JSS2 MATHEMATICS LESSON NOTES -

FIRST TERM

Scheme of Work

SN WEEK TOPICS

1 WEEK 1 WHOLE NUMBERS AND DECIMAL NUMBERS

2 WEEK 2 WHOLE NUMBERS AND DECIMAL NUMBERS

3 WEEK 3 HIGHEST COMMON FACTOR AND LOWEST

COMMON MULTIPLE

4 WEEK 4 FRACTIONS, RATIOS, AND PERCENTAGES

5 WEEK 5 HOUSEHOLD ARITHMETIC SIMPLE INTEREST,

PROFIT AND LOSS, DISCOUNT AND
COMMISSION

6 WEEK 6 APPROXIMATION AND ESTIMATION

7 WEEK 7 MIDTERM TEST

8 WEEK 8 MULTIPLICATION AND DIVISION OF DIRECTED

NUMBERS

9 WEEK 9 ALGEBRAIC EXPRESSIONS

10 WEEK 10 ALGEBRAIC FRACTIONS

 WEEK 1 & 2 WHOLE NUMBERS AND DECIMAL
NUMBERS
COMMON MULTIPLES AND FACTOR

Factors: The factor of a number is the whole number that divides the number
exactly. For example, the factors of 18 may be found as follows;

18 = 1 x 18

=2x9

=3x6

Multiples: A multiple of a number is obtained by multiplying it by any whole

number. For example, the multiples of 4 are 4, 8, 12, 16, 20, 24 etc.

A prime number is a number that can only divide itself. It has only two factors
which are 1 and itself. Examples of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19,
etc.

Example: (a) find all the factors of 18

(b) State which of these factors is even

(c) state which of these factors are prime numbers

(d) Write the first three multiples of 18

Solution

(a) Factors of 18 are 1, 2, 3, 6, 9, and 18

(b) The even numbers are 2, 6, and 18

(c) The prime numbers are 2 and 3

Example 1 Find the factor of 56

Solution:

1 × 56

2 × 28

4 × 14

2
7×8

Therefore 56 = 1, 2, 4, 7, 8, 14, 28, and 56.

Product of Prime Factors.

Example 2 Express 56 and 108 as the product of prime factors in index

notation.

Solution:

56 = 2 × 2 × 2 × 7 = 23 x 7

108 = 2 × 2 × 3 × 3 × 3 = 22 x 33

Note that the numbers must be a prime numbers

ASSIGNMENT

EXERCISE 1.1; NO 3, 5 & 9. PAGE 2

EXERCISE 1.2; NO 8, 9 & 12. PAGE 3

3
4
 WEEK 3 HIGHEST COMMON FACTOR HCF AND
LOWEST COMMON FACTOR LCM

EXAMPLE 1 Find the L C M of 18 and 24

Solution:

METHOD 1 METHOD 2

2 18 24 18 = 2 3 3

2 9 12 24 = 2 2 2 3

2 9 6 L C M = 2 2 2 3 3

3 9 3 = 72

3 3 1

1 1

L C M = 2 × 2 × 2 × 3 × 3 = 72

Example 2: Find the L C M of 72 and 90

Solution:

METHOD 1 METHOD 2

2 72 90 72 = 2 X 2X 2 X 3 X 3

2 36 45 90 = 2X 3 X3 X 5

2 18 45 LCM=2X2X2 X3X3X5

3 9 45 = 360

3 3 15

5 1 5

1 1

5
2 x 2 x 2 x3 x 3 x 5 = 360

Example 3: Find the H C F of 72 and 96

Solution: find the prime product of the number and pick the common ones

72 = 2 * 2 * 2 * 3 * 3

96 = 2 * 2 * 2 * 2 * 2 * 3

HCF=2*2*2*3

= 24

SQUARE AND SQUARE ROOT

“Squareˮ is the product of two equal terms example N * N = N²

Example 1: Find the square of 14 and 21

Solution:

14 * 14 = 196 (b) 21 * 21 = 441.

Square Root: A number that when multiplied by itself equals a given number.

Example 2: find the square root of 144

Solution:

Using a prime factor method (method 1 Factor pairs method (method 2

2 144 144 = 1 144

2 72 = 2 * 72

2 36 = 3 * 48

2 18 = 4 * 36

3 9 = 6 * 24

3 3 = 8 * 18

6
 1 = 9 * 16

Therefore 2 * 2 * 2 * 2 * 3 * 3 = 2 2 * 3 = 12 = 12 x 12

ASSESSMENT

1. Find the square of the following (a) 25, (b) 40, and (c) 132

2. Find the square root of the following (a) 6400 (b) 16900 (c) 1296

Assignment: Essential mathematics textbook for J SS 2 PAGE 10 EXERCISE 1.6

NO 2 (e, f), NO 6 (b, c)

7
 WEEK 4 FRACTIONS TYPES OF FRACTIONS, RATIO
AND PERCENTAGES

A fraction is a small part or item forming a piece of a whole. It is the quotient of

the rational numbers.

Types of Fractions

EQUIVALENT FRACTIONS Examples are 2/5 = 4/10 = 8/20 = 16/40

IMPROPER FRACTIONS Examples are 13/2, 100/7, 67/5,102/5

MIXED FRACTIONS 72/3, 91/8, 1421/2 etc.

RATIOS, DECIMALS, AND PERCENTAGE

RATIO is the relative magnitudes of two quantities (usually expressed as a

quotient).

Example 1 In a bus station, the ratio of men to women is 25 (a) what fraction of
the people are men (b) what fraction of the people

Solution:

25 = 25 = 7

The fraction of men = 2/7 (i.e. the fraction of men over the total fraction)

The fraction of women = 5/7 (i.e. the fraction of women over the total fraction)

Example 2 The population of a country is estimated to be 12 million people.

The last survey reveals that 40% were boys, 30% were girls, 10% were men
and 20% were women. Express the composition in ratio and evaluate the
estimated number of people in each category.

Solution:

40% (boys) 30% (girls) 20% (women) 10% (men)

Step 1 eliminate the % first = 40 30 20 10

Step 2 divide through by 10 to reduce the value = 4 3 2 1

8
Therefore we have 4321 = 4 + 3 + 2 + 1 = 10

Step 3 find the people for each category.

MEN = 1/10 * 12 000 000 = 1 200 000 ZERO CANCEL ZERO

WOMEN = 2/10 * 12 000 000 = 2 400 000 ZERO CANCEL ZERO

GIRLS = 3/10 12 000 000 = 3 600 000 ZERO CANCEL ZERO

BOYS = 4/10 * 12 000 000 = 4 800 000 ZERO CANCEL ZERO

THEREFORE 4 800 000 + 3 600 000 + 2 400 000 + 1 200 000 = 12 000 000.

FRACTION AND DECIMALS TO PERCENTAGES

Example 1 Express the following fraction to percentage (a) 5/12 (b) 120/500 (c)
0.009

Solution:

5/12 * 100 = 500/ 12 = 412/3

120/500 * 100 = 24%

0.009 = 9/1000 * 100 = 0.9%

ASSESSMENT

ESSENTIAL MATHEMATICS BOOK 2

EXERCISE 4.1 NO 1 I, J, K, L, NO 4 A, B, F, G PAGE 44

EXERCISE 4.3 NO 6 AND 7. PAGE 47

9
 WEEK 5 HOUSEHOLD ARITHMETIC SIMPLE
INTEREST, PROFIT AND LOSS, DISCOUNT AND
COMMISSION
PROFIT AND LOSS

When a trader buys or sells goods, the price at which he /she sells is called the
selling price while the price at which he/she buys is called the cost price.

When the good is sold at a price greater than the cost price, then the trader has
made a gain or profit. On the other hand, when the good is sold at a price less
than the cost price, then the trader has made a loss.

S.P. means selling price, and C.P. means cost price.

Profit = SP – CP %P = P/CP * 100 LOSS = CP SP

%L = L/CP 100

Example 1 A man buys a pair of shoes for Ħ3000 and sells it for Ħ3300. Find
the percentage profit.

Solution:

SP = Ħ3300, CP = Ħ3000,

P = SP – CP = Ħ3300 – Ħ3000 = 300

%P = P/CP * 100 = 300/3000 * 100 = 30000/3000

= 10%

Example 2 a market woman bought 50 oranges at a total cost of Ħ2000. She

sold each one at Ħ45. Find the percentage profit.

Solution:

CP = Ħ2000, SP = Ħ 45 500 = Ħ2250,

P = SP – CP = Ħ2250 – Ħ2000 = Ħ250

SP = P/CP * 100 = 250/2000 * 100 = 25000/2000 = 12.5%

Example 3 A dealer bought an item for Ħ6000 after three months he sold it at a
price of Ħ55000. What is the percentage loss?

10
Solution:

CP = Ħ60000, SP = 55000,

LOSS = CP – SP = Ħ60000 – Ħ55000 = Ħ5000

%LOSS = L/CP * 100 = 5000/60000 * 100 = 500000/60000

= 8.3%

Example 3A dealer bought an article for Ħ65000. Find the price he will sell it to
make a profit of 20%

Solution :

CP = Ħ65000, SP = ?, %P 20%

STEP 1 Find the % of the cost price

P = 20/100 * 65000 = 130000/100 = 13000

P = SP – CP = Ħ 13 000 = SP – CP = SP = Ħ 65000 + Ħ 13000 = Ħ 78000

SIMPLE INTEREST

SI = P * R * T/100 where P principal, R = rate, T = time and SI = simple interest

Example 1 Mr Smith saves Ħ 70000 with a bank for 3 years at the rate of 5%.

(a). calculate the interest he will receive at the end of the years

(b). calculate the simple interest for 7 years

(c). what is the total amount he will save at the end of 5 years?

Solution:

P = Ħ 7000, R = 5%, T = 3

(a). S I = P * R * T/100 = 7000 * 5 * 3/100 – 700 * 15 = Ħ 10000

(b). S I = Ħ70000 * 5 * 7/100 = Ħ 700 * 35 = Ħ 24 500

C. Ħ70000 * 5 * 5/100 = 700 * Ħ 17500

Amount = P + S I = Ħ 70000 + 17500 = Ħ 87500

11
COMMISSION AND DISCOUNT

The commission is simply a payment received for selling a good.

Example 1 An insurance company pays an agent a basic salary of Ħ5000 per

month plus a commission of 15% of all the sales above Ħ100000. Calculate his
gross earnings in a month if he sells goods to the value of Ħ1 200,000.

Solution:

Basis salary = Ħ15000, commission = 15% 0f Ħ100000

But he sold 1200000, therefore Ħ 1200000 - Ħ100000 = Ħ1100000

15% of 1100000 = Ħ165000

Basic salary + commission = Ħ5000 + Ħ165000 = Ħ180000

DISCOUNT is the amount of money taken off the price of a good to promote the
sale.

Example: Mr. Adeoye, a regular customer is given a discount of 12% on an item

that costs Ħ84500. How much does he pay?

Solution:

The item cost Ħ84500,12% of 584500

12% of 84500 – Ħ10140

He pays Ħ84500 – Ħ10140 = Ħ74360.

Example 2 A car company advertises a discount of 12.5% on all their vehicles.

How much would it cost to purchase?

(a). a Toyota car priced at Ħ650000

(b). a Volvo car priced at Ħ450000

(c). a Peugeot car priced at Ħ360000

Solution:

(a). Toyota car = 12.5% of Ħ650000 = Ħ81250

12
Therefore Ħ650000 – Ħ81250 = Ħ568750

(b). Volvo car = 12.5% of Ħ450000 = Ħ56250

Therefore Ħ450000 – Ħ56250 = Ħ393750

(c).Peugeot car = Ħ12.5% of Ħ360000 = Ħ313500

Ħ360000 – Ħ46500 = Ħ313500

ASSESSMENT ESSENTIAL MATHEMATICS BOOK FOR JSS 2

EXERCISE 6.7 NO 2A, B, C, 3, 8, 12 AND 14. PAGE 76

EXERCISE 6.8 NO 6, 14, 16 & 19. PAGE 75

EXERCISE 6.10 NO 3, 4 AND 5. PAGE 77

13
 WEEK 6 APPROXIMATION AND ESTIMATION
SIGNIFICANT AND DECIMAL PLACES
SIGNIFICANT TO ROUND OFF A NUMBER CHANGE 0,1,2,3,4 TO 0 WHILE
5,6,7,8,9 TO 1 AND ADD IT TO THE NEXT NUMBER

Example 1 round off 492.763 to (a) 3 s.f (b) 3 s.f (c) 2 d.p (d) 4 d.p

Solution:

(a). 492.763 = 49 (since the third number is 2 it has changed to zero)

(b). 492.763 = 493 (the 2 has changed to 3 because 7 has changed to 1 and
been added to 2 to become 3

(c).492.763 = 492.76 (in decimal point we count after the point).

(d). 492.763 = 492.7630(since the number is not upto 4d.p we add zero to it.)

Example 2 calculate the following and round your answer to the given degree
of accuracy

(a). 576.175 + 20.82 2 d.p)

(b). 8.52 x 0.0651 3 s.f)

Solution:

(a). 576.173 + 20.82 = 596.99

596.99 = 596.99 2 d.p)

(b). 8.52 * 0.0651 = 0.554652

0.554652 = 0.5553s.f

ASSESSMENT ESSENTIAL MATHEMATICS BOOK FOR JSS 2

EXERCISE 7.1 NO 9 A – K. PAGE 85

EXERCISE 7.3 NO 1 A R. PAGE 86

14
WEEK 8 MULTIPLICATION AND DIVISION OF DIRECTED
NUMBERS
Revision on addition and subtraction of directed numbers

Note that:

(a). + + = + OR - - = + Replacing the same signs that appear together by a

positive sign

(b). + - = - OR - + = - Replacing two different signs that appear together by

a negative sign

Example 1 Find the values of the following:

(a). + 7 + 8 (b) 13 – 6

Solution:

(a). + 7 + 8 = 7 + 8 = 15

(b). + 13 – 6 = 13 – 6 = 7

Example 2 Calculate the following (a) 25 – 3 (b) 12 – 9

Solution

(a). 25 - 3 = 25 – 3 = 22

(b). 12 - 9 = 12 + 9 = 21

MULTIPLICATION OF DIRECTED NUMBERS

RULES

(i) + * - = - OR - * + = - If different signs are multiplied the answer is

NEGATIVE.

(ii) + * + = + OR - * - = + ( If the same signs are multiplied the answer is

positive).

Example1 Simplify the following (a) 12 * 5 (b) 3 * 8

15
Solution:

(a). 12 * 5 = 60

(b). 3 * 8 = + 24

Example 2 Find the values of the following (a) 4 * 2 * 2 * 2 * 2 (b) 7 *

3 * 1 * 1 * 20

Solution:

( a). 4 * 2 * 2 * 2 * 2 = - 64 ( rules, we have equal signs to give positive

while different sign gives negative)

(b). 7 * 3 * 1 * 1 * 20 = 73 = 21 1 * 1 = 2120 = - 420

DIVISION OF DIRECTED NUMBERS

RULES

+ ÷ + = + OR - ÷ - = + ( If the sign are divided the answer is positive)

+ ÷ + = - OR - ÷ + = =- ( the sign is different the answer is negative).

Example 1 work out the following (a) 80 ÷ 10 (b) 25 ÷ 5

Solution:

(a) 80 ÷ 10 = - 8 (because the signs are different)

(b) 25 ÷ 5 = + 5(because the signs are the same)

DO THESE

Simplify the following

(b) 25 ÷ 5

(a). 3 x 5 x 2 x 15 9 (b). 8 x 11 x 9 x 5

5 x 25 x 3 2 x 33 x 3

16
ASSESSMENT ESSENTIAL BOOK FOR JSS 2

EXERCISE 10.3 N0 1, 2, 3 ATO E. PAGE121 TO 123

17
 WEEK 9 ALGEBRAIC EXPRESSIONS
To expand algebraic expressions, those expressions will have to be in brackets.
When the bracket ever moves, then any factor outside the bracket must be
multiplied by each term inside the bracket.

Example 1 Expand d(a + c)

Solution: d * a + d * c = da + dc

Example 2 Expand (y + 3 y + 4

Solution:

=Y*Y+ Y*4 +3*Y +3*4

= Y2 + 4y + 3y + 12

FACTORIZATION OF SIMPLE ALGEBRAIC EXPRESSION

Factorization is the reverse of expanding brackets. The first step in

factorization is to take any common factor which the terms are:

Example 1 Factorise 3X2 + X

Solution:

X is common to the expression

Therefore = X3X + 1

Example 2 Factorize 6y3 – 4y2 – 4y

Solution:

2y is common in the expression

Therefore 2y(3y2 2y 2

ALGEBRAIC EXPRESSION WITH FRACTIONS

Example 1 Solve X/3 + X 2/5 = 6

Solution:

18
Find the L C M = 15

5x + 3X – 6 /15 = 6

Cross multiply

= 5X + 3X – 6 = 615

8X – 6 = 90

Add 6 to both sides = 8X – 6 + 6 = 90 + 6

8X = 96 Divide both sides by 8

X = 12.

FINDING THE LOWEST COMMON FACTOR AND HIGHEST COMMON FACTOR

IN ALGEBRAIC FORM

Example 1 Find the L C M of 4xy,8xv and 10x2y

Solution:

2 4xy 8xy 10x2y

2 2xy 4xy 5x2y

2 xy 2xy 5x2y

5 xy xy x2y

X xy xy x2y

X y y xy

Y y y y

1 1 1

L C M = 2 * 2 * 2 * 2 * 5 * X * X * Y = 40X2Y

Example 2 find the H C F of 4xy, 8xy and 10x2y

Solution;

4xy = 2 * 2 * x * y

19
8xy = 2 * 2 * 2 * x * y

10x2y = 2 * 5 * x * x * y

H C F = 2 * X * Y = 2XY

ASSESSMENT ESSENTIAL BOOK FOR JSS 2

EXERCISE 11.4 NO 25 – 35. PAGE 138.

EXERCISE 11.8 NO 2 F, G & H, NO 4E&F, NO 11A, B, C & D. PAGE 143

20
 WEEK10 ALGERAIC FRACTIONS
To add or subtract fractions with different denominators, first change them to
equivalent fractions. This is done by finding the L C M of the denominators

Example1: Simplify 2a/5 + 4a/3

Solution:

The L C M is 15

6a + 20a/ 15 = 26a/15

Example 2: Simplify 3/5x + 1/2x – 1/4x

Solution:

Find the L C M of the expression = 20X

12 + 10 – 5 = 17/20X

SIMPLIFYING FRACTION

Example 1: Reduce 25X4Y3/35X3y3

solution

Divide through by 5X3Y3(the common factor )

= 5x/7

Example 2 Reduce 8X3Y2/6X3Z

Solution:

Divide through by 2X 3(the common factor)

4Y2/3Z

21
 MULTIPLICATION AND DIVISION OF ALGEBRAIC FRACTION

Example 1 Simplify X2/7 * 4/X2

Solution : X – 2/7 * 4/ X 2 X2 divides themselves

= 4/7

Example 2 Simplify 6X2/11y ÷ 18X/33Y2

Solution

6X2/11y * 33Y2/18X (division sign change to multiplication)

= XY

FRACTIONS WITH BRACKETS

Example1 simplify the following (a) 2x + 5/4 + 2X – 3 /4 (b) 7X – 2 /4 + X – 4 /6

Solution:

(a). The L C M is 4

2X + 5 + ( 2X – 3/4

4x + 2/4 = 22X + 1/4

= 2X + 1/2

(b). 7X – 2 /4 + X – 4 /6

Solution:

The L C M = 12

21x – 6 + 2X – 6 – 8 / 12

= 23X – 14/ 12

22
ASSESSMENT ESSENTIAL BOOK FOR JSS 2

EXERCISE 12.4 NO 7, 13, 19 & 20. PAGE 149

EXERCISE 12.6 NO 1, 2, 3 & 4. PAGE 151

23
 JSS2 MATHEMATICS LESSON NOTES -
SECOND TERM

Scheme of Work

SN WEEK TOPICS

1 WEEK 1 REVIEW OF FIRST TERM WORK, EXPANDING

AND FACTORIZING ALGEBRAIC EXPRESSIONS,
SOLVING QUADRATIC EQUATIONS.

2 WEEK 2 WORD PROBLEMS ON ALGEBRAIC FRACTION

3 WEEK 3 LINEAR INEQUALITIES

4 WEEK 4 LINEAR INEQUALITIES IN ONE VARIABLE,

GRAPHICAL PRESENTATIONS OF SOLUTION
OF LINEAR INEQUALITIES

5 WEEK 5 GRAPHS OF LINEAR EQUATIONS IN TWO

VARIABLES

6 WEEK 6 PLANE FIGURES OR SHAPES, IDENTIFICATION

OF PLANE SHAPES WITH THEIR PROPERTIES

7 WEEK 7 REVIEW OF THE FIRST HALF TERMSʼS AND

PERIODIC TEST

8 WEEK 8 SCALE DRAWING OF LENGTH AND DISTANCES

9 WEEK 9 QUANTITATIVE APTITUDE PROBLEMS

10 WEEK 10 REVISION OF THE SECOND HALF TERMʼS

WORK
 WEEK 2 WORD PROBLEM LEADING TO ALGEBRAIC
EXPRESSION
Sometimes you may be asked to solve a problem given in words. To do this you

need to convert the words into an algebraic equation and then solve it. The

following points will show you what to do.

a. Read the equation carefully and then decide what the unknown number is

b. Where necessary, change all the unit of measurement to the same unit

c. Use a letter to represent the unknown

d. Use the information provided to write the required equation

e. Solve the equation as usual

f. Use the solution obtained to answer the questions in words

g. You can check your answer as usual.

Example 1 Think of a number, add 5 to it, and multiply the result by 3, the

answer is 36. What is the number?

Example 2The sum of a number and 9 is multiplied by 2 and the answer is 8.

Find the number

Example 3 The smallest of three consecutive odd numbers is n if their sum is

27. Find the three numbers.

2
Solution:

Note the rules above. Follow the steps.

1. Let the number be X, add 5 to it, = X + 5, and multiply by 3

3X5, the answer is 36

Open the bracket 3X5 = 36 [use the 3 to multiply the equation]

3x + 15 = 36 [subtract 15 from both sides]

3X + 15 – 15 = 36 – 15

3X = 21[divide both side by 3

3X/3 = 21/3

X=7

Check

375

3X1236

2. Let X be the number, X + 9, multiply by 2,

2X + 9, the answer is 8

2X + 9 = 8 [open the bracket]

2X 18 = 8

Collect the like term

2X = 8 + 18 [the sign we change to positive]

2x = 10

Divide both sides by 2

2X/2 = 10/2

3
X = 5

Check

259

24 = 8

3. Let the first number be n, the second number be n+2, and the third number
be n+4.

Note that odd numbers are 1, 3, 5, and 7, the difference between each number
is 2.

NN2N4 = 27

Collect the like term

3N + 6 = 27 [subtract 6 from both sides]

3N = 27 – 6

3N = 21

Divide both sides by 3

3N/3 = 21/3

N=7

N 7, N2 72 = 9, N + 4 = 7 + 4 = 11

Therefore the consecutive numbers are 7, 9, and 11

EVALUATION Translate the following statement into algebraic equations and

then solve them

a. Think of a number, and 5, the result is 10. What is the number?

b. Two market women shared a basket full of oranges to sell them. The share
of the first woman is twice the other. Suppose there are 300 oranges in the
basket. What is the share of each woman?

c. The sum of three consecutive even numbers is 60. Find the numbers.

4
ASSIGNMENT PAGE 159 EXERCISE 13.5 NO 4, 6,8,15, AND 18

5
 WEEK 3 LINEAR INEQUALITIES
Inequality is an algebraic expression formed by replacing the equal sign of an
equation with an inequality symbol. e.g. 7X + 5 = 16 (equation while 5X + 7 > 9
(inequality).

The following are commonly used in an inequality symbol.

SYMBOL MEANING REPRESENTATION

< LESS THAN

> GREATER THAN

≤ LESS THAN OR EQUAL TO

≥ GREATER THAN OR EQUAL TO

We often use inequality in our everyday life. We can write them as algebraic
statements. For example, if the speed of a car is 250km/h or less, we can write
this as S ≤ 250, where s represents speed.

GRAPH OF INEQUALITIES

A linear inequality has no square or higher power of the unknown. In other

words, the power of the unknown is 1.

Example 1 2x > 15 is a linear inequality in one variable. X.

Solution:

2X > 10 [divide both sides by 2

2X/2 > 10/2

X = 5.

1 2 3 4 5 6 7 8

The empty circle at the end of the arrow shows that 5 is not included in the
range

COMBINING INEQUALITIES

6
When combining inequalities (sometimes an unknown quantity obeys more
than one inequality), these inequalities may be combined as one statement, the
smallest number must be written first followed by the unknown, and finally the
largest number, and vice-versa. For example, the diagram below shows that X
can take any value from 2 to 3.

6 5 4 3 2 1 0 1 2 3 4

Hence the inequalities are X ≥ 2 and X < 3

But X ≥ 2 in reverse is written as a ≤ X and X < 3 can be combined as a single

inequality as follows

2 ≤ X ≤ 3.

Example1

6 5 4 3 2 1 0 1 2 3 4 5 6

X ≥ 1, X≤5

1 ≤ x ≤ 5

Example 2

5 4 3 2 1 0 1 2

X > 5, X ≤ 2

5 < X ≤ 2

EVALUATION PAGE 167 EXERCISE 14.1 NO 1 I, J, K, L, NO 3 J, K, L, NO 5 E,

F, G, H.

PAGE 169 EXERCISE 14.2 NO 1A,B,C,D

ASSIGNMENT PAGE 169 EXERCISE 14.2 NO 2 ( A TO H

7
 WEEK 4 SOLVING INEQUALITIES
RULES IN SOLVING INEQUALITIES.

a. An inequality remains true when the same quantity is added to, or

subtracted from both sides.

b. An inequality remains true when both sides are multiplied or divided by

the same positive quantity.

c. An inequality remains true when both sides are multiplied or divided by a

negative quantity provided the inequality sign is reversed.

Example 1 Find the greatest possible value of an X that satisfies the inequality
8 + 2X > 3 + 5X. if X is an integer.

Solution:

8 + 2X > 3 + 5X

Subtract 8 from both sides

2X > 3 – 8 + 5X

2X > 5 + 5X

Subtract 5X from both sides

2X – 5X > 5 = 3x > 5

Divide both sides by 3 and also reverse the inequality

X < 12/3 [rule three]

The greatest integer value of X is 1.

Example2 Find the smallest integer value of X that satisfies the inequality 7X –
2 ≥ 5X – 6

Solution:

7X – 2 ≥ 5X – 6

Add 2 to both sides

7X ≥ 5X – 6 + 2

8
7X ≥ 5X – 4

Subtract 5X from both sides

7X – 5X ≥ 4 = 2x ≥ 4 , Divide both side by 2

X ≥ 2

WORD PROBLEMS LEADING TO INEQUALITIES

Example 1: David is X years old. In 4 years, his age will still be less than 12
years. (a) Write this information in an inequality in X. (b) Find the maximum age
of David to the nearest whole numbers.

Solution:

In 4 years, David ageʼs will be X + 4 years.

If at that time his age is less than 12, then

X + 4 < 12

Subtract 4 from both sides

X < 12 – 4

X < 8.

The maximum age of David is 7 years.

Example 2 A man had #X, out of this, he used #1000 to pay his house rent.
The amount he had left was not more than 3500. (a) Write this information in an
inequality in X (b) Solve for X.

Solution:

The man used #1000 to pay his house rent out of #X, so the amount left is #(x –
1000.

This amount is not more than 3500

X – 1000 ≤ 500

Add 1000 to both sides

X ≤ 500 + 1000 = X ≤ 1500

9
Hence the man had less than or equal to #1500

EVALUATION PAGE 172 EXERCISE 14.3 N0 3K,L,M,N AND O, PAGE 173 EX
14.4 N0 1, 2 AND 3

ASSIGNMENT PAGE 172 AND 173 EX 14.3 AND 14.4 N0 5, AND 6, 7, 8 AND 9 IN
EX 14.4

10
 WEEK 5 GRAPH OF LINEAR EQUATION

There are several ways to graph a straight line given its equation.

Let's quickly refresh our memories on equations of straight lines:

Slope Intercept Form Point Slope Form Horizontal Lines Vertical Lines

y = 3 (or any number) x = 2 (or any

when stated in "y=" when graphing, put horizontal lines have number)
form, it quickly gives this equation into "y = a slope of zero - they vertical lines have no
the slope, m, and " form to easily read have "run", but no slope (it does not
where the graphing information. "rise" -- all of the y exist) - they have
the line crosses the values are 3. "rise", but no "run"
y-axis, b, called the --all of the x values
y-intercept. are 2.

Graphing Tidbits:

Before graphing a The x-coordinate may

If a point lies on a line, be sure that your be called the
line, its coordinates equation starts with abscissa.
make the equation "y=".
true. The y-coordinate may
To graph 6x + 2y = 8 be called the
2,1 in on the rewrite the equation: ordinate.
line y = 2x - 3 2y = 6x 8
because 1 = 22 - 3 y = 3x + 4
Now graph the line
using either the
slope-intercept
method or the chart
method.

11
How Do You Graph a Linear Equation by
Making a Table?
Note:

Graphing a function? It would be really helpful if you had a table of values that
fit your equation. You could plot those values on a coordinate plane and
connect the points to make your graph. See it all in this tutorial!

Graphing in the Coordinate Plane

o o

How Do You Plot Points in the Coordinate Plane?

Knowing how to plot ordered pairs is an essential part of graphing functions. In

this tutorial, you'll see how to take an ordered pair and plot it on the coordinate
plane. Take a look!

What is an Ordered Pair?

Ordered pairs are a fundamental part of graphing. Ordered pairs make up

functions on a graph, and very often, you need to plot ordered pairs to see
what the graph of a function looks like. This tutorial will introduce you to
ordered pairs!

What is the Origin?

The coordinate plane has two axes: the horizontal and vertical axes. These two
axes intersect one another at a point called the origin.

12
 ■

What is the X-coordinate?

Ordered pairs are a crucial part of graphing, but you need to know how

Identifying Linear Equations

What's the Standard Form of a Linear Equation?

A linear equation can be written in many different forms, and each of them is
quite useful! One of these is a standard form. Watch this tutorial and learn the
standard form for a linear equation!

Further Exploration

· Working With Graphs

How Do You Check if a Point is on a Line If You Have a Graph?

Wonder if a point is part of a line? You could take that equation and graph it.
Then use the graph to get your answer! Watch how in this tutorial.

· Finding Slopes

13
 WEEK 6 PLANE SHAPES AND THEIR PROPERTIES

Identify Plane Shapes and Solid Shapes: Overview

Geometry and spatial relationships are a part of children's daily lives.
Understanding an object's position in space and learning the vocabulary to
describe the position and give directions are important. Simple terms like
above, below, left, right, or between, enable children to order and describe
the world around them. They can apply these terms as they describe planes
and solid shapes in the classroom.

Most of the objects that we encounter can be associated with basic shapes. A
closed, two-dimensional, or flat figure is called a plane shape. Different plane
shapes have different attributes, such as the number of sides or corners. A
side is a straight line that makes part of the shape, and a corner is where two
sides meet. In this chapter, children will learn to identify, describe, sort, and
classify plane shapes by these attributes.

Although children are familiar with the most common shapes, up until now they
may not have been able to verbalize what distinguishes a square from a
rectangle or a circle from a triangle. They will learn to describe shapes in terms
of their sides and corners. A triangle is a shape with three sides and three
corners. A rectangle is a shape with four sides and four corners. They may
notice that opposite sides are the same length. A square is a rectangle in which
all four sides are of equal length. A circle is a round shape that has no sides or
corners. These attributes, as well as size, can be used to sort and classify
shapes.

PROPERTIES OF PLANE SHAPES

Square

14
A square has four sides, but not just any four sides. A square's four sides are
all the same length. A square with one-inch sides is smaller than a square with
three-inch sides because one is less than three. A square also has four
corners. Opposite sides are equal, and the diagonal bisects at right angles. It
has four lines of symmetry.

Rectangle
A rectangle has two equal sides of one length and two equal sides of a
different length. A rectangle is like a stretched square. Both figures have four
corners, but no longer four equal sides for the rectangle. Write their findings on
the board under headings "square," "rectangle" and "both." The diagonals are
equal, and they bisect each other. it has two lines of symmetry.

Triangle
Two sides are equal, the base angles are equal (isosceles triangle). All the three
sides are equal, all the angles are equal which is 60, and it has three lines of
symmetry. (equilateral triangles)

Circle
Give each child a piece of string. Ask them to make circles with the string on
their desks. Discuss how many sides and corners a circle has: none. Let each
child pick a piece of construction paper. Fold it in half and show them how to
trim the edges; open it up and it's a circle. For homework, tell the class to take
home their circle, find unneeded items that are circles, and glue them on the
construction paper. The next day post the artistic circles on the bulletin board.

15
 WEEK 8 SCALE DRAWING
Since it is not always possible to draw on paper the actual size of real-life
objects such as the real size of a car, or an airplane, we need scale drawings to
represent the size like the one you see below of a van.

In real life, the length of this van may measure 240 inches. However, the length
of a copy or print paper that you could use to draw this van is a little bit less
than 12 inches

Since 240/12 = 20, you will need about 20 sheets of copy paper to draw the
length of the actual size of the van

To use just one sheet, you could then use 1 inch on your drawing to represent
20 inches on the real-life object

You can write this situation as 120 or 1/20 or 1 to 20

Notice that the first number always refers to the length of the drawing on
paper and the second number refers to the length of a real-life object

Example #1

16
 Suppose a problem tells you that the length of a vehicle is drawn to scale. The
scale of the drawing is 120

If the length of the drawing of the vehicle on paper is 12 inches, how long is
the vehicle in real life?

Set up a proportion that will look like this:

Do a cross-product by multiplying the numerator of one fraction by the

denominator of the other fraction

We get :

Length of drawing × 20 = Real length × 1

Since the length of the drawing = 12, we get:

12 × 20 = Real length × 1

240 inches = Real length

Example #2

The scale drawing of this tree is 1500

If the height of the tree on paper is 20 inches, what is the height of the tree in
real life?

17
Set up a proportion like this:

Do a cross-product by multiplying the numerator of one fraction by the

denominator of the other fraction

We get :

Height of drawing × 500 = Real height × 1

Since the height of the drawing = 20, we get:

20 × 500 = Real length × 1

10000 inches = Real height

18
 JSS2 MATHEMATICS LESSON NOTES -
THIRD TERM

Scheme of Work

SN WEEK TOPICS

1 WEEK 1 REVISION OF THE SECOND TERMʼS

EXAMINATION

2 WEEK 2 ANGLES AND POLYGON

3 WEEK 3 ANGLES OF ELEVATION AND DEPRESSION

4 WEEK 4 BEARING AND DISTANCES

5 WEEK 5 STATISTICS - DATA PRESENTATION

6 WEEK 6 STATISTICS CONTINUED

7 WEEK 7 REVIEW OF FIRST HALF TERMʼS WORK AND

PERIODIC TEST

8 WEEK 8 PROBABILITY

9 WEEK 9 PROBABILITY CONTINUED

10 WEEK 10 REVIEW OF THIRD TERMʼS WORK AND

PERIODIC TEST
 WEEK 2 ANGLES AND POLYGON
TYPES AND PROPERTIES OF ANGLES

Straight Angles

Angles which measure exactly 180° (degrees) are straight. Therefore, straight
angles are straight lines. Angles are represented by the sign ϴ, called theta.
That is, for straight angles, ϴ 180°.

Right Angles

Angles which measure exactly 90° are right angles, that is, ϴ = 90°.

Obtuse Angles

Obtuse angles are those which are greater than 90° but less than 180°, that is,
90° < ϴ < 180°.

Acute Angles

Acute angles are angles that are greater than 0° but less than 90°, that is, 0° <
ϴ < 90°.

2
Reflex Angles

Reflex angles are angles that are greater than 180° but less than 360°, that is,
180° < ϴ < 360°.

Adjacent Angles

Two angles that share the same vertex (center, usually represented by 0 and
have a common side (line) are called adjacent angles.

Complementary Angles

Complementary angles are two angles which when summed equals 90°.

Note: A and B, are ‘angle Aʼ and ‘angle Bʼ respectively.

Supplementary Angles

Supplementary angles are two angles which when summed equals 180°.

3
Vertically Opposite Angles

Vertically opposite angles are the angles opposite to each other when two
straight lines intersect. Their defining property is that vertically opposite angles
are equal in magnitude.

Corresponding Angles

When two parallel lines are crossed by a line called the transversal, the angles
formed which are in corresponding positions, are called corresponding angles.
Corresponding angles are equal in magnitude.

POLYGON A part of a triangle and quadrilateral, we have other polygons which

are also named according to the number of sides they have.

Examples are pentagon = 5, Hexagon = 6. It can be regular or irregular. A

polygon is said to be regular when all the sides and angles are equal. An
irregular polygon has neither of the sides or angles equal.

4
TYPES OF POLYGON Regular Polygons)

Regular polygons have all sides, and all angles equal.

5
Size of Internal Angles

To find the size of the internal angles of a regular polygon with ‘nʼ sides, use
the formula:

For example, the size of the interior angles of the pentagon (five sides) above
is:

The sum of all the interior angles of a polygon with ‘nʼ sides is found using the
formula:

(n – 2 x 180°

Therefore, the sum of all the interior angles of the pentagon above is:

5 – 2 x 180° = 3 x 180° = 540°

Size of Exterior Angles

Interior and Exterior angles are measured on the same line, that is, they add up
to 180°.

Therefore, the size of an exterior angle = 180° – Interior angle.

For example, the size of the external angle of the pentagon above is:

Since, interior angle = 108°

Then, exterior angle = 180° – Interior angle

180° – 108° = 72°

Below is a list of the names and the number of sides, of some of the most
popular polygons.

Name of Polygon Number of Sides

Equilateral Triangle 3

Quadrilateral 4

Pentagon 5

Hexagon 6

Heptagon 7

6
Octagon 8

Nonagon 9

Decagon 10

DO THESE

A. Calculate the total internal angle of an octagon

B. The size of each angle of a regular octagon

C. The total internal of a decagon

D. The size of each angle of a regular decagon.

ASSIGNMENT

PAGE 233 EXERCISE 18.3 QUESTION 1 TO 10 OF ESSENTIAL MATHEMATICS

FOR JUNIOR SECONDARY SCHOOLS BOOK 2

7
 WEEK 3 ANGLES OF ELEVATION AND DEPRESSION
What is the angle of elevation?
The angle of elevation is the angle between a horizontal line from the observer
and the line of sight to an object that is above the horizontal line.

In the diagram below, AB is the horizontal line. q is the angle of elevation from
the observer at A to the object at C.

B A

What is the angle of depression?

The angle of depression is the angle between a horizontal line from the
observer and the line of sight to an object that is below the horizontal line.

In the diagram below, PQ is the horizontal line. q is the angle of depression

from the observer at P to the object at R.

P Q

8
How to solve word problems that involve an angle of elevation or
depression?
Step 1 Sketch the situation.
Step 2 Mark the given angle of elevation or depression.
Step 3 Use trigonometry to find the required missing length

Example:

Two poles on the horizontal ground are 60 m apart. The shorter pole is 3 m
high. The angle of depression of the top of the shorter pole from the top of the
longer pole is 20˚. Sketch a diagram to represent the situation.

Solution:

Step 1: Draw two vertical lines to represent the shorter pole and the longer
pole.

Step 2: Draw a line from the top of the longer pole to the top of the shorter
pole. This is the line of sight).

Step 3: Draw a horizontal line to the top of the pole and mark the angle of
depression.

Example:

9
A man who is 2 m tall stands on horizontal ground 30 m from a tree. The angle
of elevation of the top of the tree from his eyes is 28˚. Estimate the height of
the tree.

Solution:

Let the height of the tree be h. Sketch a diagram to represent the situation.

tan 28˚ =

h – 2 = 30 tan 28˚

h = 30 ´ 0.5317 + 2 ← tan 28˚ = 0.5317

= 17.951

The height of the tree is approximately 17.95 m.

10
 WEEK 4 BEARING AND DISTANCES
Bearings are measured from the north in the clockwise direction

11
 WEEK 5 & 6 STATISTICS - DATA PRESENTATION

FREQUENCY TABLE

EXAMPLE 1 The following figures show the number of children per family in a
sample of 40 households. 1, 2, 4, 3, 5, 3, 8, 3, 2, 3, 4, 5, 6, 5, 4, 2, 1, 3, 2, 4, 5, 3,
8, 7, 6, 5, 4, 5, 7, 6, 3, 8, 6, 3, 5, 7, 5, 4, 3.

a. Use a tally mark to prepare a frequency table for the data

b. What is the highest frequency of the number of children per family?

Solution

NUMBER TALLY FREQUENCY

1 // 2

2 ///// 5

3 //// //// 9

4 //// / 6

5 //// /// 8

6 //// 4

7 /// 3

8 /// 3

40

12
Example 2 In a further mathematics test the following marks were obtained by
a group of students 85, 75, 95, 80, 75, 80, 90, 84, 95, 84, 85, 80, 80, 75, 80, 75,
80, 84, 81, 80, 75, 90, 80.

Use a tally mark to prepare a frequency table for this data.

Solution:

NUMBER TALLY FREQUENCY

75 //// 5

80 //// 9

81 / 1

84 /// 3

85 // 2

90 /// 3

95 // 2

a. How many students took part in the test? 25 students

b. Which mark had the highest frequency? 80 marks

AVERAGE, MEAN MEDIAN AND MODE

Average is a single value used to represent a set of numbers (i.e all values in as
et data)

The most commonly used statistic is average.

MEDIAN = THE NUMBER IN THE MIDDLE AFTER THE ARRANGEMENT OF THE

DATA.

13
MODE IS THE VALUE THAT OCCURS MOST FREQUENTLY.

EXAMPLES Calculate the mean, median, and mode of the following data

a. 45, 50, 55, 54, 48, 53, 50, 55

b. 38, 35, 36, 30.8, 34.7, 37.9, 33.1

c. 3, 0,4,7, 0, 5, 3, 4, 0, 3, 6, 5, 5,4, 6, 5

Solution:
Mean = 4550 55 54 48 53 50 55
8
= 410/8

= 51.25

Median = 45, 48, 50, 50, 53, 54, 55,55

50 + 53
2
= 52

Mode = 50 and 55

Average

The average or mean of a set of numbers is defined by the formula:

Example

Bar Charts

14
A Bar chart is a series of rectangular bars of the same width, drawn vertically
or horizontally, with an equal space between them, with the height of each bar
being a depiction of the data it is representing.

Example

The table below lists several models of Blackberry cellular phones and the
amount of each that an electronic store has in stock. Draw a vertical and
horizontal bar chart to represent the data.

Blackberry Phones Stock Amount

Blackberry Curve 8310 75

Blackberry Curve 8320 100

Blackberry Pearl 8100 50

Blackberry Bold 9650 200

Blackberry Bold 9000 150

Blackberry Curve 8520 125

15
Pie Charts

A Pie chart is a circular diagram divided into sectors, with the size of each
sector representing the magnitude of data it is depicting. Each sector of a pie
chart can either be displayed in percentages (note all sectors must add up to
100% or as an angle (note all sectors must add up to 360o).

16
Example

The table below lists some of the most popular football clubs and the number
of students at a given institution that supports each. Use a Pie chart to
represent the information given in the table.

Football Clubs Number of Students

Chelsea 50

Manchester United 200

Barcelona 350

Real Madrid 150

Inter Milan 25

Arsenal 100

Liverpool 40

AC Milan 75

17
The Pie Chart above depicts each sector as percentages. To calculate the
percentages for each sector use the formula below:
% of a sector = Number of students x 100
Total number of students

So, to calculate the percentage of Chelsea fans:

% of Chelsea fans = 50 x 100
990

% of Chelsea fans = 5%

For Pie charts that depict each sector as angles, the angles for each sector are
found using the formula below:
The angle of a sector = Number of students x
360
Total number of students

So, to calculate the angle of the Chelsea sector:

Angle of Chelsea sector = 50 x 360
990

Angle of Chelsea sector = 18 de

18
Note: In most cases, the questions set on Pie charts require those drawn
depicting sectors in percentages.

Line Graphs

Line graphs are mostly used in depicting trends, and as such, values are in
most cases plotted against time. A line graph is drawn by connecting a line to
consecutive values, with a circle/point made at each value being depicted.

Example

The table below lists the amount of Toyota motor vehicles produced in April
over the period 2000 2010.

Year Number of Toyota Motor Vehicles Produced

2000 220,382

2001 260,879

2002 213,546

2003 238,890

2004 227,678

2005 245,376

2006 240,224

2007 224,100

2008 258,100

2009 248,024

2010 249,123

19
20
 WEEK 8 & 9 PROBABILITY

Probability is the measure of the likelihood that an event will occur. Probability
is quantified as a number between 0 and 1 (where 0 indicates impossibility and
1 indicates certainty). The higher the probability of an event, the more certain
we are that the event will occur. A simple example is the tossing of a fair
(unbiased) coin. Since the coin is unbiased, the two outcomes ("head" and
"tail") are equally probable; the probability of "head" equals the probability of
"tail." Since no other outcome is possible, the probability is 1/2 (or 50% of
either "head" or "tail". In other words, the probability of "head" is 1 out of 2
outcomes, and the probability of "tail" is also 1 out of 2 outcomes.

The probability of an event A is written as , , or .24 This

mathematical definition of probability can extend to infinite sample spaces, and
even uncountable sample spaces, using the concept of a measure.

The opposite or complement of an event A is the event [not A (that is, the
event of A not occurring), often denoted as , ,
or ; its probability is given by P(not A = 1 − PA. As an
example, the chance of not rolling a six on a six-sided die is 1 – (chance of
rolling a six) . If two events A and B occur on a single performance
of an experiment, this is called the intersection of A and B, denoted as

Independent events

If two events, A and B are independent then the joint probability is

For example, if two coins are flipped the chance of both being heads is
26

21
Mutually exclusive events

If either event A or event B occurs on a single performance of an experiment

this is called the union of the events A and B denoted as . If two
events are mutually exclusive then the probability of either occurring is

For example, the chance of rolling a 1 or 2 on a six-sided die is

Not mutually exclusive events

If the events are not mutually exclusive then

For example, when drawing a single card at random from a regular deck of
cards, the chance of getting a heart or a face card J, Q, K (or one that is both)
is , because of the 52 cards of deck 13 are hearts, 12 are face
cards, and 3 are both: here the possibilities included in the "3 that are both" are
included in each of the "13 hearts" and the "12 face cards" but should only be
counted once.

Event Probability

not A

A or B

22
A and B

A given B

PROBABILITY SCALE AND TERMS

EVENT An event is something that happens. For example, tossing a coin or

throwing a die is an event.

OUTCOME An outcome is the result of an event. For example, if you toss a

coin, you will either get a Head or a Tail. This means there are 2 possible
outcomes.

IMPOSSIBLE An impossible event will happen if given a probability of 0.

UNLIKELY When the probability tends towards 0, then there is less chance that
an event will happen.

LIKELY When the probability tends towards 1, then there is a likely chance that
is 5050

APPLICATION

EXAMPLES

Each of the following numbers is written on a piece of paper and then put in a
bag. 3, 4, 6, 3, 5, 7, 5, 10, 5, 12, 7, 8, 9,7, 5, 3, 9, 6, 6, 11, 12, 11, 5

What is the probability of a picking random

23
I. An odd number

II. An even number

SOLUTION

Picking an odd number is 3, 5, 7, 11, successful outcome is 3, 3, 3, 5, 5, 5, 5, 5,

7, 7,7, 9, 9, 11, 11 = 15 outcome.

Pro. Of odd no = 15/24 = 5/8

Picking an even number is 4, 4, 6, 6, 6, 8, 10, 12, 12 = 9 outcome

Pro. Of even number is 9/24 = 3/8

Example 2 There are 7 red balls, 8 white balls, and 5 blue balls in a box. Find
the probability that the ball is

a. White

b. Red

c. Blue or red

d. Neither red nor white

e. green

Solution:

a. Total number of balls = 7 8 5

= 20

White ball = 8, pro. Selecting a white ball is = 8/20 = 2/5

b. Number of red balls = 7

24
Pro. Selecting a red ball = 7

7/20

c. Number of blue and red balls = 5 7 = 12

Pro. Selecting a blue or red ball = 12/ 20

3/5

d. If the ball is neither red nor white, then it must be blue. Pro. Of selecting
a blue ball = 5/20 = ¼

e. There are no green balls therefore the pro. Of green is 0

Example 3 A card is selected from a well-shuffled standard pack of 52 cards.

What is the probability of getting,

a. A diamond

b. A queen

c. An ace

d. A red card

e. The ace of spades

f. Any card other than an ace

NOTE A PACK OF CARDS ARE IN 4 SUITS. EACH SUIT HAS 13 DIAMONDS, 13

HEARTS, 13 SPADES, 13 CLUBS. THE DIAMOND AND THE HEART ARE BOTH
RED WHILE THE CLUB AND THE SPADE ARE BLUE. THE SIZE OF NUMBERS ON
THE CARD ARE A 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.

25
WHERE A = ACE, Q = QUEEN, K = KING, J = JACK. THERE ARE 12 PICTURE
CARDS, NAMELY; 4 KINGS, 4 QUEENS, 4 JACKS.

Solution

Total number of possible outcomes = 52

(a). pro. Diamond) = 13/52

=¼

(b). pro Queen = 4/52

= 1/13

(c). pro. Ace 4/52

= 1/13

(e). pro. (red card) = 26/52

=½

(f). pro. (any card other than an ace) = 1 1/13

= 12/ 13

DO THESE

1. A bag contains the following: 90 blue balls, 3 red balls, 50 yellow balls,
57 brown balls, and 100 green balls. What is the probability of picking at
random:

I. A blue ball

II. A yellow ball

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III. A brown ball

IV. A red balls

V. A white ball

VI. A green ball

1. A card is selected from a well-shuffled standard pack of 52 cards. What

is the probability of getting:

I. A club

II. The ace of diamond

III. A jack of hearts

IV. A diamond or a spade

2. A die has six faces numbered 1 to 6. If the die is rolled once, find the
probability of:

a. Obtaining the number 6

b. Obtaining the number 10

c. Not obtaining the number 6

d. Obtaining the numbers 1, 2, 3, 4, 5, or 6

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ASSIGNMENT

PAGE 308 EXERCISE 24.3 NO 1, 3, 4, 5, 6, 7, AND 8

WEEK 10 REVIEW OF THIRD TERMʼS WORK AND

PERIODIC TEST

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