9a 2 + 30a + 25

(4) (5x − 3 y ) (5x − 3 y )

= 5x(5x − 3 y ) − 3 y(5x − 3 y )

= 25x 2 − 15xy − 15xy + 9 y 2 = 25x 2 − 30 xy + 9 y 2

(5) 3(x + 3) (x + 2) = 3(x + 3) (x + 2)

= 3x(x + 2) + 3(x + 2)


= 3 x 2 + 2 x + 3x + 6 

= 3 x 2 + 5x + 6 
= 3x 2 + 15 x + 18

(6) 2 ( x − 5 )( x + 1) = 2 ( x − 5 )( x + 1) 

= 2x(x + 1) − 5(x + 1)

(
= 2 x 2 + x − 5x − 5 )
= 2 x 2 − 8 x − 10

22
(7) 5 ( x + 3)( x − 2 ) = 5 ( x + 3)( x − 2 ) 

= 5x(x − 2) + 3(x − 2)


= 5 x 2 − 2 x + 3x − 6 

= 5 x2 + x − 6 
= 5 x 2 + 5 x − 30

(8) 4 ( x + 8 )( x + 3) = 4 ( x + 8 )( x + 3) 

= 4x(x + 3) + 8(x + 3)


= 4 x 2 + 3x + 8x + 24 

= 4 x 2 + 11x + 24  = 4 x 2 + 44 x + 96

Exercise1.5.3 Expansion and Simplification of Algebraic Expressions

(a) Expand and simplify the following

(1) ( x − 3)( x + 3) (2) ( 4x + 5 y )( 4x − 5 y )

(3) ( 6x + 5 y )( 6x − 5 y ) (4) (5x − 2)(5x + 2) (5) ( x + 9)( x − 9)
(6) ( 7 x − 4 y )( 7 x + 4 y ) (7) (8x − 3)(8x + 3) (8) ( 6x − 4 y )( 6x + 4 y )
(9) (3x − 2 y) (3x + 2 y) (10) (16x − y )(16x + y )
(b) Expand and simplify the following

(1) (11x + 4)( 2x + 7) (2) (3x + 4 y ) (3x + 5 y )

(3) (x + 7) (2x + 7) − 4(2x + 1) (3x + 5) (4) 5(x + 3) (x − 4)

(5) 3(x + 8) (x + 4) + 7(x − 1) (x + 1) (6) (3x − 1) (x + 2) + (x + 7) (x − 4)

(7) (x + 4)(x + 5) − (3x − 2) (x + 4) (8) 2(x − 1)(x − 5) + 3(x − 4) (x − 3)

(9) 2(x + 8)(x + 2) + 3(x + 4) (x − 5) (10) 3(x − 5)(x + 5) + 4(x − 4) (x + 4)

23
1.6.0 Factorization

This is the reverse of expansion. Here we introduce brackets.

Factorize ac + ad , Solution ac + ad = a(c + d )

Looking at the two expressions, we see that a is common so bring it out and introduce a bracket.
Then we write the others in the bracket.

Example

Factorize the following expressions:

(1) 2x + 8 y = 2(x + 4 y ) (2) 16x + 24 y = 8(2x + 3 y )

(3) 6 x 2 − 5x = x(6 x − 5) (4) 5x + 5 y = 5(x + y )

(5) 7 x − 21 = 7(x − 3)

Exercise 1.6.0

Factorize the following expressions

(a) x 2 − 15 x (b) 5 x 2 − 10 x (c) − 2x 2 + 6x

(d) x(2x − 3) + 4(2x − 3) (e) x(x − 9) + y(x − 9) (f) 3(x − 4) + y(x − 4)

(g) 2x(x − 5) + 6(x − 5) (h) − x(x + 3) + 9(x + 3) (i) 6(x + 4) + x(x + 4)

(j) 3x(x + 4) + 5(x + 4)

1.6.1 Factorization of four term expressions

Factorize ax + ay + bx + by

ax + ay + bx + by = a(x + y ) + b(x + y )

= (a + b) (x + y)
After the first factorization, we see that x + y is common, so we put this x + y in one bracket
and a + b also in another

24
Example 1

Factorize the following:

(1) 7 xy + y 2 − 7 x − y

Now 7 xy + y 2 = y(7 x + y )

− 7 x − y = −1(7 x + y )

:. 7 xy + y 2 − 7 x − y = y(7 x + y ) − 1(7 x + y )

= ( y − 1) (7 x + y )

We first put 7 x + y in one bracket then y − 1

(2) ac − bc + ad − bd = c ( a − b ) + d ( a − b ) = (a + d ) (a − b)

(3) ax − xy − ay + y 2 = x(a − y ) − y(a − y ) = (x − y ) (a − y )

(4) ax − ay + bx − by = a(x − y ) + b(x − y ) = (a + b) (x − y )

(5) 2rx + 2ry + 3cx + 3cy = 2r ( x + y ) + 3c ( x + y ) = ( 2r + 3c ) ( x + y )

(6) 5x 2 + 10 x + 3x + 6 = 5x(x + 2) + 3(x + 2) = (5x + 3) (x + 2)

(7) 6 x 2 − 2 x + 15x − 5 = 2 x(3x + 1) + 5(3x − 1) = (2 x + 5) (3x − 1)

(8) x 2 + 3x + 3x + 9 = x(x + 3) + 3(x + 3) = (x + 3) (x + 3) = ( x + 3)2

(9) x 2 − 7 x + 3x − 21 = x(x − 7) + 3(x − 7 ) = ( x + 3) ( x − 7 )

(10) x 2 + 3x − x − 3 = x(x + 3) − 1(x + 3) = (x + 3) (x − 1)

25
Exercise 1.6.1

Factorize the following expressions:

(1) 7 xy − 7 x − y 2 + y (2) ac − bd − bc + ad (3) ax + y 2 − xy − ay

(4) 5 y + 1 − 5 xy − x (5) a 2 + 2ab + 4b 2 + 2ab (6) 12ac + bd + 3ad + 4bc

(7) x 2 − xz − xy + yz (8) 3x 2 − xb + 3 yb − 9 xy (9) 2m − 2n + mx − nx

(10) 6am + 15an + 2bm + 5bn (11) ax − ay − bx + by (12) a 2 + 2a + b 2

(13) x 2 − 2 xy + y 2 (14) 9 x 2 − 3xy − 3xy + y 2 (15) x 2 + 2 x + 3x + 6

(16) x 2 + 4 x + 8 x + 32 (17) x 2 − 3 x − 20 x + 60 (18) 2 x2 + 8x + x + 4

(19) 4x 2 + 4x − 9x − 9 (20) 7 x 2 + 35 x − 2 x − 10 (21) x 2 − 4x + 2x − 8

(22) x 2 − 3x + 7 x − 21 (23) 3x 2 + 6 x − x − 2 (24) x 2 − x + 3x − 3

(25) 2 x 2 + 2 x − 10 x − 10 (26) x 2 + 5 x − 9 x − 45 (27) 24 x 2 − 14 x + 36 x − 21

(28) x 2 − x − 5x + 5 (29) x 2 + 3 x + 9 x + 27 (30) 4x 2 + 2x + 2x + 1

1.6.2 Factorization of Quadratic Expressions

A quadratic expression is an expression of the form ax 2 + bx + c .

We shall consider various types.

Type I:

x 2 + bx + c . In this type a = 1 i.e.

The coefficient of x 2 is one.

Steps

Look for the factors of the constant term (c) there will be two of them such that when we add we
get b.

Replace those two factors for b and then factorize completely.

26
Example 1

Factorize the following:

(1) x 2 + 8 x + 15 :

Find factors of 15 so that when we add we get 8: they are 3 and 5.

x 2 + 8 x + 15 = x 2 + 3x + 5x + 15 = x(x + 3) + 5(x + 3) = ( x + 5)( x + 3)

Short cut

For this type. i.e.

x 2 + bx + c = (x + factor 1) (x + factor 2)

:. x 2 + 8x + 15 = (x + 3) (x + 5)

(2) x 2 − 4x − 2 1 :. Factors : − 7, 3

Short cut: x 2 − 4 x − 21 = (x − 7 ) (x + 3)

Normal route

x 2 − 4 x − 21 = x 2 − 7 x + 3x − 21 = x(x − 7) + 3(x − 7) = (x + 3) (x − 7)

NB: the order in which we write the answer is not important

i.e. (x − 7) (x + 3) = (x + 3) (x − 7)

(3) x 2 + 2 x − 3 :. Factors are –1 and 3

x 2 + 2 x − 3 = (x − 1) (x + 3)

(4) x 2 + 3x − 10 factors : 5 , –2

x 2 + 3x − 10 = (x + 5) (x − 2)

(5) x 2 + 5x + 6 factors of 6 : 2 , 3

x 2 + 5x + 6 = (x + 2) (x + 3)

(6) x 2 − 5 x + 6 : Here factors: −2 , − 3

x 2 − 5 x + 6 = (x − 2) (x − 3)

27
(7) x 2 + 5 x − 6 : factors are 6 , − 1

x2 + 5x − 6 = ( x + 6)( x −1)

(8) x 2 − 5 x − 6 : factors are −6 , 1

x2 − 5x − 6 = ( x − 6) ( x + 1)

(9) x 2 − 8x + 15 = (x − 3) (x − 5)

(10) x 2 + 10 x + 25 , factor are 5 , 5

x 2 + 10 x + 25 = (x + 5) (x + 5) = ( x + 5)
2

Exercise 1.6.2

Factorize the following expressions:

(1) x 2 + 12 x + 32 (2) x 2 − 10 x + 25

(3) x 2 − 23 x + 60 (4) x 2 + x − 42

(5) x 2 + 17 x − 84 (6) x 2 − 4 x − 45

(7) x 2 + 4 x − 45 (8) x 2 + 6x − 7

(9) x 2 + 14 x + 40 (10) x 2 − 6x + 5

1.6.3 Type II : ax 2 + bx

With this there is no constant term.

Factorize the following

(1) 2 x 2 + 5x = x(2 x + 5) (2) 3 x 2 + 6 x = 3 x( x + 2 )

(3) 8 x 2 + 4x = 4 x(2 x + 1) (4) 7 x 2 + 21 = 7 x(x + 3)

(5) 25x 2 − 50 = 25x(x − 2)

28
Exercise 1.6.3

Factorize the following

(1) 8 x 2 − 14 x (2) 21x 2 + 14 x (3) x 2 + 8x

(4) 6 x 2 + 15 x (5) 3 x 2 + 10 x

1.6.4 Type III: Difference of Two Squares

Examples : a 2 − b 2 , 4x 2 − 9

This type is also under a special factorization called difference of two squares.

Expand (a + b) (a − b)

Solution

(a + b) (a − b) = a 2 − ab + ab − b 2

Therefore (a + b) (a − b) = a − b
2 2

The right hand side is what we call difference of two squares,

To factorize m 2 − n 2

It is very simple. The answer contains m and n. one with positive sign and the other with
negative sign.

m2 − n 2 = (m + n) (m − n)

Example

Factorize the following:

(1) x 2 − 32 = (x + 3) (x − 3) (2) (2 x )2 − 5 2 = (2 x + 5) (2 x − 5)
(3) (5 x )2 − 42 = (5 x + 4) (5 x − 4) (4) (3x )2 − 5 2 = (3x + 5) (3x − 5)

(5) (9 x )2 − (4 y )2 = (9 x + 4 y ) (9 x − 4 y )

4 x 2 − 100 = (2 x ) − (10) = (2 x − 10) (2 x + 10)

2 2
(6)

(7) (
8x 2 − 32 = 8 x 2 − 4 = 8 x 2 − 2 2 ) ( ) = 8(x + 2) (x − 2)

(8) (
3x 2 − 27 = 3 x 2 − 9 ) ( )
= 3 x 2 − 32 = 3(x + 3) (x − 3)
29
(9) (x − 5)2 − 4 2 = (x − 5 + 4) (x − 5 − 4) = (x − 1) (x − 9)

81 − ( x + 3) = 92 − ( x + 3) = ( 9 + x + 3) ( 9 − ( x + 3) )
2 2
(10)

= (12 + x) (9 − x − 3) = (12 + x) (6 − x)

Exercise 1.6.4

Factorize the following expressions:

(1) 9 x 2 − 64 (2) 16 − 25 x 2 (3) x 2 − 49

(4) (x + 3)2 − 16 (5) (x − 3)2 − 25 (6) 4 x 2 − 81

(7) 16 x 2 − 25 y 2 (8) 81x 2 − 25m 2 (9) a 2 − 4 (10) 2 x 2 − 32

1.6.5 Factorization of the General Quadratic Expression

ax 2 + bx + c

Steps 1 : multiply a  c = ac

2. find all the factors of ac.

3. select two of them whose sum is b and the product is ac.

4. Replace these two factors for b and then factorize

Factorize the following

(1) 5 x 2 + 13x + 6 , 5  6 = 30

Factors are 3 and 10

: 5 x 2 + 13x + 6 = 5 x 2 + 3x + 10 x + 6

= x(5x + 3) + 2(5x + 3)

= (x + 2) (5x + 3)

(2) 6 x 2 + 13x − 5 : 15 −2 = −30

15 , − 2
30
 6 x 2 + 13x − 5 = 6 x 2 + 15 x − 2 x − 5

= 3x(2x + 5) − 1(2x + 5)

= (3x − 1) (2 x + 5)

(3) 5x 2 − 9x + 4 : −5 −4 = 20

Factors − 5, − 4

5x 2 − 9x + 4 = 5x 2 − 5x − 4 x + 4

= 5x(x − 1) − 4(x − 1)

= (5x − 4) (x − 1)

(4) 35x 2 + 12 x + 1 factors of 35 : 5, 7

35x 2 + 12 x + 1 = 35x 2 + 7 x + 5x + 1 , = 7 x ( 5x + 1) + 1( 5x + 1) ,

= ( 7 x + 1)( 5x + 1)

(5) 5x 2 + 9 x + 4 = 20 : 5, 4

5x 2 + 9 x + 4 = 5x 2 + 5x + 4 x + 4 , = 5x(x + 1) + 4(x + 1) , = ( 5x + 4)( x + 1)

(6) 6 x 2 − 13x + 5 : factor are − 10, − 3

6 x 2 − 13x + 5 = 6 x 2 − 10 x − 3x + 5

= 2x(3x − 5) − 1(3x − 5)

= ( 2 x −1)( 3x − 5)

(7) 5 x 2 + 13x − 6 :. Factor are 15, −2

= 5 x 2 + 15 x − 2 x − 6

= 5x(x + 3) − 2(x + 3) = ( 5x − 2)( x + 3)

31
Exercise 1.6.5

Factorize the following:

(a) 2x 2 − 9x + 4 (b) 10 x 2 − 53x + 15 (c) 6 x 2 + 32 x + 26

(d) 10 x 2 + 53x + 15 (e) 3x 2 − x − 14 (f) 24 x 2 − 22 x − 30

(g) 7 x 2 + 33x − 10 (h) 4x 2 − 5x − 9

1.7.0 Change of Subject

For a relation such as V = U + ab , we say that V is the subject of the formula since V is on one
side of the relation .

To make any of the variables the subject means changing the equation such that the variable in
question will be on one side of the equation.

Example 1

(1) Make a the subject if V = u + at

Solution

V = u + at

V − u = at

V − u at V −u
= :. a =
t t t

(2) Make m the subject if. f = mg − kv

Solution

F = mg − kv , F + kv = mg

F + kv mg F + kv
= Divide by g , m =
g g g

32
 rh
(3) If x = make r the subject
R−r

x ( R − r ) = rh : cross multiply

Rx − rx = rh

Rx = rh + rx

Rx = r (h + x) factorize r.

Rx r (h + x )
= Divide by h + x
h+ x h + x

Rx
=r
h+x

(4) Make q the subject if xt − q = xq + t

Solution

xt − t = xq + q

xt − t = q(x + 1) Divide by x + 1

xt − t q(x + 1 )
=
x +1 x + 1

xt − t
=q
x +1

(5) Given that p(x −1) = q + x = p , make p the subject.

Solution

px − p = q + x − p

px − p + p = q + x

px = q + x

q+x
p=
x

33
Exercise 1.7.0

Transpose each of the following to make the letters in brackets the subject:

1
(1) V 2 = U 2 + 2as (a ) (2) V = r 2 h (h )
3

(3) x(a + b) = a − b (a) (4) yz + xz = xy (z )

w x
(5) = (x ) (6) 25 = n(a + L) (L)
y x− y

(7) A = r (r + h) (h) (8) xt − q = xq + t (t )

x−a i
(9) a(w − p ) = bp ( p) (10) = (q )
p q

1.7.1 Non Linear Form

This type may consist of fractions, squares or square roots.

If there are fractions, we multiply through by the L.C.M (usually by the product of the
denominators)

Example 2

1 1 1
Make y the subject in the following relation: = +
x y z

Solution

Method I

Multiply through by the LCM : xyz

1 1 1

xyz  = xyz  + xyz 
 x  y z

yz = xz + xy , yz − zy = xz , y(z − x ) = xz , Divide by z − x

xz
y=
z−x
34
Method II

1 1 1 1 1 1
= + − = (find LCM between x and y)
x y z , x z y

z−x 1 xz y xz
= = , y= turn it upside down)
xz y , z−x 1 z−x

1 1
(2) If = b − , make c the subject.
a c

Solution

1 b 1 1 ab − 1
= − (send c to left side) , = (simplifying RHS)
c 1 a c a

c a a
= :. c=
1 ab − 1 , ab − 1

a c b d
In general, if = then = you can just turn the two fraction.
b d a c

1 1 1
(3) Make u the subject = +
f v u

v− f 1 fv
= , :. u =
fv u v− f

When a square crosses the = sign, it becomes square root.

If x 2 = k then x =  k

Similarly square root becomes square

If x = p then x = p 2

35
Examples 3

(1) Make b the subject if a 2 + b 2 = c 2

Solution

b2 = c2 − a2

:. b = c2 − a2

(2) If y = k lm Make m the subject

Solution

square both sides.

(
y 2 = k lm )2

y 2 = k 2 lm divide by k 2 l

y 2 k 2 lm y2
= :. m =
k 2l k 2l k 2l

(3) If v 2 = u 2 + 2aL make u the subject

v 2 − 2aL = u 2

v 2 − 2aL = u

y2 − d
(4) = x + c make y the subject
k

Solution

y 2 − d = k ( x + c ) multiply by k.

36
 y 2 − d = (k ( x + c) ) square it.
2

y 2 − d = k 2 (x + c )
2

y 2 = d + k 2 (x + c )
2

y = d + k 2 (x + c )
2

km2 l
(5) if h = , make m the subject
g2

Solution

hg 2 km2 l
hg 2 = km2 l multiply by g 2 , = Divide by kl
kl kl

hg 2
= m 2 (find square root)
kl

hg 2
=m
kl

M
(6) Given that P = + K , find an expression for R in term of P, M and K.
R2

Method I

M M 1 R2
P= 2 +K, P−K = 2 , = , (inverting it)
R R P−K M

M M
= R2 , =R
P−K P−K

Method II

M M
P= 2
+ K , P − K = 2 (cross multiply) , (P − K )R 2 = M Divide by P − K
R R

(P − K )R 2 =
M
, R2 =
M
, :. R=
M
P − K P−K P−K P−K

37
 1 K 2 + P2
(7) Given that = , make K the subject
n hg

Solution

hg
= K 2 + P2 (square it)
n
2
 hg  h2 g 2 h2 g 2
  = K +P
2 2
2
− P2 = K 2 , 2
− P2 = K
 n  n n

(8) Make r the subject if

1
(
V = h 3r 2 + h 2
6
)
Solution

(
6V = h 3r 2 + h 2 , ) 6V
h
= 3r 2 + h 2 (Divide by h )

6V 1  6V 
− h 2 = 3r 2 ,  − h2  = r 2 (Divide by 3)
h 3  h 

1  6V 
 − h2  = r
3  h 

3t + 2 P 3h(P + t )
(9) Given that = , make U the subject
U L

U L
= (invert it )
3t + 2 P 3h ( P + t )

L(3t + 2 P )
U= multiply by 3t + 2P
3h(P + t )

x−K U +a
(10) Make x the subject if =
U x+K

38
Solution

(x − K ) (x + K ) = U (U + a) Cross multiply

x 2 + xK − xK − K 2 = U 2 + Ua

x 2 − K 2 = U 2 + Ua , x 2 = U 2 + ua + K 2 ,:. x = U 2 + ua + K 2

Exercise 1.7.1 Change of subject

Make the variables in the bracket the subject:

1 1 1 1 1 1
(1) = + (V ) (2) = y − (z ) (3) T=K (g )
F U V x z g

x−a 1 x a
(4) L = h 2 + k 2 (k ) (5) + = − (x )
P q q P

y2 − d
(6) = x+c (y) (7) F = (Mx − e )
2
(x )
K

as + t x−K
(8) C= (s ) (9) P = 2+ (x )
K − 2x x+c

a−x
(10) 6k − 5m = 25m 2 + 6 2 (m) (11) y=k (a )
a−c

1 1 1 a z 3x − t
(12) = + (t ) (13) C= − (t) (14) M = (x )
p t2 q t d 3x + r

u K
(15 M (x − u ) = b(x + v) (x ) (16) x= h +  (K )
3 r 

a a+K 2x − y 4x + y
(17) = (a ) (18) = (x )
a−b a−c x + 2 y 2x − y

39
1.8.0 Linear Equations

A linear equation is an equation of the form ax + b = 0 where a and b are real numbers and x is a
variable.

Examples

(i) 2x + 17 = 0 (ii) 16x − 3 = 0

Linear equations have only one solution

Solving linear equations:

(1) If there are fractions, multiply through by the LCM.

(2) Expand brackets if any

(3) Collect all like terms together

(4) Then solve for x.

Solve for x if 2x + 17 = 0

Solution

2 x = −17 , 2x = −17 (send 17 to the other side of = )

2 x − 17 1
= , x = −8
2 2 2

2x
(2) Given that = −4 , find x.
7

Solution

2x 2x
= −4 7  = −4  7 (multiply through by 7)
7 , 7

2 x − 28
2 x = −28 , = , x = − 14
2 2

40
(3) Solve for x. if 7 x − 1 = 2x + 4

Solution

5x 5
7 x − 1 = 2 x + 4 , 7 x − 2 x = 4 + 1 , 5x = 5 , = x =1
5 5 ,

(4) Find x if 4(2x − 1) = 8(3x + 1)

Solution

4(2x − 1) = 8(3x + 1)

8x − 4 = 24x + 8 expanding

− 4 − 8 = 24x − 8x

− 12 = 16 x Divide by 16

− 12 16 x −3 −3
= =x :. x =
18 16 4 4

(5) Solve for x if 5(x + 3) − 4(3x + 2) = 20

Solution

5(x + 3) − 4(3x + 2) = 20 , 5x + 15 − 12x − 8 = 20

5x − 12x + 15 − 8 = 20 , − 7 x + 7 = 20 , − 7 x = 20 − 7 , − 7 x = 13

− 13 6
:. x = = −1
7 7

(6) Solve for x: 23(x − 3) − 4(2x − 5) − 6x = 32x − 5 − (3x − 7)

Solution

23(x − 3) − 4(2x − 5) − 6x = 32x − 5 − (3x − 7)

23x − 9 − 8x + 20 − 6 x = 32 x − 5 − 3x + 7

41
 2− 5x + 11 − 6x = 3− x + 2

− 10x + 22 − 6x = −3x + 6 , − 10x − 6x + 3x = 6 − 22

16
− 13x = −16 , : . x =
13

Exercise 1.8.0 Linear Equations

Solve for x:

(1) 2(9x + 4) − 8x + 17 = 0 (2) 3(7 x −1) − (2x −15) = 0

(3) 2(3x −1) − 4(x + 1) + 4(3x −1) = −5 (4) 6 − 5x = 2 x + 1

(5) 8x + 1 + 3(3x − 5) = 17 (6) 4 − 3x = 8 − 9x

(7) x − 4 = 7(x + 2) (8) 2(x + 4) = 3(x − 15)

(9) 2x − 1 − (3x − 7) = 0 (10) 3(3x − 1) − (2x + 4) − 3x = 17

1.8.1 Equations involving fractions

If the equations are of the form

a c
= then ad = bc
b d

This is called cross multiplication

Remember you can only cross multiply if there are only two fractions.

Example

(1) Solve for x if

5x − 1 − 3
=
2 4

Solution

5x − 1 − 3
= ,  4 ( 5x −1) = −3  2 ,  20 x − 4 = −6
2 4

2 1
20x = −6 + 4 , 20 x = −2 , x=− =−
20 10
42
(2) Solve for x

x+4 x−4
=
5 3

Solution

x+4 x−4
= ,  3 ( x + 4) = 5 ( x − 4)
5 3

3x + 12 = 5x − 20 ,  12 + 20 = 5 x − 3x

32 2 x
= ,  16 = x , :. x = 16
2 2

(3) Solve for x.

2 x + 5 3x − 7
=
4 2

Solution

38
2(2x + 5) = 4(3x − 7) ,  4 x + 10 = 12 x − 28 ,  38 = 8x ,  = x
8

19
 x=
4

4(2 x − 1)
(4) Solve for x. = −1
3

Solution

1
4(2x − 1) = −3 ,  8 x − 4 = −3 ,  8 x = −3 + 4 ,  8 x = 1 ,  x =
8

43
 10 x + 5
(5) Solve for x. =6
4

Solution

19
10x + 5 = 24 ,  10 x = 24 − 5 ,  10 x = 19 ,  x =
10

Exercise 1.8.1

Solve for x.

6 − 3x 2x + 5 x + 2 1 3
(1) = −2 (2) = (3) =
4 3 7 x+4 x−4

x−4 5 7 6 3
(4) =5 (5) = (6) =
x+2 3x + 2 5 x − 2 3x − 2 7

8 5 1 1
(7) − =0 (8) =
3x − 2 2 x + 5 6 x 17

4(2 x − 1) 3(3x − 7 ) 3(2 x − 1) 2(2 x + 1)

(9) = (10) − =0
5 7 4 5

1.8.2 General Fractions

Solve for x.

2x + 5 x + 2
(1) = +4 (multiply by 21)
3 7

 2x + 5   x + 2
21  = 21  + 4  21
 3   7 

 7 ( 2 x + 5) = 3 ( x + 2) + 84 ,  14 x + 35 = 3 x + 6 + 84

 14 x − 3x = 90 − 35 ,  11x = 55 ,  x=5

8x + 3 3 7 x − 1
(2) + = (multiply by 20)
4 5 4

44
  8x + 3  3  7x −1
20  + 20  = 20 
 4  5  4 

 5 (8x + 3) + 4 (3) = 5 ( 7 x −1) ,  40 x + 15 + 12 = 35 x − 5

−32 2
 40 x − 35 x = −5 − 27 ,  5 x = − 32 ,  x= = −6
5 5

x + 3 2x − 1
(3) Solve for x in the following , − = x +1
3 6

( x + 3) − 6 ( 2 x − 1) = 6
 6 ( x + 1)
3 6

 2 ( x + 3) − ( 2 x −1) = 6 x + 6 ,  2x + 6 − 2x +1 = 6x + 6

1
 7 − 6 = 6x ,  1 = 6x ,  x=
6

1
(4) Solve for x in the following: (3x − 1) − 5 (3x − 2) = 1 .
4 7 4

Solution

1 5 1
 28  ( 3x − 1) − 28  ( 3x − 2 ) =  28 ,  7 (3x −1) − 20 (3x − 2) = 7
4 7 4

 21x − 7 − 60 x + 40 = 7 ,  − 39 x + 33 = 7 ,  − 39 x = 7 − 33 ,  − 39 x = −26 ,

26 2
:. x= =
39 3

x 2x
(1) If + = 14 , find x.
2 3

x  2x  84
 6 + 6   = 6 14 ,  3x + 4 x = 84 ,  7 x = 84 ,  x= = 12
2  3  7

45
Exercise 1.8.1 Linear Equations

Solve for x if:

1
(1) 2 x − 5 − 3(2 x − 5) − 5 x = 1 (2 x − 1 − (3x − 4)
4 5

2x − 3 1 7x − 1 x + 1 2x − 1
(2) + = (3) = + 7x + 5
7 2 4 2 5

2 x − 1 3x + 1 3x − 1 3x − 1 6 1 3
(4) + = 1 − 2x (5) + = 3 − 6x (6) − =
4 5 7 5 3x − 1 5 7

3 − 8x 3 7 x − 1
(7) − = (8) 4(3x − 1) − 5(x + 2) = 7(2x + 5) − (3x − 16)
4 5 3

2x − 1 x − 1 2 x + 5 3x − 4 x
(9) − + 15 = 0 (10) − =
3 4 4 5 2

1.9.0 Simultaneous Equations

Consider the statement below:

Find two numbers whose sum is 10.

Solution

Let the numbers be x and y. then x + y = 10 . The solution is an infinite one. But when we add the
statement if the difference between them is 8, find the numbers.

Then the equations become:

x + y = 10 ….. (1)

x− y =8 ….. (2)

Solving such equations is the main subject of this topic.

Simultaneous equations is an equation of the form

a1 x + b1 y = c1 ….. (1)

a2 x + b2 y = c2 ….. (2)

We shall consider two methods under this section. The first method is called elimination method
and the second one is called substitution method.

46
Elimination Method

a1 x + b1 y = c1 ….. (1)

a2 x + b2 y = c2 ….. (2)

If we want to eliminate y, then we make the coefficients of y in both equations become the same
and likewise x.

To solve for x

Step 1: Multiply equation (1) by b2

a1b2 x + b1b2 y = c1b2 ….. (3)

Multiply equation (2) by b1

a2b1 x + b1b2 y = c2b1 ….. (4)

Now: equation (4) – equation (3)

c2 b1 − c1b2
a2b1 x − a1b2 x = c2b1 − c1b2 , ( a2b1 − a1b2 ) x = c2b1 − c1b2 , :. x=
a 2 b1 − a1b2

When we substitute the value of x into any of the equations, we can obtain the value of y. we can
also eliminate x as follows:

a1 x + b1 y = c1 …… (1)

a2 x + b2 y = c2 …… (2)

Equation (1) xa2

a1a2 x + a2b1 y = c1a2 …. (3)

Equation (2) x a1

a2 a1 x + a1b2 y = c2 a1 …. (4)

equation (4) – equation (3)

a1c2 − a2 c1
a1b2 y − a2b1 y = a2c1 , (a1b2 − a2b1 )y = a1c2 − a2c1 ,  y=
a1b2 − a2 b1

47
Example 1

Solve the following simultaneous equations using elimination method:

(1) x + 7y = 5 ….. (1)

x − 7y = 9 ….. (2)

equation (1) + equation (2):

14
2 x = 14 ,:. x = = 7 , Putting x = 7 into (1) , we have 7 + 7 y = 5 , 7 y = 5 − 7
2

−2
7 y = −2  y=
7

(2) x + 8 y = 34 ….. (1)

x − 8y = 6 ….. (2)

Add (1) and (2)

2 x = 40 , :. x = 20

Put x = 20 into equation (1)

14 7
20 + 8 y = 34 , 8 y = 14 , y= =
8 4

(3) x + 5 y = −13 …. (1)

2x − y = 7 …. (2)

Equation (2)  5

10 x − 5 y = 35 …. (3),

x + 5 y = −13 …. (1)

(1) + (3) : 11x = 22 , :. x = 2

Put x = 2 into (1)

2 + 5 y = −13 , 5 y = −15 , :. y = −3

48
(4) 5x + 2 y = 9 …… (1)

9 x − 7 y = −5 …… (2)

Equation (1) 7

35 x + 14 y = 63 ….. (3)

Equation (2)  2

18 x − 14 y = −10 …… (4),

Equation (3) + equation (4)

53 x = 53 :. x = 1, Put x = 1 into (1)

5(1) + 2 y = 9 , 5 + 2 y = 9 , 2 y = 9 − 5 , 2 y = 4 , :. y=2

(5) 4 x − y = −1 …. (1)

2 x − 5 y = −1 …. (2)

Equation (1)  (− 5)

− 20 x + 5 y = 5 …. (3)

Add equation (2) and (3)

4 2
− 18x = 4 , :. x = − =−
18 9

2  2
Put x=− into (1), 4  −  − y = −1
9  9

−8 1
− y = −1 , − 8 − 9 y = −9 , − 8 + 9 = 9y , 1 = 9y :. y =
9 9

(6) 4 x + 3 y = 6 …… (1)

3x + 4 y = 1 ……. (2)

49
Equation (1)  4 :

16 x + 12 y = 24 ….. (3)

Equation (2)  3

9 x + 12 y = 3 …. (4)

Equation (3) – equation (4)

7 x = 21 :. x = 3

Put x = 3 into (2)

3(3) + 4 y = 1

9 + 4y = 1

4y = 1− 9

4 y = −8 : . y = −2

(7) Find 2 p + 3q if

2 p − 3q = 4 and 3 p + 2q = 19

Solution

2 p − 3q = 4 …. (1)

3 p + 2q = 19 … (2)

Equation (1)  2

4 p − 6q = 8 … (3)

Equation (3) + equation (4)

13 p = 65

:. p = 5

From equation (1)

2(5) − 3q = 4 , 10 − 4 = 3q , 6 = 3q , 2=q

50
Hence 2 p + 3q = 2(5) + 3(2) , = 10 + 6 = 16

3 2
(8) Given that x + y = 12 and
5 3

1
x + y = 14 , find x + y
2

Solution

Equation(1) 15 :

3  2 
15 x  + 15 y  = 15  12
5  3 

9 x + 10 y = 180 ….. (3)

x + 2 y = 28 ….. (4)

equation (4)  −5

− 5 x − 10 y = −140 …. (5)

equation (3) + equation (5)

4 x = 40 , :. x = 10

Put x = 10 into (3)

9(10) + 10 y = 180 , 90 + 10 y = 180 , 10 y = 180 − 90

10 y = 90 , :. y = 9 , Hence x + y = 10 + 9 = 19

4 5 1 5 3
(9) Solve for x and y given that x + y = 1 and x − y = .
3 6 3 12 2

4  5 
6 x  + 6 y = 6
3  6 

8x + 5 y = 6 ….. (1)

1  5  3
12   x  − 12   y  = 12   
3   12  2

4x − 5 y = 8 ….. (2)
51
 Add (1) and (2): 12 x = 24 : .x = 2

from equation (1)

8(2) + 5 y = 6 , 16 + 5 y = 6 , 5 y = 6 − 16 5 y = −10

y = −2 :. x = 2 and y = −2

Exercise 1.9.1

Solve the Simultaneous equations below using elimination method

(1) 5 x − y = 6 and x + 2 y = 3 (2) 5 x − 4 y = 8 and 2 x + y = 11

(3) 3x + 4 y = 11 and 2 x + 3 y = 9 (4) x + 3 y = 14 and 2 x − y = 7

1 1 x y
(5) 2 x + 5 y = 20 and 3x − 2 y = −46 (6) x + y = 9 and − = −5
2 7 7 2

x y x y
(7) − = 1 and − = 1 (8) 2 x + 5 y = 16 and 10 x − 3 y = −4
2 3 4 9

(9) 5m − 6n = 12 and 2m + 9n = 20 (10) 5 x + 2 y = 9 and 9 x − 7 y = −5

(11) 3x − y = 5 and 5 x + 3 y = −8 (12) 2m + 3n = −4 and 3m + 2n = −6

(13) 2a − 3b = 5 and 2a − 5b = −1 (14) 2 x + 3 y = 14 and 4 x − 5 y = 17

y 3 3a − 2b 6a − b
(15) 2x − = 5 and x + y = −1 (16) = 9 and =9
4 4 2 2

(17) 4 x − 5 y = 30 and 4 x − 2 y = 24 (18) 3m + 2n = −6 and 2m − 3n = −4

(19) x + 5 y = −13 and 2 x − y = 7 (20) 2 x + y = 7 and x + 3 y = 1

(21) 3x + 5 y = 21 and 7 x − 2 y = 8 (22) x + 6 y = 3 and 2 x + 3 y = 15

52
1.9.2 Substitution method

With this method, we make either x or y the subject and then substitute into the other equation

Example 2

Solve the simultaneous equations:

(1) x + 7y = 5 … (1)

x − 7 y = −9 …. (2)

From (1) x = 5 − 7 y ….. (3)

Put (3) into (2)

5 − 7 y − 7 y = −9 , − 14 y = −9 − 5 , − 14 y = −14 , :. y = 1

Put y = 1 into (3), x = 5 − 7 (1) , = 5 − 7 ,= 2

:. x = 2 and y = 1

(2) 3x − y = 5 …. (1)

5 x + 3 y = −8 … (2)

From (1) make y the subject

3x − 5 = y … (3) , Put (3) into (2)

5x + 3(3x − 5) = −8 , 5x + 9x − 15 = −8

7 1
14x = 7 , :. x= =
14 2

1 1 3
Put x = into (3) , y = 3  − 5 = −5
2 2 , 2

3 − 10 − 7 1 −7
= = :. x = and y =
2 2 2 2

53
(3) y = 4x − 2 …. (1)

y = −3 x + 5 …. (2)

Equation (1) and (2) we have

4x − 2 = −3x + 5 , 4x + 3x = 5 + 2 , 7x = 7 , x =1

Put x = 1 into (1)

y = 4(1) − 2 , = 4−2, = 2 : .x = 1 and y = 2

(4) 5 x − 6 y = 12 …… (1)

2 x + 9 y = 20 ….. (2)

20 − 9 y
From (2) 2 x = 20 − 9 y then x= ….. (3)
2

Put (3) into (1)

5
(20 − 9 y ) − 6 y = 12 , 5(20 − 9 y ) − 12 y = 24 , 100 − 45 y − 12 y = 24
2

76 1 4
− 57 y = 24 − 100 , − 57 y = −76 , y= =1 , =
57 3 3

4 4 20
Put y = into (1), 5 x − 6  = 12 , 5x − 8 = 12 , 5x = 12 + 8 = 20 , x= =4
3 3 5

4
:. x = 4 and y =
3

54
1.10.0 Word Problems

In this section, we shall learn how to translate words into equations and then solve them. This
section will concentrate on problems in fractions, linear equations and simultaneous equations.

We shall start with fractions and linear equations

2
(1) After spending of her pocket money, Joyce realized she has spent GH ¢18 find her
5
pocket money

Let the pocket money be x

2
then x = 18 :. 2 x = 18  5
5

18  5
x= = GH¢45
2

Thus the pocket money = GH¢45

2
(2) After spending of her pocket money, Joyce realized she has GH¢18 left. Find her
5
pocket money.

Here we shall find the fraction left

2 3
Fraction left = 1 − =
5 5

3
Thus x = 18
5

5 18
3 x = 5  18 , :. x= = GH¢30
3

You need to know if the question relates to the money left or spent.

2 3 1
(3) A man spent of his monthly salary on food , of the remainder on rent and of
11 8 9
what still remained on books.

(a) what fraction of his income is left?

(b) If he still has GH¢ 55.00 left, find his monthly income.

Solution
55
 2 2 9
(a) fraction on food = ; fraction left after food = 1 − =
11 11 11

3 9 3 9 27
rent of =  =
8 11 8 11 88

9 27
fraction left after rent −
11 88

72 − 27 45
=
88 88

1 45 5
book =  =
9 88 88

45 5 40 5
fraction left after books = − = =
88 88 88 11

(b) Let total income = x

5
then x = 55
11

:. 5 x = 11 55

605
x= = 121.0
5

Thus his monthly income = ¢121.00

9 2
(4) A famer uses of a field for growing Cassava. He uses of the remainder for growing
16 7
corn.

(a) What fraction of the field is used for growing corn?

(b) Find the fraction of the field left for, growing other things

(c) If the fraction left is equivalent of 50 hectares find the total field in hectares.

Solution

56
 9
(a) Cassava =
16

9 7
Fraction left = 1 − =
16 16

2 2 7 1
Corn = of remainder =  =
7 7 10 8

7 1 5
(b) Fraction left = − =
16 8 16

(c) Let the whole field in hectares be x

5 50  16
Then x = 50  5 x = 50  16 , x= = 160
16 , 5

field = 160 Hectares.

1 1 1
(5) A man spends of his monthly salary on rent, on food and on clothes. If he has
9 2 4
GH¢20 left at the end of the month, how much does he earn?

Solution

1 1 1
Fraction spent = + +
9 2 4

4 + 18 + 9 31
=
36 36

31 5
Fraction left = 1 − =
36 36

Let the income = x then

5 720
x = 20  5 x = 36  20 , x= = ¢144.00
36 , 5

1 2
(6) A man gave of his social security benefit to his aged mother, of the remainder to
3 3
1
his father, of what still remained to his nephew and the rest which was GH¢4,000 to
6
his wife. Find the amount he received as social security benefit.

57
 Solution

1
Father =
3

1 2
Remainder = 1 − =
3 3

2
Aged mother = of remainder
3

2 2 4
=  =
3 3 9

2 4 6−4 2
Remainder = − = =
3 9 9 9

1
Nephew = of the amount left
6

1 2 1
=  =
6 9 27

2 1 6 −1 5
Remainder = − = =
9 27 27 27

5
Wife took the rest =
27

Let the total amount = x

5 27  4, 000
x = 4, 000 . , 5 x = 27  4, 000 ,:. x =
27 5

= GH¢21, 600

3
(7) An employee spent 3 of his salary on books, of the remainder on accommodation
13 7
3
and of what still left on other expenses. If he still has GH¢4 left, find his salary
10

58
Solution

3
Books =
13

3 10
Remainder = 1 − =
13 13

3
Accommodation = of remainder
7

3 10 30
=  =
7 13 91

10 30 70 − 30 40
Remainder = − = =
13 91 91 91

3 40 12
Others =  =
10 91 91

40 12 28
Remainder = − =
91 91 91

28
Hence x=4
91

91 4
:. x = = GH¢13
28

1 1
(8) A widow received of her husband’s estate and each of her 3 sons received of the
3 3
remainder. If the widow and one of her sons received a total of GH¢6,000 from the
estate, what was the amount of the estate?

Solution

1
Widow =
3

1 2
Remainder : 1 − =
3 3

1 2 1 2 2
One son = of =  =
3 3 3 3 9

59
 1 2 3+ 2 5
Widow + one son = + = =
3 9 9 9

5
Hence x = 6, 000
9

9  6, 000
:. 5 x = 9  6, 000 , x= = ¢10,800
5

The value of the estate is ¢10,800.

1 2
(9) of the mangoes obtained from school farm were given to form 4 class, of the
3 3
1 1
remainder to form 3 class, of what still remained to form 2 class and of what still
6 2
remained to form 1 class. If the teachers shared the rest which were 40 mangoes, how
many mangoes were there?

Solution

1
Form 4 =
3

1 2
Remainder = 1 − =
3 3

2 2 2 4
Form 3 = of remainder =  =
3 3 3 9

2 4 6−4 2
Reminder = − = =
3 9 9 9

1 1 2 1
Form 2 = of remainder =  =
6 6 9 27

2 1 6 −1 5
Remainder = − = =
9 27 27 27

1 1 5 5
Form1 = of rest =  =
2 2 27 54

5 5 5
Remainder = − =
27 54 54

5 54  40
:. x = 40 : .x = = 432 Oranges
54 5

60
(10) When 4 is added to a quarter of a certain number, the result is the same as subtracting 8
from the number, find the number.

Solution

Let x be the number

1
First part of the statement: 4 + x
4

Second part: x − 8

Equating the two,

1
4+ x = x −8
4

48
16 + x = 4x − 32 , 16 + 32 = 4x − x , 48 = 3 x , = x x = 16
3 ,

Hence the number is 16

(11) Two consecutive odd integers are such that the greater added to twice the smaller gives
65 find the integers.

Remember the odd positive integers are

1,3, 5, 7, 9

You see that they differ by 2 consecutively.

Let the first one be x. then the second one is x + 2

Then we have x + 2 + 2x = 65

3x = 65 − 2 , 3 x = 63 , :. x = 21

The numbers are 21 and 23

(12) My father is three times as old as I am now. If the sum of our ages is 64 years, find our
ages.

61
Solution

Son x, father 3x

64
Sum of ages , x + 3x = 64 ,:. 4 x = 64 , x = = 16
4

Son = 16yrs

Father = 316

= 48 years.

(13) Faraday the great is 29 years now and his daughter Nana Ama Nyarko is 7 years. In how
many years’ from now will faraday be exactly three times as old as Nana?

Solution

Let that year be x from now

Faraday will be 29 + x by then

Nana will be 7 + x

We want 29 + x = 3(7 + x) , 29 + x = 21 + 3x , 29 − 21 = 3x − x

8 = 2x , x=4

This will happen in 4 year’s time.

(14) John is twice as old as Mary if John’s age was three times Mary’s age 5 years ago, find
their ages now.

Solution

John’s age: 2x Mary’s age: x

Five years ago.

Mary’s age = x − 5

John’s age = 2x − 5

:. 2x − 5 = 3(x − 5) , 2 x − 5 = 3 x − 15 , − 5 + 15 = 3x − 2x , 10 = x

Hence Mary = 10 years now

John = 2 10 = 20 years now.

62
Exercise 1.10.0

(1) One number is 10 less than another number find the numbers if their sum is 52

(2) If 3 is added to a certain number, the result is 15 less than three times the number.
find the number

(3) Find a number such that if you add 4 and divide the result by 3, you get the same
result as adding 13 and dividing the result by 5

(4) Joan has ¢5 more than Jane. Together they have ¢40. How much has each?

(5) A father is 7 times as old as his daughter. In five years time he will be 4 times as old
as his daughter will be then. Find their ages now.

(6) John’s mother is 5 times as old as John is. Three years ago, she was 9 times his age.
Find their ages now.

(7) A boy is two years younger than his sister. Three years ago, the sum of their ages was
48. What is the boy’s age?

(8) Today is Kwaku’s 12th birthday and his father’s 40th birthday. How many years from
today will Kwaku’s father be twice as old as Kwaku is at that time?

(9) 1 of the mangoes obtained from a School Farm were given to the form 4 class, 2
3 3

of the remainder to the Form 3 class, 1 of what still remained to the Form 2 and
6
1 of what still remained to the Form 1 class. If the teachers shared the rest, which
2
were 40 mangoes, how many mangoes were there altogether?

(10) If one number exceeds another number by 13 and the larger number is 3 times the
2
smaller number, then find the larger number.

(11) Of the number of Labour Cards issued to some artisans by Obuasi Labour Officer,
1 were issued to Carpenters, 1 of what remained to Drivers, 1 of what still
3 3 4
remained to Masons and the remainder, being 15, to Welders. How many Masons
were issued with Labour Cards by the office?

63
 (12) Three pencils an eraser cost ¢1200 and two pencils and an eraser cost ¢1000. Find the
cost of 15 pencils and 12 erasers.

(13) Three packets of sugar and 4 tins of milk cost ¢635 and 4 packets of sugar and 3 tins
of milk cost ¢695. Find the cost of sugar and a milk.

(14) Tickets for a film show were sold at ¢4500 to the general public and ¢3,750 to
students. 400 people attended the show and ¢1,680,000 was collected in tickets sales.
How many tickets were sold to the general public?

(15) Find two numbers such that, if 18 is added to the first number it becomes twice the
second number and, if 6 is added to the second number, it becomes three times the
first number.

(16) Admission costs to a theatre are adults ¢5, children ¢2 if 1000 people paid to enter the
theatre and the total receipts were ¢3800, who many adults attended?

1.11.0 Ratio

a 3
This is another way of expressing fractions for example a : b = . So 3 : 5 =
b 5

(1) A father is 50yrs and his son is 20yrs express their ages as a ratio.

Father: son = 50: 20

= 5: 2 i.e. dividing by 10

Simplifying Ratio

We can divide or multiply through by factor or common multiples respectively.

Example 1

Simplify the following ratios

(1) 20: 16: 8 (2) 150: 170: 80

(3) 25: 30: 70 (4) 80: 100: 150

Solution

(1) 20: 16: 8 = 5: 4: 2 (divide by 4)

(2) 150: 170: 80 = 15: 17: 8 (divide by 10)

(3) 25: 30: 70 = 5: 6: 14 (4) 80: 100: 150 = 8: 10: 15

64
Example 2

Simplify the following ratios:

1 2 2 2 6 1 3 1 1 1 5
(1) :3: (2) : : (3) : : (4) : :
4 3 3 5 7 5 10 2 8 2 12

Solution

1 2 1 2
(1) : 3 : = 12  : 12  3 : 12  i.e. multipling by 12(LCM)
4 3 4 3

= 3: 36: 8

2 2 6 2 2 6

(2) : : = 175  : 175  : 175  ( LCM = 175)
3 5 7 3 5 7

= 35 2 : 21(2) : 15(6)

= 70 : 42 : 90

= 35 : 21 : 45

1 3 1 1 3 1
(3) : : = 10  : 10  : 10  (LCM = 10)
5 10 2 5 10 2

= 2 : 3 : 5]

1 1 5 1 1 5
(4) : : = 24  : 24  : 24  (LCM 24)
8 2 12 8 2 12

= 3 : 12 : 10

a c
If a : b = c : d then =
b d

We can therefore express ratios as equations and solve where possible

Example 3

Find the value of x in the following ratios:

(1) 2 : 7 = x : 28 (2) 3 : 5 = 15 : x (3) 3x : 32 = 9 : 16

(4) (x + 1) : 4 = (2x − 1) : 7

65
Solution

2 x 2
(1) 2 : 7 = x : 28  =   28 = x , 8= x
7 28 7

3 15 5 15
(2) 3 : 5 = 15 : x = :. =  3x = 5 15 x= = 25
5 x 3

3x 9 288
(3) 3x : 32 = 9 : 16  =  3x 16 = 9  32  48x = 288 x = =6
32 16 48

x +1 2x −1
(4) (x + 1) : 4 = (2x − 1) : 7  =  7 ( x + 1) = 4 ( 2 x −1)
4 7

 7 x + 7 = 8x − 4  7 + 4 = 8x − 7 x :. 11 = x

Total Ratio

Let A, B, C share x in the ratio a : b : c

Total ratio = a + b + c

a a b
A’s share = x = x B’s share = x
a+b+c , total , total

c
C’s share = x
total

Example 4

(1) Divide ¢250,000 in the ratio 12 : 3 : 10

Solution

Total ration = 12 + 3 + 10 = 25

12
1st share =  250,000 = 120,000
25

3
2nd share =  250,000 = 30,000
25

10
3rd share =  250,000 = 10,000
25

66
(2) A rod 270cm long is divided in the ratio 2 : 3 : 4 find the length of the longest part

Solution

Total ratio = 2 + 3 + 4 = 9

4
Longest part  2 7 0 30 = 120 cm
9

1 3 7
(3) Three boy’s shared 1008 oranges in the ratio : : find the share of the boy who
5 10 10
got the least number of oranges

Solution

1 3 7
: : =2:3:7 Total ratio = 2 + 3 + 7 = 12
5 10 10 ,

2
Least number =  1008 = 168 oranges
12

1
(4) Divide 510 exercise books amongst X, Y and Z such that z has 1 times as much as Y and
2
Y had 3 times as much as X. finds each share

1 3
Z = 1 Y = y y = 3x
2 2 ,

Thus x has the least find an expression for Z and y in terms x

3 3
Z= y = (3x ) = 9 x
2 , 2 2

9 9
Hence x : y : z = x : 3x : x divide by x , = 1 : 3 : =2:6:9
2 2

2
Total ratio = 2 + 6 + 9+ = 17 , X share =  510 = 60 books;
17

6
Y share =  510 = 180 books;
17

9
Z share =  510 = 270 books.
17

67
(5) Adwoba and Esi formed a company and agreed that their profit will be shared in the ratio
of 4:5 respectively. At the end of the year, Esi received $500 more than Adwoba. How
much profit did they make in the year?

Solution

Method1

Total ratio is 4 + 5 = 9

Let the total profit be x then

4 5
Adwoba’s share is x , Esi’s share is x Since the difference between them is 500, then
9 9
we have:

5 4
x − x = 500
9 9
5 x − 4 x = 9  500 ,
x = 4500

Method 2

Adwoba: Esi = 4:5

Differnce between the ratios is 5 − 4 = 1. Let x be the total profit, then we have

1
x = 500
9
x = 9  500 = 4500

68
Exercise 1.11.0 Ratios

(1) Amina, Esi and Afua shared ¢170m in the ratio 19 : 29 : 37 respectively. Find each
person’s share

1 3 1
(2) Three boys shared ¢150m in the ratio : : . Find the difference between the least
5 10 2
and the highest share.

(3) Divide ¢75m in the ratio 2 : 3 : 5

1 3 7
(4) When 720 apples are shared in the ratio : : find each person’s share.
5 10 10

1 1 1
(5) Three students shared ¢169,000 in the ratio : : find the least amount.
2 3 4

In the following questions, find the value of x.

(6) 5 : (3x + 2) = 7 :(5x − 2)

(7) (2x −1) : 8 = (3x +1) : 4

(8) (x + 3) : (3x − 4) = 2 : 3
(9) 6 : (3x − 1) = 1 : 5

(10) 8 : (x + 4) = 6 : (x − 4)

(11) 1 : (x − 1) = 1 : (2x − 1)

1.12.1. Proportions

Two variables x and y are said to be proportional to each other if the ratios of any two
corresponding variables are the same.

Example1

(1) A man is paid GH¢50.00 for 10 days’ work. Find his pay in

(a) 1 day (b) 20 days (c) 5 days

Solution

69
 (a) 10 days = 50

1
:. 1 day =  50 = GH¢5
10

(b) 20 days = 20  5 = GH¢100

(c) 5 days = 5  5 = GH¢25

(2) 15 boys take 60 days to do a piece of work, how many boys will be required to complete
the work in 20 days if they work at the same rate?

Solution

60 days = 15 boys

60
20 days =  15
20

= 45 boys

This type of questions requires logic. If 15 boys will take 60 days and we want to reduce the
number of days, then we must require more men to do this.

(3) Three books of the same kind cost ¢189,000. Find the cost of 5 such books

Solution

5
3 = 189,000 , :. 5=  189,000 = ¢315,000
3

(4) A piece of land has enough grass to feed 15 Cows for 4 days. How long will it lasts for 6
Cows?

Solution

15
15 Cows = 4days , : . 6 cows =  4 = 10 days
6

Certain types of ratios consist of three quantities. We shall consider such questions

70
Example 2

(1) If 15 men working independently and at the same rate manufacture 27 bags in an hour,
how many bags would be manufactured by 45 men working independently and at the
same rate in 40 minutes?

Solution

Men bags Time

15 27 60 minutes

45 - 40 minutes

15 men = 27 bags

45 new time 4 5 8 40
45 men =  27  , =  27  = 54 bags
15 old time 1 5 60

1
(2) If a worker can pack of a carton of fruits in 15 minutes and there are 40 workers in a
6
2
factory, how many cartons should be packed in the factory in 1 hours?
3

Solution

2 2
1 hours = 60 minutes  60 = 100 minutes
3 3

Workers Cartons Time

1
1 15 minutes
6

40 - 100 minutes

1
1= cartons
6

71
 40 1 new time
:. 40 workers =  
1 6 old time
4
4 0 2 0 1 100 400 4
=   , = = 44 Cartons
1 6 3 1 5 3 9 9

(3) Five teachers can mark 26 scripts in 3 days, how many scripts can 15 teachers mark in
150 days

Solution

Teachers Scripts Days

5 26 3

15 - 150

5 = 26

1 5 3 150
:. 15 =  26  = 26  150 , 3900 scripts
5 1 3

(4) A contractor laying 200 metres of drain pipe requires 12 men working for 10 days. What
length in metres would be laid by 15 men in 9 days assuming that they all work at same
rate?

Length Men Days

200 12 10

- 15 9

12 men = 200

15 new time
15 men =  200 
12 old time

15 93
=  2 0 0 5  = 15  5  3 , = 225 metres
1 2 4 10

72
Proportion
In this section, either a fraction of work or the total number of days required to do a piece of
work by a person (or group of persons) will be given and you will be required to use the given
information to perform certain calculations. You will need the information below:
a
1) If a person can perform a piece of work in number of days, to find the fraction of work
b
that can be done in a day, you just turn the given fraction upside down. Thus the fraction
b
of work that can be done in a day is
a

Examples.
(a) A person will take 7 days to complete a work, find the fraction of the work that can
1
be done in a day . He can do of the work in a day.
7
1
(b) What fraction of a work can be done in a day by a man if he will use 17 days to
2
35
complete the whole work? Here the fraction to improper to get . Hence the
2
2
fraction of the work that can be done in a day is .
35

c
2) Also if the fraction of work that can be done in a day by a person is , then the total
d
d
number of days required to perform the whole work is .
c

For example,

1
(c ) The fraction of work that can be done in a day by a man is then It will take 7 days to
7
complete the whole work.
2 15 1
(d) If the fraction of work to be done in a day is , then it will take him or 7 days
15 2 2
to complete the whole work.

We now take complete look at a question in this section.

3) A father and a son working together can do a piece of work in 5 days. If the father alone
1
can do the work in 6 days, how long will it take the son to do this work?
2
1
The fraction of work that can be done by both is .
5

73
 2
The fraction of work that can be done by the father alone is
. Then the fraction of the
13
1 2 13 − 10 3
work that can be done by the son alone is − = = . Then the son alone will
5 13 65 65
65 2
take days or 21 days.
3 3

4) Alhassan, Ato and Atsu, working together can do a piece of work in days. If
Alhassan alone can do it in days and Ato alone can do it in 5 days, how long will it take Atsu
alone to do the work?

Solution
5
The fraction of work that can be done in day by all the three is .
9
2
The fraction of work to be done Alhassan is and the fraction of work to be done by
9
1
Ato is .
5
2 1 10 + 9 19
Alhassan and Ato together can do + = = in a day so, we now find what
9 5 45 45
5 19 25 − 19 2
Atsu can do in a day. So the fraction of work that Atsu can do is: − = = .
9 45 45 45
45 1
Then the number of days that he will use is or7 days
2 2

5) Faraday alone can do a piece of work in 5 days and his wife Paulina also can do it 8 days.
If they work together, how long will it take them to complete the job?

We first find the fraction of the work that can be done in a day by each .
1 1
Fraction in a day by Faraday is and fraction that Paulina can do in a day is .
5 8
Working together, the fraction that can be done in a day is obtained by adding the two.
1 1 8 + 5 13 40 1
The fraction is + = = . Hence the number of days required is or 3 .
5 8 40 40 13 13

74
Exercise 1.12.0 Proportion

(1) A contractor laying 500 metres of drain pipe requires 20 men working for 10 days. What
length in metres would be laid by 50 men in 5days?

(2) 25 teachers can mark 200 scripts in 50 days. If 5 teachers are required to mark 600 scripts
how long will it take them if we assume they work at the same rate?

(3) If a worker can pack 15 cartons of fruits in 5 minutes and there are 50 workers available,
1
how many cartons will they pack in 2 hours?
4

(4) If 5 men can paint 75 houses in 20 days, how long can 15 men take to paint 410 houses?

(5) Three masons can build 25 rooms in 4 days. If there are 20 masons, how many rooms can
be built in 36 days?

1
(6) If a worker can pack cartons of fruits in 20 minutes and there are 50 workers, how
6
many cartons can be packed in 6 hours?

(7) 15 carpenters can make 27 tables in an hour, how many tables can 75 men make in 20
minutes?

(8) 27 students can pick 945 oranges in 20 minutes. If there are 30 students, how many
oranges can be picked in 1 hr?

(9) 16 workers can make 60 baskets in 75 minutes. How many baskets can 48 workers make
1
in 7 hours?
2

(10) A typist can type 25 pages in 10 minutes if there are 26 typists and they work at the same
2
rate, how many pages can they type in 1 hours?
3

75
1.13.0 Percentages

Under percentages, we shall first consider equivalent fractions. Consider the fractions below

2 20
and . The two are said to be equivalent since
5 50

20 2
Since =
50 5

A fraction whose denominator is 100 is called a percentage

x 1
= x% where % =
100 100
20 35 125
For example, = 20% , = 35% ,, = 125%
100 100 100

1.13.1 Changing Fractions to Percentages

To change a fraction to percentage, we multiply the fraction by 100%

Example 1
Change the following fractions to percentages.
2 7 1 15 1
(1) (2) (3) 2 (4) (5) 12
5 8 4 2 3

Solution
2 2
(1) =  1 0 0 % 20 = 2  20% = 40%
5 5
7 7 700
(2) =  100% = % = 87.5%
8 8 8
1 9 9 900
(3) 2 = =  100% = % = 225%
4 4 4 4
15 15
(4) =  100% = 15  50% = 750%
2 2
1 37 37 3, 700
(5) 12 = = 100% = % = 1, 233.3%
3 3 3 3

76
Exercise 1.13.1
Convert the following fractions to percentages.
1 3 1 6
(1) 3 (2) 16 (3) 12 (4)
2 4 5 7
1 15 2 1
(5) (6) (7) 15 (8) 17
9 7 5 4
1 3
(9) 3 (10) 18
7 4

1.13.2 Changing Decimals Fractions to Percentages

To change a decimal fraction to a percentage we multiply by 100% .

Example1. Change the following to percentages.

(1) 0.25 (2) 16.25 (3) 123.738 (4) 0.005
(5) 1.675

Solution
(1) 0.25 = 0.25 × 100% = 25%
(2) 16.25 = 16.25 × 100% = 1625%
(3) 123.738 = 123.738 × 100% 12373.8%
(4) 0.005 = 0.005 × 100% = 0.5%
(5) 1.675 = 1.675 × 100% = 167.5%

Thus if we want to change a fraction to a percentage, we can change the fraction to decimals
before multiplying by 100%.

Example 2
2
(1) = 0.4 = 0.4  100% = 40%
5
8
(2) = 0.5333 = 0.5333 × 100% = 53.33%
15

77
 3
(3) 15 = 15.75 = 15.75 × 100% = 1575%
4
We can also change from percentages to fractions or decimals

Example 3
Change the following percentages to decimals
(1) 75% (2) 143% (3) 1573% (4) 8%

Solution
75 143
(1) 75% = = 0.75 (2) 143% = = 1.43
100 100
1573 8
(3) 1573 = = 15.73 (4) 8% = = 0.08
100 100

Example 4
Change the following percentages to fractions:
(1) 255% (2) 17% (3) 75% (4) 0.5%
(5) 875

Solution
255 11 17
(1) 255% = =2 (2) 17% =
100 20 100
75 3 0.5 0.5  10 5 1
(3) 75% = = (4) 0.5% = = = =
100 4 100 100  10 100 20
875 75 3
(5) 875% = =8 =8
100 100 4

Exercise 1.13.2
Change the following percentages to fractions
1 3
(1) 12 % (2) 125% (3) 0.25% (4) 15 %
2 4
1
(5) 17 % (6) 80% (7) 90% (8) 25%
2
(9) 85% (10) 755%

78
1.13.4 Applications of Percentages
Under percentages, we shall consider profit and loss, simple interest and other general questions
on percentages.

General questions under percentages

(1) In a class of 50 students, 35 are males. Express the number of males as a percentage of
the class populations.

Solution
35 7
Fractions of male = =
50 10
7
Percentage of male =  100% = 70%
10
1
(2) 510 exercise books were shared among schools X, Y, and Z such that Z has 1 times as
2
much as Y and Y has 3 times as much as X. express X’s share as a percentage of the total.

Solution
We need to find X’ share first

Method I
x + y + z = 510 ………. (1)
1 3 2z
z =1 y ,  z= y,  = y ………….. (2)
2 2 3
and y = 3x ………. (3)
put (3) into (2) to get:
2z 9x
= 3x ,  z= …………. (4)
3 2
Substituting these into equation (1), we have
9
x + 3x + x = 510 , 2x + 6x + 9x = 2  510 , 17 x = 1020
2
1020
:. x = = 60
17
79
 60 600
Hence x’s share as a percentage of total  100 % = % = 11.76%
510 51

Method II
We will use ratio to find x’s share:
x y z
9
x 3x x
2
9
X : Y : Z = x : 3x : x i.e. dividing by x, we have:
2
9
= 1: 3: multiply by 2
2
=2:6:9
2
total ratio = 2 + 6 + 9 = 17 , x’ share =  510 = 60
17
60
X’s as a percentage, we have  100% = 11.76%
510

(3) An employer pays three workers X, Y and Z a total of GH¢ 610.00 a month X is paid
125% of the amount Y is said and 80% of the amount Z is paid. How much does X receive
a month?

Method I
X + Y + Z = 610 ………… (1)
x = 125% of y
125 5 4
= y x= y, :. y = x ………… (2)
100 4 5
80 4 5x
x = 89% of z, = 2, =  2 , :. z= ………… (3)
100 5 4
Putting these into (1), we have
4 5
x+ x + x = 610
5 4
4 5
20 x + 20  x + 20  x = 20  610
5 4

80
20 x + 16 x + 25 x = 12, 200 , 61x = 12, 200
12, 200
:. x = = 200 Thus x is paid GH¢200
61

Method II
We can use ratio.
Write all in terms of x
x y z
4 5
x x x
5 4
4 5
X : Y : Z = x: x : x (Divide by x)
5 4
4 5
=1: : (Multiply by 20)
5 4
4 5
= 20 :  20 :  20
5 4
= 20 : 6 : 25
Total ratio = 20 + 16 + 25 = 61
20
x’s share =  6,100, 000
61
= 20  100,000
= ¢2,000,000

(4) If the price of petrol is increased by 20% and then later decreased by 20%. What is the
effect of price of petrol?

Solution
Since we do not know the original price, we can choose any value for it.
Let the initial price = 100
20
Increase of 20% =  100 = 20
100
New price 120

81
 Decrease of 20% the new price = 20% of 120
20
 120 = 24
10 0
Resulting price = 120 – 24 = 96
This is a net decrease of 100 − 96 = 4
change
To express this as a percentage =  100%
original
4
=  100% = 4%
100
Thus there was a decrease of 4%

(5) Nana withdraws 15% of the amount in her savings account. If she must deposit
GH¢45.00 to bring the amount in the savings account back up to the original amount,
find the original amount.

Solution
Let x be the amount in her savings account.
15% of x = 45
15
 x = 45
100
15x = 45 100
    300
4,500
x= = GH¢300

15

1.13.5 Profit and Loss

Profit = selling price – cost price
Loss = cost price – selling price
Profit
Profit% =  100%
cost price
loss
Loss % = 100%
cost price

82
Example 1
(1) An article which costs GH¢25 was sold for GH¢30, find:
(a) the profit (b) the profit percent

Solution
(a) Cost price (C.P) = 25
Selling price (S.P) = 30
Profit = S.P – C.P = 30 – 25 = 5
Profit
(b) Profit% =  100%
C.P
5
=  100%
25
= 20%

(2) Goods, that cost GH¢50 was sold for GH¢30, find
(a) the loss (b) the loss percent

Solution
Cost price, C.P = 50
Selling price S.P = 30
(a) loss = 50 – 30 = 20
loss
(b) loss percent =  100%
C.P
20
 100% , = 40%
50

Sometimes, the profit or loss percent will be given and you will be required to find the selling
price.

Example 2
(1) Goods that cost ¢50 was sold at a profit of 20%, find the selling price.

Solution

83
 Method I
Profit = 20% of 50
20
=  50 = 10 Thus selling price 50 + 10 = 60
100

Method II
Profit = 20% of C.P
To find the selling price add Profit % to 100% = (20 + 100)% = 120%
Hence S.P = 120% of 50
120
 50 = ¢60
10 0
If is a loss percent, subtract from 100%

(2) A book which cost ¢40 was sold at a loss of 15%. Find the selling price.

Solution
Method I
Loss = 15% of 40
15
 40 = ¢6
100
Selling price = 40 – 6 = ¢34

Method II
Selling price = (100 − 15)% of C.P
= 85% of 40
85 3400
=  40 ,=
100 100
= ¢34

(3) A phone which cost ¢400 was sold at a profit of 25% find
(a) the profit (b) the selling price.
Solution
(a) Profit = 25% of 400

84
 25
=  40 0 = 25  4 = ¢100
10 0
(b) selling price = 400 + 100 = ¢500

(4) An article which cost ¢600 was sold at a loss of 5%. Find the
(a) loss (b) selling price

Solution
5
(a) loss = 5% of 600 =  600 = 5  6 = ¢30
100
(b) selling price = 600 – 30 = ¢570

There is a third type where the profit or loss percent will be given and you will be given the
selling price. You will be required to find the cost price.

Example 3
(1) A book was sold at GH¢240 at a profit of 20% find the cost price.

Method I
Let the cost prices = x
Then profit = 20% of x
20
=  x = 0.2 x
100
Selling price = x + 0.2x =1.2x

 1.2 x = 240

240
x=  x = GH ¢200
1.2

Method II
Selling price = 20% of cost price.
Let the cost price = x

85
 120
Then 120% of x = 240 , x = 240 , 1.2x = 240
100
240
x= ,  x = GH ¢200
1.2

(2) The selling price of pens is GH¢1.15 at a profit of 15%. Find the cost price.
Solution
(a) (100 + 15)% = 115%, 115% of x = 1.15
115 1.15
 x = 1.15 , 1.15x = 1.15 , x= = ¢1.00
100 1.15

(3) A calculator was sold at GH¢17 at a loss of 15%. Find the cost price.
Solution
Selling price = (100 – 15)% of cost price
= 85% of cost price
85
Hence 85% of x = 17 ,  x = 17
100
17
0.85x = 17, x= = GH¢20
0.85

(4) A book was sold at GH¢5.00 at a loss of 20%. Find the selling price.
Solution
Selling price = (100 – 20)% = 80%
80% = 5.00
100
100% =  5 = ¢6.25
80

Exercise 1.13.5 Profit and loss

(1) Find the profit or loss and profit or loss percent:
(a) The cost price of an item is GH¢15.00 and selling price GH¢18.00;
(b) The cost price of an item is GH¢21.00 and selling prices GH¢25.00;
(c) The cost price of an item is GH¢18.00 and selling price is GH¢25.00;
(d) The cost price of an item is GH¢35.00 and selling price is GH¢30.00.

86
(2) Find the selling price of an item, if:
(a) The cost price is GH¢18.00 at a profit of 10%;
(b) The cost price is GH¢25.00 at a profit of 50%;
(c) The cost price is GH¢40.00 at a loss of 25%;
(d) The cost price is GH¢50.00 at a loss of 15%;
(e) The cost price is GH¢75.00 at a loss of 30%.

(3) Find the cost price of an item, if:

(a) The selling price is GH¢15.00 at a profit of 10%;
(b) The selling price is GH¢25.00 at a profit of 25%;
(c) The selling price is GH¢60.00 at a loss of 40%;
(d) The selling price is GH¢12.00 at a loss of 20% .

1.13.6 Simple Interest

Principal is the amount which a lender gives to the borrower. If the interest on the principal is
fixed, then we have simple interest. For example, one can borrow ¢500 and will be asked to pay
¢10 at the end of each month as interest. Since the person is paying ¢10 at the end of each month,
we say that the ¢10 interest per month is a simple interest.

Formula for simple interest

Let P – Principal, R – Rate, T – Time, then
P  R T
Interest (I) =
100
Amount = Principal + Interest
Example
(1) Find the simple interest on GH¢50 at 12% per annum for 5 years
Solution
P = 50 , R = 12, T=5
P  R T 50  12  5
I= = = GH¢30
100 100
Interest = GH¢30

(2) Find the simple interest on GH¢150 at 18% per annum for 8 months.

87
 Solution
Since the rate = 18% per year, change the time to year.
8 2
8 months = = yrs
12 3
2
Time T = , Principal, P = 150, Rate, R = 18%
3
P  R T 150 18  2 5400
I= = = = GH¢18
100 100  3 300

(3) At what rate can GH¢200 yield an interest of GH¢40 in 5 years.

Solution
Principal, P = 200, Interest, I = 40, Time, T = 5, Rate R = ?
P  R T 200  R  5
I= :. 40 =  40 = 10 R :. R = 4
100 100
Thus the rate of interest = 4% per annum

(4) In what time can GH¢250 yield an interest of GH¢20 at the rate of 5% per annum.
Solution
Principal, P = 250, Interest I = 20, Rate, R = 5%, Time T = ?
P  R T 250  5  T
I=  20 =
100 100
2000
 2000 = 1250 T ,  =T
1250
 T = 1.6yrs

Exercise 1.13.6 Simple Interest

(1) Find the simple interest on GH¢500 for each of the following:
(a) 3 years at 15% per annum;
1
(b) 8 years at 12 % per annum;
2
1
(c) 15 months at 6 % per annum;
2
1
(d) 9 month at 3 % per annum;
2

88
(2) In what time can GH¢750 yield:
(a) GH¢25 interest at 15% per annum;
(b) GH¢45 interest at 25% per annum;
(c) GH¢150 interest at 10% per annum.
(3) Find the rate at which GH¢150 can yield:
(a) interest of GH¢5 in 8 years;
(b) GH¢15 interest in 2 years;
(c) GH¢25 interest in 3 years.

1.14. Compound Interest Theory

Interest is the sum paid as compensation to those who lend money and who thus give up the
opportunity of using it themselves and, therefore, expect to be reimbursed.
If we let
P be the principal, i.e. the initial sum or investment;
r be rate of interest per a given period;
t be the number of time periods.
Then, we have:
Simple interest (i):
i = P r t .
The total amount A is the principal plus the accrued interest i.e. A = P + i .

Example
The simple interest on GHC 500.00 for 2 years at 3% is:
3
i = P r t = 500   2 = GHC 30 .
100
When interest is added to the principal at the end of each period, it is converted into principal.
The total amount due at the end of the period is called the compound amount.
Compound interest is the difference between this compound amount and the original principal.
If the initial investments is P, after one period time, it becomes P (1 + r ) . After two periods time,
it is P (1 + r )(1 + r ) = P (1 + r ) .
2

The amount of money accumulated at the end of n compounding periods is

A = P (1 + r ) .
n

89
Example
The compounded amount of GHC 500.00 after 5 years at 5% is

A = P (1 + r ) = 500 (1 + 0.05 )
n 5
= GHC 638.30 .

Example
The compounded amount of GHC 3,000.00 after 4 years at 3% interest each half-year,
compounded semi-annually is

A = P (1 + r ) = 3, 000 (1 + 0.03)
n 8
= GHC 3,800 .

Since there are 8 periods of six months each i.e. semi-annually, n = 4  2 = 8 .

Modification of Compounded Amount

A principal P compounded m times a year for n years at the annual interest rate r grows to an
amount:
mn
 r
A = P 1 + 
 m
Example
The compounded amount of GHC 500.00 quarterly, after 5 years at 5% interest per annum is
mn 54
 r  0.05 
= 500 (1 + 0.0125 ) = GHC 641.02 .
20
A = P 1 +  = 500 1 + 
 m  4 

Equivalent Rate of Interest

Interest rates resulting in equal amounts in the same length of time on the same principal are
called equivalent rates.
Let i be the effective rate equivalent to a given nominal rate r, compounded m times a year.
Since equivalent rates result in equal amounts in the same length of time, we have:
m
 r
P (1 + i ) = P 1 + 
 m
m
 r
 1 + i = 1 + 
 m
m
 r
 i = 1 +  − 1
 m

90
Example
The equivalent of 8% compounded quarterly is obtained by writing
r 0.08
m=4 , = = 0.02. Therefore we have,
m 4
m
 r
(1 + 0.02 )
4
i = 1 +  − 1 = − 1 = 0.0824
 m
The effective rate of interest is 8.24%.

 Activity 4.5
(a) Find the (i) simple and (ii) compound
interest at 8% converted quarterly of GHC 2,000 invested for 1 34 years?

(b) Find the effective rate equivalent to 6% nominally converted

(i) semi-annually (ii) quarterly (iii) monthly (iv) bi-monthly.
(c) If money is invested for 8 years, what should be the rate of interest for this money to (i)
treble (ii) quintuple?

[Hint: For (i), you may do this by writing 3P = P (1 + r ) , and solve for r.]
3

91
 CHAPTER TWO
MATHEMATICAL REASONING AND SET THEORY

Inductive and Deduction reasoning

Introduction
Mathematics has a well defined language and rules of operation. It provides one of the best
methods of dealing with complex analytical problems. Mathematics deals with abstract general
models, which is concerned with their internal logic and not their application. Mathematics is a
tool which can be used by each and everyone in his or her own field of study or application.
Every idea developed in Mathematics need not necessarily have a direct application to the
science under consideration but it can be relevant to the development of some principle.

Mathematics enables us to see the structure of an economic or social science model in a clear,
precise and compact form. In any such study, mathematics begins with the empirical content of
the problem, makes use of deductive reasoning and arrives at empirical conclusions or evidence
whose validity depends on the original empirical statement. Mathematics is not committed in the
choice of the model but it is very much necessary in discovering how the model fits the problem
under consideration. Reasoning inductively, or deductively, is an essential element in all
mathematical work.

What is Inductive Reasoning?

Inductive reasoning consists of conclusion drawn on the basis of observations.

What is Deductive Reasoning?

Deductive Reasoning proceeds from assumptions and not from observations?

Induction is often the road to some discovery. In the course of inductive reasoning, one guesses,
makes observations and may be led to certain general truths or principles which cannot be
accepted outright. They must be proved. Proofs are more often arrived at through deduction
reasoning. The deductive arguments, used in proofs must be consistent and convincing.

Mathematical work requires the use of both deductive and inductive reasoning. Pure deduction is
impossible is social science which deals with a certain aspect of reality. Judgments on reality can
be based on induction and hence on what may be called intuition in the arts or social sciences.

The Logic of Reasoning

Mathematical logic is the art of reasoning which helps us to systematize the principles of valid
reasoning and which may be used to judge the correctness of a claim of reasoning. Mathematical
reasoning is formal i.e. it is based on the form of the sequence of the statements. It gives us a
mathematical theory as a logical system.

Philosophers believe that mathematics is a part of logic whereas mathematicians feel the
opposite. In either case, the importance of logic cannot be over emphasized. Logical reasoning is
a must in modern society. Intuition and accidental results may be prominent at times but even in
such case, logic can still play an important role. The ability to reason logically reflects
92
intelligence. This ability can be improved upon and nurtured, though sometimes a genius may
have it as a natural gift.

Real world observations which are sometimes undefined or defined motives of the behaving
units, give rise to propositions. The complex relationships in the behaviours of human beings and
of nations along with these propositions lead to predictions, recommendations and various types
of suggestions. A consideration of economic phenomena, for example, in terms of variables like
prices, production levels, taxes etc. although subject to severe limitations and practical
difficulties, is helpful in giving a systematic formulation for various activities of the economic
system. Logic brings economic analysis and mathematical reasoning together to help draw
conclusions form a set of postulates. Logical reasoning, used along with the characteristics of the
economic system can be useful in understanding complex economic phenomena more clearly.

Statements and Truth Tables

Statement:
A statement (also called an assertion or a proposition), either written or oral, is a sentence which
is either true or false. Note that a statement cannot be true and false at the same time.

The statement ‘Kofi is a young man’ has a truth value. The truth value of a statement is its
truthfulness (T) or falsehood (F). The truth value of a statement is a relation between the subject
(in this case, Kofi) and the predicate (in this case, Young man)

A statement is usually denoted by lower case letters like p, q, r, etc.

Compound (or composite) statements:

A statement which is the combination of two or more simple statements, combined by
connectives to give compound or composite statements.

The truth value of a statement is completely determined by the truth value of each of its sub-
statements and the way they are connected.

If we write all the possible truth values or a given statement for all possible combinations of truth
values of its components, we get a device known as the truth table which is useful in considering
the validity of the statement.

Types of Statements
Conjunction p  q
Conjunction is the joining of two statements p and q by the connective ‘and’. The statement ‘p
and q’ written p  q is called a conjunction of p and q.

If statement p is true and statement q is true, then p  q is true. Otherwise p  q is false.

93
Truth table for ‘ p  q ’
p q pq
T T T i.e. p true, q true  p  q true;
T F F p true, q false  p  q false;
F T F p false, q true  p  q false;
F F F p false, q false  p  q false.

Disjunction p  q
Disjunction is the joining of two statements by the connective ‘or’ used in the inductive since.
p  p ( ie., p or q ) is called the inductive disjunction of statements p and q.
If both p, q are false, p  q is false; otherwise p  q is true. There are four cases shown in the
truth tables.

Truth table for ‘ p  q ’

p q pq
T T T
T F T
F T T
F F F

Example 1.1
Given the following statement:

p : Rent is high and q: price of foodstuffs is low

then,
p  q : Rent is high and price of foodstuffs is low;
p  q : Rent is high or price of foodstuffs is low.

Negation ~ p
Conjunction and disjunction are binary compositions, i.e. combining two statements to give a
new one. Negation just turns a statement into a new one; it is unary. Negation consists of adding
to or removing from a sentence (or statement) the word “not” to form a new statement.
If p is a statement, ~ p (i.e. not p) is a statement such that

If p is true, then ~ p is false.

If p is false, then ~ p is true.

The truth value of the negation of a statement is just the opposite of the statement.

94
Example 1.2
p: rent is high ~ p : Rent is not high
p: sometimes rent is high ~ p : Rent is never high
p: some rents are not high ~ p : all Rents are high
p: Rent and foodstuff are high ~ p : rent or foodstuff is not high.

Conditional p  q
A conditional statement has two components: the antecedent or hypothesis and the consequent.

If p, q are statements the p  q (read ‘p implies q’ or, ‘if p then q’) is called an implication or a
conditional. p  q is false when p is true and q is false. Otherwise it is true.

p q pq
T T T
T F F
F T T
F F T

Converse q  p
The implication q  p (i.e. ‘if q the p’) is called converse of the implication p  q

Inverse ~ p ~ q
The implication ~ p  ~ q (i.e. ‘If not p then not q’) is called the inverse of the
implication p  q .

Contra positive ~ q ~ p
The implication ~ q  ~ p (i.e. ‘If not q then not p’) is called contra positive of the
implication p  q .
The contra positive method of proofs begins by assuming the false result to be truth and arrives
at a contradiction. This is the famous ‘reduction ad absurdum’ method of proving results in an
indirect manner.

Equivalent statements 
Equivalent statements are true together. They have identical truth tables.

Example 1.3
(i) p ~ ( ~ p ) (ii) ~ ( p  q )  ( ~ p ~ q )
(iii) ( p  q)  p (iv) ( p  q)  ( p  q)
(v) ( p  q)  p (vi) ( p  q)  ( p  q)
Example 1.3
95
Show that:
The statement p  q and its contrapositive ~ q  ~ p are equivalent.
The converse q  p and the inverse ~ q  ~ p are equivalent.

These equivalent statements can be verified from the truth table as follows:

p q ~p ~q pq q p ~ p ~ q ~ q ~ p
T T F F T T T T
T F F T F T T F
F T T F T F F T
F F T T T T T T

Example 1.4
Given the statements: p: Price is high , q: Demand is low ;
Form (i) Implication (ii) Converse (iii) Inverse (iv) Contrapositive
statements using p and q.

Implication : If price is high then demand is low

Converse : If demand is low then price is high
Inverse : If price is not high then demand is not low
Contrapositive : If demand is not low then price is not high

Tautology
A statement logically true in all possible cases is a tautology.

The following are instances of tautology: p  p , p   q  ( p  q ) 

Contradiction
A statement that is logically false in all possible cases is a contradiction.

The following are instances of a contradiction: p ~ p , p ~ p .

Necessary and Sufficient Conditions

In the implication p  q :
(i) If statement p is satisfied, the statement q is true. Statement p is said to be a
sufficient condition for q because the knowledge that p is true is sufficient for
knowing q is true.

(ii) p cannot be true unless q is satisfied. Statement q is said to be a necessary condition

for p. q may be necessary for p but may not be sufficient for p.

96
 e.g. Attending lectures in Numeracy Skills may be necessary to pass the
examinations but may not be sufficient for it.

If q is both necessary and sufficient condition for p, the converse of the original implication is
true. Thus for a necessary and sufficient condition both the implication and its converse must be
true i.e. p  q and q  p . For this purpose we use the phrase “If and only if” in forming a
statement. We thus write p  q to mean ‘p is true if and only if q is true.

Example 1.5
Consider the statement:
Only if a student is active he should be the Class Representative.

This means: If a student is the Class Representative, then he should be active. Being active is a
necessary condition to being a leader but not a sufficient condition.

Exercise
1) Prove that the inverse and the contra positive of an implication are not
equivalent. (Hint: Use truth tables).

2) Prove that the converse and the contra positive of an implication are not
equivalent. (Hint. Use truth tables).

3) State the inverse, converse and contra positive of the following

implication: If x is an integer then 2x is an even integer.

4) Given the statement:

p: The per capita Income is low q: Ghana is poor.
Give verbal representations of the following symbolic forms:
(i) pq (ii) pq (iii) p ~ q
(iv) ~ pq (v) ~ ( p  q )  ( p ~ q )

5) Consider the following statement:

‘If I pass the Numeracy Skills examinations, then I shall invite you to a
party’.

Judge the following conditions if this statement is true or false:

(i) I pass the Numeracy Skills examinations and invite you to a party.
(ii) I pass the Numeracy Skills examinations and do not invite you to a
party.
(iii) I do not pass the Numeracy Skills examinations and invite you to
party.
(iv) I do not pass the Numeracy Skills examinations and do not invite
you to a party.
97
6) Comment on the statement:
A professor is good if and only if he imparts knowledge to his students.

7) Comment on the statement:

If either labour or management is stubborn, then the strike will be settled if and
only if the government obtains an injunction, but troops are not sent to the factory.
If we use the symbols:
L: Labour is stubborn
M: Management is stubborn
S: Strike will be settled
G: Government obtains an injunction
R: Troops are sent to the factory.

(a) Write out he statement in symbolic form

(b) By a truth value analysis, determine whether the above statement is true or
false under each of the follow assumptions:
(i) Labour is stubborn, management is not, the strike will be settled,
the government obtains an injunction and troops are sent to the factory
(ii) Both lobour and management are stubborn, the strike will not be
settled, the government obtains an injunction and troops are sent to the
factory.
(iii) If the government obtained an injunction the troop will be sent to
the factory and the strike will not and settled.

8) Comment on the validity of the following statement:

(i) If our exports do not exceed faster than our imports, then foreign
exchange will centime to flow out. Foreign exchange stops flowing
not. Thus exports have increased faster than imports.

(ii) Either businessmen are honest or there is a parallel economy.

There is no parallel economy. Therefore businessmen honest.
(iii) Either the country should import more goods or it should stat
manufacturing goods. The country is not at all thinking of import
therefore the country is going to start manufacturing goods.

98
Definition of a Set and some Examples

Introduction
In mathematics, the idea of a set is fundamental and all mathematical objects and constructions
have a basis on set theory. In the social sciences and the arts, we encounter sets of data, sets of
items produced, sets of outcomes of decisions etc.

Words like group, family, association, club, flock, church, spectators, choir, team, etc. are often
used to convey the idea of a set in everyday life. Enumeration and measurement lead to numbers
and sets whose study gives us a good insight into the depths of a subject under study.

Definition of a set
A set is a collection of any type of objects, items or numbers usually connected with grouping.

The constituents or members of a set are called elements of that set. The elements may either be
listed by enclosing them in curly brackets   or they may be described by a statement of the
properties of the elements within the set.

A set is known by its elements. A set is well defined if and only if it can be decided that a given
object is an element of the set. Capital (or upper case) letters A, B, C etc. are usually employed
to denote sets. Elements of sets are usually denoted by lower case letters, a, b, c e.tc. or shown by
description.

Example 2.1
A = 2, 4, 6, 8 is a set of even numbers 2, 4, 6, 8 . We may say A is a set of all even numbers
between 1 and 9 ; or 2 , 4 , 6 , 8 are element of A.
Example 2.1 gives the Roster or tabulation method to describe a set. It consists of listing each
element of the set within the brace or brackets.

Another method is the descriptive phrase method which consists of placing a phrase describing
the elements of the set within the brace. Thus A = {even numbers between 1 and 9}. This
method may be used when there is a large number of elements or when all the elements cannot
be named.

A third method known as the rule or set builder method consists of enclosing within the brace or
brackets, a general element and describing it.

Thus we have:
A =  x x is an even integer, 1  x  9 or A =  x is even integer, 1  x  9 .

A set is determined by its elements and not by the method of description. The set A described in
any of these three different ways, remain the same.

The algebra of set provides you with laws or rules which can be used to prove important results
in any field of application.
99
Diagrammatic Representation of Sets
Sets may be represented by shapes of enclosed curves commonly known as Venn diagrams.
Venn diagrams were first introduced in set theory by an English Mathematician known as John
Venn in the early 16th Century. In a Venn diagram, every element in a set is represented by a
point.

Example 2.2
Let V be the set of vowels in the alphabet.

Then
V = a, e, i, o, u
The Venn diagram shows how the vowels which are the elements of the set V belong to the set
V.

a e
i
o u

Types of Sets
Finite and Infinite Sets
A set is finite, if it has a finite number of elements. The elements of such a set can be enumerated
or counted by a finite number. The number of elements in a finite set A is denoted by n ( A) .
Here n is a finite positive integer.

As the name suggests, an infinite set is a set that has infinite number of elements. The elements
of such a set cannot be counted by a finite number.

Example 3.1
Let us see some examples of finite and infinite sets:
(1) F = a , b , c , 0 , 1, 2 . Here F is finite since n ( F ) = 6
(2) S =  x x is grain of sand on the sea shore .
Here S is infinite, since grains of sand cannot be counted or enumerated.
(3) N = n : n is a natural number
= 1, 2, 3, 
:. The set N is an infinite set.

100
Null Set or Empty Set
The null set is the set which contains no elements. A null set is denoted by  or  
A =  x x is married bachelor = 
B =  x x is a planet on earth = 
C =  x x is a Senior High School that is also a university = 

In each of the above examples, there are no elements in them, thus they are all empty or null sets.

Simple Set
A simple set is a set containing only are element. It is also called a singleton or Unit Set.

Example of simple sets

A = a
B = 0
Note that the set B contains a simple meant 0. The set B is not a null set.

Universal Set
The Universal set is the totality of elements under consideration. It is the set of all items relevant
to a particular application. A universal set is denoted by U or  . Its representation on a Venn
diagram is a rectangular box.

Examples of Universal Set

U = a , b , c , , z the set of letters of the English alphabet.
U = head, tail , the set obtained by the throw of a coin.
U = all students taking the Numeracy skills course .

Disjoint Sets
Two Set A and B are said to be disjoint if they have no elements in common.
Example
If A = { all odd numbers} and B = {all even numbers}
then A and B are disjoint.
101
Membership of a Set
If an element x belongs to a set A, we write x  A which is read as ‘x is an element of A’.

If an element x does not belong to a set A, we write x  A , which is read as ‘x is not an element
of A’.

Subset
A subset of a set A is a set which consists of some or all of the elements in A. The set B is a
subset of A if every element of B is also an element of A and we write B  A .

This means that x  B  x  A .

A set B is a proper subset of A if and only if

B  A and B  A and B  

Note that
• Every set is a subset of itself i.e. A  A
• Every set is a subset of the Universal set with respect to the application under
consideration.

Equal or Identical Sets

If every element of set A is also an element of set B, then sets A and B are said to be equal and
we write A = B

Example
If A = 2, 4, 6, 8 and B = 4, 6, 2, 8 then A = B

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Example
Given that A = 2, 4, 6, 8 and B = 2 , 2 , 4 , 4 , 6 , 8 , 8 then A = B since set A contains 4
distinct elements which are the same distinct elements in set B.

However n ( A)  n ( B )

Equivalent Sets
Two sets A and B are said to be equivalent if there is a one-to-one correspondence, between the
element of set A and the of set B. the order in which the element appear in each set is immaterial
It may be noted however. That a one-to-one correspondence exits between set A and B if it is
possible to associate the elements of set A with the elements of set B in such a way that each
element of set A is associates with exactly one elements in se B. we express equivalence
symbolically by writing A  B or A  B .
Equivalent sets have the same number of elements i.e. n ( A) = n ( B )

Example
If A = ( a , b , c ) and B =  ,  ,   , then A and B are equivalent sets.
Since n ( A) = n ( B ) = 3

Note that two sets are equivalent if they have the same of elements and no element is repeated
when the members of the sets are listed.

Operations with Sets: Union, Intersection, Set Difference and

Symmetric Set Difference
Union of the two Sets

The union of the two sets A and B is the set containing all elements belonging to A or B or to
both A and B.
A  B =  x x  A, x  B or  x  both A or B 
Diagrammatically we have

A B

The shaded area in the Venn diagram

represents A  B which shows the set of
items that are elements of at least one of the
sets A and B
A B

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Example 4.1
Let A = a, b, c, d  and B = c, d , e . Find A  B

Solution A B
A  B = a, b, c, d , e a b c
d e

Note that every element in either A or B or both are represented only once in the union.

Intersection
The intersection of two sets A and B is the set containing all elements belong to both
sets A and B

Symbolically, we write
A  B =  x x  A and x  B

Diagrammatically, we have
A B

A B

Complement of a Set
If A is a subset of the Universal set U, then the collection of element belonging to U but not
belonging A is called the complement of A.

We denote this diagrammatically as

The shaded portion in the Venn diagram depicts the complement of A.

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We thus note that the complement of a Set A to a universal set U is the set of all elements of U
which are not elements of A.

The complement of A is written A

A =  x x  A
A is also written as Ac or A

Example
Let U =  x x is any letter in the alphabet be a universal set and
C =  x x is any consonant in the alphabet . Then the complement of C is the set containing the
vowels.
i.e. C = a , e , i , o , u .

Note that if A has complement A then A has a complement A. i.e. ( A ) = A

Two sets are complements of each other if all the elements of the two, together constitute the
universal set.

i.e. C  C = U

The complement of the universal set U contains no element.

i.e.  = U

Similarly
 = U

Example
If U = 1, 2, 3, 4, 5 and A = 1, 3, 4 then
A = 2, 5

Set Difference
The difference of two sets A and B is the set of all elements belonging to A but not to B. written,
symbolically as
A − B =  x x  A , x  B

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Illustrated on a Venn diagram as:

A B

A–B

Example
If A = a, b, c, d  and B = c, d , e, f 
Then A − B = a, b and B − A = e, f 

A B
a c
a c e
A–B e
A b b d f B–A
d f

Symmetric Difference
We define the symmetric difference of two sets A and B are the set of elements either in or B but
not in both A and B. we write the symmetric difference symbolically as A B
i.e. A  B =  x x  A  B, x  A  B

This is depicted on a Venn diagram as:

The shaded portion is the symmetric difference A  B

A B

A–B B–A

A B

Laws of Operations in Set Algebra

Every law is associated with a dual the principle of duality is that, if in any true statement
involving only union, intersection or complementation, ,  are interchangeable and U and 
are interchangeable, then a true statement results.

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These laws can be verified by means of Venn diagrams.

The Idempotent Laws

These laws are also referred to as inclusion or tautology.

Let A be any set. Then

A  A = A and A  A = A

The commutative Laws

Let A and B be any two sets, then

A  B = B  A and A  B = B  A

The Associative Laws

Let A, B and C be any three sets, then
( A  B )  C = A  ( B  C ) and ( A  B )  C = A  ( B  C )

The Distributive laws

Let A and B be any three sets, then

A  ( B  C ) = ( A  B )  ( A  C ) and A  ( B  C ) = ( A  B )  ( A  C )

The Identity Laws

A   =  , A   =  and A   =  , A   = A
Where A is any subset of the Universal set 

Complement laws

A  A =  , ( A) = A
and
A  A =  ,  = ,   =  where  is the universal set.

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De Morgan’s Laws
( A  B ) = A  B and ( A  B ) = A  B

Solving Two- and Three- Set Problems

Set theory can be applied to solve basic real problems. In this section we will illustrate problems
with which set theory can solve and how to obtain solutions to such problems. We will however,
limit our discussions to two- and three- set problems.

Cardinality of set
The cardinality of set is the number of elements that constitute the set. If A is a set then its
cardinality is denoted by n ( A) .

Example
If A = a, b, c, d , e then n ( A) = 5 .
Also, if  is the empty set then n ( ) = 0 .

Cardinally of the set A  B

Let n ( A) be the number of elements in A and n ( B ) the number of elements in B then the
number of element in A  B is

n ( A  B ) = n ( A) + n ( B ) − n ( A  B )

If A and B are disjoint sets,

n ( A  B ) = n ( A) + n ( B )

We can extend this formula to three sets:

n ( A  B  C ) = n ( A) + n ( B ) − n (C ) −n ( A  B ) − n ( B  C ) − n (C  A) +n ( A  B  C )

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Two and Three Sets Illustration
The intersection and union of three sets gives rise to several interesting results. We consider
here, some Venn diagrams for two and three sets.

Note that A  B  C is the same as ( A  B )  C or A  ( B  C ) since we get the same result by

joining the union of any two of A, B, C to the third (see Association laws in Section 5).

An element in A  B  C belongs to at least one of A , B , C.

A B
A B

C
C
A  B C A  B C

Similarly A  B  C is the same as A  ( B  C ) or ( A  B )  C thus an element in A  B  C

is common to all three sets A , B , C.

Example
Given that A = 1, 2 ,3 , 4 , B = 1,3 ,5 ,7 and C = 1,5 ,6 ,8 ,
then A  B  C = 1, 2 ,3 , 4 ,5 ,6 ,7 ,8 and A  B  C = 1

A  ( B  C ) is the intersection of A with the union of B and C.

( A  B )  ( A  C ) is the union of A  B and A  C . We obtain the same diagram as for

A(B C) :
A B

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:. A  ( B  C ) = ( A  B )  ( A  C )

The shades portion in the Venn diagram is the complement of A  B  C . It is denoted

( A  B  C ) . This is the same as A  B  C
A
B

Solving set Problems

Example : Two Sets Problem

A firm has 40 workers working in its factory premises, 30 working in its office and 20 working
in both places. How many workers are there in the firm? How many are working exclusively in
the (i) factory? (ii) office?

Solution
Let F represent those working in the factory;
O represent those working in the office.

Then we have
n ( F ) = 40, n (O ) = 30 and n ( F  O ) = 20

:. n ( F  O ) = n ( F ) + n (O ) − n ( F  O ) = 40 + 30 − 20 = 50

The number working in the factory only in

n ( F only ) = n ( F − O ) = n ( F ) − n ( F  O )
= 40 − 20 = 20

Similarly, the number working in the office only is

n ( O only ) = n ( O − F ) = n ( O ) − n ( F  O ) = 30 − 20 = 10

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Example 2: Two-Set Problem
In a class 64% of the students have taken political Science (P) and 56% students have taken
History (H). How many students have taken both courses?

Solution
The number of students in Universal set in assumed to be 100%

:. n ( P  H ) = 100

n ( P ) = 64 n ( H ) = 56

Let x represent the percentage of students taking

x both course.
:. n ( P  H ) = x

Now
n ( P  H ) = n ( P) + n ( H ) − n ( P  H )

 100 = 64 + 56 − x

Making x the subject, we obtain x = 20.

Examples: Three – set problem

In a survey of 100 students, it was found the 50 used the Balme Library, 40 had their own library
and 30 borrowed books. Of these, 20 used both Balme Library and their own library. 15 used
their own library and borrowed books and 10 used Balme library and borrowed books. How
many students used al three sources of books?

Solution
Let
A represent those students who use Balme Library;
B represent those students who use their own library;
C represent those students who borrow books..

Let x present students who use all the three sources.

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Then we have n = ( A) = 50 n n ( B ) = 40

n( A B C) = x 20 – x
n ( A  B  C ) = 100
x
n ( A) = 50 , n ( B ) = 40 , n ( C ) = 30
n ( A  B ) = 20 , n ( B  C ) = 15 , n ( A  C ) = 10 10 – x
15 – x

n ( C ) = 30
Now n ( A  B  C ) = n ( A) + n ( B ) + n (C )
−n ( A  B ) − n ( A  C ) − n ( B  C )
+n ( A  B  C )

:. we have

100 = 50 + 40 + 30 − 20 −10 −15 + x

:. x = 25

Exercise
1. Comment on the validity of the arguments that follows and construct the truth table. If the
commodity market is perfect, the prices of units of a particular commodity are equal. But
it always happen that prices for such units are not equal. Therefore the commodity market
is not perfect.

2. Give that A = 0 ,1, 2 , 3 ,  . Assume x is a positive integer. State whether the

following statements are true or false.
(i) 10  A (ii) 101 A (iii) 0.5  A (iv) 1.5  A
(v) 0  A (vi) −10  A (vii) 0 = 0 (viii)    0

3. In a certain examination 32% students failed in Economics, 30% in Political

Science, 46% in History, 12% in Economics and Political Science 9% in Political
Science and History, 10% in Economics and History and 3% in all the three, how
many students passed in all the three subjects? How many students failed in exactly
one subject?

4. Shade the following in a Venn diagram.

(i) B  (ii) A  B (iii) ( A  B ) (iv) ( A − B) (v) A  B (vi) A  B
(vii) A  B (viii) A − B

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2.2.3 Two Set Venn Diagram

Consider the diagram below

U

A B

a b c

a belongs to only A
b belongs to both A and B
c belongs to only B
d does not belong to neither A nor B

Formula
n (u ) = a + b + c + d
n (u ) = n ( A) + n ( B ) − n ( A  B ) + n ( A  B)

Examples
(1) In a class of 40 students, 30 read Mathematics and 20 read English. If each student reads at
least one subject, find the number of students that reads (a) both subjects (b) only Mathematics
(c) only English
number in set, n(u) = 40,
mathematics (m) English (E)

n(m) = 30, n(E) = 20

30 + 20 = 50 but the universal set is 40 hence intersection = 50 – 40 = 10

u = 40

E = 20
M = 30

20 10 10

(a) both subjects = 10 (b) only Mathematics = 20 (c) only English = 10

(2) In a class of 36 students, 19 read biology, 16 read chemistry and 5 were not allowed to read
both subjects, find how many students, read
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 (i) both subjects (ii) only one subject

Solution
n(u) = 36, Biology, n(B) = 19 Chemistry n(C) = 16
n ( C  B) = 5
Now 19 + 16 + 5 = 40
:. both subjects = 40 – 36 = 4

u = 36

B = 19
C = 16

12 4 15

(i) both subjects = 4

(ii) one subject = 12 + 15 = 27

(3) In a class, 30 students read English, 25 read mathematics, 5 read both subjects and 6 were not
allowed to read both subjects. How many students are in the class?

Solution

B = 25
C = 30

25 5 20

n ( u ) = 25 + 5 + 20 + 6 = 56

There are 56 students in the class

(4) In a survey of a town, it was found that 65% of the people surveyed watched the news on
television 40% read a newspaper and 25% read a news paper and watched the news on

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 television. What percentage of the people surveyed neither watched the news on television
or read a newspaper?

Solution
Let T – television N – newspaper U = universal set
N(u) = 100% since percentage add up to 100
N(T) = 65% n(N) = 40% n (T  N ) = 25%

n ( u ) = 100

n(N) = 40
N(T) = 65

40 25 15

40 + 25 +15 + a = 100
80 + a = 100
= 20%

Exercise 2.2.3 Two Set Problem

(1) In a class, 50 students read math, 60 read English, 10 read both subjects and 5 were not
allowed to read math nor english subjects. Find the number of students in the class

(2) In a class of 120 students, 40 read history and 70 read geography. If 30 students were not
allowed to read any of the two subjects, find the number of students who read both
subjects.

(3) In a class of 75 students, 30 read math and 40 read English. If the number of students who
read only one subjects is 50 find the number of students who
(i) read both subjects
(ii) do to read both subjects.

(4) In a class of 50 students, 30 offer economics, 17 offer Government and 7 offer neither
Economics nor Government. How many students offer both subjects?

(5) In a class of 20 students, 16 play soccer 12 play hockey and 2 do not play any of the two
games. How many students play only soccer?

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 CHAPTER THREE
INTRODUCTION TO PROBABILITY

History and the concept to Probability

The development of a mathematical theory of probability began in the 17th Century when the
French Noble Antoine Gombauld , known as the Chevalier De Mere, raised certain questions
about games of chance. He was puzzled about the chance of obtaining two sixes at least once in
24 rolls of a pair of dice. De Mere posed the question to a young French mathematician Blaise
Pascal who solved the problem. Following this, Pascal discussed this and other puzzles raised by
De Mere with the famous French mathematician Pierre de Fermat. In the course of their
correspondence, the mathematical theory of probability was born. Given its origin in games of
chance, it is not surprising that the first methods developed for measuring probabilities were
appropriate for gambling situations and most examples used to explain the concepts of
probability involve games of chance. The theory of probability is a mathematical theory, which
deals with laws of chance. It is an abstract discipline, which is used as a model in order to make
deductions about events, which possibly may occur in actual or an imagined fashion.

Decisions we take in life are governed by probability. Chance and luck play an important role in
our life. Events such as births, marriages, deaths etc. depend substantially on chance.

The Concept of Probability

One may try to measure the extent of chance or probability of any event as is often done in
gambling, weather forecasts, business etc. A right decision may mean, for instance, the choice of
the best possible bet, irrespective of the ultimate actual results. Sometimes, experience can help a
person make fairly accurate decisions and predictions.
Any attempt to investigate a phenomenon is likely to have to reckon with the elements of
probability.
In everyday conversation, the term probability is used loosely. Statements of the following types
are common:
• It is likely to rain today
• It will probably rain today
• There is fair chance of it raining today.

In each of the above instance, a sense of probability measure is conveyed. They all convey some
level or degree of confidence or belief in the outcome of the event.

Definition of Probability
Probability is a mathematical measure that quantifies the degree of confidence in the occurrence
of an event. If we toss a coin, the event, ‘head’ or the event ‘tail’ may occur. Both of these
possibilities as assumed to be equally likely when the coin is fair or balanced.

116
 1
We say that the probability or chance of a head occurring is 50 percent or and the probability
2
1
of a tail occurring is allow .
2
If we roll a fair die, any of the six faces (1 , 2 , 3 , 4 , 5 , 6) is equally likely to show up. The
probability of any of these events is 1 .
6
We shall express the probability of an event A as a ratio denoted P ( A) .

Some Related Definitions

Event:
An event is said to have occurred in an experiment if the result of the experiment is a specified
outcome E, say. If this specified outcome is observed when the experiment is performed, then E
is said to have occurred.

Experiment:
A procedure that leads to outcomes.

Trial:
A procedure or a single performance of an experiment is called a. trial. Its primary objective is to
collect statistical information.

Exhaustive events:
A set of events is exhaustive of all possible events associated with a trial of an experiment are
included in the set. The set consisting of all possible events associated with a trial of an
experiment is called the sample space, denoted S.

The set of events {head, tail} exhausts all possibilities in the toss of a thin coin since nothing also
can occur. In the roll of a die, {1 , 2 , 3 , 4 , 5 , 6} is an exhaustive set of events.

Favourable events:
Such cases as result in the happening of an event are said to be favourable to that event.

Elementary or Simple Event:

Such events that contain only a single outcome are called elementary or simple events.

Impossible Events:
An event that can never occur is an impossible event.

Certain Events:
An event which is certain to occur is a certain or sure event.

The occurrence of the number 8 when a die is rolled is an impossible event.

The death of a person is a certain event.

117
Complementary Events:
Two events, E1 , E2 are complementary, if whenever E1 occurs, E2 does not; whenever E2
occurs, E1 does not.
Occurrence of head and tail in the toss of a coin are complementary events. The complement of
an event E is E  (also denoted as E c or E )

Subjective, Classical and Frequentist Definitions of Probability

Subjective Probability
As the name suggest, it is quantifying a probability measure in subjective terms. Subjective
probabilities though comparable, is not unique and thus no reliability measures are possible. Two
individuals may quantify the occurrence of an event hampering subjectively which may lead to
different scenarios.
Let as consider the following statements that are examples of subjective probability
(i) It is likely Ghana will win the 2010 World Cup in South Africa;
(ii) It is probable, Ghana will win the 2010 World Cup Tournament in South Africa;
(iii) Ghana, has a fair chance of winning the 2010 World Cup football tournament in
South Africa.

In each of the three statements, probability is quantified subjectively. Each showing some degree
of confidence of the chance or likelihood of the event occurring. In making or quantifying
probability subjectively, one needs to depend on his or her experience to make a fairly accurate
decision or prediction.
Different persons may assign different probability values to the same event. This normally
happens in marketing, managerial and economics related situations.

Classical Definition of Probability

If an event E can occur in m cases out of a total of n possible equally likely outcomes, then the
classical definition of probability of the event E is defined as
m
p(E) = (3.1)
n
If we can find:
n, the total number of outcomes;
m, the number of favourable outcomes;

m
Then the expression in (3.1) is simply, the ratio . Note that, here n − m outcomes are
n
favourable to E 
i.e. to the nonoccurrence of E and E  are complementary events.
n−m m
P ( E) = = 1− (3.2)
n n
From (3.1) and (3.2) we can see that
P ( E ) + P ( E ) = 1 (3.3)
or. P ( E ) = 1 − P ( E ) (3.4)

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Example
What is the probability of obtaining a six’ when a fair die is rolled once?

Solution
Number of possible outcomes: n = 6
Number of favourable outcomes: m = 1
1
:. P ( E ) = P ( a six ) =
6

Example
What is the probability of obtaining a white ball from an urn containing 2 white and 3 red balls?
All balls are indistinguishable except for colour and selection is randomly done.

Solution
Number of possible outcome: n = 5 (five balls)
Number of favourable outcomes: m = 2 (two white balls)
2
:. P ( E ) = P ( white ) =
5

The Frequentist Probability

In the case of the classical definition of probability, all possibilities of the occurrences are
considered well in advance. These events are equally likely and mutually exclusive.

With this type of probability, which is known as the frequentist probability or empirical or
posteriori probability, determined experimentally rather than theoretically. It is concerned with
problems in which past experiences are influential. To arrive at the conclusions, a large number
of cases must be observed.

If m is the number of successes in event A in n trials and if the sequence of relative frequencies
m
of event A denoted f A = obtained for larger and larger values of n, approaches a limit P ( A)
n
i.e. when a large number of trials is made, we define the frequentist probability of the event A as
m
p ( A ) = lim
n → n

Mutually Exclusive Events

Two events are said to be mutually exclusive when they cannot occur simultaneously. If one of
the events occurs, the other does not and vice-versa.
Example
Suppose an experiment consists of selecting a student in the University of Ghana who is in Level
100. Let
Event A: the student is a male;
Event B: the student is a female;

119
 Event C: the students is aged under 25 years.
It is clear that, Events A and B are mutually exclusive. They cannot happen simultaneously i.e.
no one student can be chosen who is a male and is a female at the same time. However events A
and C are not mutually excuses. They could happen together (as could events B and C).

Independent and Conditional Events

An independent event is one where the probability of the event occurring is not affected by other
events. A conditional event is one whose probability is so affected. It seems reasonable to
suppose, for example, that the probability that a student taking Numeracy Skills course is
independent of the probability of the student being left-handed. That is, the probability of a
student taking Numeracy Skills is independent of whether he or she left-handed or right-handed.
The probability of a student taking Numeracy Skills, however may well be conditional on which
type of course they are taking in the University of Ghana. That is, the probability of a student
taking Numeracy Skills is affected by the type of course they are taking.
You shall shortly see why it is important to distinguish between conditional and independent
events.

Sets and Probability

Events can be treated as subsets of an outcome space (or sample space). All relations in Set
Theory apply without change to events. We shall, from now on, consider an event to be a
collection of points which is a subset of the outcome space, S, of the experiment.
Let A , B and C be events associated with an experiment  . We can use set notations to depict
the following compound events:

Event Description Set Notation

Event that A and B occurs A B
Event that A or B occurs A B
Event that at least one of A , B or C occurs A  B C
Event that A does not occur A
Event that exactly one of A or B occurs ( A  B)  ( A  B )
Event that A or B and C occurs A(B C)
Event that A and B or C occurs A(B C)

Rules of Probability
Probability Axioms
Axioms are rules that govern probability computations.

Axiom 1
An impossible event is that which equals the empty set,  . The probability of an impossible
event is zero.
i.e. P ( ) = 0 .

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Axiom 2
A certain (or sure) event equals the outcome space, S. The probability of a certain event is 1. i.e.
P (S ) = 1.

Axiom 3
For any event E , 0  P ( E )  1 .

Axiom 4
The complement E  of an event E , is the collection of all sample points in S, but not in E.
P ( E ) + P ( E) = 1.
:. P ( E ) = 1 − P ( E ) .

Compound Events
Most events occurring in practice are compound i.e. combinations of two or more events.
Let E1 and E2 be two events. Then we define
1. Union event: E1  E2 or E1 + E2 i.e. either E1 or E2 or both E1 , E2 occur.
2. Intersection event: E1  E2 or E1E2 i.e. both E1 and E2 occur.
3. Inclusion event: E1  E2 i.e. if E1 occurs, then so does E2
4. Difference event: E1 − E2 or E1 E2 i.e. if E1 occurs then E2 does not occur
5. Mutually exclusive events; E1  E2 =  i.e. E1 and E2 are mutually exclusive if the
event E1 , E2 contains no sample point. i.e. P ( E1E2 ) = 0
6. Independent events: Two events E1 , E2 are said to be independent if
P ( E1 E2 ) = P ( E1 ) and P ( E2 E1 ) = P ( E2 ) .

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Diagrammatic Representation of Events
The use of Venn diagrams to represent events makes computations much easier.

Two mutually exclusive events Two compatible events

E1 E2 E1 E2

E1 or E2 occurs E1 or E2 but not both occurs

E1 E2 E1 E2

E1 and E2 occurs E1 but not E2 occurs

E1 E2 E1 E2

E2 but not E1 occurs Neither E1 nor E2 occurs

E1 E2

E1 E2

The Fundamental Rules of Probability

The Addition Rule

(i) The probability of the union E1 or E2 i.e. ( E1 + E2 ) is given by
P ( E1 + E2 ) = P ( E1 ) + P ( E2 ) − P ( E1E2 ) .
(ii) If E1 , E2 are mutually exclusive events, P ( E1E2 ) = 0
:. P ( E1 + E2 ) = P ( E1 ) + P ( E2 ) .

Example
What is the probability of obtaining a 1 or 4 in a throw of a fair die once?

Solution
Let E1 represent the event that 1 occurs and
122
 E2 represent the event that 4 occurs.
Then
1 1
P ( E1 ) = and P ( E2 ) = .
6 6

Therefore, the required probability is

P ( E1 or E2 ) = P ( E1 + E2 ) = ( E1 ) + P ( E2 )
since occurrence of a 1 or a 4 are mutually exclusive
1 1 2 1
= + = = .
6 6 6 3

Example
What is the probability of obtaining a 2 or an even number in a throw of a fair die once?

Solution
Let E1 = ‘event of a 2’;
E2 = ‘event of an even number’
Then
1
P ( E1 ) = ;
6
E2 = 2 , 4 , 6
3 1
:. P ( E2 ) = = ;
6 2
The event E1 E2 is the event of a ‘2 and an even number’.
i.e. E1E2 = 2
1
:. P ( E1 E2 ) = ;
6

Required probability is
P ( E1 or E2 ) = P ( E1 + E2 ) = ( E1 ) + P ( E2 ) − P ( E1E2 )
1 1 1 1
= + − = .
6 2 6 2

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The Multiplication Rule of Probability
The probability of the intersection event E1 E2 is given by
P ( E1 E2 ) = P ( E1 )  P ( E2 E1 )
= P ( E2 )  P ( E1 E2 )
Where P ( E2 E1 ) is the conditional probability of E2 given that E1 has already occurred.
P ( E1 E2 ) is similarly defined [conditional probability computations are illustrated in the next
section].

If E1 and E2 are independent then

P ( E1E2 ) = P ( E1 )  P ( E2 )

Example
A boy is to receive a prize if he can throw 2 sixes on 2 tosses of a fair die. What are his chances
of winning the prize?
Solution
Let E1 = ‘1st throw is a six’
E2 = ‘2nd throw is a six’

1 1
Then P ( E1 ) = , P ( E2 ) =
6 6
1 1 1
:. P ( E1 and E2 ) = P ( E1 E2 ) =  = .
6 6 36

Note that, here, we assumed that events E1 and E2 are independent.

Conditional (or Dependence) Probability

Introduction
When two or more events are independent, computations involving composite events amongst
them is fairly simple. However when these events are not all independent they are said to be
dependent or conditional. Computations involving conditional events use some basic definition
as we will be discussing.

Definition
The conditional probability of event E1 , given that event E2 has already occurred, is defined by
P ( E1 and E2 )
P ( E1 E2 ) = ;
P ( E2 )
Also, the probability of E2 , given E1 is

124
 P ( E1 and E2 )
P ( E2 E1 ) = .
P ( E1 )

Example
Consider the following data obtained form 300 level 100 students at the University of
Ghana.

Activity Participation
Gender Yes No Total
Male 120 80 200
Female 60 40 100
Total 180 120 300

Let E1 = 'Yes' , E2 = 'males' , E11 = 'No' , E21 = 'Females' .

180 200 120
Then P ( E1 ) = , P ( E2 ) = , P ( E1E2 ) = .
300 300 300
P ( E1 E2 ) 120 300 3
:. P ( E1 E2 ) = = = .
P ( E2 ) 200 300 5
P ( E1 E2 ) 120 300 2
Also, P ( E2 E1 ) = = = .
P ( E1 ) 180 300 3

Illustrative Examples of Probability

Example
What is the probability of obtaining double sixes in two rolls of a die (or one roll of two dice
simultaneously)?

Solution
Let E1 be the event of obtaining a six in the 1st roll;
E2 be the event of obtaining a six in the 2nd roll.

Note that E1 and E2 are independent.

: . P ( E1E2 ) = P ( E1 )  P ( E2 )
1 1 1
=  =
6 6 36

Example
What is the probability of obtaining four heads in a single toss of 4 coins? Note that this
experiment is similar to 4 tosses of one coin simultaneously.

Solution
Let E1 be the event of a head on the ith coin ; i = 1 , 2 , 3 , 4.
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Then
1
P ( E1 ) = P ( E2 ) = P ( E3 ) = P ( E4 ) =
2
Now the required probability is
P ( E1E2 E3 E4 ) = P ( E1 ) P ( E2 ) P ( E3 ) P ( E4 ) , since the events Ei s are independent.
1 1 1 1
=   
2 2 2 2
4
1 1
=   = .
2 6

Example
What is the probability of getting 2 aces in two draws from a deck of playing cards?
(i) If the first card is replaced before the second draw;
(ii) If the first card is not replaced before the second draw.

Solution
(i) If the first card is replaced before the second draw, then

4 4 1
P ( E1 E2 ) = P ( E1 ) P ( E2 ) =  = .
52 52 169

(ii) If the first card is not replaced before the second draw, then

P ( E1 E2 ) = P ( E1 )  P ( E1 E2 )
4 3 1
=  = .
52 51 121

Remark
In the first solution, the events E1 and E2 are considered independent i.e. the outcome of the
second draw is not affected by the outcome in the first draw since the card drawn in the first
draw is replaced before the second draw.

In the second solution, the event E1 and E2 are not independent since the outcome of the first
draw affects the outcome in the second draw. For instance, in a batch of 52 playing cards there
are 4 aces, therefore if the first draw results in an ace, then in the second draw we will be left
with 51 cards with only three aces, since the first card drawn is not replaced before the second
draw. These events are therefore conditional (or dependent).

Example
From an urn containing 3 white (W) and 2 black (B) balls, two balls are drawn one after the other
without replacement. What is the probability that the first ball drawn is white and the second
black?

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Solution
P (WB ) = P (W ) P ( B W )
3 2 6 3
=  = =
5 4 20 10

Example
From a pack of 52 playing cards, a card is drawn and found to be a face card. What is the
probability that it is a Queen?

Solution
P ( Queen ) 4 52 1
P ( Queen Face Card ) = = = .
P ( Face Card ) 12 52 3

Exercise
(1) What is the probability of obtaining a six at least once in two rolls of a die?

(2) What is the probability of obtaining two kings in drawing two cards from a
shuffled duct of playing cards?

(3) If two dice are rolled, what is the probability that a sum of 7 will be obtained?

(4) What is the probability of getting a sum of

(i) 9 (ii) 10 (iii) 11
when 3 dice are thrown?

(5) The following table refers to 250 employees of the University of Ghana classified by
Sex and by Opinion on a proposal to emphasize fringe benefits rather than wage
increases in an impending contract negotiation with their Union TEWU:

Opinion
Sex In Favour Neutral Oppose Total
Male 90 20 40 150
Female 30 10 60 100
Total 120 30 100 250

Calculate the probability that an employee selected from this group will be:
(a) a female opposed to the proposal;
(b) neutral;
(c) opposed to the proposal given that the employee selected is a female;
(d) either a male or opposed to the proposal.
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 CHAPTER FOUR
Statistical Display of Data: Tables, Graphs and Charts

Data, Data Types and Frequency Distributions

Introduction
The ability to present the data that a student is working with is very important for the
purpose of providing a vivid picture of what the data is saying. Various tables and
graphs have been developed for giving this vivid picture of what the data is saying.
The main illustrative tools used in this regard are bar and pie charts for qualitative
data and frequency distribution, the histogram, the frequency polygon and the
cumulative frequency curve for quantitative data.

Primary/Raw Data
What is Data?
First we need to understand what data is. Primary data is data that has been collected
but which has not been organized numerically e.g. the set of 90 Level 100 students
collected from an alphabetical listing of all Level 100 University of Ghana students’
records. Similarly the set of 100 primary school pupils collected from the University
of Ghana Basic Schools is an example of raw data.

Data Types
There are two main types of data. They are Qualitative and Quantitative data.
▪ Categorical or qualitative data have values that can only be placed into
mutually exclusive categories, such as “yes” and “no.”
▪ Numerical or quantitative data have values that represent quantities.

Figure 4.1 represents the types of data and examples of each.

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Figure 1.1: Data and Data Types

Data

Categorical Numerical

Discrete Continuous

Exercise

Give examples of four types of raw data you would collect from your town or locality
and identify whether they are categorical or numerical. In the case of a numerical
example, say whether they are discrete or continuous.
(a)……………………………………………………………………..
(b)……………………………………………………………………..
(c)……………………………………………………………………..
(d)……………………………………………………………..………

Arrays
When the raw numerical data are arranged in either ascending or descending order of
magnitude they are called arrays e.g. 2, 5, 10, 12 (ascending) or 12, 10, 5, 2
(descending).
Ascending order of magnitude means that you are starting with the smallest number.
Descending order of magnitude is when you start with the highest number first.

129
Range
The difference between the largest and smallest values in the data set is called the
range of the data. For example in the array above the largest number is 12 and the
smallest is 2, therefore, the range is 12 – 2 = 10.

Frequency Distribution for Categorical Data

In organising categorical data into a frequency distribution, we record for each
category, the number of times it occurs in the data set, i.e. the frequency of each
category. A table depicting all categories and their corresponding frequency is
known as a Frequency Distribution, Frequency Table or Summary Table. Table 4.1
is an example of a Summary Table.

Table 1.1 Faculty Classification by a sample of 100 students of the University

of Ghana, Legon

Faculty # of Students (f)

Arts 25
Social Studies 38
Science 20
Engineering Sciences 5
Agricultural Science 10
Total = N 100

Frequency Distribution for Numerical Data

What is a Frequency Distribution?
When we need to summarize large masses of raw data it is often useful to distribute
the data into classes or categories and to determine the number of individuals
belonging to each class/category, called the class frequency. A table of data made up
of classes/categories together with corresponding class frequencies is called a
frequency distribution or frequency table.
Extensive data in an ungrouped form cannot be as useful as grouped data in a form of
a frequency table

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 Table 1.2 Marks obtained by 100 students of the University of Ghana, Legon
in UGRC 120.

Marks # of Students (f)

20 – 29 5
30 – 39 9
40 – 49 19
50 – 59 25
60 – 69 20
70 – 79 12
80 – 89 10
Total = N 100

The Data, which is arranged and summarized in Table 1.2, are called grouped data.
Although the grouping tends to destroy much of the original detail of the data, its
advantage is that a clear overall picture is obtained and certain valid relationships are
made evident.
To form a frequency distribution from raw data, we tally each observation with either
a mark or stoke; every fifth mark or stoke across the first four marks or strokes
giving us a frequency of five at a time. This facilitates the counting process. The first
three classes of Table 4.2 may be obtained by tallying as follows:

Marks Tally # of Students (f)

20 – 29 //// 5
30 – 39 //// //// 9
40 – 49 //// //// //// //// 19

Class Intervals and Class Limits

In Table 4.2, 20 – 29 is called the class interval (made up of all the values from 20
to 29). The values 20 and 29 are called class limits. The value 20 is the lower class
limit and 29 is the upper class limit. This indicates the first number and the last
number in a class respectively.

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 Class Boundaries or Exact Limits
If grades are recorded to the nearest whole number, the class interval 20 – 29 in
theory includes all the values from19.5 to 29.5. These are called the class boundaries
or exact limits. Sometimes class boundaries are used to symbolize the classes. Thus
the class boundaries in Table 4.2 could be indicated as follows: 19.5 – 29.5 , 29.5 –
39.5 , 39.5 – 49.5 , etc.

Exercise
Develop a table of class boundaries or exact limits of the data in Table 1.2.
Marks obtained by 100 students of the University of Ghana, Legon in UGRC 120

Marks Class Boundaries No. of Students (f)

20 – 29 5
30 – 39 9
40 – 49 19
50 – 59 25
60 – 69 20
70 – 79 12
80 – 89 10
Total = N 100

Class Width or size of Class Interval (c)

The width of a class interval is the difference between the lower and upper class
boundaries and is often called the class width or the size of the class interval. If all
the class intervals of a frequency table have equal widths, the common width is
denoted by “c”. The c is also equal to the difference between any two successive
lower class limits or any two successive upper class limits. The class width may also
be obtained for any particular class by subtracting the lower class boundary from the
upper class boundary.

Class Mark/Class Midpoint

The class mark is the midpoint of the class interval. It is the sum of the lower and
upper class limits divided by 2. It is also called the class midpoint. It is denoted by x.

132
Exercise
Develop a table of class midpoints for Table 1.2

Marks obtained by 100 students of the University of Ghana, Legon in UGRC 120

Marks # of Students (f) Class Midpoints (x)

20 – 29 5
30 – 39 9
40 – 49 19
50 – 59 25
60 – 69 20
70 – 79 12
80 – 89 10
Total = N 100

Relative Frequency Distribution

In a frequency distribution, if the frequency in each class interval is converted into a proportion
by dividing each by the total frequency, we get a series of proportions called relative
frequencies. A distribution presented with relative frequencies, is called a relative frequency
distribution. The sum of all relative frequencies in a distribution is 1.

Percentage Relative Frequency Distribution

When all relative frequencies are multiplied by 100, they are converted to percentage relative
frequency. The ensuing distribution is thus called the Percentage Relative Frequency
distribution. The sum of all percentage relative frequencies in a distribution is 100.

Stem-and-Leaf Displays
Another way of presenting frequency distributions for numerical data is the Stem-and-leaf
display or plot. In this Section, I will introduce to you how to display data in a stem-and-
leaf plot and also discuss some of its advantages and disadvantages.

A stem-and-leaf display organizes data into groups (called stems) so that the values
within each group (the leaves) branch out to the right on each row. The values in the stem
are arranged vertically in ascending order from top to bottom. Each data value is split into
133
 two values, one part called the stem and the other part called the leaf. The normal practice
of splitting a numeric value is that, all digits preceding the last digit is taken as the stem
and the last digit of the number is taken as the leaf. E.g. the number 25 has a stem of 2
and a leaf of 5, the number 128 will then have a stem value of 12 and a leaf value of 8.
We note that the definition we give to splitting the values in a particular data set must be
the same for all values in the data set.

Example 2.1
Obtain a stem-and-leaf display for the data below:
16 17 17 18 18 18 19 19 19 20
20 21 22 22 25 27 32 38 42 45

A cursory look at the data indicates that the values are in tens, twenties, thirties and forties. We
will therefore suggest the stems to be the first digits of the numbers and the last digits constitute
the leaves. We therefore have:
Stem Leaf
1 6 7 7 8 8 8 9 9 9
2 0 0 1 2 2 5 7
3 2 8
4 2 5

Exercise
Develop a stem-and-leaf display for each of the following sets of data. In each case
tell us your definition of the stem and the leaf:
(a) 10.5 , 11. 1 , 11. 2 , 11. 3 , 11. 5 , 12.0 , 12.2 , 12. 3 , 15.7 , 16.2
(b) 120 , 123 , 98 , 99 , 110 , 157 , 143 , 152 , 135 , 149.
The Stem-and-leaf display has the advantage of retaining the actual values in the display as well
as shows the distribution of the values. This makes it the preferred choice when organizing a data
set that contains only few observations.
However, when the number of observations in the data set is large, it is cumbersome to display
and therefore is less preferred to other tabular forms.

One- and Multi- Dimensional Tables

The Construction of a Frequency Table/Distribution
The following are a helpful guide for the construction of a frequency
table/distribution
▪ Given the specified class interval (C.I.), and starting from the lowest or highest
value, write out the classes with each class distinct from each other.

134
 ▪ Distribute the individual values into each class/category. This is done by putting
a slash adjacent to the class. This process is called tallying, as discussed in
Section 1.
▪ Count the number of slashes to give the total number of cases for each class.
This is called the class frequency denoted by (f).
▪ Indicate the total number of cases for the whole table. This is denoted by N. The
total of the frequencies must equal the total number of cases in the raw data.

Example 3.1
The following are the grades of 50 students in an examination:
13 47 27 55 41 58 35 58 48 53 58 22 55 32 45 48 54 78 66 58
42 35 18 57 30 72 57 81 33 63 54 79 47 64 36 45 51 24 79 26
33 60 18 68 35 20 68 36 60 55
Use a C.I = 10
The lowest value is 13 and the highest is 82. We will consider class intervals: 10 – 19
, 20 – 29 , 30 – 39 , etc. We therefore, construct the frequency table as follows:

Class Tally # of cases (f)

10 – 19 /// 3
20 – 29 //// 5
30 – 39 //// //// 9
40 – 49 //// //// 9
50 – 59 //// //// // 12
60 – 69 //// // 7
70 – 79 //// 4
80 - 89 / 1

Total Frequency N = 50

Using Technology
Excel is a spreadsheet which is a software application that comes together with every
computer. Excel has the capability of analyzing data even for larger sets of data.

135
 Microsoft Excel Terms
When you use Microsoft Excel, you place the data you have collected in
▪
worksheets.
▪ The intersections of the columns and rows of worksheets form boxes called cells.
▪ If you want to refer to a group of cells that forms a contiguous rectangular area, you
can use a cell range.
▪ Worksheets exist inside a workbook, a collection of worksheets and other types of
sheets, including chart sheets that help visualize data.
Figure 4.1 is an illustration of the Microsoft Worksheet when Excel is opened on your
computer.

Figure 3.1: Microsoft Worksheet

To be able to perform basic statistical analyses that we shall be discussing in Excel,

you have to install the Excel Analysis Tool Pack.

136
 Installation of the Analysis Tool Pack

Step 1:
Open Excel. Click on the
Office button and click on
Excel Options as indicated
in the Figure.

Step 2:
Select Add-Ins and click
OK as indicated in the
Figure.

137
Step 3:
Make sure ‘Add-ins’ is
selected in the Manage:
dialogue box and click
Go… as indicated in the
Figure.

Step 4:
Select ‘Analysis ToolPak’ and
‘Analysis ToolPak – VBA’ in the
Add-Ins available: dialogue box and
click OK as indicated in the Figure.

Step 5:
When the Analysis ToolPak Add-In has been installed, click on Data, the last Menu bar shows Data 138
Analysis. You are now ready to explore the wide variety of statistical analysis that Excel offers.
Example 3.2
Use the data in Example 3.2 to construct a frequency table in Excel.
Enter the data in the worksheet in one column as indicated below:

The bins are the upper end points (or upper boundaries of the intervals).

Step 1:
Click on ‘Data’ , then ‘Data
Analysis’ in the task pane located at
the uppermost parts of the worksheet.
Select Histogram in the ‘Data
Analysis’ dialogue box and click OK
as indicated in the Figure.

139
Step 2:
Input the cell range of the data by
placing the cursor in the ‘Input
Range: box and clicking and
holding to select the cells
containing the marks.
Also input the cell range of the
Bins in the ‘Bin Range:’ box.
Select ‘New Worksheet Ply:’ to
store the result in a new worksheet
or select ‘Output Range:’ and
click any cell within the same
worksheet to store the result there.
Click on OK.

Step 3:
Delete the last row and change ‘Bins’
to ‘Class Intervals’. Change the first
column values appropriately to show
the class intervals.

140
 This is the ensuing Frequency
distribution.

Multi-Dimensional Tables
When we present data to show more than one characteristic or groups of items, then
we refer to it as Multi-dimensional tables. We present an illustration of a multi-
dimensional table as an example.

Example 3.3
The table below is the result of a survey into cinema attendance habits of adult
factory workers:
Table 3.2: Cinema attendance among Adult Adult Factory Workers in Tema, March 2010:

Single Married
Cinema Attendance
Under 30 years Over 30 years Under 30 years Over 30 years

Less than once a month 122 374 1400 1880

1 – 4 times a month 1046 202 288 115
More than 4 times a month 882 24 112 15

Total 2050 600 1800 2010

Frequently, table construction involves deciding which characteristic or attribute

should be taken as primary and which as secondary. For Table 3.2, we might ask
whether it would be improved if “under 30 years” and “ 30 years and over had been
the main column headings and “single’ and “married” the sub-headings. The answer
depends on the purpose of the table. If the activities of age groups are to be
compared, it is best left as it stands. But if a comparison between men of different
marital status is required, the change would bring an improvement.

141
 Exercise
In a` survey of 10, 048 persons, of whom 2, 703 were children under 16 years of age,
605 retired men and 2, 212 housewives, the following information was obtained. Of
those above age 16, there were 2, 720 males 16 and under 60, and 2, 931 females of
this age group.
There were 227 males between 60 and 65 and 482 males above 65 and 270 and 515
females in these age ranges respectively.
Construct tables showing the distribution of the distribution of the sample by work
status and the adult population by age and sex. Use percentages to help in making
comparisons. Give you comment on the data.

Charts: Simple Bar, Pie, Multiple and Composite (or Stacked) Bar
In this Section, we will learn to present these frequency distributions in the form of graphs and
charts. We will specifically restrict ourselves to graphical presentation of qualitative or count
data. These charts include:
▪ Simple Bar Chart;
▪ Pie Chart;
▪ Multiple Bar Chart;
▪ Composite Bar Chart;

Simple Bar Chart

In a bar chart, a bar shows each category, the length of which represents the amount,
frequency or percentage of values falling into a category.

Example 4.1
Draw a bar chart for the data in Table 1.1

142
Table 1.1 Faculty Classification by a sample of 100 students of the University
of Ghana, Legon

Faculty # of Students (f)

Arts 25
Social Studies 38
Science 20
Engineering Sciences 5
Agricultural Science 10
Total = N 100

143
 Pie Chart
The pie chart is a circle broken up into slices that represent categories. The size of each
slice of the pie varies according to the percentage in each category.

Example 4.2
Draw a pie chart for the data in Table 1.1

Multiple Bar Chart

In a multiple bar chart, bars are drawn adjoining themselves to show each category, sub
categories. It is usually employed when we want to display multi-dimensional chart.

Example 4.3
Draw a multiple bar chart for the data in Table 3.2:

144
Exercise
Use the multi-dimensional table you found in Activity 3.1 to draw a multiple bar chart.

Composite (or Stacked) Bar Chart

In a composite (or Stacked) bar chart, bars are drawn to show each category, with sub
categories stacked on top of each other in each of the bars. Similar to the multiple bar
chart, it is usually employed when we want to display multi-dimensional chart.

Example 4.4
Draw a composite bar chart for the data in Table 3.2:

145
Histogram and Cumulative Frequency Curve
We will discuss how to organise numerical data into histograms and cumulative frequency
curves.

The Histogram
A histogram is a set of adjoining vertical bars whose areas are proportional to the frequencies
represented by the bars. A histogram is drawn by taking the class intervals on the horizontal
axis (or x-axis) and the frequencies on the vertical axis (y-axis). When the class intervals are
of equal width, the height of a bar corresponds to the frequency of that class.
The class intervals with the highest frequency will have the highest bar and the highest
concentration of observations. This class as we shall see in Unit 5 is called the modal class.

Example 5.1
Draw a histogram for the data in Example 3.1:

146
Frequency and Cumulative Frequency Polygon
▪ A polygon is formed by having the midpoint of each class represent the data in that class
and then connecting the sequence of midpoints at their respective class percentages.
▪ The cumulative percentage polygon, or ogive, displays the variable of interest along the
X axis, and the cumulative percentages along the Y axis.

147
 Advantages of the Frequency Polygon
Most people prefer the frequency polygon to the histogram because
(a) They think that it gives a clear picture of the shape of the contours;
(b) Also, the frequency polygon makes it possible to make comparisons to some
baseline.

The Cumulative Frequency

Cumulative frequency is the total number of cases that are added up higher or lower
than a particular score or value. For example a Professor might want to know what
number or percentage of the final examination scores was higher than 50. To answer
this question we use a graph of the cumulative frequency distribution. To do this we
cumulate the frequencies to see how many items or cases lie more than or less than a
particular value. A cumulative frequency distribution shows the relation between a
class interval and the frequency at or below its upper or lower class boundaries.
The cumulative frequency distribution is constructed by adding the frequencies of
scores in any class interval to the frequencies of all the class intervals below it on the
scale of measurement.

148
 Frequency Distribution of Examination Scores with Cumulative
Frequency

Example
Marks obtained by 100 students of the University of Ghana, Legon in UGRC 120

Class Exact Class No. of Class CF CF

interval Limits Students (f) Midpoint (x) Less than More than
20 – 29 19.5 – 29.5 5 24.5 5 100
30 – 39 29.5 – 39.5 9 34.5 14 95
40 – 49 39.5 – 49.5 19 44.5 33 86
50 – 59 49.5 – 59.5 25 54.5 58 67
60 – 69 59.5 – 69.5 20 64.5 78 42
70 – 79 69.5 – 79.5 12 74.5 90 22
80 – 89 79.5 – 89.5 10 84.5 100 10
Total = N 100

The “less than” cumulative frequency is computed based on the upper class
boundaries/exact limits and the “more than” cumulative frequency is computed based
on the lower class boundaries/exact limits.
This indicates that 14 students scored less 39.5, 33 students scored less than 49.5, 86
students scored more than 39.5, and 67 students scored more than 49.5.

A graph showing the cumulative frequency “less than” or “more than” any exact
limit is called a cumulative frequency polygon or ogive.

149
 In plotting the graph of the cumulative frequency the exact limits are used rather than
the class midpoints. The graph of the cumulative frequency allows researchers to
make graphic approximation and arithmetic interpolation of values not shown in the
cumulative distribution. For example from the graph of the cumulative frequency,
about 20 students have marks of 60 and more. This is done by drawing a line
perpendicular to the horizontal axis at (100% - 20%) = 80% and extend the line till it
intersects the “more than” curve. At this point of intersection a line is drawn
perpendicular to the vertical axis to intersect the vertical axis. At this point on the
vertical axis the value is seen to be approximately 63.5 marks.

Relevance/Appropriateness of use of Graphs and Charts

Introduction
In this section we finalize the discussion on this unit with the basic objectives of
constructing graphical displays, Chart norms and relevance and appropriateness of the use
of these graphical displays. I do not in any way, intend to tend you into a ‘Guru’ in
graphical displays but going through this section will help you appreciate what goes into
the consideration of such graphical displays and subsequently impart to you some
knowledge which will enable you to be in a better position to draw more appropriate
graphs.

Statistical results can be presented in an appealing and vivid form with the help of charts.
An appropriate chart can give a clear, truthful and easily understandable picture of the
facts contained in the data. A carelessly drawn inappropriate chart can mislead and waste
the labour in undertaking such venture.

150
 The drawing of an effective chart requires talent, imagination and grasp of the subject
under consideration.

Objectives of Constructing Charts

The following have been identified as objectives of constructing charts. Bearing these in mind
help us to draw quality useful charts;
▪ To get a quick view of the data. For understanding and remembering the nature and trend
of the data.
▪ To compare with other data and charts.
▪ To analyze the data, for further study and planning.
▪ To check the accuracy of certain computational results. Incompatibilities can be noticed
from the trend in the charts.
▪ To make a step by step visual study of the development of the issue(s) imbedded in the
data.
▪ To present the data in an alternate form.

Principles of Excellent Graphs

The following principles for constructing graphs need to be adhered to in other to draw excellent
graphs.
▪ The graph should not distort the data.
▪ The graph should not contain unnecessary adornments (sometimes referred to as chart
junk).
▪ The scale on the vertical axis should begin at zero.
▪ All axes should be properly labeled.
▪ The graph should contain a title.
▪ The simplest possible graph should be used for a given set of data.

Chart Norms
A chart should be planned so as to possess the following fundamental features:
▪ Simplicity and Neatness: The chart should not include unnecessary information and
should deal with the relevant data for the relevant purpose only. The form should be
attractive and easily understandable.
▪ Size and Proportion: It should be neither too large nor too small. The size depends on the
data and other details. The proportion of its height to width should be reasonable. Often,
it is taken as 1 : 1.5.
▪ Accuracy: Conscious and unconscious misrepresentation should be avoided.
▪ Emphasis: Colours, heavy, light and dotted lines may be suitably used. The materials
should be chosen carefully and skilfully.

151
Construction of a Chart
The above norms should be kept in mind when constructing a chart. With proper scales on the x-
axis and y-axis, it is possible to show useful information in a convenient form. A title, possibly a
subtitle, scale headings and proper labelling should accompany every chart drawn.

Exercise

1. A recent survey showed that the typical American car owner spends GH¢2, 950 per year
on operating expenses. Below is a breakdown of the various expenditure items. Draw an
appropriate chart to portray the data and summarize your findings in a brief report.

Expenditure Item Amount (in GH¢)

Fuel 603

Interest on car loan 279

Repairs 930

Insurance and license 646

Depreciation 492

Total 2,950

2. The Ministry of Health reported that in the year 2007, the distribution of Ghanaian cancer
patients by age was as follows:

Age Group 0 – 19 20 – 39 40 – 59 60+

Percentage of Patients 12.2 13.6 33.1 41.1

Draw a pie chart to represent the data.

152
3. The Registrar of the University of Ghana has 16 late applications of the Matured
Entrance Examinations (MEE) for admission into the humanities programme in the
university next academic year. The composite MEE scores of these applicants were:

27 27 27 28 27 25 25 28
26 28 26 28 31 30 26 26

The MEE scores are to be organized into a frequency distribution.

(a) How many classes would you recommend?
(b) What class intervals would you suggest?
(c) What lower limit would you recommend for the first class?
(d) Organize the scores into a frequency distribution and determine the relative
frequency distribution.
(e) Comment on the shape of the frequency distribution.

4. ECOBANK is studying the number of times their Automatic Teller Machine (ATM),
located at Legon Campus, is used each day. The following is the number of times it was
used during each of the last 30 days. Develop a stem-and-leaf display for the data.
Summarize the data on the number of times the ATM was used: How many times was the
ATM used on a typical day? What are the largest and the smallest number of times the
ATM was used? Around what values did the number of times the ATM was used tend to
cluster?

83 64 84 76 84 54 75 59 70 61
63 80 84 73 68 52 65 90 52 77
95 36 78 61 59 84 95 47 87 60

5. A large retailer is studying the lead time (elapsed time between when an order is placed
and when it is filled) for a sample of recent orders. The lead times are reported in days:

Lead Time (days) Frequency

0 up to 5 6

5 up to 10 7

10 up to 15 12

15 up to 20 8

20 up to 25 7

(a) How many orders were studied?

(b) What is the midpoint of the first class?
153
(c) Draw a histogram for the data.
(d) Draw a frequency polygon for the data.
(e) Interpret the lead times using the two charts.
(f) How many orders were filled in less than 10 days? In less than 15 days?
(g) Convert the frequency distribution into a less-than cumulative frequency
distribution.
(h) Develop a less-than cumulative frequency polygon.
(i) About 60 percent of the orders were filled in less than how many days?

154
 CHAPTER FIVE
SUMMARY STATISTICS

Measures of Central Location

Introduction
Quantitative data arranged in the form of frequency distributions generally exhibit
certain common characteristics: they have the tendency to concentrate at certain
values, usually somewhere near the centre of the distribution. This tendency of
observations to cluster near the central portion of the distribution is known as
Central Tendency and can be measured statistically.
A measure of a central tendency or an average is regarded as the most representative
value of the given data or observed values. This is because it is determined at the
point where the concentration of the values is the greatest, i.e., the frequency is
highest on a distribution scale of measurement. A central tendency is thus a precise
yet simple expression representing a series of divergent individuals, or in other
words, it is the consolidated essence of a complex distribution.

Central tendencies serve the following purposes:

1. They provide a condensed or consolidated formulation of a large quantity of

numerical data which ordinarily cannot be easily interpreted. For example, it
might not just be possible to memorise the individual performances of students in
an examination, but quite simple to remember their average performance.

2. They afford us a basis for comparison with other similar grouped data. For
example, it is impossible to compare the weight of individual babies born in a
particular hospital with every baby born in another hospital, but it is possible to
compare the mean weights of babies born in the two hospitals.

The following types of central tendencies will be discussed:

▪ The arithmetic mean;

▪ The median
▪ The mode.

These are the central tendencies that are in common use.

155
Notations and Symbols
Computations in statistics make use of basic knowledge of index notation and the
summation notation. We will learn how to develop and use index and summation
notations.

Index or Subscript Notation

The index notation is a symbolic terminology that statisticians use to represent the
variables they use in computations. To develop symbols for a set of variables we
have to start with some assumptions. Let us say we want to let the symbol X j (read
X sub j) denote or represent any of the N values X1 , X 2 , X 3 , . . . , X N that is assumed
by a variable X. The letter j in X j , which can stand for any of the numbers 1 , 2 , 3 , .
. . , N is called an index or subscript. Any other small letter of the alphabet other than
j for example i, k, l can also be used as an index or subscript.

Summation Notation
If we now use the knowledge of the index notation we can develop the summation
N
notation. The symbol Xj =1
j is used to denote/represent the sum of all the X j
N
variables from j = 1 to j = N i.e. X
j =1
j = X1 + X 2 + ... + X N . Without loss of

generality, this sum can be written simply as  X . The symbol  is the Greek
capital letter called sigma, which statisticians use to denote ‘sum up values’.

Example 1.1
Expand the following:
N 6 5 4
1. Xj
j =2
2. Xj
j =4
3.  ( X j + 2)
j =3
4. ( X
j =2
j − 4)

Solution
N
1. X
j =2
j = X 2 + X 3 + ... + X N

6
2. X
j =4
j = X4 + X5 + X6

156
 5
3. ( X
j =3
j + 2) = ( X 3 + 2) + ( X 4 + 2) + ( X 5 + 2)

4
4. ( X
j =2
j − 4) = ( X 2 − 4) + ( X 3 − 4) + ( X 4 − 4)

Example 1.2
Write the following in symbols:
5. X1 + X 2 + ... + X N 6. X 4 + X 5 + X 6 7. ( X 3 + 5) + ( X 4 + 5) + ( X 5 + 5)

Solution
N
5. X1 + X 2 + ... + X N =  X j
j =1

6
6. X 4 + X 5 + X 6 =  X j
j =4
5
7. ( X 3 + 5) + ( X 4 + 5) + ( X 5 + 5) =  ( X j + 5)
j =3

Exercise
Expand the following:
N
N 6 8
1. X
j =4
j 2. X j =3
j 3. X
j =5
j

Write the following in sigma notations:

4. X 2 + X 3 + ... + X N =

5. X 5 + X 6 + X 7 + X 8 =

6. ( X 5 − 3) + ( X 6 − 3) + ( X 7 − 3) =

We will now use our ability to expand an expression and write in symbols to begin
computations involving measures of central location.

157
Measures of Central Location
As stated earlier in the introduction of this section, a measure of central location or
average is that value which is typical or representative of a set of data. The value
tends to lie centrally within a set data arranged according to magnitude.
Such measures of central tendency serve two purposes:
(a) It is a concise, brief and economic description of a mass of data and
(b) It is also a simple measure that represents all the measures in a sample and it
enables us to compare two or more distributions.
Several types of measures of central location/tendency exist. The most
common is the arithmetic mean. Others are the median, the mode, the
geometric mean and the harmonic mean. Each of them has advantages and
disadvantages depending on the data and the intended purpose of the data.

The Arithmetic Mean

The arithmetic mean is what the ordinary man calls the average. The arithmetic mean
or simply the “mean” is calculated by adding up all the values/variables and divide
the total by the number of values/variables. Therefore the mean of a set of N
variables: X1 , X 2 , X 3 , . . . , X N is denoted by X (read X bar) and defined as follows:

X =
 X = X1 + X 2 +...+ X N
N N
Example 1.3
Raw Data
Find the mean of the values 4 , 5 , 2 , 3 , 10.

X =
 X = 4+5+ 2+3+10 24 = 4.8
N 4 5
Arithmetic Mean for Grouped Data
Where the data has been grouped in a frequency table the calculation is different.
If X is a variable having values x1 , x2 , x3 , ... , xk occurring with respective
frequencies f1 , f 2 , f3 , . . . , f k , then the arithmetic mean of the given data may be
obtained by the formula:

f1x1 + f 2 x2 + ... + f k xk  fx  fx
X = = =
f1 + f 2 + ... + f k f N
158
Example 1.4
Grouped Frequency Data
Find the arithmetic mean of the following data:
2 , 2 , 2 , 3 , 3 , 4 , 4 , 4 , 5 , 5

Solution
We may obtain the mean using the formula (5.1)

32 + 23 +34 + 25 34

X = = = 3.4
3+ 2+3+ 2 10
Quite often, we have data in the form of frequency distributions in which there are
regular class intervals. If we assume that each class interval can be represented by its
midpoint, then the formula in (5.1) can easily be applied and the procedure can be
simplified appropriately as illustrated in the following example.

Example 1.5
The table below is a grouped frequency distribution of ages of 34 children. Use it to
find the mean age of the children.
Age f x fx
1-2 5 2 10
4-6 6 5 30
7-9 10 8 80
10-12 4 11 44
13-15 6 14 84
16-18 3 17 51
N = 34  fx = 299

X =
 fx = 299 = 8.8
N 34
This method is called the long method and involves a lot of tedious calculations.
When you are confronted with large values in the class intervals, the formula in (5.1)
makes the computation quite tedious. To avoid this, there is the use of two short
methods I will introduce: (i) The Assumed Mean method and (ii) The Coding
method.

159
Exercise
Use the following frequency distribution to find the arithmetic mean:

Marks Frequency

10 – 19 2
20 – 29 3
30 – 39 8
40 – 49 5
50 -59 2

The Assumed Mean Method

If A is any guessed or assumed mean (any value of x) then let d = x − A be the
deviations of x from A then X is computed as follows:

X = A +
 fd (5.2)
N
Example 1.5
Using the data in Example 1.4 and an Assumed mean A = 11, find the mean age of
the children.

Solution

Age f x d = x-A fd
1–3 5 2 -9 - 45
4–6 6 5 -6 - 36
7–9 10 8 -3 - 30
10 – 12 4 11 0 0
13 – 15 6 14 +3 18
16 – 18 3 17 +6 18
N= 34 Σfd = - 75

160
 Using (5.2), we have:

X = A +
 fd
N
= 11 + (- 75)/34
= 11 – 2.2
= 8.8

The Coding Method

Instead of using X in formula (5.1) we create a new variable
x− A
u=
c
Where c is the equal class interval sizes of the data provided and A is any Assumed
mean value which is any of the X values picked up from the data. Usually, a value
close to the middle of the data is taken.

Then u =
 fu =  fu
f N

To find the arithmetic mean of the data, we use the formula:

X = A + cu (5.3)

Example 1.6
Use the coding formula to obtain the mean for the following data:

Exact Class Midpoints (x) Frequency (f)

Boundaries

0 – 10 5 12
10 – 20 15 15
20 – 30 25 28
30 – 40 35 25
40 – 50 45 20

100

161
 Solution
x − 25
Here, c = 10 , we take A = 25. Therefore u =
10
So we have:

Exact Class Midpoints (x) Frequency (f) x − 25 fu

Boundaries u=
10

0 – 10 5 12 -2 -24
10 – 20 15 15 -1 -15
20 – 30 25 28 0 0
30 – 40 35 25 1 25
40 – 50 45 20 2 40

100 26

Therefore:

u=
 fu = 26 = 0.26
 f 100
To find the mean, we use the formula (5.3):
X = A + cu = 25 + 10  0.26 = 27.6

When the class intervals are of equal sizes then the Assumed Mean or the Coding
method is advisable, but when the class intervals are of unequal size the long method
is recommended. The mean is used when the greatest reliability of the data is
required. We also use it when the distribution is reasonably symmetrical and when
subsequent statistical calculations are to be made.

Exercise
(a) Using an assumed mean of 34.5, compute the arithmetic mean for the data in
Activity 1.2.
(b) Use the coding method to obtain the mean for the data in Activity 1.2

162
Properties of the Arithmetic Mean
(i) The mean is affected by all the values in the data set, which is good so far as it
makes the mean a very representative central tendency. It may however not be a
good representative of the data if there are large deviations from the central value.
(ii) It is easy to compute and is capable of algebraic manipulations.
(iii)It is determinate i.e. unique.
(iv) It is a typical representative for all the values in the data set.
(v) Its value may be substituted for each value in the data set without changing the
total: i.e.  fx =  f x = N x .
(vi) The algebraic sum of the deviations of a set of numbers from their arithmetic
mean is zero.
i.e.  f ( x − x ) =  f x − f x = N x − N x = 0 .

Advantages of the Mean

The arithmetic mean is the most widely used measure of central tendency
(i) Its definition is clear and precise. It corresponds to the centre of gravity of the
observations.
(ii) It is simple to understand and easy to compute.
(iii)It uses each and every item in its computation
(iv) It has a determinate value and is rigidly defined.
(v) It can be subjected to further algebraic treatment and advanced statistical theory is
based on it.
(vi) It can be found even if only the sum of the values is known and the individual
values are not known.
(vii) It provides a good standard of comparison since extreme values can cancel
each other out when the number of observations is large.

Disadvantages of the Mean

(i) It may be influenced greatly by unrepresentative values. In such cases the
representative character of the mean is lost.
(ii) It gives greater importance to larger values and less importance to smaller values.
i.e. it has an upward bias.
(iii)It cannot be computed if one or two values are missing in the data set.
(iv) It cannot be located just by inspection like the median and the mode.
(v) It may conceal facts and may lead to distorted conclusions.

The Median
Raw (Ungrouped) Data
The median of a set of raw data arranged in order of magnitude is the middle value
when the number of values is odd, or it is the mean of the two middle values, when
the number of values is even.

163
 For example the median for the following values: 3 , 4 , 6 , 8 , 10 , 12 , 14 is 8 , this
value is the 4th ranked value.
th
 7 +1
Note that 4th ranked value =   ranked value.
 2 
Also, finding the median for the following data: 5 , 7 , 9 , 11 , 12 , 15 , 18 , 20 , then
11 + 12
we have the median as = 11.5
2
which is the average of the two middle observations.
The median is defined as that value which divides a distribution so that an equal
number of items or values occur on either side of it. Note also that the median cannot
be found for disorderly data. We must first arrange the data in order of magnitude.
th
 n +1
Generally, if there are n observations, the median is the value of the   ranked
 2 
item.
▪ Rule 1: If the ranked item results in a whole number, then the median is value that occurs
at that ranked value.
▪ Rule 2: If the ranked item results in a fractional half (2.5, 3.5, etc), then the median is
equal to the average of the corresponding ranked values within which the result falls.

Exercise
Determine the median for the following ungrouped data:
(a) 10, 5 , 13 , 12 , 15 , 17 , 21 , 7 , 9
(b) 10 , 8 , 7 , 15 , 21 , 22 , 23 , 14 , 8 , 20

Median for Grouped Data

When data has been grouped into a frequency table we cannot determine the median
using the method used for raw data. To determine the median for the age distribution
of children in a frequency table, we first find n 2 of the cases and working from
either end of the table, we interpolate to determine the value of the median at n 2 . If
we decide to start from the lower end of the table, the median is computed from the
following expression:
 n 2 − ( f ) 
Median = L1 +  1
c (5.4)
 f med 

Where
L1 = Lower class boundary of class in which median falls
164
 n = Total frequency
(Σf)1 = Sum of frequencies in classes below the median
class
fmed = Frequency of median class
c = Class width for the table

Example

Find the median for the table of age distribution of 34 children below:

Age Frequency (f) less than cumulative

frequency
1–3 5 5
4–6 6 11
7–9 10 21
10 – 12 4 25
13 – 15 6 31
16 – 18 3 34
n = 34

Step 1
Find n 2 of cases = 34/2 = 17. This means we need to count 17 cases out of 34 to
locate the class in which the median would lie.
Step 2
The lowest class (1 – 3) has 5 cases; therefore the median would not lie in that class.
The next class (4 – 6) has 6 cases and when this is added to the 5 in the previous
class we have cumulatively 11 cases. This means the median would not lie in the (4 –
6) class. The next class (7 – 9) has 10 cases. If we add the 10 cases to the 11 we
have, we would get 21 and this is more than the 17 cases we are looking for. This
means we cannot take all the 10 cases lying in this class. This means that the median
would lie in the (7 – 9) class and it becomes the median class.
Step 3
Having determined the median class we can now compute the median using the
expression given as follows:
L1 = Lower class boundary of class in which median falls: = 6.5

165
 N = Total frequency:= 34
Σf1 = Sum of frequencies in classes below median class: =11
fmed = Frequency of median class: =10
c = Class width for the table:= 3

Step 4
Substituting in the expression, we have:
 N / 2 − ( f ) 
Median = L1 +  1
c
 f med 

 34 
 2 − 11 
Median = 6.5 +  3
 10 
 
6
= 6.5 +  3 = 6.5 + 1.8 = 8.3
10

Estimating Median by Graph

The median may also be located using a graph of the cumulative frequency (ogive).
This is done as follows:
▪ Draw a perpendicular to the vertical axis at n 2
▪ Extend the perpendicular till it intersects the less than line graph
▪ From the point of intersection draw another perpendicular to intercept the
horizontal axis.
▪ The point of intersection on the horizontal axis is the value of the median.

166
Estimating Median from Graph of Cumulative Frequency

Median value

Uses of the Median

The median is used when the distributions are badly skewed. It is also used when a
quick and calculated average is required. Finally when extreme cases are likely to
affect the mean disproportionately the median is not useful.

Properties of the Median

(i) Geometrically, the median divides a histogram or a frequency curve into
two parts with equal areas.
(ii) It is unaffected by the magnitude of extreme deviations from the average.
(iii) It may be located even when the observations in the data cannot be
measured quantitatively, so long as they can be ranked or arranged in
order of magnitude.

Advantages of the Median

(i) It can easily be understood and its computation is simple.
(ii) It can be obtained even with incomplete data. It is only concerned with
only a few central observations.
(iii) It balances the number of observations in a distribution. Quite useful in
describing scores, ratios and grades.

167
 (iv) It is useful in the case of skewed distributions like those of incomes and
prices.
(v) It can be used for qualitative data.
(vi) In the case of open-ended classes, the median can be calculated but the
mean cannot.
(vii) It can easily be determined graphically. (See Subsection: Estimating the
median from graphs)

Disadvantages of the Median

(i) Re-arrangement of data values is necessary to compute the median.
(ii) The median is not easily capable of algebraic manipulations. As such it is
not used much in advanced statistical studies.
(iii) The empirical formula for the median based on interpolation may not
always give correct results.
(iv) It ignores significant extreme values.
(v) Weighting cannot be used in the case of the median, as is the case with the
mean. The scope of operations with the median is narrowed.
(vi) It cannot be computed as exactly as the mean.

Practical Applications of the Median Concept

One of the most important practical applications of the concept of the median is to be
found in road construction. It is used to divide the road into two equal halves or into
dual carriageways whereby a road median is created in the middle of the road using
concrete or hedges.
Examples of such dual carriageways include the following:
• Accra-Tema Motorway
• Nkrumah Circle - Obetsebi-Lamptey
• Obetsebi-Lamptey – Mallam
• Nkrumah Circle - La

Exercise
List three examples of data you can compute the median from.
(a)-------------------------------------------
(b) ------------------------------------------
(c) ------------------------------------------

168
The Mode
The term ’mode’ literally means ‘norm’ or ‘fashion’. The mode is thus a typical
measure of central tendency inasmuch as it is the most probable value in a given data
set. This concept is often in the minds of many people when they speak about
averages. The modal shoe size is the size that is bought by many people, the modal
customer is so called ‘average’ customer.
The mode of a set of raw data is that value which occurs with the greatest frequency
i.e. the most common value, or it is the value around which the values tend to
concentrate: For example the mode of the values 3 , 4 , 5 , 5 , 6 , 7 is 5. The values 2 ,
4 , 6 , 8 , 10 , 12 have no mode, but the mode for the following set of values 2 , 3 , 4 ,
4 , 4 , 5 , 7 , 7 , 7 , 9 is 4 and 7 (bimodal).

Mode from a Grouped Data

For data in a frequency table, the mode is obtained as follows
 1 
Mode = L1 +    c (5.5)
 1 +  2 
Where
L1 = Lower class boundary of modal class
∆1 = Difference between frequency of modal class and frequency of next lower class.
∆2 = Difference between frequency of modal class and frequency of next higher
class.
c = Class width.

Example
From the table of age distribution of 34 children the following values are computed
L1 = 6.5 ∆1 = 4
∆2 = 6 C =3

169
 Age Distribution of 34 Children

Age f

1–3 5

4–6 6

7–9 10

10 – 12 4

13 – 15 6

16 – 18 3

N= 34

 1 
Mode = L1 +    c
 1 +  2 
 4 
= 6.5 +    3
4 + 6
= 6.5 + 1.2
= 7.7

Estimating the Mode from the Mean and Median

An empirical, approximate relationship has been shown to exist between the mean,
median and mode:
Mode 3Median – 2Mean (5.6)
Therefore, we can compute the mode if we know the values of the mean and the
median. The mode obtained by this method need not necessarily be the same as that
found by using any other method.

Example
Given that for particular distribution, the mean = 125.7 and the
median = 124.9, find the mode.
Mode = 3Median − 2Mean
= 3 124.9 − 2 125.7 = 123.3

170
Estimating the Mode Graphically
It is possible to locate the mode from a histogram of a given distribution. This is
done by drawing two diagonally intersecting lines joining each upper corner of the
modal class bar to each opposite upper corner of the adjacent bars. The
perpendicular drawn from the point of intersection on the horizontal axis gives the
modal value on the horizontal axis as depicted in the figure below:

Advantages of the Mode

(i) It is simply defined and computed. When graphed, it can be located almost
by inspection.
(ii) It is a popular central tendency in the sense that it is the one that most
people use without being aware of it e.g. when they speak of the average
number of car accidents, machine breakdowns etc.
(iii) Extreme values have no effect on the computation of the mode and it can
be calculated when complete data is not available.
(iv) It is the most typical in the sense that it denotes the most probable value in
the data values.

Disadvantages of the Mode

(i) In the case of bimodal or multimodal distributions, it is not possible to
pinpoint any one value as the central tendency.
(ii) It is not rigidly defined and thus cannot be called an ideal central
tendency.
(iii) It is often indeterminate when the distribution is highly irregular.

171
 (iv) It is not based on all observations in the data and hence is not an ideal
measure of central tendency. It lays too much emphasis on the modal
group and thus may not be fully representative of the whole data set.
(v) It is not easily amenable to algebraic manipulations.

Practical Applications of the Concept of Mode

The concept of the mode or the most typical value, finds practical application in
various aspects of daily life. The concept finds practical application in the work of
the following:
• The Motor Traffic and Transport Unit (MTTU) of the police
• Manufacturers of ready to wear clothes
• Manufacturers of shoes
• Manufacturers of hats and belts

The Motor Transport and Traffic Unit (MTTU) finds the concept of the mode of
practical use in the control of traffic. By studying the frequency of the use of streets,
they are able to determine which roads and streets need traffic officers at specific
times of the day i.e. during the morning and evening rush hours in urban areas
between 7 and 9 am and between 5 and 7 pm.
The manufacturers of ready to wear clothes as well as shoes need to know the sizes
of clothes and shoes that people wear most so as to set their machines to cut such
sizes to be sewn and made for the consumers. Shoe sizes are indeed modal sizes.
This is true of all ready to wear clothes for which sizes like: small (S), medium (M),
large (L), extra large (XL) and extra-extra large (XXL).
The manufacturers of hats and belts need to know the modal size/circumference of
the head and waist. Hats are thus made to specific head sizes and belts are cut to
specific lengths.

Other Measures of Location Related to the Median

Although they are not measures of central tendency, there are certain other measures,
which are related to the median because they too are “positional measures”. These
are the quartiles, quintiles, deciles and percentiles.

Quartiles
There are three quartiles: Q1, Q2, Q3 and they divide a distribution into four equal
parts.
The first quartile, Q , is the value for which 25% of the observations are smaller
1
and 75% are larger. Q is the same as the median (50% are smaller, 50% are larger).
2
Only 25% of the values are greater than the third quartile.

172
 We find a quartile by determining the value in the appropriate position in the ranked
data, where;
First quartile position: Q = (n+1)/4 ranked value
1
Second quartile position: Q = (n+1)/2 ranked value
2
Third quartile position: Q = 3(n+1)/4 ranked value
3
where n is the number of observed values.

Guidelines in Locating the Quartiles

▪ Rule 1: If the result is a whole number, then the quartile is equal to that ranked value.
▪ Rule 2: If the result is a fractional half (2.5, 3.5, etc), then the quartile is equal to the
average of the corresponding ranked values within which the result falls.
▪ Rule 3: If the result is neither a whole number or a fractional half, you round the result to
the nearest integer and select that ranked value.

Example
Find Q1 using the sampled data in an ordered array:

11 12 13 16 16 17 18 21 22

Solution
First, note that n = 9.
Q1 = is in the (9+1)/4 = 2.5 ranked value of the ranked data, so use the value half way
nd rd
between the 2 and 3 ranked values,
so Q1 = 12.5

Exercise
Using the Example, compute Q2 and Q3.

Locating the Quartiles from Grouped Data

For grouped data, we use a formula similar to that of finding the Median in grouped
data:

173
  n 
 i − (  f )Qi 
Qi = LQi +  4 c
 f Qi 
 

where Qi is the i th Quartile value;

LDi is the lower class boundary the i th Quartile class;
n is the total frequency;
( f ) Qi
is the cumulative frequency before the i th Quartile class;

f Qi is the frequency of the i th Quartile class;

c is the size of the i th Quartile class.

Deciles
There are nine (9) deciles: D1, D2, D3 ,. . . , and D9 which divide a distribution into
ten equal parts.
The ith Decile in a raw data set is the
i ( n + 1)
th

Di = ranked value, where i = 1, 2 , 3 , . . . , 9.

10

Locating the Deciles from Grouped Data

For grouped data, we use a formula similar to that of the Median:

 n 
 i 10 − (  f ) Di 
Di = LDi +  c
 f Di 
 

where Di is the i th Decile value;

n is the total frequency;
LDi is the lower class boundary the i th Decile class;
( f ) Di
is the cumulative frequency before the i th Decile class;

f Di is the frequency of the i th Decile class;

c is the size of the i th Decile class.

Note that
D5 = Q2 = Median
174
 Percentiles
There are 99 percentiles that divide a distribution into 100 equal parts. The percentiles
to note include: P25 , P50 and P75.
The ith Percentile in a raw data set is the
i ( n + 1)
th

Pi = ranked value, where i = 1, 2 , 3 , . . . , 99.

100
Note that:
Q1 = P25
Q2 = D5 = P50
Q3 = P75
The P50 is the same as the 2nd Quartile (Q2) and the 5th Decile (D5).
The P75 is the same as the 3rd quartile (Q3)

Comparison of the Mean, Mode and Median

The mean, mode and median are the most commonly used measure of central
tendency. The mean is measure with which most people are vaguely familiar. It is the
most frequently used measure of central tendency.
The three measures of central tendency embody completely different ideas. The
mean is the sum of the values divided by the number of values/items. The median is
the value, which divides a series or distribution so that half or more of the items are
equal to or less than it, and half or more of the items are equal or greater than it. The
mode is the value around which the items tend to concentrate.
There is however no best measure of central tendency. No one of the three measures
is the one that should always be used under all conditions. The measure to use
depends on the idea we want to convey and the nature of the data. The mean uses
more information than the median because all the values are used in computing the
mean, whereas the median only uses the relative position of the values. For example
we would look at the following values 72, 81, 86, 69, 57. If the highest value had
been 126 rather than 86, the median would have been unaffected but the mean would
have greatly increased. The mean is therefore affected by changes in extreme values
whereas the median will be unaffected unless the value of the middle case is also
changed. In the example above as long as 72 remains the third value (in ranking) the
median will be unchanged. Generally the mean is a more stable measure than the
median because it varies less from sample to sample i.e. sample median tends to
differ from one sample to another more than the means. A rule of thumb however is
that when in doubt use the mean in preference to the median.
The median does not depend on extreme values and when data has many extreme
values, the mean may give misleading results under some circumstances. For this
reason when a distribution has many extreme vales the median will generally be

175
 more appropriate than the mean. The mode is used when the greatest estimate or
value of central tendency or when the most typical value is required.

The Normal Distribution

When the mean, mode and median values in a distribution all coincide in exactly the
middle of the distribution, we describe the distribution as a normal distribution. A
normal distribution has certain characteristics, Mean = Mode = Median, the
distribution is symmetrical, which means that
• The shape of the curve is bell-shaped
• Mean = Mode = Median
• The Left Hand Side (Curve) = Right Hand Side (Curve), when folded onto each
other.
• The distribution is fairly distributed on both the Right Hand Side (RHS) or the
positive side and the Left Hand Side (LHS) or the negative side.
• In an examination where the performance of the students is normal, it means
there are about the same percentage of students who have performed very well
and those who have performed very poorly.
• The performance of the majority of students is average.

The Normal or Symmetrical Distribution

176
Skewness
The tendency of a distribution to depart from normal is called a skew. Skewness
refers to the symmetry or the lack of symmetry in the shape of a frequency
distribution. This characteristic is of particular importance in judging how typical
certain measures of central tendency are. There are two types of skew, namely a
positive and a negative skew. When a distribution is positively skewed (skewed to
the right) it means that when we compare the mean, mode and median, the mean is
higher than the median and the median is higher than the mode
i.e. Mean > Median > Mode.
This means that if an examination performance is positively skewed then the
majority of students performed very poorly. We would describe the income
distribution of Ghanaian workers, which is generally accepted to be low to be
positively skewed or skewed to the right.

When a distribution is negatively skewed (skewed to left) it means that when we

compare the mean, mode and median, the mode is higher than the median and the
median is higher than the mean i.e. Mean < Median < Mode. This means that if an
examination performance is negatively skewed then the majority of students
performed very well. If most students secure distinction in an examination, then the
performance of the students would be described as negatively skewed or skewed to
the left.

177
Measures of Dispersion or Variability
Introduction
In many cases in social science research, the focus of attention is on measures of
central tendency e.g. we may want to compare various pressure groups with respect
to average attendance at meetings and income levels. We may also wish to obtain
measures of homogeneity of these groups.
In social science research, notions like concentration of power imply that dispersion
or homogeneity may be of theoretical interest. Even when comparing two groups in
terms of averages, we may still need to know something about the spread or
dispersion within each group. We would realize that each group is extremely
homogeneous with regard to income; in such a case a given difference between the
means of the groups would not be as important or indicative as would be the case if
each group were more homogeneous.
A comparisons limited only to the means would carry no hint of the fact that groups
differ in regard to dispersion. Although two sets of data may have similar averages,
they may differ considerably with respect to the spread or dispersion of the
individual observations. Measures of dispersion describe the variation in numerical
observations. Various measures of dispersion or variation exist. The most important
include the range, the mean deviation and the standard deviation.

The Range
The simplest and at times the least satisfactory measure of dispersion is the range. It
is the difference between the largest and the smallest values of a series or scores e.g.
the range of the data set 2 , 5 , 8 , 10 , 12 is 12 - 2 = 10. The range is sometimes
given by simply quoting the smallest and the largest values.

If the data is grouped into a frequency table, the range is the difference between the
upper class boundary of the highest class and the lower class boundary of the lowest
class e.g. the range for the following table:

178
Age Distribution of 34 Children

Age f

1-3 5

4-6 6

7-9 10

10-12 4

13-15 6

16-8 3

N=34

The range for the above table: 18.5 – (0.5) = 18.5 - 0.5 = 18.

A second method for computing the range for grouped data is the difference between
the class mark (mid point) of the highest and lowest classes. For the table above the
range would be 17- 2 = 15
The second method tends to eliminate extreme cases to some extent.
The problem with the use of the range is that it is based on only two cases and they
are the extreme cases. Since extreme cases are likely to be the rare or unusual cases
in most empirical situations, it is not advisable to base a measure of variability on
them e.g. if a village has a population of one millionaire and any 50 people are
randomly chosen for a study, the millionaire may or may not be included in the
sample. If the millionaire is included the range of the incomes will be very large and
thus misleading as a measure of dispersion. The range will not tell us anything about
the dispersion of values between the two extreme values. Also the range will not will
ordinarily be greater for large samples than small ones simply because in large
samples we have a better chance of including the most extreme cases. For those
reasons, the range is not ordinarily used in social science research except at the most
exploratory levels.

179
Quartile Deviation and Semi-Inter-Quartile Range
The quartile deviation, denoted by QD is a type of range but instead of representing
difference between extreme values. It is defined as the distance between the first and
third quartiles.
QD = Q3 – Q1
Half of this measure is called the semi-inter-quartile range. Usually Q1 and Q3 will
vary less from one sample to another than most extreme cases; the quartile deviation
is a far more stable a measure of spread than the range. However it does not take
advantage of all available information. As a result it does not include the values
below Q1, between Q1 and Q3 or above Q3. To take care of the variability among the
middle cases and what is happening at the extremes of the distribution, we turn to the
mean deviation and the standard deviation.

The Mean Deviation

The mean deviation takes into consideration the values of all items in a distribution.
To compute the MD, we take the deviations of each value from the mean and take
the mean of the deviations. It is computed from the following expression:

MD =
 X −X
N
Where the sign (which are two vertical lines) is the absolute value of the
deviation of X − X (the absolute value of a number is the number without its
associated sign and is indicated by two vertical lines placed around the number) e.g.
−5 = 5 , +5 = 5 , −0.6 = 0.6 . The absolute value of a number is the number
without regard to the sign preceding it.

Example
Compute the MD for the following data: 72 , 81 , 86 , 69 , 57

Solution

X=
X =
365
= 73
N 5

180
 X X −X

72 1
81 8
86 13
69 4
57 16

42

MD =
 X −X =
42
= 8.4
N 5

This means that on average, the values differ from the mean in absolute terms by 8.4
The problem with the mean deviation is that, it has some serious limitations. These
are (a) that absolute values are not easily manipulated algebraically (b) the mean
deviation is not easily interpreted theoretically and it does not lead to simple
mathematical results.

The Standard Deviation

The standard deviation is the most useful and frequently used measure of dispersion.
It is defined as the square root of the arithmetic mean of the squared deviations of
observations from the mean. There are two formulas for the standard deviation i.e.
for the sample and for the population.

The Population Standard deviation

The population standard deviation, denoted  , of a set of N numbers
X1 , X 2 , X 3 , . . . , X N is given by the following expression:

( X − )
2

 =
N

Where  =
X is the population mean;
N
N is the population size.
Using the raw data constituting the population of values: 72 , 81 , 86 , 69 , 57, then
the population standard deviation  is computed as follows:

181
 X ( X − ) (X − )
2

72 -1 1
81 8 64
86 13 169
69 -4 16
57 -16 256
506

( X −  )
2
506
= = = 101.2 = 10.06
N 5

The Sample Standard deviation

The sample standard deviation, s of a set n numbers X1 , X 2 , X 3 , ... , X n is given by
the following expression:

( X − X )
2

s=
n −1

Where X =
X is the sample mean;
n
n is the sample size.

The standard deviation is used a good deal in experimental work as well as in social
science research. Its importance lies in the fact that statistically it is the most reliable
measure of dispersion, i.e. it varies less from one sample to another. It would
therefore be regarded as the most accurate estimate of the dispersion of the
population. When it is necessary to distinguish between the standard deviation of a
population from the standard deviation of a sample drawn from the same population
the symbol s is used for the sample standard deviation and sigma (σ) is used for the
population standard deviation. Note that the formulas for their computations differ in
terms of the denominator.

182
 The Variance
The variance of a set of data is defined as the square of the standard deviation.

The Population Variance

( X −  )
2

 2
=
N

Where σ2 = Population standard deviation

The Sample Variance

( X − X )
2

S =
2

n −1
where S2 = Sample standard deviation

Methods for Computing the Standard Deviation

To avoid computing the standard deviation by having to compute the mean and the
deviations from the mean we use the following expression:

( X )
2

X 2
−
N
 =
N
for the population standard deviation and

( X )
2

X 2
−
n
s=
n −1
for the sample standard deviation.
183
 This means that for a set of raw data, we need only the values and the square of each
value.

Example
The population standard deviation σ, for the following data set:
72 , 81 , 86 , 69 , 57
is computed as follows:

X X2
72 5184
81 6561
86 7396
69 4761
57 3249
365 27,151

( X )
2
3652
X 2
−
N
27,151 −
5
 = = = 10.059
N 5

Finding the Standard Deviation from Grouped Data

For data that is grouped into a frequency distribution the standard deviation is
computed from the following expression:

(  fX )
2

 fX 2
−
N
=
N

Where N =  f is the total frequency;

f is the class frequency.

184
 Example

Class
Marks Frequency Midpoint X2 fX fX2
(f) (X)
60 – 62 5 61 3,721 305 18,605
63 – 65 18 64 4,096 1,152 73,728
66 – 68 42 67 4,489 2,814 188,538
69 – 71 27 70 4,900 1,890 132,300
72 – 74 8 73 5,329 584 42,632
100 6,745 455,803

(  fX )
2
67452
 fX 2 −
N
455803 −
100
= = = 8.5275 = 2.92
N 100

Using the Assumed Mean Method to Calculate the Standard

deviation (Short Method)
Introduction
To use this method, we need to make an initial assumption. The assumption we make
is as follows:
If we assume that d = X – A are the deviations of X from an arbitrary constant A, then
the standard deviation can be computed from the following expression:

(  fd )
2

 fd − 2

s =
f
 f −1

185
 Using data from the previous table of the distribution of 34 children we develop the
following working sheet:

Class
Marks Frequency Midpoint d = X- A fd fd2
(f) (X)
60 – 62 5 61 -6 -30 180
63 – 65 18 64 -3 -54 162
66 – 68 42 67 0 0 0
69 – 71 27 70 3 81 243
72 – 74 8 73 6 48 288
100 45 873

From the table above, we have the following sums:

Σf =100 , Σfd = 45 , Σfd2 = 873

(  fd )
2

 fd −2
873 −
452
s =
f = 100 = 8.6136 = 2.93
 f −1 100 − 1

Finding the Standard Deviation Using the Coding Method

(Short method)
Where the data is grouped with frequencies whose class intervals have equal size “c”
we have d = cu or X = A + cu and the standard deviation is computed from the
following:

(  fu )
2

 fu −2

s = c
f
 f −1
where u =
( X − A) .
c

186
 This method is recommended for group data when class interval sizes are equal. It is
called the coding method.

Class
Marks Frequency Midpoint
u =
( X − 67 ) fu fu2
(f) (X)
3
60 – 62 5 61 -2 -10 20
63 – 65 18 64 -1 -18 18
66 – 68 42 67 0 0 0
69 – 71 27 70 1 27 27
72 – 74 8 73 2 16 32
100 45 97

From the table above, we have the following sums:

c=3 , Σf =100 , Σfu = 15 and Σfu2 = 97

(  fu )
2

 fu − 2
97 −
152
s = c
f = 3 100 = 3 0.9571 = 2.93
 f −1 100 − 1

Uses of the Standard Deviation

Among the uses of the standard deviation are the following
• When the statistics has the greatest stability or reliability.
• When it is desired that an extreme deviation should have proportionally greater
effect on the measure of deviation.

Exercise
1. Find the standard deviation of the following sets of data:
(a) 3 , 6 , 2 , 1 , 7 , 5
(b) 3.2 , 4.6 , 2.8 , 5.2 , 4.4

187
 2. The following is the age distribution of 40 students from the University of Ghana:
Age distribution of 40 Students

Age (f)
18-22 8
23 -27 7
28-32 10
33-37 5
38-42 7
43-47 3
N=40

(a) Find the range and the mean deviation for the table:
(b) Compute the standard deviation for the age distribution for the students.

Some Standardized Measures and their Uses

The Coefficient of Variation (CV)
The coefficient of variation is the standard deviation divided by the mean, multiplied by 100.
It is always expressed as a percentage (%). It shows variation relative to mean.

The CV can be used to compare two or more sets of data measured in different units or when
their means are significantly different.

Mathematically stated, the CV is obtained by the formula:

s
CV =  100%
X

Example
Suppose that two stocks, A and B have the following summary statistics:
Stock A:
Average price last year = GHC 50
Standard deviation = GHC 5
Stock B:
Average price last year = GHC 100
Standard deviation = GHC 5
Which of these two stocks’ price is less variable?

188
 Solution
 s  GHC 5
CVA =   100% = 100% = 10%
X GHC 50

 s  GHC 5
CVB =   100% = 100% = 5%
X GHC 100
From the two CVs computed, you can observe that the two stocks have the same standard
deviation but stock B is less variable compared to its price.

Exercise
The points of two football teams in ten league seasons are given below. Which of the two
teams are more consistent?
Team A: 32 28 47 63 71 39 10 60 96 14
Team B: 19 31 48 53 67 90 10 62 40 80
[Hint: Compute the CV of each team and compare the results]

The Standard Score (or Z-Score)

To compute the Z-score of a data value, subtract the mean and divide by the standard
deviation. The Z-score is the number of standard deviations a data value is from the mean.

A data value is considered an extreme outlier if its Z-score is less than -3.0 or greater than
+3.0. The larger the absolute value of the Z-score, the farther the data value is from the mean.

Mathematically stated, the CV is obtained by the formula:

X −X
Z =
s
where X is the data value;
X is the mean value;
s is the standard deviation.

Example
Suppose the mean math SAT score is 490, with a standard deviation of 100.
Compute the z-score for a test score of 620.

Solution

X −X 620 − 490 130

Z = = = = 1.3
s 100 100 189
 A score of 620 is 1.3 standard deviations above the mean and would not be
considered an outlier.

Exercise
The average performance in Mathematics in school A is 80 with a standard deviation
of 10. That of school B is 70 and 15 respectively. Adjei who is in school A, scored
75 while his friend Obonto, in school B, scored 65. Compare their relative
performance in Mathematics.

Measures of Symmetry, Skewness and Kurtosis

Here we will discuss some basic measures of shape. They are: Symmetry, Skewness
and Kurtosis.

Symmetry and Skewness of a Distribution

Skewness refers to lack of symmetry in a distribution. In a symmetric distribution,
the mean, median and mode coincide. In a positively skewed (or right skewed)
distribution, the longer tail is on the right side and the mean is on the right side of the
median. In a negatively skewed (or left skewed) distribution, the longer tail is on the
left side and the mean is on the left side of the median.
In a skewed distribution, the distance between the mean and median is nearly one-
third of that between the mean and the mode.

Mathematically,
1
Mean − Median  ( Mean − Mode )
3
or 3 ( Mean − Median )  Mean − Mode

Measures of Skewness
A measure of skewness gives a numerical expression for and the direction of
asymmetry in a distribution. It gives information about the shape of the distribution
and the degree of dispersion on either side of the central value.
We consider some relative measures of skewness.
(i) Pearson’s Coefficient of Skewness
x − Mode 3 ( x − Median )
PSk = =
s s
PS k may have any value. But, usually it lies between – 1 and + 1.

190
Example
If for a given data set, it is found that:
x = 10 , Mode = 8 , s = 4 ,

then we have

PSk =
x − Mode
=
(10 − 8) = 0.5
s 4

(ii) Bowley’s Coefficient of Skewness

Q1 − 2Q2 + Q3
BSk = , − 1  BSk  +1
Q3 − Q1

Example
If for a given data set, it is found that:
Q1 = 20 , Q2 = 50 , Q3 = 80 ,

then we have
Q1 − 2Q2 + Q3 20 − 2 ( 50 ) + 80
BSk = = = 0
Q3 − Q1 80 − 20

Note that a skewness measure of zero indicates that the underlying distribution is
symmetric.

Kurtosis
Kurtosis is a measure of peakedness of a distribution. It shows the degree of
convexity of a frequency curve. If the normal curve is taken as the standard,
symmetrical, bell-shaped curve, kurtosis gives a measure of departure form normal
convexity of a distribution.

Insert diagram

191
 The normal curve is mesokurtic. It is of intermediate peakedness. The flat-topped
curve, broader than the normal curve is platykurtic. The slender, highly peaked
curve, is leptokurtic.

Measure of Kurtosis based on Percentile Values

The Percentile Coefficient of Kurtosis, k , is
1
SIQR ( Q3 − Q1 )
k = = 2
P90 − P10 P90 − P10

Exercise
The following is the age distribution of 40 students from the University of Ghana:
Age distribution of 40 Students

Age (f)
18-22 8
23 -27 7
28-32 10
33-37 5
38-42 7
43-47 3
N=40

(a) Compute a measure of skewness and commen on your result.

(b) Compute a measure of kutosis and comment on your results.

192
The 5 Number Summary Measures and the Box-and-Whisker Plot
The Five Number Summary
The five numbers that describe the spread of data are:
▪ Minimum value
▪ First Quartile (Q )
1
▪ Median (Q )
2
▪ Third Quartile (Q )
3
▪ Maximum value

The Box-and-Whisker Plot

The Box-and-Whisker Plot is a graphical display of the five number summary.

193
Exercise
The following are age distributions for a group of pupils:
14 , 8 , 9 , 5 , 17 , 12 , 20 , 7 , 6 , 2 , 4 , 7 , 8 , 9 , 8 , 17.
(a) Compare the quartile values for the distribution.
(b) Obtain a box-and-whisker plot for the distribution and use it to describe the
shape of the distribution.

Exploratory Data Analysis using Excel

General Descriptive Statistics Using Microsoft Excel
We illustrate how to obtain general descriptive statistics of a distribution using Microsoft Excel.

Example
The following are the marks obtained by 50 students in an examination:
83 64 84 76 84 54 75 59 70 61
63 80 84 73 68 52 65 90 52 77
95 36 78 61 59 84 95 47 87 60
54 79 47 64 36 45 51 24 78 26
33 60 18 68 35 58 48 53 55 45

Solution
1. Enter the data into one column of cells as shown.

194
2. Select Data.
3. Select Data Analysis.
4. Select Descriptive Statistics and click OK.

5. Select the cell range that contains the data and put into the ‘Input Range:’
6. Check ‘Labels in first row’.
7. Select ‘New worksheet Ply:’ to store the output in a new worksheet.
8. Select ‘Summary statistics’.
9. Click OK.

195
 Microsoft Excel
descriptive statistics output

Exercise
5.1 he following are ages of 8 patients in the emergency room of a certain hospital on Friday
night:
52, 48, 37, 54, 48, 15, 42, 12

i. Calculate three measures of central location for these data.

ii. Which of the measures you have calculated is the best? Give reasons

5.2 The data below gives the ages of 27 employees of a company:

41, 35, 49, 36, 34, 24, 36, 30, 40
24, 41, 52, 45, 53 23, 43, 45, 35,
37, 41, 39, 22, 23, 26, 35, 94, 22
i. Construct a stem-and-leave plot for the data. How would you describe the distribution of the
ages of these employees?
ii. Find the values of the three quartiles Q1 , Q2 , and Q3 .Where does the age of 30 fall in relation
to the ages of the employees?
iii. Find the inter-quartile range. What does this statistic measure?
iv. Construct a box plot for the above data. Identify any outlier (s)

5.3 Explain the following statement “The frequency distributions of family income, size of
business, and wages of skilled employees all tend to be skewed to the right.

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5.4 The following daily wage has been collected from two distinct groups of unskilled
workers within a company:
Daily wage (Cedis) Number from Number from
Group A Group B
8000 but under 10000 5 1
10000 but under 12000 17 19
12000 but under14000 23 25
14000 but under 16000 3 4
16000 but under 18000 1 1
18000 but under 20000 1 0

(i) Determine the coefficient of variation and a measure of skew for

each group of workers
(ii) Comment on your results
5.5 Which of the following statements represents ‘skewed to the right” and ‘skewed to the
left” and “symmetrical”?
I. Most students in an exam fail miserably
II. Performance of students in an exam is said to be average
III. Performance of a class in an exam is deemed very good
IV. The salaries of most Ghanaian workers are considered to be low
V. Most students find their first University exams difficult frequency
(more/less than).

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 CHAPTER SIX

Measures of Linear Association or Relationship

Section 1 Correlation Analysis

Introduction
Correlation analysis deals with the analysis of the relationship between two
quantitative variables. The two variables are designated as X and Y. The variable
which we wish to explain or estimate is referred to as the dependent variable and is
denoted by the symbol Y. The variable or factor from which the estimates are made is
called the independent variable and is denoted by the symbol X. The terms dependent
and independent do not necessarily imply any cause and effect relationship between
the two variables. What is meant is simply that estimates of the values of the
dependent variable (Y) may be obtained for given values of the independent variable
X. For that reason the values of Y are dependent upon the values of X. The X variable
may or may not be causing changes in the Y variable. For example if we are
estimating sales of a product from figures on advertising expenditures, sales are the
dependent variable and advertising expenditures is the independent variable. There
may or may not be a causal connection between these two factors in the sense that
changes in advertising expenditures cause changes in sales. In fact in certain
situations, the cause-effect relation may be just the opposite of what appears to be the
obvious one. For example suppose a company budgets a product’s advertising
expenditures for the next year as a flat percentage of the sales of that product during
the preceding year. Then advertising expenditures are more directly dependent on
sales than vice versa.
There are numerous illustrations of variables that can reasonably be assumed related
to one another, that is, they are correlated.
For example:
(a) consumption expenditure (Y) and income (X)
(b) personal net savings (Y) and disposable income (X)
(c) success in University (Y) and examination grades (X)
For each pair, the first named is to be estimated and thus is the dependent variable
and the second factor is the independent variable.

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Establishing Existence of Relationship
Introduction

A very key technique used in establishing existence of relationship between

variables is the use of a scatter diagram or scattergram.

Scatter Diagram
A very useful aid in studying the relationship between two variables is to plot the
data on a graph. This allows a visual examination of the extent to which the variables
are related. The chart used for this purpose is known as a scatter diagram which is a
graph on which each plotted point represents an observed pair of values of the
dependent and independent variables. The establishment of whether there exists any
relationship between any variables is done by the use of a scatter diagram. This is
illustrated by diagrams below. The diagrams below are scatter diagrams with
corresponding values of the correlation coefficients:

Each diagram represents an observed pair of values of the two variables X and Y. The
scatter diagram is usually compressed into an index called the coefficient of
correlation (r).

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Direction of Relationship
Introduction
The coefficient of correlation (r) provides the measure of the strength and direction
of the relationship, which can be used for direct comparisons. The coefficient of
correlation can be either positive or negative. The value of r ranges from –1 to +1. If
r is –1, it means there is a perfect negative relationship between the two variables.
This means that as the X variable increases, the Y variable decreases and shows an
inverse relationship. If r is +1 it means there is a perfect positive relationship
between the two variables. This means that as the X variable increases the Y variable
also increase. In the social sciences it is difficult to find two variables that are
perfectly correlated in a positive or negative direction. This is due to the fact that we
always have to take account of the effect of other intervening variables. The
influences of these other variables usually reduce the ability of social scientists to
predict to precision.

Positive Relationship

Negative Relationship

A positive r indicates that as the independent variable (X) increases e.g. as the
number of years spent on education increases, it influences an increase in the
dependent Y variable or the income people earn. A negative r indicates an inverse

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 relationship between the X and Y variables. This means that an increase in the
independent variable (X), influences a decrease in the dependent (Y) variable. Where
the r = 0 it indicates no relationship between X and Y.

Properties of the Correlation Coefficient

▪ Unit free
▪ Ranges between –1 and 1
▪ The closer to –1, the stronger the negative linear relationship
▪ The closer to 1, the stronger the positive linear relationship
▪ The closer to 0, the weaker any linear relationship

Degree of Relationship
Introduction
This involves the computation of the coefficient of correlation (r) between the X and
Y variables. Linear r is usually called Pearsonian correlation or the Pearson’s
Product-Moment correlation. Pearson’s correlation measures linearity between X and
Y. The term “product-moment” is a mathematical term, which refers to the mean of a
product. Pearson’s r equals the sum of the product of X and Y deviations from their
respective means divided by the value that sum would have if the observations fell
perfectly on a straight line.
By definition Pearson’s r is as follows:

r=
 XY
 X Y
2 2

Where X = X − X and Y = Y − Y
For purposes of computation the expression for r is given as follows:

N  XY − (  X )(  Y )
r=
 N X 2 − ( X )2   N Y 2 − ( Y )2 
       
This is the mathematical equivalent of the definitional formula but it avoids the need
to compute deviations from the mean for each pair of observations.

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Example 1
Given the following data of number of years spent in school (X) and income (Y):

X 1 3 4 6 8 9 11 14

Y 1 2 4 4 5 7 8 9

Determine the following:

(a) Is there a relationship between X and Y?
(b) Calculate the degree of relationship between X and Y.
To determine whether a relationship exist between X and Y, we sketch a scatter
diagram as follows:
Scatter Diagram of X and Y

From the configuration of the coordinates of X and Y we see that there is an upward
slope from the bottom left hand corner to the top right hand corner of the graph. This
suggests that a linear relationship exist between X and Y and that the relationship is
positive, i.e. as the independent variable (X) (i.e. number of years spent in school)
increases it influences an increase in the dependent variable (Y) (i.e. income) also
increase.

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Degree of Relationship between X and Y
This involves the computation of the coefficient of correlation between X and Y.

X Y X2 Y2 XY

1 1 1 1 1
3 2 9 4 6
4 4 16 16 16
6 4 36 16 24
8 5 64 25 40
9 7 81 49 63
11 8 121 64 88
14 9 196 81 126

56 40 524 256 364

n XY − (  X )(  Y )
r=
 n X 2 − ( X )2   n Y 2 − ( Y )2 
       

=
8(364) − (54)(40)
=
( 2,912 – 2,240 )
8(524) − (56) 2  8(256) − (40) 2   4,192 – 3,136   2,048 – 1,600
672
= = 0.977
473, 088
r = 0.977

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The Correlation Coefficient Using Microsoft Excel
Enter the given data into two columns of the data worksheet as shown below:

Click ‘Data’ and ‘Data Analysis’ in the main task pane.

Select Correlation and

Click on OK.

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Select the range of the cells
containing the data in the
worksheet and place into the
‘Input Range:’ column.

Check the box by the ‘Label

in First Row’ box.

Select ‘New Worksheet

Ply:’ if not already selected
by default.

Click on OK.

This is the output of the

Correlation Analysis.

Example 2

1 calculate the coefficient of correlation for the following ages of husbands and wives.
Husband’s age 23 27 28 28 29 30 31 33 35 36
(Y)
Wife’s age 18 20 22 27 21 29 27 29 28 29
(X)

Solution

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 Husband’s age Wife’s age (X) XY X2 Y2
(Y)
23 18 414 324 529
27 20 540 400 729
28 22 616 484 784
28 27 756 729 784
29 21 609 441 841
30 29 870 841 900
31 27 837 729 961
33 29 957 841 1089
35 28 980 784 1225
36 29 1044 841 1296

300 250 7623 6414 9138

n XY − (  X )(  Y )
r=
 n X 2 − ( X )2   n Y 2 − ( Y )2 
       
10  7623 − 300  250 76230 − 75000 1230 1230
r= = = = = 0.81
(10  6414 − 2502 ) (10  9138 − 3002 ) (1640 1380) 2263200 1504.39

Example 3 Find the coefficient of correlation for the following data.

X 10 14 18 22 26 30
Y 18 12 24 16 30 36

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 Solution
X Y Xy X2 Y2

10 18 180 100 324

14 12 168 196 144

18 24 432 324 576

22 16 352 484 256

26 30 780 676 900

30 36 1080 900 1296

120 136 2992 2680 3496

n XY − (  X )(  Y )
r=
 n X 2 − ( X )2   n Y 2 − ( Y )2 
       

6  2992 − 120 136 17952 − 16320 1632

r= = = = 0.80
(6  2680 − 1202 ) ( 6  3496 − 1362 ) 1680  2480 2041.18

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Guide to Interpreting the Correlation Coefficient
Introduction
Generally the following is a rough but useful guide to the degree of relationship
indicated by the size of the correlation coefficient.
Less than 0.20 : Indicates a slight correlation i.e. the relationship is so small
as to be negligible.
0.21 – 0.40 : Indicates a low correlation i.e. a definite relationship exists
between the two variables, but the relationship is a weak one.
0.41 – 0.70 : Indicates moderate correlation i.e. a substantial relationship
exist between the two variables.
0.71 – 0.99 : Indicates a very high correlation i.e. a very strong
relationship exist between the two variables.
1.00 : Perfect correlation i.e. an exact linear
relationship between the two variables.

Interpretation of Correlation coefficient (r)

The answer r =0.977 is indication of a very high linear correlation between X and Y.
Since the answer is positive it means that there is a very strong relationship between
X and Y such that as X (years in school) variable increases it influences an increase
in the Y(income) variable. It means that as the years spent in school increases, the
person acquires higher qualifications and this tends to make the person earn a higher
income.

Coefficient of Determination
The coefficient of determination r 2 measures the extent to which the independent
variable (X) explains the dependent variable (Y) i.e. it measures how much of the
variation in the Y variable is explained by the X variable. This also indicates the
strength of the X variable. It is usually computed as a percentage.
r 2 = (0.977)2 100
= 0.954529 x 100
= 95.45
= 95.5%
What this means is that the X variable (years in school) explains as much as 95.5%
of the variation in Y (income) i.e. the number of years spent in school accounts for
95.5% of the income people earn. This leaves only 4.5% of Y(income) accounted for
by factors other than years in school. Years in school (X) is therefore a very strong
variable in the determination of income (Y).
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Rank Correlation Analysis
Introduction
This is a special type of correlation coefficient whose computation is based on the
rankings of two variables. Rank correlation was developed by Spearman. In this
computation, the ranks of the values of the two variables are substituted for the actual
values and used for the computation. Sometimes in the social sciences it is possible
to rank the characteristics of an item but not to give it a more specific value.

Computation
Rank correlation is computed from the following expression:
6 d 2
r = 1−
n ( n 2 − 1)

Where d = difference between pairs of ranks

Example 1
Compute the coefficient of rank correlation between the marks of students in Core
Mathematics and Numeracy Skills.

Student Core Maths Numeracy d = Rank_X d2

Rank_X Skills – Rank_Y
Rank_Y

A 1 3 -2 4
B 2 2 0 0
C 3 1 2 4
D 4 6 -2 4
E 5 5 0 0
F 6 8 -2 4
G 7 4 3 9
H 8 10 -2 4
I 9 7 2 4
J 10 9 1 1

d 2
= 34

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 Substituting the values, we have:
6 d 2 6  34
r = 1− = 1−
n ( n − 1)
2
10 (102 − 1)
204
= 1− = 1 − 0.206 = 0.794
990

Example 2

1. The following table shows the rankings of 8 students in English and mathematics in
an entrance examination.

English 3 8 9 2 7 10 4 6

mathematics 9 5 10 1 8 7 3 4
Compute the rank correlation coefficient and comment on the results.
Solution
Let RE and RM denote the rankings in English and mathematics respectively.
RE RM d = RE -RM d2

3 9 -5 25

8 5 3 9

9 10 -1 1

2 1 1 1

7 8 -1 1

10 7 3 9

4 3 1 1

6 4 2 4

51

The rank correlation coefficient is given by the formula

210
 6d2
 = 1−
n(n2 − 1)
6  51 306
 = 1− = 1− = 1 − 0.607 = 0.39
8 ( 8 − 1)
2
504

There is a weak positive relation between the English and the mathematics score.

Interpretation
The value of rank correlation (rrank) range between -1 and +1. When the value of rrank
>0 then there is an agreement between the rankings in the same positive (+ve)
direction i.e. high ranks in one variable tend to be associated with high ranks in the
second variable. If the value of rrank < 0 it means the rankings in the two variables are
in disagreement.

In the example above the value of rrank = 0.79 is an indication that there is a marked
positive (+ve) relationship between ranks in Mathematics and Numeracy Skills.

Exercise
6.1. The following table shows the income of 7 people and the corresponding
number of years spent educating themselves.

Income ( ‘000 per month) 10 12 8 11 5 2 3

Years in school ( years) 5 6 4 7 2 2 3

(i) Specify which of the variables; Income and Years in school, will be
Independent (X) and Dependent (Y) with reasons.
(ii) Calculate the linear correlation coefficient between Income and Years spent
in school.
(iii) Interpret the answer you had in (ii) above.
6.2 The Marks obtained by nine students in their Mathematics and English examinations
are shown below:

211
 Mathematics (x) 78 59 65 70 90 42 35 52 58
English (y) 68 37 73 65 59 55 47 42 52

(i) Determine the value of product moment correlation coefficient for these data.
(ii) Comment on your result.

6.3 The following are measured attributes on two variables X and Y:

X 7 4 6 2 1 1 3 3 2
Y 2 4 2 5 7 6 5 4 6

(i) Draw a scatter diagram and suggest the type of relationship existing
between X and Y.
(ii) Determine the value of the product moment correlation coefficient for
these data.
(iii) Comment on your result.

6.4. The following table shows the rankings of 8 students in English and
Mathematics in an entrance examination.

English 8 3 9 2 7 10 4 6

Mathematics 9 5 10 1 8 7 3 4

(i) Calculate the degree of agreement between the rankings in English and
Mathematics.
(ii) Interpret the answer in (i) above.

6.5 Two people ranked 10 brands of beer in terms of their alcohol content.
The data below shows the rankings of the two judges.

Beer A B C D E F G H I J

Judge A 1 2 3 4 5 6 7 8 9 10

Judge B 10 9 8 7 6 5 4 3 2 1

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 (i) Using the appropriate correlation method, find the degree of agreement
between the rankings of the two judges.
(ii) Explain your answer in (i) above.
6.6 The following table shows the marks (out of a total of 10) awarded by two judges to
some contestants of a beauty pageant on the same attribute:

Contestant 1 2 3 4 5 6 7
Judge 1 3 5 6 4 7 9 8
Judge 2 4 6 7 5 8 10 9

(i) Calculate the Spearman rank correlation coefficient between the scores of Judge 1
and Judge 2.
(ii) Comment on the degree of agreement between the two judges.

213