# Fractions Continued: (a) Multiplication and Division of fractions (b) Prime numbers and factors

Subject :

Mathematics

Term :

First Term

Week:

Week  Nine

Class :

Jss 1

Previous lesson :

The pupils have previous knowledge of      Fractions continued: Addition and subtraction of fractions

Topic :

### Fractions Continued: (a) Multiplication and Division of fractions (b) Prime numbers and factors

Behavioural objectives :

At the end of the lesson, the pupils should be able to

• Mention or explain how to solve questions on multiplication and division of fractions
• say the process of solving word problems on multiplication and division of fractions
• solve simple sums on multiplication and division of fractions
• explain the process of calculating the prime numbers of figures
• express prime factors of numbers in index form

Instructional Materials :

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching :

• Class Discussion
• Group Discussion
• Explanation
• Role Modelling
• Role Delegation

Reference Materials :

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum
• Workbooks

Content :

WEEK NINE

TOPIC : MULTIPLICATION AND DIVISION

1. Simple Examples on Multiplication and Division of Fractions
2. Harder Examples
3. Word Problems.
4. Prime Numbers and Fractions

Introduction

Multiplication of fractions is simply a direct method compared to division of fractions. In multiplication, there is direct multiplication of the numerator of one fraction with the other and the denominator with the other.

Division is usually technical as there is reversal of the sign ( ÷) to multiplication sign 9x ) thereby leading to the reciprocal of the right-hand value.

A x x = A x α

B y B x y

But A ÷ x = A x y = Ay

B y B x Bx.

BODMAS

When signs are combined as a result of combination of fracitons, it is therefore important to apply some rules that will enable us know where to start from. Such guide is BODMAS. It states that when there is combination of signs, they should be taken in order of their arrangement

B= bracket

O= of

D= division

M= multiplication

S= subtraction.

Simple example

Example 1

Simplify the following;

(a) 1 3/5 x 6 (b) 4/11 of 3 2/3

(c) 3 ¾ x4/9 x 1 1/5 (d) 12/25 of (1 ¼ )2

Solution

(a) 1 3/5 x 6 (b) 4/11 of 3 2/3

= 8/5 x 6 = 4/11 x 11/3

= 8 x 6 = 4/3

5 x 1 = 1 1/3

= 48

5

= 9 3/5

c. 3 ¾ x 4/9 x 1 1/5 (d) 12/25 of ( 1 ¼ ) 2

15/4 x 4/9 x 6/5 12/25 x 5/4

15 x 6 = 12/25 x 5/4 x 5/4 =12/16 = 2

9 x 5

Example 2

Simplify (a) 7 1/5 ÷ 25 (b) 12/25 ÷9/ 10

(c) 7 7/8 ÷ 6 5/12

Solution.

(a) 7 1/5 ÷ 25 (c ) 7 7/8 ÷ 6 5/12

= 36/5 ÷ 25/1 = 63/8 ÷ 77/12

= 36/5 x 1/25 = 63/8 x12/77

= 36/125

b. 12/2 ÷ 9/10 = 9 x 3

12/25 x 10/9 2 x 11

4 x 2 = 27/22

5 x 3 = 1 5/22

= 8/15.

Example 3

Simplify

(a) 3/10 x 35/36 (b) 5 ¼ ÷ 2 /5

14/15 3 ¾

= 7 x 1 21/4 ÷ /5

2 x 12 15/4

7/24 ÷14/15 =21/4 x 5/14

15/14

7/24 x 15/14 = 15/8

15/4

= 15 = 15/8 ÷ 15/4

24 x 2

= 5 = 15 x 4

8 x 2 8 15

= 5 = 4

16 8 = ½

5/16.

EVALUATION

Simplify the following ;

1. 8 1/6 X 3 3/7

11 2/3

2. 7 3/7÷ 5/21

9 ¾ x 2/3

1.Essential mathematics for jSSI by AJS Oluwasanmipg 52

2. New General Mathematics for JSSI by MG macrae et al pg 36.

Harder examples

Example I

Simplify the following fractions

a. 5/8 x 1 3/5 b. ¾ of 3 3/7 c. 9/16 ÷3 3/8

Solution

5/8 x 1 3/5 b ¾ of 3 3/7

5/8 x 8/5 =3/4 x 24/7

5 x 8 = 3 x 24

8 x 5 4 x 7

= 1 3 x 6

7 = 18/7 = 2 4/7

c.9/16 ÷ 3 3/8

9/16 ÷ 27/8

9/16 x18/27 = 6/16 = 3/8.

Example 2.

Simplify 2 4/9 x 1 7/8 ÷ 2 1/5

Solution

2 4/9 x 1 7/8 ÷ 2 1/5

= 22/9 x 15/8 ÷ 11/5

BODMAS: application

= 22/9 x 15/8 x 15/11

5 x 5

3 x 4

= 25/12

= 2 1/12.

Example 3

Simplify 3 ¾ ÷ ( 2 1/7 of 11 2/3 – 5)

Solution

3 ¾ ÷ ( 2 1/7 of 11 2/3 – 5)

BODMAS – application ( the bracket first0

= 3 ¾ ÷ ( 15/7 of 35/5 – 5)

3 ¾ ÷ (15/7 x 35/35/1)

see BODMAS also multiplication first

3 3/3 ÷ ( 5 x 5 – 5)

15/4 ÷ ( 25 – 5)

15/4 ÷ 20/1

= 3

4 x 4

3

1 6

EVALUATION

Simplify the following

1. 9 ¾ – 1/3 ) x 4 1/3 ÷ 3 ¼
2. ( 2 ¾) 2 ÷( 3 1/3 of 2 ¾ )

III. Word problems

Example I

What is the area of a rectangle of length 12 2/3m and breadth 7 ¼ m?

Solution

7 ¼ m

12 2/3m

Area of rectangle, A = L x B

L = 12 2/3m, B= 7 ¼

Area = 12 2/3 x 7 ¼

Area = 28/3 x 29/4

Area = 19 x 29

6

Area = 551

6

Area = 91 5/6m2

Example 2

Divide the difference between 4 1/5 and 2 2/3 by 1 2/5.

Solution

Interpreting the question

= 4 1/5 – 2 2/3

1 2/5

= ( 21/5 8/3 ) ÷ 1 2/5

( 21/5 x 3/38/3 x 5/5 ) ÷ 1 2/5

( 63/1540/15) ÷ 7/5

( 63 – 40 ) ÷7/5

15

23/15 ÷ 7/5

23/15 x 5/7

23/21

= 1 2/21

Example 3

What is three-quarters of 3 3/7 ?

Solution

= three-quarter = ¾

= ¾ of 3 3/7

= ¾ x 24/7

= 3/1 x 24/7

3/1 x 6/7.

Example 4

In a school, 9/10 of the students play sports. 2/3of these play football. What fraction of the students play football.

Solution

Fraction who play sports = 9/10

Fraction that play football = 2/3 of 9/10.

2/3 x 9/10

1 x3

5

=3/5.

:.3/5 of the students play football.

Example 5

Three sisters share some money. The oldest gets 5/11 of the money. The next girl gets 7/12 of the remainder. What fraction of the money does the youngest girl get?

Solution

Let the total money; be a unit = 1

1st girl gets = 5/11 of 1 = 5/11

remainder = 1 –5/11 = 11 – 5 = 6/11

11

2nd girl gets = 7/12 of the remainder

=7/12 of 6/11

= 7/12x 6/11 = 7/22

3rd girl will get 6/117/22

= 6/11 x 2/27/22

= 12/227/22

= 12 – 7

22

=

:.the fraction of the money that the youngest girl will get =

EVALUATION

1. A boy eats ¼ of a loaf at breakfast and 5/8 of it for lunch. What fraction of the loaf is left?
2. In a class of 4/5 of the students have mathematics instrument . ¼ of these students have lost their protractors. What fraction of students in the class has protractors?

IV. PRIME NUMBERS

A prime number is a number that has only two factors, itself and 1. Some examples are 2,3,5,7,11,13,…

1 is not a prime number because it has only one factor, that is, itself unlike 2 which has itself and 1 as its factors. All prime numbers are odd numbers except 2 which is an even number.

Example 1

Write out all prime numbers between I and 30

Solution

Between 1 and 30.

Prime numbers = 2,3,5,7,119………..

1 is not a prime number because it has only one factor, that is, itself unlike 2 which has itself and 1 as its factors. All prime numbers are odd numbers except 2 which is an even number.

Example 1

Write out all prime numbers between I and 30

Solution

Between 1 and 30.

Prime numbers = 2,3,5,7,11, 13, 17,19,23,29.

II. Factors

A factor of a given number is a number which divides the given number without leaving any remainder. For instance, 10÷ 2 = 5 without a remainder, therefore, we say 5 is a factor of 10. Thus, we say that 3 is not a factor of 10.

Example 1

Find the factors of 32

Solution

32 = 1 x 32 =2 x 16 =4x 8

= 8 x 4 = 16 x 2 = 32 x 1

:. Factors of 32 = 1, 2,4,8, 16, and 32 or

Using table

32 1 x 32

2 x 16

4 x 8

Factors of 32 = 1, 2,4,8, 16 and 32

Note; A case were you have a particular number occurring two times, duplication is not allowed. Pick it once. See the example below;

Example 2

Find the factors of 144

Solution

144 1 x 144

144 2 x 72

144 3 x 48

144 4 x 36

144 6 x 24

144 8 x 18

144 9x 16

144 2 x 12

Factors of 144 = 1,2,3,4,6, 8,9,12,16, 18, 24, 36, 48, 72 and 144

from the example, 12 occurred twice, but only one was picked.

Example 3

Find the factors of 120

Solution

120 1 x 120

1. 2 x 60

120 3 x 40

120 4 x 30

120 5 x 24

120 6x 20

120 8 x15

120 10 x 12

Factors of 120 = 1,2,3,4,5,6,8, 10,12,15,20,24,30,40,60 and 120.

Prime Factors:

From our definition of prime numbers, it will be easy joining factor to it and getting the meaning of prime factors.

Prime factors of a number are the factors of the number that are prime.

To find the prime factors of a number.

1.Start by dividing the number with the lowest number that is its factor and progress in that order.

2. If you divide with a particular number, check if it can divide the new number again before moving to the next prime number.

Example 1

Express the following whole numbers as product of prime factors.

(a) 12 (b) 18 (c) 880 (d) 875.

Solution

1. 12 2 18
2. 6 3 9

2 2 3 3

1 1

12 is expressed as a product of 18 = 2 x 3 x 3

primes.

12 = 2 x 2 x 3

( c ) 2 880 5 875

2 440 5 175

2 220 5 35

2 110 7 7

5 55

11 11 875 =5 x 5x 5 x 7

1

= 880 = 2 x 2 x 2 x 2 x 5 x 11

example 2

Express 1512 as a product of prime factors.

Solution

Following the example above

2 1512

1. 756
2. 378
3. 189
4. 63
5. 21

7 27

1512 = 2 x 2 x 2 x3 x 3 x 3 x 7

EVALUATION

Express the following as product of prime numbers

1. 108
2. 216
3. 800
4. 900
5. 17325

Index Form

If we have to write the following 4, 18, 16 as a product of prime factors, it will pose no challenge

4 = 2 x 2

8 = 2 x 2 x 2

16 = 2 x 2 x 2 x 2

As their products increase, the challenge of how to write 2 or whichever number is multiplying itself will arise.

A way of writing this in a shorter form is called index form.

The general form is xn

Where x = the base, that is the multiplicative value and

n=index or power or the number of times a particular number multiplies itself.

Example 1

Express the following index

1. 3 x 3 x 3 x 3
2. 5 x 5 x 5
3. 2 x 2 x 2 x 2 x 2 x 2 x 2

Solution

(a) 3 x 3 x 3 x 3, this shows that four 3’s are to be multiplied together.

Writing index form

= X n

x = 3, n = 4

:. 3 4

(b) 5 x 5 x 5 = three 5’s in general form Xn

= x =5, n = 3

= 5 3

(c) 2 x 2 x2 x 2 x 2 x 2 x 2

= seven 2’s

= X n.

x = 2, n =7

= 2 7

As product of primes

800 …… 2 x 2 x 2 x 2 x 2 x 5 x 5

800 = 25 . 52

example 3

Express the following as a product of primes in index form.

(a) 720 (b) 1404

(a) 720 (d) 1404

720 = 2 x 2 x2 x 2 x3 x 2 x 5 1404 = 2 x 2 x 3 x 3 x3 x 13

= 24x 32 x 5 = 22 x33 x 13

1.Essential mathematics for JSSI by AJS Oluwasanmipg29, 46-51

2. New General Mathematics for JSSI by MG macrae et al pg25

WEEKEND ASSIGNMENT

1.The fractions C/D ÷ a/b is same as

(a) Ca/Db (b) Cb/Da (c ) C x a (d) a x C (e) a x b

D x b b x D D x b

2. Simplify 11/25 x 1 4/11

(a) 2/3 (b) 3/5 (c) 2/5 (d) 4/5 (e) 1 ¼

3.Find the length of a rectangle whose breadth and area are 7/20m and 8 1/5m2

(a) 23 3/7 (b) no answer ( c) 21 2/7 (d) 1 7/20 (e) 8 11/20.

4. Simplify 5 ¼ + 1 1/6 – 3 2/3

(a) 5 11/4 (b) 2 ¾ ( c) 3 1/12 (d) 1 ¾ ( e) 3 – 3/2

5. The product of prime factor of 28 is

(a) 2 x 3 x 7 (b) 2 x 4 x 7 (c ) 4 x 7 (d) 2 x 2 x 7 (e) 2 x 2 x2 x 7.

THEORY

1. Simplify 2 2/5 – 1 ¾

4/5 + ½

2. Express the following as a product of prime factors; i. 105 ii. 75 iii. 60.

Presentation

The topic is presented step by step

Step 1:

The class teacher revises the previous topics

Step 2.

He introduces the new topic

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

Conclusion

The class teacher wraps up or conclude the lesson by giving out short note to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she does the necessary corrections when and where  the needs arise.