Oasis Math 6

Now,
18 = 2 × 3 × 3

24 = 2 × 2 × 2 × 3

L.C.M. = 2 × 3 × 3 × 2 × 2 = 72

Example: 3
Find the smallest number which is exactly divisible by 12 and 20.
Solution:

The smallest number which is exactly divisible by 12 and 30 is their L.C.M.

Now, 2 12 2 30

2 6 3 15

3 5
12 = 2 × 2 × 3

15 = 3 × 5

∴ L.C.M. 2 × 3 × 2 × 5 = 60

Exercise 2.9

1. Find the L.C.M. of each of the following numbers using set of multiples method.
(a) 4 and 5 (b) 6 and 8 (c) 3 and 6 (d) 6 and 7 (e) 6 and 9

2. Find the L.C.M. of the following numbers by prime factor method.

(a) 9 and 12 (b) 12 and 16 (c) 15 and 25 (d) 22 and 33

(e) 24 and 30 (f) 9and 24 (g) 12 and 30 (h) 21 and 18

3. Solve the following problems:

(a) Find the smallest number which is exactly divisible by 10 and 15.

(b) Find the smallest number which gives remainder 0 while divisible by the number

12 and 15.

(c) Find the smallest number which is exactly divisible by 15 and 20.

õõõ

Answers (b) 24 (c) 6 (d) 42 (e) 18 (g) 60 (h) 126
(b) 48 (c) 75 (d) 66 (e) 120 (f) 72
1. (a) 20 (b) 60 (c) 60
2. (a) 36

3. (a) 30

Oasis School Mathematics – 6 45

2.9 Square and Square Roots

Let's look at the following product:

1×1 = 12 = 1 [1 is a square of 1] 2×2 = 22 = 4 [ 4 is a square of 2]

3×3 = 32 = 9 [9 is a square of 3]

4×4 = 42 = 16 [16 is a square of 4] and so on.

From the above example, it is clear that when a number is multiplied by itself, the
product obtained is called a square of that number. In other word, square number is
the product of two identical numbers. The number pattern of the square number can
be shown as follows:

14 9 16
Activity

• Finding of square and square root using Geoboard.
• Count the length of each square. • Count the area of each square

Fig. Length of unit square Area of unit square

I.
II.
III.
IV.

Obtain the relation of length and area and draw out the conclusion about square.

46 Oasis School Mathematics – 6

Activity 1

• Make a group of 3 students.
• Give 16 pieces, 20 pieces, 25 pieces of corn seeds.
• Ask them to arrange the seeds in rows and columns such that number of

seeds in each row and column is the same.
• Ask them whether it is possible to do by all students.
• Draw out the construction.

Shortcut method of finding square of the numbers likes 10, 20, 30, 200, 500 etc.

Square of 30

= (30)²

= 30 × 30 Multiply 3 and 3, 3×3=9. Write 2 zeros after 9 i.e. 900
= 900 ∴ 30 × 30 = 900
Square of 600

= (600)² Multiply 6 and 6, 6 × 6 = 36
= 600 × 600 Write 4 zeroes after 36 i.e. 360000 ∴ (600)² = 360000
= 360000

Square root

Let's look at the following examples:

1 = 1×1=12 → 1 is the square of 1.
→ 1 is the square root of 1.

4 = 2×2=22 → 4 is the square of 2.
→ 2 is the square root of 4.

9 = 3×3 = 32 → 9 is the square of 3.
→ 3 is the square root of 9.

This shows that square and square root are mutually reverse operations.

The perfect square number is the product of two same numbers. One of the same
numbers is the square root of given square number.
The square root is denoted by the radical sign ( ) .

\ 52 = 25 → ( 25) = 5. = (Numerator)²
Square of a number in fraction (Denominator)²

Example: square of 3 = (3)² = 9
5 (5)² 25

Square root of a fraction = Square root of numerator
Square root of denominator

Oasis School Mathematics – 6 47

∴ Square root of 9 = 9 = 3² = 3
16 16 4² 4

Activity 2

• Make a group of 3/4 students.
• Ask them to take 16, 9, 25 pieces of corn seed.
• Ask them to arrange in such a way that number of seeds in both row and

column is the same.
• Discuss the number of seeds in each row and in each column.
• Are they equal?
• What does the number of seeds in each row and column represent?
Discuss in the group.

Method of finding square roots

Prime factorisation method

Let's see an example to get an idea of getting square root by prime factorisation

method.
For example:
Find the square root of 324.

Here, Steps:
2 324 • Express the given number as the
2 162
3 81 product of its prime factors
3 27 • Make pairs of equal prime factors.
39 • Take one factor from each pair and

3 multiply them.
The product is required square root of
Here, 324 = 2 × 2 × 3 × 3 × 3 × 3
given number.
= 2² × 3² × 3²

\ Square root of 324 = 2 × 3 × 3 = 18

Shortcut method of finding square roots of the numbers like 100, 400, 900,
3600 etc.

Square root of 100 = 100 1 = 1, write 1 zero after 1
∴ 100 = 10.
= 10

Square root of 2500 = 2500 25 = 5² = 5, write 1 zero after 5
∴ 2500 = 50
= 50

48 Oasis School Mathematics – 6

Worked Out Examples

Example: 1 24
× 24
Find the square of 24.
Solution: 96
+ 480
Square of 24 = 24²
= 24 × 24 576
= 576

Example: 2 I know !
2² = 2
Find the square root of 324
Solution: 3² = 3 ....

Given number = 324
∴ 324 = 2 × 2 × 3 × 3 × 3 × 3
= (2 × 2) × (3 × 3) × (3 × 3)
∴ Square root of 324

= 324 = (2 × 2) × (3 × 3) × (3 × 3)

= 22 × 32 × 32
= 2 × 3 × 3 = 18

Example: 3

Find whether the number 216 is a perfect square numbers or not.
Solution:

Since the last two factors I understand ! To be a
are not in pair. perfect square number,
every prime factor should

have pair of

Example: 4

What least number must be multiplied to 162 to make it a perfect square.

Solution: 2 162
Here, 162 = 2 × 3 × 3 × 3 × 3

or, 162 = 32 × 32 × 2

Every factor has its own pair except 2.

2 must be multiplied to 162 to make it a perfect square.

Oasis School Mathematics – 6 49

Example: 5

Find the square root of 49 .
144
Solution:

Here, square root of 49 = 49
144 144

Now, 7 49 Here, 49 = 72 = 7
2

Again, 144 2×2×2×2×3×3
2 144
22 × 22 × 32
2 72
2× 2 × 3
2 36 12
49 7
2 18 144 12

39 Hence,

3



Example: 6

A square garden has 196 flowers. The number of flowers in each row and each

column is the same. Find the number of flowers in each row.
Solution:

Here, total number of flowers = 196. Since the number of flowers in each row and
column is the same, then :
The number of plants in each row.

= 196 = 2 × 2 × 7 × 7

= (2 × 2) × (7 × 7) = 2 × 7 = 14

Exercise 2.10

1. Find the value of.

(a) 6² (b) 8² (c) 12² (d) 16² (e) 25²

2. Find the square of the following numbers.

(a) 9 (b) 14 (c) 32 (d) 84 (e) 96

3. Find the square of the following numbers using shortcut method.

(a) 30 (b) 70 (c) 90 (d) 400 (e) 800 (f) 5000

4. Find the square root of the following numbers using factorization method.

(a) 16 (b) 36 (c) 121 (d) 169 (e) 196
(f) 441 (g) 576 (h) 625
(i) 2025

50 Oasis School Mathematics – 6

5. Find the square root of the following numbers using shortcut method.

(a) 400 (b) 900 (c) 1600 (d) 8100 (e) 6400

6. Identify whether the given numbers are square number or not.

(a) 144 (b) 392 (c) 128 (d) 200 (e) 576

7. Find the least number which should be multiplied to the following numbers to

make them perfect square.

(a) 32 (b) 28 (c) 300 (d) 125 (e) 882

8. Find the square root of the following numbers.

(a) 9 (b) 241 (c) 25 (d) 144
16 49 256

9. (a) Which number multiplied by itself gives the product 121?

(b) Which number multiplied by itself gives the product 1764?

(c) Area of a square room is 441 m², find the length of each side.

10. (a) A certain number of boys collected Rs. 1296 for a picnic, each boy gave as
many rupees as there were boys, find the number of boys.
(b) A general having 11025 men under him arranges them into a solid square. Find the
number of men in each row.

11. (a) 3 2 flowers are planted in a row and column of a square garden. Find the total

number of flowers in the garden.

(b) 45 students are kept in each row and a column of a square ground. Find the number

of students in the ground.

Answers

1. (a) 36 (b) 64 (c) 144 (d) 256 (e) 625 2. (a) 81 (b) 196
(e) 9216
(c) 1024 (d) 7056 (e) 640000 3. (a) 900 (b) 4900
(i) 45
(c) 8100 (d) 160000 (f) 25000000 4. (a) 4 (b) 6 (c) 11 (d) 13 (e)

14 (f) 21 (g) 24 (h) 25 5. (a) 20 (b) 30 (c) 40 (d) 90 (e) 80

6. (a) Square number (b) not a square number (c) not a square number (d) not a square number (e)

square number 7. (a) 2, (b) 7 (c) 3 (d) 5 (e) 2 8. (a) 3 (b) 3 (c) 5
4 2 7
12
(d) 16 9. (a) 11 (b) 42 (c) 21m 10. (a) 36 (b) 105 11. (a) 1024 (b) 2025

Do you know!
A square number contains 0, 1, 4, 5, 6 and 9 in ones place.

Oasis School Mathematics – 6 51

Unit

3 Integer

3.1 Introduction of Integers

We have already learnt about natural numbers. Let us consider the set of natural
numbers,
N = {1, 2, 3, 4, 5, ………}
Let's take any two natural numbers 4 and 5, find their sum and product, their
sum = 4 + 5 = 9, which is a natural number.
Their product = 4 × 5 = 20, which is also a natural number.
Hence, the sum and the product of two natural numbers is also a natural number.

Which is a natural number?
Let's see what happens in the subtraction.
4 – 5 = – 1, which is not a natural number.
In order to make the operation of subtraction meaningful negative number and zero
are introduced.
Hence,
The set of natural numbers together with their negatives including zero are called
integers.

Integers in Number Line
The set of integers is denoted by Z.
\ Z = { ..........., -4, -3, -2, -1, 0, 1, 2, 3, 4, ......... }

Z+ = {1, 2, 3, 4, ........... } • Opposite of negative
= a set of positive integers. integer is positive.
Z— = {-1, -2, -3, -4,........... }
= a set of negative integers. – (– 5) = + 5
– (– 7) = + 7 etc.

52 Oasis School Mathematics – 6

• Positive integers are shown on the right side of zero and the negative integers
on the left side of zero.

• The opposite integers + 1 and – 1 are marked at the same distances from
zero. Similarly + 2 and – 2, + 3 and – 3 are marked on the same way.

• An integer on the number line is always less than an integer on its right.
i.e. 3 > 2> 1> 0> – 1> – 2 > – 3 and so on.

• The arrow marks at the both ends of a number line represent that the numbers
are infinite.

Use of Integers in our daily Life
In many situations of our daily life, we have to use the opposites.

Profit and Loss

Profit of Rs. 40 is expressed as + 40.

Loss of Rs. 40 is expressed as – 40.

Altitude

Height of a mountain is 8250 m above sea level. Depth 750 m means
i.e. height of a mountain is + 8250 m 750 m below sea level.

Depth of the sea is 750 m. It is represented as –750 m.

Temperature

During the day of winter, maximum temperature –5º means 5º below 0ºC.
of winter is + 10ºC and minimum temperature
is – 5° C,

Summary

• Whole numbers greater than 0 are positive integers.
∴ +1, +2, +3, ... are positive integers; denoted by z+.
• Whole numbers less than 0 are negative integers.
∴ -1, -2, -3, ... are negative integers; denoted by z–.
• Positive integers, negative integers including zero are integers, denoted by z.

Worked Out Examples

Example: 1

Write all the integers between -3 to 5.
Solution:

Oasis School Mathematics – 6 53

Here,
The integers between -3 and 5 are -2, -1, 0, 1, 2, 3 and 4.

Example: 2

Find the integer which is 4 units left from the integer 3.
Solution:

From the above number line the integer which is 4 units left from 3 is -1.
It can also be obtained
3 - 4 = -1 [4 units left means - 4]

Example: 3

Pema and Harrish had to travel towards opposite direction. Pema travelled
25 km towards east and Harrish 30 km west. What is the distance between
them?
Solution:
Here, O is the point of reference (i.e. starting point). Then, distance from the starting
point O to A is 25 km and the distance from the starting point to B is 30 km.

A OB

So, the distance between A and B = AB = 25 km + 30 km = 55 km.
If we consider OA to be -30 km, then its direction compared to the direction of OA
is negative.
So, AB = OB - (-OA) = 25 – (-30) = 25 + 30 = 55 km.

Exercise 3.1

1. Draw a number line in your copy and answer the given questions.
(a) On which side a number smaller than zero lies?
(b) On which side a number greater than zero lies?
(c) On which side a number greater than the given number lies?
(d) On which side a number smaller than the given number lies?

54 Oasis School Mathematics – 6

2. Use + or – sign to represent the following.

(a) A loss of Rs. 30 (b) 40ºc above zero (c) 25ºc below zero

(d) Profit of Rs. 80 (e) Loss of Rs. 120

(f) 15000m above the sea level (g) 250m below the sea level

3. Use the given number line to compare each of the following pair of integers.
Write > or <.

-8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7 +8

(a) –1 +1 (b) –2 - 4 (c) +7 +4 (d) –7 –1 (e) –2 1
4. Draw a number line and show the number – 6 to + 6 in the number line.

5. Draw a number line and write down number lying 5 units towards the left of
each of:

(a) -5 (b) -3 (c) 2 (d) 5 (e) 3

6. Draw a number line and using it write the number which is 5 units right from
the number:

(a) 2 (b) 5 (c) – 3 (d) – 7 (e) 0

7. Draw a number line and find which integer is greater one.

(a) -5, - 7 (b) -2, 0 (c) 0, 3 (d) -3 and 7

8. In each of the following pairs, find which integer is smaller one.

(a) -2, -5 (b) 0, -3 (c) 2, -3 (d) -4, 0

9. Draw a number line and write all integers between:
(a) -2 and 3 (b) -3 and 3 (c) -5 and 0 (d) -5 and 1 (e) -6 and -1

10. Insert an appropriate sign < or > in the following circles.
(a) -2 2 (b) 0 -2 (c) 2 -2 (d) -5 5

11. In each case, arrange the given integers in ascending order.

(a) -8, 0, -5, 5, 4, -1 (b) -3, 3, -4, 0, -6, 2, 5 (c) -1, -5, -6, -8, 0, -7

12. In each case, arrange the given integers in descending order.

(a) -2, -3, 0, 5, -5, 6 (b) -1, -2, -3, 0, 5, 3 (c) 0, -2, 5, -3, -6, -1
13. Simplify:

(a) – (–5) (b) + (–6) (c) – (–20) (d) – (+8) (e) – (–4)

Answers
Consult your teacher.

Oasis School Mathematics – 6 55

3.2 Operations on integers

Addition of integers: Depending upon the sign of integers there are three cases on
the addition of integers.
When both integers are positive:

For example: (+5) + (+3)

\ (+5) + (+3) = 8 • Starting form 0, move 5 steps right.
• Again move another 3 steps.
• Final point is + 8.

When both integers are negative:

For example: (-3) + (-4)

4 steps left 3 steps left

\ (-3) + (-4) = -7 • Starting form 0, move 3 steps left.
• Again, move another 4 steps left.
• Final point is -7.

When two integers are of opposite sign:
For example: (-4) + (+6)

\ (–4) + (+6) = +2 • Starting from 0 move 4 steps left.
• Again from that point move 6 steps right.
• Final point is +2.



56 Oasis School Mathematics – 6

Again, (+7) + (-8)

\ (+7) + (–8) = -1. • Starting from 0 move 7 steps right.
• Again from that point move 8 steps left.
• Final point is –1.

From the above examples, we can conclude that

• Sum of two positive integer is positive.

• Sum of a positive and a negative integer is +ve if the greater one is + ve, and the
sum is –ve if the greater one is –ve.

Multiplication of integers

Multiplication of integers can be done by using number lines:

Multiplication of two positive integers

(+2) × (+3)

Here, (+2) × (+3) • Starting from 0 move 3 units right from the origin.
= 2 times of (+3) • Repeat the same process from 3.
= 3+3 • Final point is 6.

∴ (+2) × (+3) = +6
Multiplication of two integers having different signs

(+3) × (-4) • Starting form 0, move 4 steps left.
• Repeat the same process 3 times.
Here, (+3) × (-4) = 3 time of (–4) • Final point is -12.
= (-4) + (-4) + (–4)

∴ (+3) × (-4) = –12.

Again, (-2) × (+4)
= 2 × (– 4)
= 2 times (– 4)

Oasis School Mathematics – 6 57

= (– 4) + (– 4)
\ (– 2) × (+4) = – 8

When both integers are negative:
(-3) × (-2) = 3 × 2
= 3 times of 2.
=2+2+2
\ (– 3) × (– 2) = 6

From above three examples, we conclude that:
(a) If two integers are of the same sign (both are positive or both are negative)
then their product is a positive integer.
i.e. (+ve integer) × (+ve integer) = + ve integer.
(-ve integer) × (-ve integer) = + ve integer.

(b) If two integers are of opposite sign (one is positive and the other is negative)
then their product is a negative integer.

(+ve integer) × (-ve integer) = (-ve integer)

Note: The product of zero and an integer is zero.

Division of integers

Division is the inverse operation of multiplication. So the sign rules of division are
same as the sign rules of multiplication.
Therefore:

(+2) × (+3) = +6 → (+6) ÷ (+2) = +3 and (+6) ÷ (+3) = (+2)

(-4) × (+3) = -12 → (-12) ÷ (-4) = +3 and (-12) ÷ (+3) = (-4)

(-6) × (-3) = + 18 → (+18) ÷ (-6) = -3 and (+18) ÷ (-3) = (-6)
Look at the following examples;
(-12) ÷ (+4)

–13 –12 –11 –10 – 9 – 8 – 7 – 6 – 5 – 4 –3 –2 –1 0 1 2 3

–3 –3 –3 –3

Here, -12 = (-3) + (-3) + (-3) + (-3)
-12 = 4 × (-3)

58 Oasis School Mathematics – 6

\ (-12) ÷ (+4) = -3
Again, (-8) ÷ (-2)

Here, -8 = 2 × (-4) or, -8 = (-2) × 4 ∴ (-8) ÷ (-2) = 4
Therefore in the case of division:

(a) The quotient of two integers with the same sign is positive.
i.e. (+ve integer) ÷ (+ve integer) = (+ve) integer.
(-ve integer) ÷ (-ve integer) = (+ve integer)
(b) The quotient of two integers with opposite sign is negative.
i.e. (+ve integer) ÷ (-ve integer) = - ve integer
(-ve integer) ÷ (+ve integer) = - ve integer

Note: (–) even power = + ve, (–) odd power = -ve

Exercise 3.2

1. Draw a number line and add:

(a) (+5) + (+7) (b) + 8 + (–2) (c) (–6) + (+3)
(f) (–3) + (+6)
(d) (+7) + (+4) (e) (–5) + (–4)
(c) (–3) × (–4)
2. Draw a number line and multiply: (f) (– 2) × (+ 3)

(a) (+3) × (+2) (b) (+4) × (–2) (c) (+6) ÷ (–2)
(f) (– 12) ÷ (– 4)
(d) (–5) × (–2) (e) (–4) × (+3)
(e) (–4)4 (f) (–2)7
3. Use number line and divide:

(a) (+8) ÷ (+2) (b) (+6) ÷ (–3)

(d) (–8) ÷ (–4) (e) (–9) ÷ (+3)

4. Find the value of: (d) (–3)³
(a) (–1)5 (b) (–1)12 (c) (–2)4

5. Simplify:

(a) (+6) + (–8) + (–12) (b) (–12) × (–4) + (–8) + (+3)
(c) (–3) × (–9) – (–4) (+5) (d) (–72) ÷ (+9) + (–16) ÷ (–4)

Answers (b) + 6 (c) – 3 (d) + 11 (e) – 9 (f) + 3
(b) –8 (c) + 12 (d) + 10 (e) – 12 (f) – 6
1. (a) + 12 (b) – 2 (c) – 3 (d) + 2 (e) – 3 (f) + 3
2. (a) + 6 (b) 1 (c) 16 (d) – 27 (e) 64 (f) – 128
3. (a) + 4 (b) + 43 (c) + 47 (d) – 4
4. (a) – 1
5. (a) – 14

Oasis School Mathematics – 6 59

Objective Questions

1. The smallest number formed by the digits 2, 3, 5, 0, 1 is :

(i) 01235 (ii) 10235 (iii) 53210

2. The value of 20 + 2 {12 ÷ (6 ÷ 3)} is :

(i) 32 (ii) 132 (iii) 68

3. Value of which of the following expression is not equal to 20?

(i) 32 – 3 {16 ÷ (8 ÷ 2)}
(ii) 40 ÷ [12 – {6 + (10 ÷ 5 + 2)}]
(iii) 20 + [18 – {22 – (6 + 2 – 3)}]
4. A number is divisible by both 2 and 3 then it is divisible by:

(i) 8 (ii) 4 (iii) 6

5. The numbers 14 and 15 are :

(i) twin prime numbers (ii) co–prime numbers (iii) prime numbers

6. a, b, c, d, .................... common multiples of two numbers. If a <b, b <

c, c < d, ........... then the L.C.M. of the numbers is :

(i) a (ii) b (iii) d

7. L.C.M. of two prime numbers is :

(i) 1 (ii) their sum (iii) their product

8. The smallest number which should be multiplied to 108 to make it a cube

number is :

(i) 2 (ii) 3 (iii) 108

9. Which of the following may not be a natural number?

(i) Sum of two natural numbers (ii) Product of two natural numbers

(iii) Difference of two natural numbers
10. Which of the following may not be an integer?

(i) Sum of two integers.

(ii) Quotient of two integers.

(iii) Difference of two integers.
11. Which one of the following statement is not true?

(i) Every integer is a rational number.
(ii) Every natural number is whole number.
(iii) Every integer is a whole number.

60 Oasis School Mathematics – 6

12. Which of the following relation is not true?

(i) 5 < 7 (ii) –5 < –7 (iii) 2 > –1

13. Which of the following number is not a square number?

(i) 64 (ii) 8 (iii) 36

14. In the relation 3 × 5 = 15, which one of the following statement is not true.

(i) 15 is the multiple of 3 and 5.

(ii) 3 and 5 are factors of 15.

(iii) 3 and 5 are the multiple of 15.

15. The smallest number which is exactly divisible by 6 and 4 is:

(i) 24 (ii) 36 (iii) 12

Assessment Test Paper

Group "A" [5 × 1 = 5] Full marks: 30

1. (a) What is the smallest number formed by the digits 3, 4, 2, 0, 9?

(b) Simplify : 5 × 10 ÷ 2 + 1

(c) If M(4) be the set of multiples of 4 less than 20, then find M(4)

(d) Is 1 a rational number? Justify your answer.
2

Group "B" [7 × 2 = 14]

2. (a) Convert the given statement into Mathematical sentence and simplify:

12 is multiplied to the difference of 10 and 6.

(b) Test whether 14328 is divisible by 3 or not.

(c) Find the possible factors of 18.

(d) Find the prime factors of 24.
3. (a) Find the square root of 196.

(b) Draw the number line and add: (– 3) + (+ 5)

(c) If the cost of 1 kg orange is Rs 65.35, what is the cost of 6 kg orange?

Group "C" [3 × 4 = 12]
4. Simplify: 60 ÷[180 ÷ 2 {6 + (17 – 14)}]
5. Find the greatest number which exactly divides 24, 32 and 48
6. Find the cube root of : 1728

Oasis School Mathematics – 6 61

Unit Fractions and
Decimals
4

4.1 Equivalent fractions

1 is shaded.
2

2 is shaded.
4

3 is shaded.
6

In the above figures, same part is shaded but represented by different notation.

Hence, 1 , 2 and 3 are equivalent fractions.
2 4 6

Again,

1 is shaded.
2

2 is shaded.
4

4 is shaded.
6

In all three figures, same part is shaded but represented by different notation.

Hence, 1 , 2 and 4 are equivalent fractions.
2 4 8

Finding of equivalent fractions of given fraction.

Lets take a fraction 1 .
3

Multiply both numerator and denominator by 2, then

62 Oasis School Mathematics – 6

1×3 = 2
3×2 6

Multiply both numerator and denominator by 3, then

1×3 = 3
3×3 9

Hence, 1 , 2 and 3 are equivalent fractions.
3 6 9

Exercise 4.1

1. Write the fraction of the shaded part.

a) b)

c) d)

2. What fractions of these circles are shaded?

a) b) c)

3. Write any two equivalent fractions of given fraction.

a) 2 b) 3 c) 1 d) 2
3 4 5 7

4. Write the correct number in the gap.

a) 1 = 9 b) 2 = 6 c) 1 = 6 d) 3 = 12
3 5 4 4

5. Separate the equivalent fractions from the given list:

a) 1 , 1 , 2 , 2 b) 2 , 1 , 4
2 3 3 4 3 3 6

Answers
Consult your teacher.

Oasis School Mathematics – 6 63

4.2 Comparison of Fractions

Comparison of like fractions

Activity

• Take a rectangular piece of paper.
• Draw rectangle as shown in the figure and divide it into equal 5 parts

along its length and equal 4 parts along its breadth.
• Shade one box in the first row, 2 boxes in the second row, 3 boxes in the

third row and so on.
• Write the fraction represented by shaded part of each row.
• Compare the length of shaded part each row and arrange them in ascending

and descending order.

1 1 1 1 1 1 part is shaded
5 5 5 5 5 5

1 1 1 1 1 2 part is shaded
5 5 5 5 5 5

1 1 1 1 1 3 part is shaded
5 5 5 5 5 5

1 1 1 1 1 4 part is shaded
5 5 5 5 5 5

It is clear that, shaded region of the first stripe is the shortest and that of fourth is
the longest.

i.e. 1 < 2 < 3 < 4
5 5 5 5

From the above it is clear that :

A fraction with smaller numerator is smaller than the fraction with greater numerator,
if their denominator is the same.

Which is greater 4/5 4>2
or 2/5 ?
∴ 4/5 > 2/5

Example: compare 3 and 8
11 11

64 Oasis School Mathematics – 6

Solution: Here the denominator is the same. Since 3 < 8

∴ 3 < 8 .
11 11

Comparison of unlike fractions

Activity

• Take a rectangular piece of paper.

• Draw rectangle as shown in the figure and divide rectangle into four equal
parts.

• Divide first row into two equal parts, second row into three equal parts,
third into 4 equal parts and last row into 5 equal parts.

• Shade one part of each row.

• Write the fraction represented by shaded part of each row.

• Compare the length of shaded part.

1 1 1/2 part is shaded
1/3 part is shaded
2 2 1/4 part is shaded
11 1 1/5 part is shaded

33 3
11 11

44 44
111 11

555 55

Let's compare the length of the shaded part:

Length of the shaded part of first stripe is the longest one and that of the last one is

the shortest one.

i.e. 1 > 1 > 1 > 1 .
2 3 4 5

If the numerators of the fractions are the same, the fraction with smaller denominator
is greater than that having larger denominator.

Which is greater 2 /5 5<7
or 2 /7 ?
∴ 2/5 > 2/7

Oasis School Mathematics – 6 65

Example: Compare 4 and 4
11 7

Solution: Here the numerator of both fractions are the same.

Comparing the denominator, 11 > 7 \ 4 < 4
11 7

C omparison of unlike fractions by converting into the like fractions:

Let's compare the fractions 2 and 3
5 4

Let's convert them into the like fractions.

L.C.M. of 5 and 4 is 20.

\ 2 = 2 × 4 = 8 20 ÷ 5 = 4
5 5 4 20

3 = 3 × 5 = 15 20 ÷ 4 = 5
4 4 5 20


Since 8 < 15, ∴ 8 < 12 Compare the numerators.
20 20
2
i.e., 5 < 3
4

Activity

Comparison of fractions

• Take a rectangular sheet of paper.

• Draw a rectangle into 10 equal parts along its breadth as shown in figure.
0 1
• Initial point of first row represents 1 and the final point represents 1 .

• Divide the second row into two equal parts. Similarly, third row into three equal

parts and fourth into four equal parts and so on.

• Write the fraction represented by the divided part as shown.

• Find any two equivalent fractions. 2 2 1
f35raacntidon55s:in(i)as14cen.d..i.ng15or(diie)r. 5 3 3 4
• Compare the following .... 4 (iii) ....
•
Arrange 1 , 1 , 2 , 3 , 2 ,
2 3 3 4 5

20 11

02 12 22
33
30 13 23 44
55
40 14 42 43

05 51 25 53 54

66 Oasis School Mathematics – 6

Worked Out Examples

Example 1: 5 7
13 13
Which is greater or ?
Solution:
5 7
Given fractions, 13 and 13 . Since the denominators of both fractions are the same,

comparing numerator 7 > 5, ∴ 5 < 7 .
13 13

Example: 2 5 5
7 9
Compare the fractions and without converting them into like fractions.
Solution:

Given fractions, 5 and 5 . Since the numerator of both fractions is the same,
7 9
5 5
comparing denominators, 7 < 9 then 7 > 9 .

Example: 3 3 2
5 3
Compare the unlike fractions and .

Solution: Given fractions 3 and 2 .
5 3

L.C.M. of 5 and 3 is 15.

Now,

3 = 3 × 3 = 9
5 5 3 15

2 = 2 × 5 = 10
3 3 5 15
9 10 .
Since 9 < 10 , 15 < 15

i.e. 3 < 2 .
5 3

Example: 4 2 5 3
3 6 4
Arrange the following fractions into ascending order , ,

Solution: 2 5 3 3=3
3 6 4 6=2×3
Given fractions , ,

L.C.M. of 3, 6 and 4 is 12. 4=2×2

Now, L.C.M. = 3 × 2 × 2

= 12

Oasis School Mathematics – 6 67

2 = 2 × 4 = 8
3 3 4 12
5 5 2 10
6 = 6 × 2 = 12

3 = 3 × 3 = 9
4 4 3 12
8 9 10
Since, 8 < 9 < 10, i.e. 12 < 12 < 12

\ 2 < 3 < 5 .
3 4 6

Example: 5 7
3 5 12
4 of a pipe is in the mud; 8 in the water and is above the water. Which

portion of the pipe is the longest?

Solution:

Here, the unlike fractions are 3 , 5 and 7 .
4 8 12

To find which portion of the pipe is the longest, we have to change these fractions

into like fractions.

Here, 4 = 2 × 2 ; 8 = 2 × 2 × 2 ; 12 = 2 × 2 × 3

L.C.M. = 2 × 2 × 2 × 3 = 24

Now, 3 = 3×6 = 18 [∵ 24 ÷ 4 = 6]
4 4×6 24

5 = 5×3 = 15 [∵ 24 ÷ 8 = 3]
8 8×3 24

and 7 = 7×2 = 14 [∵ 24 ÷ 12 = 2]
12 12 × 2 24

Obviously, 18 > 15 > 14 , hence, 18 i.e. 3 is the longest portion of the pipe.
24 24 24 24 4

Exercise 4.2

1. Compare the following pairs of like fractions:

(a) 3 and 5 (b) 4 and 3 (c) 2 and 1 (d) 5 and 7
8 8 9 9 3 3 13 13

2. Compare the following unlike fractions without converting into like fractions.

(a) 2 and 2 (b) 3 and 3 (c) 1 and 1 (d) 7 and 7
3 5 7 5 4 3 13 15

68 Oasis School Mathematics – 6

3. Convert the following unlike fractions into like fractions and compare them.

(a) 2 and 4 (b) 1 and 1 (c) 3 and 1 (d) 3 and 1
3 5 5 7 5 7 8 4

4. Arrange the following fractions in ascending order.

(a) 3 , 2 and 4 (b) 3 , 5 and 7 (c) 7 , 5 and 2 (d) 1 , 3 and 5
5 5 5 4 6 18 12 6 3 2 4 8

5. Arrange the following fractions in descending order.

(a) 3 , 8 and 4 (b) 2 , 6 and 1 (c) 1 , 1 and 3 (d) 17 , 9 and 4
7 7 7 3 7 2 4 7 8 25 10 5
5
6. (a) Chameli and Juneli each was given bread, Chameli ate 6 of the bread and Juneli

ate 1 of the bread. Who ate more bread?
6

(b) Ram Bilash and Dhaniya were given to do a piece of work. Ram Bilash completed

2 of the work and Dhaniya completed 3 of the work. Who completed more parts
5 7
of the work?

(c) Lakpa spends 1 part of his monthly income on food, 1 part on fuel and 1 part on
5 4 2
education. On which items does he spend more money?

Answers

1. Consult your teacher 2. Consult your teacher.

3. (a) 10 and 12 (b) 7 and 5 (c) 21 and 5 (d) 3 and 2
15 15 35 35 35 35 8 8

4. (a) 2 < 3 < 4 (b) 7 < 3 < 5 (c) 7 < 2 < 5 (d) 1 < 5 < 3
5 5 5 18 4 6 12 3 6 2 8 4

5. (a) 8 > 4 > 3 (b) 6 > 2 > 1 (c) 3 > 1 > 1 (d) 9 > 4 > 17
7 7 7 7 3 2 8 4 7 10 5 25

6. (a) Chameli (b) Dhaniya (c) Education.

4.3 Fundamental Operations on Fractions

The four fundamental operations on fractions are addition, subtraction, multiplication
and division.

A ddition and Subtraction of Fractions

There are different ways of adding and subtracting fractions depending upon the

type of the fractions. 4
5
Addition and subtraction of like fraction

Look at the given examples,

In the given figures, 3 1 31
5 5 55
One part of stripe is and another part is .

Oasis School Mathematics – 6 69

Then 4 parts out of 5 are coloured.

∴ 3 + 1 = 4
5 5 5

Hence, while adding like fractions, simply add the numerators keeping the

denominator same.

In the given figure,

fraction represented by the shaded part = 8
9
= 93
fraction represented by the crossed part 8/9
5/9 3/9
remaining coloured part = 5
9
∴ 8 – 3 = 5
9 9 9

Hence, while subtracting the like fractions, subtract keeping the numerator same.

Sum of like fractions = sum of numerators and,
common denominators

Difference of like fractions = difference of numerators
common denominators

Look at one more example:

3 1 + 4 2 = 10 + 14 Convert mixed number into
3 3 3 improper fractions
3
10 + 14 Keep the denominator
= same and add numerator.
3

= 24 =8
3

Note:

If any one of the fractions is a mixed number, then change it into improper
fractions.

Alternative method:

3 1 + 4 2 = 3+ 1 + 4 + 2 Separate the whole part and
3 3 3 3 fractional part of mixed number.

Add whole numbers and
= 3+ 4 + 1 + 2 fractional parts separately.
3 3

= 7+ 2+1 = 7+ 3
3 3

= 7+1 = 8.

70 Oasis School Mathematics – 6

Addition and subtraction of unlike fractions

Activity

Addition of unlike fraction by paper cutting method:

1 + 1
2 3
Take a long strip of paper. Divide it into two equal parts. Each part represent 21.

11

22

Take a second strip of paper of same length and divide it into three equal parts.

111
333

Again, L.C.M. of 2 and 3 is 6. So, take another strip of paper of same length and divide
the strip into 6 equal parts.

1 1 11 1 1
6 6 66 6 6

Cut 1 and 1 from first two stripes and keep 1 and 1 together above the third strip
2 3 2 3

11
23

1 1 11 1 1
6 6 66 6 6

Compare the length of combination of two stripes with third stripes.

Position of third strip is 56.

∴ 1 + 1 = 65.
2 3

Method of adding unlike fractions

Take any two fractions 2 and 1 , they are unlike fractions:
5 4

Let's convert them into like fractions.

L.C.M. of 5 and 4 is 20.

∴ 2 + 1
5 4

= 2 × 4 + 1 + 5 Converting into like fractions
5 4 4 5

Oasis School Mathematics – 6 71

= 8 + 5 My Idea ! As we can add like
20 20 fraction, let's convert them in to

8+5 like fractions and add them.
20
=

= 13
20

Alternative method • Take L.C.M. of denominators
• Divide L.C.M by each denominators and multiply
2 + 1
5 4 the respective numerator by quotiet.
• Add or subtract as in the like fractions.
= 2×4+5×1
20

= 8+5
20

= 13
20

Let's be clear with one more example:

Add: 3 and 2
5 3

Solution:

3 + 2
5 3

= (3×3) + (2 × 5)
15
L.C.M. of 5 and 3 = 15
= 9+10
15 15 ÷ 5 = 3, then (3 × 3) and
15 ÷ 3 = 5, then (2 × 5)
= 19 = 1 145
15


Note: Follow the same method in the subtraction of fractions also.

Mathematical operation "of"

The meaning of operation "of" is same as the multiplication

example: 1 of 50 m = 1 × 50m = 25m
2 2

1 of Rs. 100 = 1 × Rs.1000 = Rs200
5 5

Similarly, 1 of 2 = 1 × 2 = 2 .
3 5 3 5 15

72 Oasis School Mathematics – 6

Worked Out Examples

Example : 1 Alternative method:

Subtract: 5 − 1 . L.C.M. of 6 and 4 is 12
6 4
Solution : 5 1 5×2 −1×3
Now, 6 − 4 = 12
L.C.M. of 6 and 4 is 12

5 − 1 = 5×2 − 1×3 = 10 − 3
6 4 6×2 4×3 12

= 10 − 3 = 7
12 12 12

= 10 − 3 = 7
12 12

Example : 2 6=2×3
4=2×2
Simplify: 8 1 + 4 3 – 2 1 ∴ L.C.M. = 2 × 2 ×3
Solution: 8 4 2
12 ÷ 6 = 2 Multiply 5 by 2
8 1 + 4 3 – 2 1 12 ÷ 4 = 3 Multiply 1 by 3
8 4 2

= 65 + 19 –– 5
8 4 2

= 65×1+19×2–5×4 = 65 + 38 – 20
8 8

= 103–20 = 83 =10 3
8 8 8

Example: 3 1
3
Salim has given 3 hours to complete a test. He finished the test in 2 3
4
hours. How much earlier did he finish the test?

Solution: 1 10
3 3
Total time he has given = 3 hours = hours

He finished the test in 2 3 hours = 11 hours.
4 4

( )∴ He finished the test 10 – 11 hours earlier.
3 4

( )∴ Required time = 10 – 11
3 4

= 10 × 4 – 11 × 3 = 40 – 33 = 7 hour.
12 12 12
7
∴ He finished the test 12 hour earlier.

Oasis School Mathematics – 6 73

Exercise 4.3

1. Add the following like fractions and reduce them into its lowest term (if necessary).

(a) 4 + 2 (b) 5 + 4 (c) 2 1 + 3 2 (d) 2 3 + 3 1
11 11 13 13 3 3 8 8

2. Subtract the following and reduce them into lowest term (if necessary).

(a) 3 − 1 (b) 8 − 3 (c) 4 2 − 1 1 (d) 7 2 − 4 5
4 4 13 13 3 3 11 11

3. Carry out the following addition and subtraction.

(a) 3 + 4 + 7 (b) 4 1 + 3 2 + 6 3 (c) 5 2 − 2 3 − 2 1
15 15 15 5 5 5 7 7 7

4. Add the following fractions and reduce them into lowest term (if necessary).

(a) 1 + 3 (b) 3 + 1 (c) 2 + 1 (d) 5 + 7
2 8 8 4 3 4 6 12

(e) 4 + 3 + 1 (f) 3 3 + 8 2 + 7 (g) 2 1 + 3 1 + 6 1
5 10 2 5 5 10 6 12 2

5. Carry out the following subtractions.

(a) 2 – 1 (b) 7 − 5 (c) 11 – 5 (d) 5 – 4 1
3 4 8 12 12 6 4

(e) 9 1 – 8 (f) 6 2 – 4 1 (g) 4 5 – 1 2
2 3 2 6 3


6. Simplify the following.

(a) 5 – 1 + 1 1 (b) 2 +1–1 (c) 1+ 1 – 1
8 4 2 13 26 2 3

(d) 2 3 + 1 1 – 17 (e) 5 1 – 6 1 + 2 5
4 2 8 2 3 6

7. (a) Bikalpa drinks 5 litres of milk. Sankalpa drinks 3 litres of milk. Find the total
8 4

quantity of milk taken by both.

(b) The sides of a triangle are 3 1 cm, 4 2 cm and 5 2 cm. Find the perimeter of the
6 3 3

triangle.

(c) Weight of three bags are 2 1 kg, 3 1 kg and 5 1 kg respectively. What is the total
2 4 6

weight of three bags.

74 Oasis School Mathematics – 6

8. (a) The sum of two fractions is 7. If one of them is 2 3 , find the other.
(b) From a cloth of 4 cut away for making
5 3 m length, a piece of 3 3 m a dress.
4 8 is

How much cloth is left?

(c) Ansu ate 1 part of cake and Adhya ate 1 part of cake. By how much part Aadhya
4 3

ate more cake than Ansu?

Answers

1. (a) 6 (b) 9 (c) 6 (d) 5 1 2. (a) 1 (b) 5 (c) 3 1
11 13 2 2 13 3

(d) 2181 3. (a) 14 (b) 14 1 (c) 1 4. (a) 7 (b) 5
15 5 8 8

(c) 1112 (d) 1152 (e) 1 3 (f) 11 3
5 4

5. (a) 5 (b) 11 (c) 112 (d) 34 (e)112 (f) 3 2 (g) 3 61
12 24 3

6. (a) 1 7 (b) 2 (c) 1 1 (d) 2 3 (e) 2
8 6 8

7. (a) 138 l (b) 13 1 cm (c) 10 11 kg 8. (a) 4 1 (b) 2 3 (c) 112
2 12 4 8

4.4 Multiplication of Fractions

Multiplication of fractions

Multiplication of a fraction by a whole number:
We know that, multiplication is the repeated addition.


3 times

\ While multiplying a fraction by a whole number, multiply the numerator by

whole number and leave the denominator same.

Hence, 5× 1 = 5×1 = 5 , 6× 3 = 6×3 = 18
4 4 4 4 4 4

7× 1 = 7 ×1 = 7
3 3 3

Oasis School Mathematics – 6 75

Activity
Write the fraction of shaded by the product of whole number and a fraction as shown.

22 2 Fraction of shaded part = three 2
11 11 11 11

= 3× 2
11

Lets be clear with an example;

Let's multiply 1 and 3 I understand! We have to multiply
2 4 numerator with numerator and
denominator with denominator.
Multiply the numerators: 1 × 3 = 3

Multiply the denominators: 2 × 4 = 8

\ 1 × 3 = 3
2 4 8

Note:

If the fraction is a mixed number, change it into improper fraction then
multiply and reduce the product into its lowest term.

Example: 4 2 × 6 3 Alternatively
3 4
4 32134×1×76 3
2 3 = 4 2 63 31
Solution: 3 4 9 6
4 × 6 27 2 3
2
Convert into the 4 1
improper fraction
= 14 × 27 = 63 12 378 31
3 4 2 36
18
= 378 = 31 1 12
12 2 6
Multiply numerators and
= 31162 denominators separately.

= 31 1
2

76 Oasis School Mathematics – 6

Worked Out Examples

Example: 1

Multiply: 3 3 2
5 5 5
(a) 3 × 4 (b) ×

Solution: Solution:

3 × 4 3 3 × 2
5 5 5

= 3 × 23 = 3×2
5 5×5

= 3 × 23 = 6
5 25

= 69 = 13 4
5 5

Example: 2 1 2 12 .
3 5
Multiply and reduce into its lowest term: 4 × 2 × 2

Solution:

4 1 × 2 2 × 2 1 = 13 × 12 × 5
3 5 2 3 5 2

42

= 133 ×× 12 × 5 = 13 × 2 = 26
5×2

Example: 3 of (i) 2 of Rs 360 (ii) 3 of 1
3 5 2
Find the value

Solution:

Here, 2 of 360 = 2 × Rs 360 = Rs 2 × 360 = Rs 240
3 3 3×1

Again, 3 of 1 = 3 × 1 = 3×1 = 3 .
5 2 5 2 5×2 10

Exercise 4.4

1. Multiply:

(a) 1 × 8 (b) 1 × 15 (c) 185× 24
2 6

Oasis School Mathematics – 6 77

2. Multiply given two fractions:

(a) 1 × 2 (b) 5 × 1145 (c) 2 5 × 2 2
2 3 7 8 3

3. Find the value of :

(a) 1 of 24 (b) 2 of 48 (c) 2 of 120
2 3 5

(d) 3 of 50 (e) 3 of 420 (f) 3 of 480
5 7 8

4. Find the value of:

(a) 1 of 2 (b) 1 of 2 (c) 2 of 1 (d) 2 of 3 1
4 3 2 3 5 3 7 2

5. Carry out the following multiplication:

(a) 7 × 8 × 5 (b) 9 × 3 × 6
10 21 7 24 18 7

(c) 5 × 5 × 24 (d) 4 1 × 4 2 ×1
9 12 25 2 3

6. a) The weight of a cucumber is 3 kg. What is the weight of 4 such cucumber.
4

b) The weight of an iron rod is 3 3 kg. What is the weight of 16 such rods?
4

c) The cost of 1 kg rice is Rs. 40 1 , what is the cost of 3 kg rice?
2

7. a) The cost of 1 kg sugar is Rs. 20 3 . What is the cost of 5 1 kg of sugar?
4 2

b) A man has to complete 3 work. But he has completed 1 part of given work,
4 2
how much work has he completed?

c) A man has Rs. 500. He gave 2 part of money to his son and 1 part to his wife.
5 10

(i) How much money does he give to his son?

(ii) How much money does he give to his wife?

Answers

1. (a) 4 (b) 2 1 (c) 6 2 2. (a) 1 (b) 2 (c) 9 2
2 3 2 3 3

3. (a) 12 (b) 32 (c) 48 (d) 30 (e) 180 (f) 180

4. (a) 1 (b) 1 (c) 2 (d) 1 5. (a) 4 (b) 3 (c) 2 (d) 24.
6 3 15 21 56 9
3 1
6. (a) 3 4 (b) 60kg (c) Rs. 121 2 (d) (i) Rs. 200 (ii) Rs. 50

7. (a) Rs. 114 1 (b) 3 (c) (i) Rs. 200 (ii) Rs. 50
8 8

78 Oasis School Mathematics – 6

Reciprocal of a Fraction

Take a fraction 3 .
5

Interchange the position of number and denominator.

Then it becomes 5 .
3
3 5
Find their product: 5 × 3 =1

Hence, 5 is the reciprocal of 3 .
3 5

The reciprocal of a fraction is obtained by interchanging its numerator and
denominator.

Hence, when the product of two fractions is 23
1 then they are reciprocal to each other. • Reciprocal of 3 is 2

1
• Reciprocal of 5 is 5

4.5 Division of a whole number by a fraction

A man has 2 litres of milk. He has to divide 1 litre of milk in each bottle. How
4
many bottles will he need?

Activity
• Take 3 apples.

• Divide each apple into two equal halves.

11 1 1 11
22
22 2 2
Oasis School Mathematics – 6 79
• Count how many 1 apples are there?
2

• There are 6 half apples.

• Hence, 3 ÷ 1 = 6
2

Method of dividing whole number by fraction
1
Lets divide 2 by 4 .

i.e. 2÷ 1 • Write whole number into fraction.
4 • Multiply the dividend with the reciprocal of the

= 2 × 4 divisor.
1 1 • Reduce into lowest term if necessary.

=8

5 ÷ 1
3

I understand!
Whole number ÷ fractional number = whole

number × reciprocal of fraction

Division of a fraction by a whole number

5 bottles of milk is to be poured equally in 4 glasses.
6

How much milk will be there in each glass?
5
To get it, we have to divide 6 by 4.

i.e. 5 ÷4
6

= 5 × 1 = 5
6 4 24

\ Each glass contains 5 bottles of milk. 4 glasses
24

Steps
• Find the reciprocal of the divisor.

• Multiply the dividend with the reciprocal of the divisor.


80 Oasis School Mathematics – 6

Division of a fraction by a fraction:
3
A boy has 4 m coloured paper. He wants to cut it into some pieces each having the

length of 1 m. How many pieces can he make?
8
3 1
To get it we have to divide 4 m by 8 m.
÷
i.e. 3 1 Steps
4 8
• Find the reciprocal of the divisor.
= 3 × 8 • Multiply the dividend by the reciprocal of the divisor.
4 1

=6

I understand!
A fractional number ÷ a fractional
number = Dividend × the reciprocal

of the divisor.

Worked Out Examples

Example: 1

Simplify the following:

(a) 35 (b) 3 (c) 2 7 ÷ 5
15 125 9 9

Solution: (b) 32 (c) 2 7 ÷ 5
15= 3÷ 125 9 9
3
= 25 ÷ 5
( a) 155= 35 ÷15 9 9
= 35 × 115
= 215 = 3 × 15 = 45 = 25 × 9
2 2 9 5
= 22 12
= 5

Oasis School Mathematics – 6 81

Example: 2

If the cost of 4 kg rice is Rs. 135 1 what is the cost of 1 kg of rice?
Solution: 2

The cost of 4 kg rice = Rs. 135 1 = Rs. 271
2 2

271

The cost of 1 kg of rice = Rs. 2
4

= 271 ÷ 4
2

= 271 × 1 = 271 = 33 7
2 2 8 8
Exam ple: 3

266 2 kg rice is to be distributed equally among 50 family of earthquake
3
victim. How much does each family receive?

Solution: 2 800
3 3
Total quantify of rice = 266 kg rice = kg rice

This rice is to be distribute equally among 50 families.

800

Each family receives 3 kg rice

50

= 800 × 1 kg = 5 1 kg rice.
3 50 3

Exercise 4.5

1. Find the reciprocal of the following:

(a) 5 (b) 1 (c) 8 (d) 3 2
6 17 9

2. Carry out division.

(a) 1 ÷ 5 (b) 1 ÷ 8 (c) 2 ÷5
3 4 3

3. Carry out division.

(c) 7 ÷ 2 (c) 5 ÷ 2 (c) 4 ÷ 3
3 5 7

4. Divide:

(c) 2 ÷ 1 (c) 3 ÷ 1 (c) 4 ÷ 1 1
5 2 7 2 5 2

82 Oasis School Mathematics – 6

5. Simplify: 5

(a) 20 (b) 15 (c) 8 (d) 7
5 2 1 10
12 3
4

35 3
8 16 5
(e) 9 (f) 9 (g) 7

6. Simplify: 14 10

(a) 2 × 9 ÷ 3 (b) 1 × 2 ÷ 4
3 4 2 4 3 5

4 2 ÷ 1 1 × 2 2 (d) 3 3 × 2 2 ÷ 2 2
(c) 3 3 7 4 3 9

7. (a) From a ribbon of length 15 meter, piece of ribbon of length 3 m is to be cut. Find
the number of such ribbon. 5

(b) From 2 litre milk cups of capacity 1 litre are to be filled, how many such cups
3
can be filled?

(c) 1 3 cloths is required to stich a shirt. How many shirt piece can be stich from the
4

bundle of cloths of length 14m.

8. (a) i12n bottles of milk is to be poured equally in 3 glasses. How much milk will be there
each glass?

(b) 3 bucket of water is to be poured equally in 5 jugs. How much water will be there
4
in each jug?

9. (a) 12 m ribbon is to be cut into some pieces each having length 1 m, how many pieces
8
can be made?

(b) 15 3 litres of milk is to be distributed into a vessel of capacity 1 litre. Find the
4 4

number of such vessel.

Answers

1. (a) 1 (b) 6 (c) 17/8 (d) 9/29 (e) 3/5 2. (a) 1/15 (b) 1/32 (c) 2/15
5
1 1 (c) 913 4. (a) 54 6 8
3. (a) 10 2 (b) 12 2 (b) 7 (c) 15

5. (a) 48 (b) 2225412( c)(c8) 32 (d) 114 (e) 214 (f) 35 (g) 6
6. (a) 1 (b) 72 7
1
(d) 4 2

7. (a) 25 (b) 6 (c) 8 8. (a) 1 (b) 230 7.(a) 4 (b) 63
6

Oasis School Mathematics – 6 83

4.6 Simplification of Fractions

To simplify expressions involving fractions, we remember the ‘BODMAS’ rule

where order of the letters of this word are as follows:

Steps ______

• R emove bar , brackets ( ), { }, [ ] in 4th in order

order by simplifying all the operations within it.(B) 2nd in order

• Perform the operation involving 'of' (O) 1st in order
• Perform the operation involving division (D) 3rd in order

• Perform the operation involving multiplication (M)

• Perform the operation involving addition (A)

• Perform the operation involving subtraction (S)

Worked Out Examples

Example : 1

Simplify:

5 1 + 2 1 × 3 1 − 1
6 2 3 6

Solution:

5 1 + 2 1 × 3 1 − 1
6 2 3 6


= 31 + 5 × 10 − 1
6 2 3 6

= 31 + 50 − 1
6 6 6

= 31 + 50 − 1
6
80
= 6

= 13 2 = 13 1 .
6 3

84 Oasis School Mathematics – 6

Example : 2 Example : 3

Simplify: Simplify : 3 × 2 1 ÷ 5 of 3 1
4 1 2 7 5 2 6 5
3 5 − 2 10 ÷ 1 5 + 3 10 Solution:

Solution: 1 2 7 3 × 2 1 ÷ 5 of 16
4 10 5 10 5 2 6 5
3 5 − 2 ÷ 1 + 3
5 16 5 16 16 
 6 of 5 = 6 × 5 = 6 

= 19 − 21 ÷ 7 + 37 3 1 5 16
5 10 5 10 = 5 2 6 5
× 2 ÷ ×
19 21 5 37
= 5 − 10 × 7 + 10 = 3 × 2 1 ÷ 16
5 2 6
19 3 37 3 × 5 × 63 = 9
= 5 − 2 + 10 3 5 6 5 2 16 16
= 5 × 2 × 16

= 19 × 2 − 3 × 5 + 37 = 9
10 16

= 38 − 15 + 37 60 = 6.
10 = 10

Example: 4

{ ( )}Simplify:1 − 1 − 1 + 1 − 1 
6  2 3 4 8 

Solution:
{ ( )}Here,11 1 1 1 
6 −  2 − 3 + 4 − 8 

{ ( )} =1 −  1 − 1 + 2 −1 
6  2 3 8 

{ } = 1 −  1 − 1 + 1 
6  2 3 8 

{ } = 16 − 1 − 8+3 
 2 24 

= 1 - 1 − 11 
6  2 24 


= 1 − 12 − 11
6  24 

= 16 − 1 = 4 −1 = 3 = 1
24 24 24 8

Oasis School Mathematics – 6 85

Example : 5

Make mathematical expressions and simplify:

The sum of 3 and 5 is divided by the difference of 1 and 1 .
4 8 2 3
Solution:

( ) ( )= 3 + 5 ÷ 1 – 1
4 8 2 3

( ) ( )= 3 3
4 + 8 ÷ 1×3–1×2
6

( ) ( )= 6+8 5 ÷ 3–2 = 11 ÷ 1
6 8 6

= 11 ×6 = 11×3 = 33 =8 1
8 4 4 4

Exercise 4.6

1. Simplify :

(a) 3 + 1 × 2 (b) 2 × 2 1 − 1 (c) 20 × 5 ÷ 40
4 2 5 3 2 2 27 8 81

(d) 1 2 × 2 1 + 3 × 3 1 (e) 1 3 × 9 + 4 ÷ 32 (f) 1 2 × 2 1 –1125 ÷ 125
5 7 5 3 4 14 5 35 5 2

2. Simplify:

(a) 3 + 2 of 5 (b) 4 × 1 ÷ 3 of 2
7 5 7 5 2 2 5

3. Simplify:

( ) ( ) ( )(a)4×3 5 × 7 2 2 × 1 1 1 3 × 3
7 4 ÷ 7 10 (b) 7 3 × 4 5

( ) { ( )}(c)14 3 3 20 3 1 − 2 1 + 2 − 1
5 + 4 ÷ 16 ÷ 21 (d) 3 3 3 3

( ) { ( )}(e)31– 2 ÷ 3 + 1 3 – 3
3 3 10 2 5

4. Simplify:

{ ( )}(a)6 1 − 2 ×1 1 + 6 ÷1 1 { ( )}(b) 1 ÷ 1 ÷ 2 1 1 − 1
2 3 2 2 2 2 2 4

{ ( )}(c)3 + 5 1 + 2 − 1 ÷ 2 × 1 
4 2  4 5 5 15 10 

5. (a) Multiply the sum of 1 and 1 by 4 divided by 2.
2 3

(b) Multiply the difference of 2 1 and 1 1 by the sum of 2 and 1 .
3 2 3 3

86 Oasis School Mathematics – 6

(c) Divide the sum of 5 and 1 by the product of 2 and 1 1 .
8 4 3 8

6. (a) A man had Rs. 250. He spent 1 part of his money to buy a copy and 2 part to
5 5

buy a pen. How much money is left with him?

(b) Ashok has Rs 2000. He spends 1 past of his money to buy books and 2 part of his
5 5
money to buy cloths. Find how much money is left with him?

Answers

1. (a) 19 (b) 1 1 (c) 15 (d) 5 (e) 2 (f) -5 2. (a) 5 (b) 2
20 6 16 7 3
2
3. (a) 6 (b) 3 15 (c) 6 (d) 3 (e) 2 1 (f) 1 7 4. (a) 1 1 (b) 1150
7 3 9 2
5 1 1
(c) 2 5. (a) 5/12 (b) 6 (c) 1 6 (d) 6 5 6. (a) Rs. 100 (b) Rs. 800.

4.7 Decimals

Take a piece of paper. 0.1

Divide it into 10 equal parts.

Each part represents 1
10

In decimal, each part represents 0.1. or one tenths.

Again, Here two full 1
stripes of paper are 0.3
shaded and 3 parts out
of 10 of third stripe are
shaded.
∴ Shaded part represents 2.3.

2.3 Remember !

1 = 0.1 = one-tenths
10
Whole number part Decimal part
Decimal point 2 = 0.2 = two-tenths
10

Again, 3 = 0.3 = three-tenths
Here 1 whole is divided into 100 equal parts. 10

4 = 0.4 four-tenths etc.
10

Oasis School Mathematics – 6 87

Each part represents 1 = 0.01 1 or, 0.01
100 10

= one-hundredths .

2 parts represent 2 = 0.02
100

= two-hundredths

3 parts represent 1300= 0.03
= three-hundredths

Similarly,

1 = 0.001 (one-thousandth) Remember !
1000
2 = 0.02, = (two-hundredths)
2 100
1000 = 0.002 (two-thousandths)
3 = 0.03, = (three-hundredths)
100
92 = 0.092, 456 = 0.456 etc.
1000 1000 15 = 0.15 = (fifteen-hundredths), etc.
100
How to read the decimal numbers?

Write Read

2.5 Two point five
7.06 Seven point zero six
8.23 Eight point two three

Conversion of fraction into decimal

We already know that,

1 = 0.1, 1 = 0.2, 310= 0.3, etc. 5 20 0.4
10 20 20
×
1 = 0.01, 2 = 0.02, 65 = 0.65, etc.
100 100 100

1 = 0.001, 2 = 0.002, 432 = 0.432, etc.
1000 1000 1000

When the denominator is other number

while converting 2 into decimal, I understand !
5 I have to make denomina-

5 we have to divide 2 by 5 tor 10, 100, 100 ect.

2 = 0.4
5
2 2 2 4
Alternatively, 5 = 5 × 2 = 10 = 0.4

88 Oasis School Mathematics – 6

Conversion of Decimal into fraction

As we know that

0.1 = 1
10

0.8 = 180, 0.15 = 15
100

We get an idea that, to convert decimal into fraction,

• Count the number of decimal places in the decimal

• Ignore the decimal point and write all the digits of the
numerator of the fraction.

• Write as many zero after 1 in the denominator as these were deci-
mal places in the fraction. Reduce the fraction in to its lowest
there.

Now, 0.65 = 65 13 = 13 .
10020 20

Review Exercise

1. Express the following fractions into decimal numbers.

(a) 6 (b) 170 (c) 8 (d) 56 (e) 1
10 100 1000
100

(f) 13 (g) 3110 (h) 4 9 (i) 12 1
1000 100 1000

2. Change the denominator in 10 or 100 and convert given fractions into decimal.

(a) 1 (b) 2 (c) 2130 (d) 17
2 5 50

3. Convert the following decimals into fractions.

(a) 0.3 (b) 4.6 (c) 16.5 (d) 0.95
(e) 0.385 (f) 0.008 (g) 6.045

4. Expand the number and write in words.

(a) 0.8 (b) 0.09 (c) 4.75 (d) 12.374 (e) 6.321

Answers

1. (a) 0.6 (b) 0.7 (c) 0.08 (d) 0.56 (e) 0.001 (f) 0.013 (g) 3.1 (h) 4.09

(i) 12.003 2. (a) 0.5 (b) 0.4 (c) 0.65 (d) 0.34 3. (a) 3 (b) 46 (c) 165
10 10 10
95 385 8 6045
(d) 100 (e) 1000 (f) 1000 (g) 1000 4. Consult your teacher.

Oasis School Mathematics – 6 89

Multiplication of decimals by a whole numbers

Let's learn multiplication of a decimal number by a whole number.

Example: • Perform the multiplication as if we are multiplying
7.432 the two whole numbers.
× 6
44.592 • Put the decimal point in the product to get as many
decimal places in the multiplicand.

Example: As there are 3 digits after the
decimal point in multiplicand, I
Multiply: 9.034 × 6 have to put decimal after 3 digits
9.034 counting from the right in the
× 6
54.204 product.

Worked Out Examples

Example: 1 I have to shift
Multiply: 0.374 × 100 decimal point two

Solution: steps right.

0.374 × 100

= 37.4

Example: 2

A packet contains 16.5 kg of rice. How much rice is contained in 10 such
packets?
Solution:

Rice contained in one packet = 16.5 kg.
Rice contained in 10 packets= 16.5 kg × 10
= (16.5 × 10) kg
= 165kg

Example: 3

The cost of 1 m cloth is Rs. 75.35. Find the cost of 5.4 m of cloth.

Solution:

90 Oasis School Mathematics – 6

Cost of 1 m of cloth = Rs. 75.35
Cost of 5.4 m of cloth = Rs. (75.35 × 5.4)

75.35 ← 2 (Decimal places)
× 5.4 ← 1 (Decimal place)
← 3 (Decimal places)
30140
+376750

406.890


(Rs 75.35 × 5.4) = Rs 406.89
∴ The cost of 5.4m cloth is Rs 406.89

Exercise 4.7

1. Find the product in each of the following.

(a) 0.6 × 10 (b) 0.07 × 10 (c) 4.25 × 10

(e) 0.009 × 10 (f) 35.75 × 100 (g) 0.9 × 100

(i) 6.09 × 100 (j) 0.735 × 100 (k) 0.5 × 1000

(m) 2.05 × 1000 (n) 625.006 × 1000

2. Find the following products.

(a) 0.5 × 2 (b) 0.6 × 4 (c) 0.05 × 9 (d) 0.06 × 12
(h) 0.545 × 15
(e) 42.5 × 4 (f) 55.8 × 12 (g) 0.65 × 8

3. Find the following products.

(a) 0.8 × 0.5 (b) 0.35 × 0.7 (c) 5.55 × 0.6

(d) 5.957 × 0.8

4. (a) A litre of petrol costs Rs. 72.35. Find the cost of 10 litres of petrol.
(b) A bag contains 225.5 kg. of rice. How much rice is contained in 100 such bags?
(c) One kg of oranges costs Rs. 45.75. Find the cost of 9 kg. of oranges.
(d) The cost of one metre of cloth is Rs 55.50. Find the cost of 35 metres of cloth.
(e) The cost of 1 m of cloth is Rs. 85.25. Find the cost of 4.75 m of cloth.

5. Area of a rectangle = l × b and the area of a square is l². Using these formula
find the area of given rectangles and squares.

(a) A square having a side 3.5 cm

Oasis School Mathematics – 6 91

(b) A square having a side of 4.6cm
(c) Rectangle having l = 4.6 cm, b = 3.2cm
(d) A rectangle having l = 5.2cm, b = 2.5cm

Answers

1. (a) 6 (b) 0.7 (c) 42.5 (d) 0.09

(e) 3575 (f) 90 (g) 609 (h) 73.5

(i) 500 (j) 2050 (k) 625006

2. (a) 1 (b) 2.4 (c) 0.45 (d) 0.72

(e) 170 (f) 669.6 (g) 5.2 (h) 8.175

3. (a) 0.4 (b) 0.245 (c) 3.33 (d) 4.7656

4. (a) Rs. 723.5 (b) 22550 kg. (c) Rs. 411.75 (d) Rs. 1942.5

(e) Rs. 404.9375 5. (a) 12.25 cm2 (b) 21.16 cm2 (c) 14.72 cm2 (d) 13cm2

4.8 Division of Decimals

Division of decimals

Division of decimals by 10, 100 and 1000:

We already know that
2
2 ÷ 10 = 10 = 0.2 (two - tenths) Shift decimal point one step left.
Shift decimal point two steps left.
3 ÷ 10 = 3 = 0.3 (three- tenths)
10

5 ÷ 100 = 5 = 0.05 (five - hundredths)
100

12 ÷ 100 = 12 = 0.12 (twelve - hundredths)
100

Division of decimals by a whole number

Let's learn the division of a decimal number by a whole number,

Divide 12.645 by 5

Solution:

5 ) 12.645 ( 2.529

- 10 Division of decimal number by a whole num-
26 ber is same as the division of a whole number
- 25 by a whole number. Just we keep decimal after
14 dividing the whole number.

- 10

45 \ 12.645 ÷ 5 = 2.529
- 45

0

92 Oasis School Mathematics – 6

Division of decimal number by another decimal number

Let's learn the division of a decimal number by another decimal number.
Example :

Dividing 5.22 ÷ 0.9

Here, 5.22 = 5.22 × 10 = 52.2
0.9 0.9 × 10 9
• Make divisor a whole number multiplying both
Now, 9 ) 52.2 ( 5.8 numerator and denominator by appropriate
- 45 number.

72 • Divide as in the division of decimal number by
a whole number.
- 72

0
\ 52.2 ÷ 9 = 5.8

Worked Out Examples

Example: 1 Steps:

Divide 8.55 by 5. • D ivide first the whole number part.
Solution: Here are 8 ones. So, 8÷5 = 1, remainder 3.

• P ut the decimal point in the quotient. Also bring
Example: 2 down 5 and the new dividend in 35 and 35÷5 = 7

Divide 0.355 by 5. • Again, bring down 5 and 5÷5 = 1. Write 1 in the
Solution: hundredth place.

5 0 . 3 5 5 0.071 Since 3 < 5 put zero in the quotient.
–35
5
–5
0

∴ 0.355 ÷ 5 = 0.071

Oasis School Mathematics – 6 93

Example: 3

Divide 3.51 by 0.003

Solution:

3.51 ÷ 0.003 = 3.51 = 3.51×1000 = 3510
0.003 0.003×1000 5

Next divide 3510 by 3

∴ 3.51 ÷ 0.003 = 1170

Example: 4 • While dividing a number by 10, shift decimal
point one step left.
4.543 ÷ 10 = 0.4543

36.18 ÷ 10 = 3.618 • While dividing a number by 100, shift decimal

23.16 ÷ 100 = 0.2316 point two steps left.

5614.34 ÷ 1000 = 5.61434 • While dividing a number by 1000, shift decimal
point three steps left.

Exercise 4.8

1. Divide:

(a) 25 ÷ 10 (b) 3 ÷ 10 (c) 1.54 ÷ 10 (d) 32.4 ÷ 10 (e) 426.4 ÷ 10

2. Divide: (b) 43 ÷ 100 (c) 18.2 ÷ 100
(a) 3 ÷ 100

(d) 304.8 ÷ 100 (e) 637.42 ÷ 100

3. Divide: (b) 23 ÷ 1000 (c) 426 ÷ 1000
(a) 7 ÷ 1000
(d) 18.2 ÷ 1000 (e) 374.5 ÷ 1000

4. Divide: (b) 0.35 ÷ 7 (c) 1.25 ÷ 3
(a) 0.63 ÷ 3 (e) 0.637 ÷ 7
(d) 3.43 ÷ 7

5. Divide:

(a) 0.64 ÷ 0.4 (b) 0.85 ÷ 0.5 (c) 8.64 ÷ 0.24
(d) 0.222 ÷ 6 (e) 8.88 ÷ 0.12

94 Oasis School Mathematics – 6