Math 7

Example: 2

Using prime factorisation method, find the L.C.M. of 12, 18 and 24.

Solution:

Here, 12 2 18 2 24
2 6 39 2 12
3 26
2 3
3
Now,

12 = 2 × 2 × 3 L.C.M.
18 = 2 × 3 × 3 = 2 × 3 × 2 × 2 × 3 → Remaining in second
24 = 2 × 2 × 2 × 3
Common Common from Remaining in
L.C.M. = 2 × 3 × 2 × 2 × 3 = 72 from all first and third third

Example: 3

Using division method, find the L.C.M. of 20, 25, and 40.
Solution: Here,

2 20, 25, 40
2 10, 25, 20
5 5, 25, 10
1, 5, 2

∴ L.C.M. = 2 × 2 × 5 × 1 × 5 × 2 = 200

Example: 4

Find the least number which when divided by 27, 36 and 48 will leave the remainder
3 in each case.

Solution: Here,

2 27, 36, 48

2 27, 18, 24

3 27, 9, 12

3 9, 3, 4

3, 1, 4

L.C.M. of 27, 36 and 48 is 2 × 2 × 3 × 3 × 3 × 4 = 432

Since the remainder is 3,

required number = 432 + 3 = 435.

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Example: 5

Three bells ring at intervals of 15, 20 and 30 minutes respectively. If they all ring at 7
a.m., when will they ring together again?

Solution: Here, 2 15, 20, 30
5 15, 10, 15
3 3, 2, 3

1, 2, 1

Here, L.C.M. of 15, 20 and 30 is 2 × 5 × 3 × 1 × 2 × 1 = 60

i.e. Three bells ring together again after 60 minutes.

∴ They ring at 8 a.m.

Example: 6

Find the least number which when increased by 8 is divisible by 21, 28, and 45.

Solution: Here,

3 21, 28, 45
7 7, 28, 15
1, 4, 15

∴ L.C.M. of 21, 28, and 45 is 3 × 7 × 4 × 15 = 1260

∴ Required number is 1260 – 8 = 1252.

Exercise 11.4

1. List three common multiples of 8 and 12 and take the smallest one.

2. Find the L.C.M. of each of the following numbers by set of multiples method:

(a) 3, 6 (b) 4, 8, 12, (c) 6, 9 (d) 10, 15 (e) 12, 16, 24

3. Using prime factors method, find the L.C.M. of the following.

(a) 15, 25 (b) 12, 15 (c) 24, 30 (d) 16, 24, 32 (e) 20, 30, 40

(f) 48, 64, 36 (g) 36, 45, 54 (h) 15, 21, 30 (i) 15, 45, 60

4. Find the L.C.M. of the following using division method:

(a) 18, 27 (b) 21, 35 (c) 15, 18 (d) 24, 32 (e) 24, 32, 36

(f) 28, 42, 56 (g) 40, 48, 70 (h) 70, 90, 120 (i) 96, 120, 216

5. (a) Find the least number which is exactly divisible by 18 and 24.
(b) Find the smallest number which is exactly divisible by 18, 27 and 36.

146 Oasis School Mathematics – 7

(c) Find the least number which when divided by 24, 18 and 15 leaves no remainder.

(d) Find the smallest number which is exactly divisible by 24, 30 and 40.

6. (a) Find the smallest number which when divided by 5, 12 and 15 leaves the same
remainder 2.

(b) Find the least number which when divided by 12, 16 and 18 will leave in each
a remainder 5.

(c) Find the least number which being increased by 7 will be exactly divisible by
24, 36 and 48.

(d) Find the least number from which 6 may subtract so that the remainder will be
exactly divisible by 30, 45 and 90.

7. (a) Three bells ring at intervals of 10, 15 and 20 minutes respectively. If they all
ring together at 10 A.M. When will they ring together again?

(b) Find the least capacity of a drum which can be filled by the jugs of capacity
36l, 24l and 18l in exact number of times.

(c) Three measuring rods of length 25cm, 15cm and 30cm are used to measure the
benches in the class room. What is the length of the shortest bench which can
be measured by each of the rods in exact number of times?

1A.n sw24e,r 48, 72, Smallest common multiple = 24 2.(a) 6 (b) 24 (c) 18 (d) 30 (e) 48
(g) 540 (h) 210 (i) 180
3. (a) 75 (b) 60 (c) 120 (d) 96 (e) 120 (f) 576 (g) 1680 (h) 2520 (i) 4320
(c) 137 (d) 96
4. (a) 54 (b) 105 (c) 90 (d) 96 (e) 288 (f) 168

5. (a) 72 (b) 108 (c) 360 (d) 120 6. (a) 62 (b) 149

7. (a) 11 A.M. (b) 72 litres (c) 1.5 m

11.5 Relation between H.C.F. and L.C.M. of Two Numbers

Let us consider any two numbers 12 and 16.

Here, 2 12 2 16

26 2 8
3 2 4
2

12 = 2×2×3

16 = 2×2×2×2

∴ H.C.F. = 2 × 2 = 4

L.C.M = 2 × 2 × 2 × 2 × 3 = 48

Now, L.C.M. × H.C.F = 48 × 4 = 192

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Product of two numbers = 12 × 16 = 192

This shows that product of two numbers = H.C.F. × L.C.M

or, H.C.F = Product of two numbers .....................(i)
L.C.M

L.C.M = Product of two numbers .....................(ii)
H.C.F.

One unknown number = H.C.F × L.C.M ...................................(iii)
Given number

Worked Out Examples

Example: 1

The H.C.F. and L.C.M. of any two numbers are 720 and 18 respectively. If one of the
numbers is 90,find the other number.

Solution:

Here, H.C.F. = 18, L.C.M. = 720.

We have, One number = 90, Other number = ?

H.C.F × L.C.M
Other number =

∴ Given number
= 18 × 720

90
= 144

Other number = 144

Example: 2

The product of two numbers having H.C.F 15 is 2700. Find the L.C.M.
Solution:

Here, product of two numbers = 2700

We have, L.C.M. = Product of two numbers = 2700 = 180
15
H.C.F.

Exercise 11.5

1. (a) The H.C.F. and L.C.M. of any two numbers are 6 and 36 respectively. If one
number is 18 find the other number.

(b) The H.C.F. and L.C.M. of 'x' and 210 are 30 and 2310 respectively. Find the

148 Oasis School Mathematics – 7

value of x.

2. (a) If the H.C.F. of 45 and 60 is 15. Find their L.C.M.
(b) The product of two numbers having H.C.F. 14 is 1260. Find their L.C.M.

(c) The product of two numbers is 448 and their L.C.M. is 112. Find their H.C.F.

1A.n sw(ae)r 12  (b) 90 (c) 4
(b) 330 2. (a) 180

Self Practice Materials

I. H.C.F. of two co–prime numbers is 1, and L.C.M. of two co–prime numbers is their
product. Using this concept, complete the following table.

Numbers H.C.F. L.C.M.

3 and 5

4 and 9

3 and 13

13 and 17

2 and 5

II. Use the given number line to find the L.C.M. of given numbers.

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

(iii) 3 and 5 (iv) 3 and 6
(i) 2 and 5 (ii) 3 and 4

III. Fill in the blanks of given table and find the relation between H.C.F. and L.C.M.

Numbers H.C.F. L.C.M. H.C.F. × L.C.M. Product of two Numbers
6, 10 2 30 60 60
12, 30 ....... ....... ....... .......
18, 27 ....... ....... ....... .......
15, 45 ....... ....... ....... .......
Conclusion:

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11.6 Square and Square Roots

Let us 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 words, square number
is the product of two identical numbers. The number pattern of the square number
can be shown as follows.


1 4 9 16
Let us 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.

I. Method of finding square roots

(a) Prime factorisation method
The steps of finding square root of a perfect square number by prime factorization

method are:

150 Oasis School Mathematics – 7

For example:

Find the square root of 324. Steps

Here, • Express the given number as the product of its

2 324 prime factors.

2 162 • Make pairs of equal prime factors.
3 81
• Take one factor from each pair and multiply them.
3 27

39 The product is required square root of given number.
3

Here, 324 = 2 × 2 × 3 × 3 × 3 × 3

= 2² × 3² × 3²

∴ Square root of 324 = 2 × 3 × 3 = 18

(b) Division method

This method is useful to find the square root of larger numbers. This method is also
useful to find the square root of decimal number also.

Let us look at this example with steps of finding square root.

Here, Steps

1 3 68 64 192 • Starting from the place of unit make a number pair
+1 1 using bar mark just over each pair.

29 268 • Divide first pair(or a number) 3 by a number so as
+9 -261 to get a perfect square equal to or less than 3. i.e. 1.

382 764 • Square root of 1 is 1, so take 1 as the divisor and

+2 -764 quotient.

384 × • Write their product 1 just below 3 and subtract

from 3 which is 2.
∴ Square root = 192
• Add another 1 on the divisor, then divisor becomes

2. It is trial divisor.

• Bring down another pair 68 on the dividend, then
new dividend is 268.

• Write a number on divisor side i.e. 9 so that 29×9 is
just less than 268.

• Continue this process until the remainder is zero.

• Quotient is the square root of given number.

Division method is useful to find the square root of the decimal numbers.
For example:
Find the square root of 9.7344.

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For example:
Find the square root of 9.7344.

3 9. 73 4 4 3.12
+ 3 -9
61 73
1 - 61
622 1 2 4 4
2 - 1 2 4 4
624 ×
∴ Square root of 9.7344 is 3.12.

Note:
8×8 = 64 ⇒ 64 = 8
0.8 ×0.8 = 0.64 ⇒ 0.64 = 0.8
i.e. Square root of a number having 2 decimal place has one decimal place.
Square root of a number having 4 decimal place has two decimal place.
For. e.g., if the square root of 256 is 16, then square root of 2.56 is 1.6.
If the square root of 144 is 12 then square root of 0.0144 is 0.12.

Worked Out Examples

Example: 1 18 (b) 56 5 6
× 18 × 56
Find the square of: 1 4 4 3 3 6
(a) 18 + 2800
Solution: + 180 3 1 3 6

3 2 4

(a) Square of 18 = (18)² = 324 (b) Square of 56 = (56)² = 3136

Example: 2

Using prime factorisation method, find the square root of 400.

Solution: 2 400
Here, 2 200
400 = 2×2×2×2×5×5 2 100
= 2²×2²×5² 2 50
∴ Square root of 400 = 2×2×5 =20 5 25
5

152 Oasis School Mathematics – 7

Example: 3

Find the square root of 105625 using division method.
Solution:

3 10 56 25 325

+3 9

62 156
+2 -124

645 3225

+5 -3225
×
650

∴ Square root of 105625 is 325.

Example: 4 Alternative Method

Find the square root of 29.16. 29.16 = 2190106
Solution:
∴ 29.16 = 2190106
5 29 . 16 5.4
+5 -25 Here, 5 29 16 54
104 416
+4 -416 +5 -25
108 ×
104 416
∴ 29.16 = 5.4 +4 -416
108 ×
Example: 5
and 1 100 10
Simplify: 63 × 27 × 21
Solution: +1 1
63 × 27 × 21 20 00
+0 - 0
= 3²×7 × 3²×3 × 3×7 20 ×

∴ 2916 = 2916 = 54 = 5.4
100 100 10

= 3 7× 3 3 × 3×7

= 9 7×3×3×7 3 63 3 27 3 21
= 9 7²×3² 3 21 39 7
= 9×7×3
7 3

= 189

Note: a = a and a × b = ab
b b


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Example: 6

Which prime factor of the number 972 is pairless? By what least number 972 is to be
multiplied to make it a perfect square?

Solution: 2 972
Here, 2 486
9 7 2 = 2 × 2 × 3 × 3 × 3 × 3 × 3 3 243
= 2 ²× 3 ²× 3 ²× 3 3 81
∴ Prime factor 3 of 972 is pairless. 3 27

3

Hence, 972 is to be multiplied by 3 to make it a perfect square.

Example: 7

Find the least number that has to be subtracted from 1531 to make a perfect square.

Solution:

Here, 3 15 31 39

+3 -9
69 631
+9 -621
78 10

∴ 10 should be subtracted from 1531 to make perfect square.

Example: 8

Find the least number that must be added to 2906 so that sum will be a perfect square.
Solution:

Here, 5 29 06 53
+5 -25
103 4 0 6
+3 - 3 0 9

106 97

The whole number just greater than 53 is 54. Its square = (54)² − 2916.

∴ 2916 – 2906 = 10 must be added to 2906 so that the sum will be a perfect square.

154 Oasis School Mathematics – 7

Exercise 11.6

1. Find the square of the following numbers.

(a) 18 (b) 24 (c) 117 (d) 201
(g) 550
(e) 345 (f) 400

2. Find the square root of the following numbers using prime factorization method.

(a) 225 (b) 324 (c) 625 (d) 676 (e) 729

(f) 484 (g) 900 (h) 784 (i) 1296 (j) 1764

3. Simplify: (b) 26245 (c) 767296 (d) 5 × 20
(a) 144
(f) 18 × 50 × 16 (g) 4 8 × 3 250 × 2 80
(e) 2 3 ×5 27

(h) 15 × 8 × 30 (i) 300 × 75 × 400

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

(a) 5184 (b) 7744 (c) 16384

(d) 6561 (e) 8464 (f) 2809

(g) 13689 (h) 855625 (i) 169744

5. Find the square root of the following decimal numbers.

(a) 0.64 (b) 1.44 (c) 4.41

(d) 6.25 (e) 0.0729 (f) 0.0256

6. Find the square root of the following decimal numbers using division method.

(a) 1.5376 (b) 0.9216 (c) 1.0404

(d) 9.7344 (e) 152.5225 (f) 530.3809

7. (a) Some students are arranged in equal number of rows and columns. If the number of
rows is 29, find the total number of students.

(b) In a square garden, 24 flowers are planted in each row and column. Find the total
number of flowers in the garden.

(c) 35 students of a class collect as much money as their number to make a fund. Find
the amount of money collected altogether.

8. (a) A garden has as many rows of flowers as there are flowers in each row. If there are
1024 flowers altogether, find the number of flowers in each row.

(b) Rs. 900 is distributed among some students so that each of them gets equal sum.
If the money received by each student is equal to the number of students, find the
number of students.

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(c) 1089 marbles are arranged in square shape. Find the number of marbles in each row.

9. (a) Find which prime factor of 1250 is unpaired. What least number is to be multiplied
to 1250 to make it a perfect square?

(b) Find the smallest number which should be multiplied to 343 to make it a perfect
square?

(c) Find the least number which is to be multiplied to 726 to make its product a perfect
square.

10. (a) Find the smallest number by which 162 is to be divided so that the quotient is a
perfect square.

(b) Find the least number by which 363 is divided to make it a perfect square.

11. (a) Find the smallest number that has to be subtracted from 735 to make it a perfect
square.

(b) Which least number should be subtracted from 2320 to make it a perfect square?

(c) Find the least number that should be added to 310 to make the sum perfect square.

(d) Which least number should be added to 700 to make it a perfect square?

Answer (b) 576 
1. (a) 324 (c) 13,689 (d) 40,401 (e) 1,19,025 (f) 1,60,000

(g) 30,2500 2. (a) 15 (b) 18 (c) 25 (d) 26 (e) 27 (f) 22
(g) 30 (h) 28 (i) 36 (j) 42 3. (a) 12 (b) 185 26
(d) 10 (e) 90 (f) 120 (g) 9600 (h) 60 (i) 3000 (c) 27



4. (a) 72 (b) 88 (c) 128 (d) 81 (e) 92 (f) 53 (g) 117

(h) 925 (i) 412 5. (a) 0.8 (b) 1.2

(c) 2.1 (d) 2.5 (e) 0.27 (f) 0.16 6. (a) 1.24

(b) 0.96 (c) 1.02 (d) 3.12 (e) 12.35 (f) 23.03

7. (a) 841 (b) 576 (c) Rs. 1225 8. (a) 32 (b) 30 (c) 33

9. (a) 2, 2 (b) 7 (c) 6 10. (a) 2 (b) 3

11. (a) 6 (b) 16 (c) 14 (d) 29

Quick way of squaring a number ending with 5.

• Take a number ending with 5. Say 35.
• Add 1 on the first digit of the number. 3 + 1 = 4
• Multiply this number with the first digit of the given number. 4 × 3 = 12.
• Write 25 after this number. 1225.
∴ 35 × 35 = 1225.
Try this with other numbers also.

156 Oasis School Mathematics – 7

11.7 Cube and Cube Root

Let's look at the following examples:
1 = 1×1×1 = 13
⇒ 1 is cube of 1.

⇒ 1 is the cube root of 1.

8 = 2 × 2 × 2 = 2³ ⇒ 8 is the cube of 2.
⇒ 2 is the cube root of 8.
27 = 3×3×3 = 3³ ⇒ 27 is the cube of 3.
⇒ 3 is the cube root of 27.

If a number is multiplied by itself three times; the resulting number is the cube of
the given number.

Cube root of the given number is a number which is multiplied three times to
produce the given number.

Cube root of a given number is denoted by the symbol 3 or, ( )1/3
Thus, 23 = 8 ⇒ 3 8 = 2

3³ = 27 ⇒ 3 27 =3
4³ = 64 ⇒ 3 64 = 4 and so on.

Method of finding cube roots
Generally we use the prime factor method to find the cube root of the number.

Let's be clear with an example, find the cube root of 216.

Here, 2 216 Steps:
2 108 • Express the given number as the
2 54
3 27 product of its prime factors.
3 9 • Make a group of three equal factors.
3 • Take one factor from each group

and multiply them.
216 = (2 × 2 × 2) × (3 × 3 × 3) = 23 × 33 • The resulting number is the cube

∴ Cube roof of 216 = 3 216 = 3 23×33 root of the given number.
= 2 × 3 = 6.

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Worked Out Examples

Example: 1

Find the cube of : (a) 8 (b) 12
Solution:

Cube of 8 = (8)³ = 8 × 8 × 8 = 512

Cube of 12 = (12)³ = 12 × 12 × 12 = 1728

Example: 2

Find the cube root of 1728.

Solution: 2 1728 Here, 1728 = 2×2×2×2×2×2×3×3×3
Here, 2 864
2 432 = 2³ × 2³ × 3³
2 216
Example: 3 2 108 ∴ Cube root of 1728 = 2 × 2 × 3 = 12
2 54
3 27
3 9
3

What smallest number should be multiplied to 1372 to make it a perfect cube?

Solution: Here, 2 1372
Here, 1372 = 2×2×7×7×7 2 686
= 2² × 7³ 7 343
To make it a perfect cube there should be 2³. 7 49
7

∴ Given number should be multiplied by 2 to make it a perfect cube.

Example: 4

The volume of a cubical box is 512 cm³. Find the length of each side of the cube.

Solution: 2 512
2 256
Volume of the cubical box = 512 cm³ 2 128
2 64
l × l × l = 512 cm³ [ ∵ volume of the cube = l³ ] 2 32
2 16
l³ = 512 cm³ 2 8
2 4
l = 3 512 cm³
2

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Now, ∴ l = 3 2×2×2×2×2×2×2×2×2 cm³

= 3 2³×2³×2³ cm³

= 2 × 2 × 2 cm

= 8 cm

∴ Length of each side = 8 cm.

Exercise 11.7

1. Find the cube of the following numbers.

(a) 6 (b) 7 (c) 12 (d) 20 (e) 30 (f) 35

2. Find the cube root of the following numbers.

(a) 64 (b) 125 (c) 343 (d) 1728 (e) 5832

3. (a) Find the cube root of square of 8.

(b) Find the cube of square root of 25.

(c) Find the product of square root of 36 and cube root of 125.

4. (a) What smallest number should be multiplied to 54 to make it a perfect cube?
(b) What least number should be multiplied to 576 to make it a perfect cube?

5. (a) By which least number 54 should be divided to make the quotient a perfect cube?
(b) By which smallest number 256 be divided to make it a perfect cube?

6. (a) A cubical cistern contains 3375 m³ of water. What is the length of the edge of the
cistern?

(b) A cubical water tank is 25 cm long. What is its capacity?

(c) Find the length of a cubical tank containing 343 cm³ of water.



Answer (b) 343 (c) 1728 (d) 8000 (e) 27000 (f) 42875
1. (a) 216 (b) 5 (c) 7 (d) 12 (e) 18
2. (a) 4 (b) 125 (c) 30
3. (a) 4 (b) 3
4. (a) 4 (b) 4 6. (a) 15 m (b) 15625 cm³ (c) 7 cm
5. (a) 2

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Unit Operations on
12 Integers

12.1 Integers

Introduction

We have already learnt about natural numbers. Let us consider the set of natural
numbers.

N = {1, 2, 3, 4, 5, ......... }

Let's be clear with an example.

3 is a natural number, 5 is a natural number.

3 + 5= 8 is a natural number.

3 × 5 = 15 is also a natural number.

Natural numbers are closed under addition and multiplication. i.e. sum and prod-
uct of two natural numbers is also a natural number.

Let's see in the case of subtraction.

(3 – 5) = – 2 which is not a natural number.

(6 – 6) = 0 which is not natural number.

i.e. natural numbers may not always be closed under subtraction. So. in order to
make the operation of subtraction meaningful, negative numbers and zero are
introduced.

The set of natural numbers together with their negatives including zero are called
integers.

The set of integers is denoted by Z.

∴ Z = { ..........., –4, –3, –2, –1, 0, 1, 2, 3, 4, ......... }

Z+ = {1, 2, 3, 4, ........... }

= a set of positive integers.

Z– = {–1, –2, –3, –4,........... }

= a set of negative integers.

160 Oasis School Mathematics – 7

Integers on the number line: Integers are closed under addition, subtraction and
multiplication.



-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 678

Negative integers Zero Positive integers



While showing the integers on the number line.
• Positive integers are shown on the right side of zero and 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.
• The arrow marks at the both ends of a number line represent that the inte-

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

Note: –3 < – 2 ⇒ 3 > 2.

Remember !

• In the number line of integers, every number on the right is greater than
all the numbers on its left and vice versa.

• The whole number 0 is neither a positive integer nor a negative integer, as
the negative of zero is zero.

Absolute value of integers

In the given figure, B, C, D are the points which are 1 unit, 2 units, 3 units right from
the point A.

G F E A BC D E

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

i.e. AB = 1 unit.

AC = 2 units.

AD = 3 units.

Again, E, F, G are the points which are 1 unit, 2 units, 3 units left from the point A.

i.e. AE = –1 unit.

AF = –2 units.

AG = –3 units .

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Here, AB = 1 unit AE = –1 unit they are equal in magnitude only the direction is
opposite.

∴ |1| = |–1| = 1 units.

AC = 2 units. AF = – 2 units. |2| = |–2| = 2 units.

AD = 3 units: AG = –3 units. |3| = |–3| = 3 units.

We can say that absolute value of AB and AE, AC and AF, AD and AG are equal.

Thus the absolute (modulus) of an integer is its numerical value which is always
positive.

Thus, the absolute value of an integer is the distance of that integer from 0 irre-
spective of the direction.

∴ |3| = 3

|–3| = 3

Remember !

• The smallest positive integer is 1.
• The greatest negative integer is –1.
• There is no greatest positive integer and smallest negative integer.
• The integer on the right is always greater than the one on the left.
• Absolute value of any integer is positive. It represents the distance of the

integer from origin.

Worked Out Examples

Example: 1

Write all the integers between –3 and 5.
Solution:

-4 -3 -2 -1 0 1 2 34 5

Here, from the number line, the integers between –3 and 5 are –2, –1, 0, 1, 2, 3, 4.

Example: 2

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

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

From the above number line the integer which is 4 units left from 3 is –1.

162 Oasis School Mathematics – 7

It can also be obtained

3 – 4 = –1 [4 units left means – 4]

Example: 3

A city bus went 30 km east and another bus went 40 km west from the same city. Find
the distance between the two buses.

Solution: 30 km east
OB
A 40 km west

Distance between two buses
= (30 + 40) km

= 70 km

Exercise 12.1

1. Using a number line, answer the following questions:

(a) On which side from 0, a number less than zero lies?

(b) On which side from 0, a number greater than zero lies?

(c) How many integers are there between –3 and 2?

(d) How many integers are there between – 2 and 1?

2. Write all the integers between.

(a) –2 and 5 (b) 0 and 3. (c) –6 and –3 (d) –3 and 4.

3. Using a number line, answer the following.

(a) Write an integer which is 3 units right from –2.

(b) Write an integer which is 4 units left from 1.

(c) Write an integer which is 5 units right from –3.

(d) Write an integer which is 10 units left from –1.

(e) Write the integers which are 3 units right and 5 units left from –1.

(f) Write the integers which are 4 units right and 3 units left from –6.

4. Write the opposite of each of the following.

(a) + 7 (b) –2 (c) +10 (d) Moving 20 m left.

(e) Ascending 30 m up. (f) Rising of temperature by 15°C.

5. (a) Insert the appropriate sign < or > in the following blanks.

Oasis School Mathematics – 7 163

(a) –5 ..... –2 (b) 2 ..... 3 (c) –3 ...... –6

(d) 2 ..... – 2 (e) –1 ..... 3
6. Arrange the given integers in ascending order.

(a) –5, –6, 3, 0, 2, –2 (b) –3, –2, –4, 1, 0, 2
7. Arrange the given integers in descending order.

(a) 3, 4, –2, –3, 0, 1, 2 (b) –6, –5, 6, 5, 2, –2
8. Find the absolute value of each of the following integers.

(a) +2 (b) –5 (c) –7 (d) +5
9. Simplify:

(a) |–7| + |2| (b) |–9| + |5| – |–7| (c) |–2| |3| – |–3||–2|

(d) (|5| – |–3|) × |–6| (e) (|–3| + |2|) (|–4| – |–2|)

10. (a) Point A is 10 units left form C and point B is 5 units right from C. Find the distance
between A and B.

(b) From a village, a man ascends 15 km north and from the same village another man
descends 10 km south. What is the vertical distance between them?

Answer  (b) 1, 2 (c) -5, -4
1. Consult your teacher. 2. (a) -1, 0, 1, 2, 3, 4

(d) -2, -1, 0, 1, 2, 3 3. Consult your teacher. 4. (a) -7 (b) +2

(c) -10 (d) 20 m right (e) descending 30 m down. (f) falling of temperature by 15°C.

5. Consult your teacher. 6. (a) -6 < -5 <-2 < 0 < 2 < 3 (b) -4 < -3 < -2 < 0 < 1< 2

7. (a) 4 > 3 > 2 > 1 > 0 > -2 > -3 (b) 6 > 5 > 2 > -2 > -5 > -6

8. (a) 2 (b) 5 (c) 7 (d) 5 9. (a) 9 (b) 7 (c) 0

(d) 12 (e) 10 10. (a) 15 units (b) 25 km

12.2 Operation on Integers

Integers are closed under addition, subtraction, multiplication and division.
i.e. (i) an integer + an integer = an integer.
an integer – an integer = an integer.
a n integer × an integer = an integer.

I. Addition of integers:
Depending upon the sign of integers there are three cases on the addition of integers.

164 Oasis School Mathematics – 7

a) When both integers are positive:

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

5 steps right 3 steps right

-5 -4 -3 -2 -1 0 1 23 45 67 89
+8

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

∴ (+5) + (+3) = 8
b) When both integers are negative:
For example:(–3) + (–4)

4 steps left 3 steps left 0 1 2 34 5

-7 -6 -5 -4 -3 -2 -1
–7

• Starting from 0, move steps left.
• Again, move another 4 steps left.
• Final point is –7.

∴ (–3) + (–4) = –7
c) When two integers are of opposite sign:
For example:(–4) + (+6)

4 units left 6 units right

-7 -6 -5 -4 -3 -2 -1 01 2 34 5 67
+2

• Starting from 0 move 4 steps left.
• Again from that point move 6 steps right.
• Final point is +2.


∴ (–4) + (6) = +2

Again, (+7) + (–8)

∴ (+7) + (–8) = –1.

Oasis School Mathematics – 7 165

7 units right

8 units left

-4 -3 -2 -1 01 2 34 56 7

• Starting form 0 move 7 steps right.
• Again from that point move 8 steps left.
• Final point is –1.

Additive inverse of an integer

Look at the given number line.

-3 -2 -1 0 1 2 3

–1 is 1 unit left from 0. i.e. –1 and 1 are additive inverse of each other.
1 is 1 unit right form 0. –2 and 2 are additive inverse of each other.
–2 is 2 units left from 0. ∴ –2 is the additive inverse of 2.
2 is 2 units right from 0. 2 is the additive inverse of –2.
Here,
–1 + 1 = 0, –2 + 2 = 0

– 3 + 3 = 0 and so on,

Hence, the two integers are called additive inverse of each other if their sum is zero.

Note: • For every positive integer, there exists a negative integer at the same
distance from zero in the opposite direction. These two integers are
called the opposites of each other. These two integers are also called
additive inverse of each other.

• Sum of an integer and its additive inverse is zero.

II. Subtraction of integers

Look at an example:

(+5) – (+3) = +2 and (+5) + (–3) = +2.

In this case, first operation is subtraction of integers and the second is addition of
integers. Result of both operation is the same.

It means that subtraction of +3 from +5 is same as addition of +5 with (–3).

i.e. subtraction of an integer is the addition of its additive inverse.

Therefore, subtraction is the inverse operation of addition.

Let's look at an example: (+5) – (+8)

Here, (+5) – (+8) = (+5) + (–8) [∵ Subtraction is the addition of its inverse]

166 Oasis School Mathematics – 7

-5 -4 -3 -2 -1 0 1 23 4 56 7 8
-3

∴ (+5) – (+8) = –3
Again, look at another example:
–6 – (+4)
Here, –6 – (+4) means –6 + (–4)

-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-10

∴ –6 – (+4) = – 10

Properties of addition of integers

(i) Closure property:
The sum of two integers is also an integer.
∴ If a ∈ Z, b ∈ Z then a + b ∈ Z.
For example:
(+5) + (+4) = + 9 (an integer)
(–6) + (+4) = –2 (an integer)
(ii) Commutative property:
The sum of two integers remain the same, even if their places are interchanged.
i.e. if a ∈ Z, b ∈ Z then a + b = b + a.
For example:
(–3) + (+7) = (+7) + (–3) = +4.
(–4) + (–3) = (–3) + (–4) = –7
(iii) Associative property:
The sum of three integers remains unchanged if the order in which they are grouped

is altered.
i.e. if a, b, c ∈ Z then,
a + (b + c) = (a + b) + c
For example,
(−3) + (0 + 7) = (−3 + 0) + (+7)
or, (−3) + (+ 7) = (−3) + (+7)

∴ (+ 4) = (+4)

(iv) Existence of additive inverse:

Oasis School Mathematics – 7 167

For every integer, there exists an integer such that their sum is zero. Here each inte-
ger is said to be additive inverse of each other.

i.e. if a ∈ Z then there exists –a ∈ Z such that a + (–a) = (–a) + a = 0
and – a is additive inverse of 'a'

a is additive inverse of '–a'.

For example,

additive inverse of (+7) is (–7) and additive inverse of (-10) is (+10)

(v) Identity law:

For all integers there exists an integer zero such that sum of an integer and zero is
given integer itself.

If a ∈ Z then a + 0 = 0 + a = a.

Note: • While adding two positive integers, add the integers and the sum
holds positive sign. e.g. (+5) + (+7) = +12

• While adding two negative integers, add the integers, the sum holds
negative sign e.g. (–3) + (–2) = –5

• While adding a positive integer with a negative integer, subtract
smaller integer from the larger. The difference holds the sign of the
larger integer.

e.g. (–3) + (+2) = –1, (–7) + (+10) = 3.


Exercise 12.2

1. Answer the following questions.
(a) What is the additive inverse of –3?

(b) What is the sum of an integer and its additive inverse?

(c) If 'x' and 'y' are any two integers write the commutative property of the integers
taking x and y.

(d) If 'x', 'y' and 'z' are any three integers. Write the associative property of the i n -
tegers taking 'x', 'y' and 'z'.

2. Add the following integers by using number lines.

(a) (+3) + (+4) (b) (+7) + (–4) (c) (–6) + (+4)

(d) (–6) + (–3) (e) (+2) + (+4) (f) (–5) + (–3)

3. Subtract the following integers by using number lines.

(a) (+5) – (+3) (b) (+8) – (–4) (c) (–7) – (–3)
(d) (–9) – (+6)
(e) (–2) – (–3) (f) (–5) – (+2)

168 Oasis School Mathematics – 7

4. Verify the given commutative law:

(a) (+7) + (–3) = (–3) + (+7) (b) (+2) + (–5) = (–5) + (+2)

5. Verify the given associative law:

(a) [(+2) + (+5)] + (–3) = (+2) + [(+5) + (–3)].

(b) [(+9) + (–2)] + (+6) = (+9) + [(–2) + (+6)]

6. Simplify:

(a) (+4) + (–3) – (–6) (b) (+8) + (–6) + (+5) (c) (+3) – (+5) – (–8)

(d) (+9) – (–8) + (+6) – (–7) (e) –(–10) + (+4) – (–6) + (–8)

7. If a = (+8), b = (–6), c = (–12), find the value of: (d) a – b – c
(a) a – b + c (b) a + b + c (c) a + b – c

Answer (c) -2 (d) -9 (e) +6
1. Consult your teacher: 2. (a) +7 (b) +3 (e) +12

(f) -8 3. (a) +2 (b) +12 (c) -4

(d) -15 (e) +1 (f) -7 4. Consult your teacher.

5. Consult your teacher. 6. (a) +7 (b) +7 (c) +6 (d) +30

7. (a) +2 (b) -10 (c) +14 (d) +26

12.3 Multiplication and Division of Integers

(a) Multiplication of integers

Multiplication of integers can be done by using number lines:

Multiplication of two positive integers(+2) × (+3)

3 units 3 units

–5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
6

Here, (+2) × (+3) Starting from 0, jump
= 2 times of +3 = 3 + 3 3 steps right twice.
Therefore: (+2) × (+3) = +6

I. Multiplication of two integers having different signs.

(+3) × (–4)

Here, (+3) × (–4) = 3 times of (–4)

= (–4) + (–4) + (–4)

∴ (+3) × (–4) = –12.

Oasis School Mathematics – 7 169

• Starting from 0, move 4 steps left.
• Repeat the same process 3 times.

• Final point is –12.

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


∴ (–2) × (+4) = – 8

II. 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 nega-

tive) then their product is a negative integer.
(+ve integer) × (–ve integer) = (–ve integer)
Note: The product of zero and an integer is zero.

170 Oasis School Mathematics – 7

(b) Division of integers
Division is the inverse operation of multiplication. So the sign rules of division are

same as the sign rule 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)

Here, –12 = (–3) + (–3) + (–3) + (–3)

–12 = 4 × (–3)
∴ (–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

Oasis School Mathematics – 7 171

Exercise 12.3

1. Using number line, find the product of:

(a) (+3) × (+2) (b) (+3) × (–2) (c) (–2) × (+4) (d) (–3) × (–2)
(e) (+2) × (+5) (f) (–5) × (+3) (g) (–4) × (–3) (h) (+4) × (–3)

2. Using number line, find the quotient of:

(a) (+8) ÷ (+2) (b) (+6) ÷ (–3) (c) (–10) ÷ (–2)

(d) (–8) ÷ (+4) (e) (–12) ÷ (+4) (f) (+9) ÷ (–3)

(g) (–16) ÷ (–2) (h) (+12) ÷ (+3)

3. Simplify:

(a) (–15) × (+10) (b) (–8) × (+10) (c) (+9) × ( –6) (d) (+15) × (+7)

(e) (–30) ÷ (+6) (f) (+20) ÷ (–2) (g) (–35) ÷ (–5) (h) (+30) ÷ (+5)

4. Simplify:

(a) (+5) × (–3) × (–2) (b) (+6) × (–2) × (–5)

(c) (+10) × (+4) × (+2) (d) (–12) × (–2) × (–3)

(e) (–9) × (+3) × (–2) (f) (–7) × (–4) × (+5)

(g) (–8) × (+2) × (+10) (h) (+5) × (+5) × (+4) × (–7)

5. (a) The product of two integers is (–27). If one integer is +9, find the other.

(b) The product of two integers is (+20). If one integer is –4, find the other.

(c) What integer should be multiplied to +4 to make it + 24?

(d) What integer should be multiplied to (–6) to make it + 18?



Answer (b) -6 (c) -8 (d) +6 (e) +10 (f) -15
1. (a) +6, (h) -12 2. (a) +4 (b) -2 (c) +5 (d) -2

(g) +12

(e) -3 (f) -3 (g) +8 (h) +4 3. (a) -150 (b) -80 (c) -54
(d) +105 (e) -5 (f) -10 (g) +7 (h) +6
(f) +140
4. (a) + 30 (b) +60 (c) +80 (d) -72 (e) +54 (d) -3

(g) -160 (h) -700 5. (a) -3 (b) -5 (c) +6

172 Oasis School Mathematics – 7

12.4 Order of Operations for Simplification

Some mathematical problems contain mixed operations addition, subtraction, mul-
tiplication, division and different brackets. When more than one operation is in-
volved, we have to use the following rules for the simplification of given expression
of integers.

This rule is known as 'BODMAS'

Steps:
Remove bar — , brackets ( ), { }, [ ] in order. By simplifying all the operations
within it or by simplifying, all the operations within it. (B)
Perform the operation involving 'of' (O)
Perform the operation involving division (D)
Perform the operation involving multiplication (M)
Perform the operation involving addition (A)
Perform the operation involving subtraction (S)

Worked Out Examples Example: 2

Example: 1 Simplify: [20 + {8 + (5–2)}] – 25
Solution:
Simplify: 18–4×5+6÷2 Here,
Solution: [20 + {8 + (5 – 2)}] – 25
Here, 18 – 4 × 5 + 6 ÷ 2 = [20 + {8 + 3}] – 25 [Operation of ( )]
= 18 – 4 × 5 + 3 [Operation of division] = [20 + 11] – 25 [Operation of { } ]
= 18 – 20 + 3 [Operation of multiplication] = 31 – 25 [Operation of [ ] ]
= 21 – 20 [Operation of addition] = 6 [Operation of subtraction]
= 1 [Operation of subtraction]
Example: 4
Example: 3
Simplify: 10–{10 – (10 ÷ 2×4+2)}
Simplify: 60 ÷ 3 [150 ÷ 2 {6 + 3 (17 – 14)}] Solution: Here,
Solution: Here,
60 ÷ 3 [150 ÷ 2 {6 + 3 (17 – 14)}] 10 – {10 – (10 ÷2 × 4 + 2)}
= 60 ÷ 3 [150 ÷ 2 {6 + 3 of 3}] = 10 – {10 – (10 ÷ 8+2) }
= 60 ÷ 3 [150 ÷ 2 {6 + 9}] = 10 – { 10 – (10 ÷ 10) }
= 60 ÷ 3 [150 ÷ 2 of 15] = 10 – {10 – 1}
= 60 ÷ 3 [ 150 ÷ 30] = 10 – {10–1}
= 60 ÷ 3 of 5 = 10 – 9
= 60 ÷ 15 =1

=4

Oasis School Mathematics – 7 173

Example: 5

Convert the given statement into mathematical expression and simplify:
5 times the difference of 18 and 10 is divided by the product of 4 and 2.
Solution:
Here, 5 times the difference of 18 and 10.

= 5 (18 – 10)

Again, product of 4 and 2 = 4 × 2

∴Required Mathematical expression,

= {5 (18 – 10)} ÷ (4 × 2)

= { 5 × 8 } ÷ 8

= 40 ÷ 8 = 5

Exercise 12.4

1. Simplify the following:

(a) 15 – 5 + 14 (b) 3 × 10 + 6 × 3 – 8 × 5

(c) 20 ÷ 5 – 3 × 5 + 30 ÷ 6 (d) 3 × 8 ÷ 4 – 2 × 6 ÷ 2

(e) 20 + 3 × 8 ÷ 4 – 12 (f) 115 + 51 ÷ 17 × 5 – 80

(g) 15 + 2 × (–3) – 10 ÷ (–5) (h) (–2) × 6 – (–12) ÷ 4 + 16 ÷ 4

(i) (5 – 11) × (–2) + 32 ÷ 8 × 2



2. Simplify:

(a) 5 + [7 – {4 + (5–2)}] (b) 10 – [20 ÷ {20 – 5(4–1)}]

(c) 17 – {9 + 2 (3 – 1)} (d) 9 – [7 + {4 – (5 – 2)}]

(e) 14 ÷ [ 3 + 16 ÷ {2 + 8 ÷ (6 – 2)}] (f) 5 [{84 – (8 × 8)} ÷ 4] – 20

(g) 32 – [90 – 5 {9 – (14 – 12)}] ÷ 5 (h) 3 {24 + 8 ÷ (6 – 4)} – 60

(i) 56 ÷ 8 – { (51 + 7 × 2) – (5 × 11 – 5)} (j) 39 – 7 {28 ÷ (17 – 10) + 1 }

(k) 10 + [3 × { 6 + (16 × 2 + 7)} ÷ 5] (l) 28 ÷ [5 + 16 ÷ { 4 + 8 ÷ (4 – )}]

3. Simplify:
(a) 15 + (20 – 8 + 4) (b) 60 + (16 × 4 – 12 - 6 + 9 × 4)
(c) 18 + [ 16 – {4 + (3 + 6 - 4 )}] (d) 40 – [28 – {24 – (20 – 8 - 4 )}]
(e) 15 [30 ÷ {7 × 4 – 8 × 2 – (6 – 3 + 1 )} ]
(f) 20 [160 ÷ {6 × 5 – 15 ÷ 3 – (8 – 7 - 4 )}]

174 Oasis School Mathematics – 7

4. Convert the following statement into mathematical expression and simplify:
(a) 50 is added to the product of 5 and 3.
(b) 13 is subtracted from 4 times the difference between 9 and 5.
(c) 8 times the difference of 8 and 3 is divided by the product of 2 and 5.
(d) 6 times the sum of 7 and 3 is divided by the sum of 8 and 2.
(e) The sum of 90 and 30 is divided by 12 and the quotient is again divided by the product

of 2 and 5.

Answer (b) 8 (c) -6 (d) 0 (e) 14 (f) 50 (g) 11
(i) 20 2. (a) 5 (b) 6 (c) 4 (d) 1 (e) 2
1. (a) 24 (g) 21 (h) 24 (i) -8 (j) 4 (k) 37 (l) 4
(h) - 5 (b) 64 (c) 25 (d) 20 (e) 45 (f) 160
(f) 5 (b) 3 (c) 4 (d) 6 (e) 1
3. a) 23
4. (a) 65

Self Practice Material

Start with the number given and perform the operations in order shown by arrows.
(a)

Operation ÷ 2 × 6 + 20 – 49

Answers 10

(b) ÷ 2 (10-6) × (-3-2) + (5+4) ÷ 49
Operation
Answers 40

Do You Know!

The word integer is derived from integrity indicating that negative numbers were universally
accepted.

Oasis School Mathematics – 7 175

Magic Square

Let us learn a method popularly known as cup method to
generate the magic squares of order 22, 23, 24 .....

Magic Square of Order = 22 = 4.

Materials required:

(a) Cardboard cut into a square (b) White sheet of paper

(c) Sketch pens (d) Ruler

Method:

• Take a sheet of chart paper and draw a square on it.

• Divide the square into 16 square boxes as shown.

• Pick any set of 16 consecutive natural numbers and arrange them as shown i.e. write
first 8 natural numbers as open cup.

• Write the next 8 natural numbers as a cup upside down.

• Invert the middle two rows.
Find the sum of the numbers along all rows, columns and diagonals.

5 12 16 17 5 12 16 17
6 11 15 18 18 15 11 6
7 10 14 19 19 14 10 7
8 9 13 20 8 9 13 20

The sum is 50 in each case.
Observe the given figure and get the idea to get the magic square of order 23 = 8.

11 26 34 35 43 58 66 67 11 26 34 35 43 58 66 67
12 25 33 36 44 57 65 68 12 25 33 36 44 57 65 68
13 24 32 37 45 56 64 69 69 64 56 45 37 32 24 30
14 23 31 38 46 55 63 70 70 63 55 46 38 31 23 40
15 22 30 39 47 54 62 71 71 62 54 47 39 30 22 15
16 21 29 40 48 53 61 72 72 61 53 48 40 29 21 16
17 20 28 41 49 52 60 73 17 20 28 41 49 52 60 73
18 19 27 42 50 51 59 74 18 19 27 42 50 51 59 74

Keep the position of first two and last two rows same.
Invert the middle four rows.

176 Oasis School Mathematics – 7

Unit

13 Rational Numbers

All the counting numbers starting from 1 are called natural numbers. e.g. 1, 2, 3,
..................... , etc.

All the positive numbers with zero are called whole numbers. e.g. 0, 1, 2, 3, ...............,
etc.

All positive and negative numbers including 'zero' are called integers e.g. ............
–3, – 2, –1, 0, 1, 2, ............, etc.

Here are some operations on integers.
Let us consider any two integers 1 and 2.

(i) 1 + 2 = 3 = 2 + 1 (3 is also a integer).

(ii) 1 – 2 = –1 and 2 – 1 = 1 (both –1 and 1 are also integers).

(iii) 1 × 2 = 2 = 2 × 1 (2 is an integer)

(iv) 1÷ 2 = 1 and 2 ÷ 1 = 12 ( 1 is not an integer but 2 is an integer.)
2 2

Thus, when an integer is divided by another integer the quotient is not always an integer.

Hence, to make the quotient of two integers meaningful, rational number is introduced.

Any numbers as a number of the form p where p and q are integers and q ≠ 0 are
q

called rational numbers. The set of rational numbers is denoted by Q.

All the natural numbers are whole numbers and all the integers are also rational
numbers.

Positive and Negative Rational Numbers

As the numerator and denominator of a rational number should be integers, they

can be either positive or negative integers. Thus, there are two types of rational

numbers, they are:

(i) Positive rational numbers: 1 , 3 , 6 etc., are examples of positive rational
2 5 7
numbers.

(ii) Negative rational numbers: -1 , 3 , -76, etc. are negative rational numbers.
2 5

Oasis School Mathematics – 7 177

Mixed Rational Numbers

3 4 is a mixed number. It is also a rational number since 19 is in the form of pq. So,
5 5
mixed number is a rational number.

Terminating and Non–terminating Rational Numbers

When a rational number is decimalized, the decimal so obtained may be terminating

or non–terminating recurring.

For e.g.,

• 1 = 0.5( Terminating decimal) • 1 = 0.25 (Terminating decimal)
2 4

• 1 = 0.333 (Non–terminating recurring decimal)
3

• 2 = 0.666 (Non–terminating recurring decimal)
3

22 = 3.142857142857... (Non terminating recurring)
7

Irrational Number

The numbers which are not rational are called irrational numbers. 2, 3, 5, etc. are

the examples of irrational number.

When irrational numbers are decimalized, the decimals are non–terminating non–
a
recurring. So, they cannot be expressed in b form.

4 = 2 (2 is a rational number)

3 = 0.375 (The decimal is terminating 3 is a rational number.)
8 8

And 2 = 1.41421 ... (decimal is non–terminating, non–recurring. It is an irrational number.)

3 = 1.732 ........ (decimal is non–terminating non–recurring, it is an irrational number).

π = c = 3.1428... (Non terminating non recurring)
d

Remember !

• Rational numbers are terminating or non terminating recurring decimal.
• Irrational numbers are non terminating and non recurring decimal.

178 Oasis School Mathematics – 7

Worked Out Examples

Example: 1

Examine whether following numbers are rational or irrational.

(a) 2 (b) 0.5 (c) 23 (d) 36 (e) – 3

Solution:

(a) Here, 2 = 1.4142134….
This is a non–terminating, non–recurring decimal.

So, 2 is an irrational number.

(b) This is a terminating decimal. So, 0.5 is a rational number.

(c) 2 = 0.6666.
3

This is a non–terminating recurring decimal.

So, it is a rational number.

(d) 36 = 6. As the square root of 36 gives a perfect whole number, it is a rational
number.

(e) –3 is a rational number.

Exercise 13.1

1. Are the following numbers rational? Write with reasons.

(i) 0 (ii) 3 (iii) 16 (iv) 10 (v) 12

2. Which of the following numbers are irrational. List out rational and irrational
numbers separately.

(i) 2 (ii) 49 (iii) 25 (iv) 3 27
36

(v) 64 (vi) 3 54 (vii) 3 64 (viii) 1000

3. Insert any two rational numbers between each of the following pairs of rational
numbers.

(i) 2 and 3 (ii) 4 and 5

4. Define following terms with suitable examples:

(i) Natural numbers and whole numbers

(ii) Integers

(iii) Rational numbers and irrational numbers

(iv) Terminating and non terminating decimals
(v) Recurring and non–recurring decimals

Answer
Consult your teacher.

Oasis School Mathematics – 7 179

Unit Fractions and
14 Decimals

14.1 Fractions

Review

We have already learnt about fractions in earlier classes. Let’s review some of its
basic terms.

A fraction is a number representing a part of a whole.

It is written as a , where 'a' is the numerator and 'b' is the denominator.
b

Diagram Fraction Meaning Numerator Denominator

3 3 8
8 3 parts out of 8 equal parts

Proper fractions

A proper fraction is a fraction, in which the numerator is less than the denominator.

2 , 152, 163, etc. are proper fractions.
3

Improper fraction

An improper fraction is a fraction in which the numerator is greater than the

denominator:

5 , 7 , 3 , etc. are improper fractions.
3 2 2

Mixed fraction

A fraction which contains both whole number and proper fraction is called a mixed

fraction.

In a fraction 4 2 , 4 is a whole number and 2 is a proper fraction. Hence 4 2 is a mixed
3 3 3
fraction.

Like fractions

Two or more fractions with the same denominators are called like fractions:

180 Oasis School Mathematics – 7

1 , 2 , 2 etc. are like fractions.
5 5 5

Unlike fractions

Two or more fractions with different denominators are called unlike fractions:

2 , 3 , 1 , etc., are unlike fractions.
7 5 2

Equivalent fractions:

The fractions having the same value are called equivalent fractions:

1
2

2
4

3
6



Shaded parts in each figure represent the equal part.

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

Conversion of improper fraction into mixed fraction

To convert improper fraction into mixed fraction, divide numerator by denominator.

The quotient so obtained gives the whole number part and remainder gives the

numerator part of the fraction.

Remainder
∴ Mixed fraction = Quotient Denominator

Example: E xpress 13 into mixed fraction.
4

Here, 4 13 Hence, 13 = 3 1
3 4 4

–12

1

Conversion of mixed number into improper fraction

To convert mixed number into improper fraction, multiply the whole number by
the denominator and add the numerator to the product.

∴ Improper fraction = Whole Number × Denominator + Numerator
Denominator
3
Example: Covert 2 5 into improper fraction.

Here, 2 3 = 2×5+3
5 5

= 153

Oasis School Mathematics – 7 181

Formation of equivalent fractions

(a) M ultiplying its numerator and denominator by a common number

(b) dividing its numerator and denominator by a common number

Let's find the equivalent fractions of

Now, 1 = 1 × 2 = 2
2 2 2 4

1 = 1 × 3 = 3
2 2 3 6

∴ 1 , 2 , 3 , etc. are equivalent fractions.
2 4 6

Similarly, to find equivalent fraction of

182 = 182÷÷22 = 4
6

182 = 182÷÷44 = 2
3

4 , 2 , etc. are equivalent fractions .
6 3

Conversion of Unlike Fractions into Like Fractions

To convert the unlike fractions into like fractions, proceed as under:

Let us convert the fractions • FindtheL.C.M.ofthedenominators
of all the given fractions.
3 and 5 into like fractions.
4 6 • Divide the L.C.M. by each
denominator of given fractions
4 = 2 × 2 ; 6 = 2 × 3 and multiply the numerator and
denominator of the given fractions
∴ L.C.M. = 2 × 2 × 3 = 12 with their respective quotient.

12 ÷ 4 = 3, 12 ÷ 6 = 2

Now, 3 = 3 × 3 = 192 • The fractions so obtained are
4 4 × 3 equivalent fractions such that the

and 5 = 5 × 2 = 10 denominators are equal to the
6 6 × 2 12 LCM of denominators of the given
fractions.
Hence, 192 and 1120 are like fractions.

Comparison of Fractions

To compare fractions, follow the following steps:

Example: 1 • Convert unlike fractions into like
fractions.
Compare the fractions 2 and 3 .
3 4 • Compare numerator of like
fractions.
L.C.M. of 3 and 4 is 12.

2 = 2 × 4 = 8 and • Fraction having greater
3 3 4 12 numerator is the greater one.

182 Oasis School Mathematics – 7

3 = 3 × 3 = 8
4 4 3 12

Since 9 > 8 then 3 > 2
12 12 4 3

Worked Out Examples

Example: 1

Arrange the fractions 3 , 5 and 7 in ascending order.
4 6 18
Solution:

Find the L.C.M. of 4, 6 and 18.

4 = 2 × 2; 6 = 2 × 3; 18 = 2 × 3 × 3

∴ L.C.M. = 2 × 2 × 3 × 3 = 36

Now, 43 = 3×9 = 3276 ; 5 = 5×6 = 3306 ; 7 = 7×2 = 14
4×9 6 6×6 18 18 ×2 36

Since, 3146 < 3276 < 3306 then, 178 < 3 < 5
4 6

So, the fractions in ascending order are 178, 3 and 5 .
4 6

Exercise 14.1

1. Mention whether the given fractions are proper, improper or mixed.

(a) 8 (b) 5 (c) 4 5 (d) 1 (e) 15
5 8 7 57

(f) 2 (g) 3 2 (h) 17
5 5 5

2. Express the following fractions as improper fractions.

(a) 3 4 (b) 4 6 (c) 8 5 (d) 12 4 (e) 11 3
5 9 7 5 9

3. Express the following fractions as mixed fractions.

(a) 27 (b) 65 (c) 85 (d) 500 (e) 45
7 12 13 7 14

4. Fill in the blanks with correct number.

(a) 3 = 2...7. (b) ... = 2 (c) 3..5.. = 5 (d) 5 = 2...5. (e) 45 = ...
5 24 3 6 7 75 5

5. Write any three equivalent fractions of each of the following fractions.

(a) 8 (b) 5 (c) 1 (d) 48 (e) 24
9 7 2 64 36

Oasis School Mathematics – 7 183

6. Convert the following unlike fractions into like fractions.

(a) 3 , 5 , 7 (b) 3 , 2 , 8 (c) 7 , 5 , 2 (d) 1 , 5 , 3
4 8 12 5 5 15 9 6 3 2 8 4

7. Arrange the following fractions in ascending order.

(a) 2 , 6 , 1 (b) 5 , 3 , 3 (c) 3 , 185, 5
3 7 2 8 16 4 5 6

8. Arrange the following fractions into descending order.

(a) 172, 5 , 2 (b) 3 , 5 , 7 (c) 5 , 172, 3
6 3 4 6 8 8 4

Answer 

1. (a) Improper (b) Proper (c) Mixed (d) Proper (e) Improper (f) Proper

(g) Mixed (h) Improper 2. (a) 19 (b) 492 (c) 61 (d) 64 (e) 102
5 7 5 9

3. (a) 3 6 (b) 5 5 (c) 6 7 (d) 71 3 (e) 3134 4. (a) 45
7 12 13 7

(b) 16 (c) 42 (d) 35 (e) 3 5. (a) 16 , 2274 , 32 (b) 1104 , 15 , 20
18 36 21 28

(c) 63, 84, 5 (d) 24 , 1162 , 6 (e) 1182 , 69, 2 6. (a) 1248 , 15 , 14 (b) 195 , 6 , 8 (c) 14 , 1185 , 12
10 32 8 3 24 24 15 15 18 18

(d) 84, 85, 6 7. (a) 21, 23, 6 (b) 3 , 58, 3 (c) 185 , 35, 5 8. (a) 56, 23, 172 (b) 78, 65, 3
8 7 16 4 6 4

(c) 34, 58, 7
12

14.2 Fundamental Operations on Fractions

(a) Addition and Subtraction of Fractions

Addition and subtraction of like fractions

When adding two or more like fractions or when subtracting two like fractions, we
add or subtract them just by adding or subtracting their numerators and reduce the
fraction obtained in lowest term.

(i) 3 4 + 2 3 + 1 4 (ii) 152 – 112 – 122
5 5 5

= 159 + 153 + 9 = 5–1–2
5 12

= 19+13+9 = 5–3
5 12

= 451 = 8 1 = 122
5

= 1
6

184 Oasis School Mathematics – 7

Addition and subtraction of unlike fractions

When adding or subtracting two or more unlike fractions, first change them into
like fractions such that denominators are equal to the L.C.M. then add or subtract
their numerator.

For examples:

Example of addition and subtraction of unlike fractions:

For example: 2 + 5 2 4, 8, 12 Alternative Method
3 4

Here, L.C.M. of 3 and 4 is 12 2 2, 4, 6 2 5
3 4
Now, 2 + 5 2 1, 2, 3 +
3 4 LCM = 2 × 2 × 1 × 2 × 3 = 24
2 4 5 3 2×4+5×3
= 3 × 4 + 4 × 3 = 12

= 8 + 1152 = 8+15
12 12

= 8+15 = 1223
12
11
= 23 = 1 12
12

= 11121


Note: I f the resultant fraction is an improper fraction, convert the improper

fraction into mixed fraction.



Addition of fraction follows the following properties

Closure property:

The sum of two rational numbers a and b is also a rational i.e. a + b = c where c is a

rational.

Example: 1 + 1 = 185, which is also a rational.
3 5

Commutative property:

If a and b are any two rational numbers then we can get a + b = b + a.

Example: 1 + 2 = 2 + 1
2 3 3 2

Associative property:

Let a, b and c are any three rational numbers, then, (a + b) + c = a + ( b + c)

Example:

1 1 1 5 1 1123 and 1 1 1 1 172 = 13
( ) ( ) 2+3 + 4 = 6 + 4 = 2 + 3 + 4 = 2 + 12

∴ ( 1 + 1 ) + (1 = 1 + 1 + 1 )
2 3 2 3 4
4

Oasis School Mathematics – 7 185

Additive Identity:
If a is any rational number then a + 0 = 0 + a = a, where 0 is another rational number.

Example:

1 + 0 = 1
Also, 3 3
∴
0+ 1 = 1
3 3
1 1 1
3 + 0 = 0+ 3 = 3

Hence, 0 is additive identity.

Additive inverse:

For every rational number say a, there exists another rational number – a such that
their sum is zero i.e. a + ( – a) = 0.

Example:

1 + (– 1 ) = 1 – 1
2 2 2 2

= 1–21 – 0 =0
2

Hence, (– 1 ) is additive inverse of 1 .
2 2

Worked Out Examples

Example: 1

Add: 2 + 1 + 5
3 6 12
Solution:
Alternative Method

L.C.M. of 3, 6 and 12 is 12. 2 + 1 + 152
3 6
∴ 2 + 1 + 5
3 6 12 2×4+1×2+5×1
= 15
= 2 × 4 + 1 × 2 + 5 × 1
3 4 6 2 12 1 8+ 2 + 5
= 12
182 + 122 + 5
= 12 = 1125

= 8+2+5 = 5
12 4

= 15 = 1 1
12 4

= 5
4

= 1 1
4

186 Oasis School Mathematics – 7

Example: 2

Simplify: 1 1 + 2 3 – 1152
4 8

Solution:

= 1 1 +2 3 – 1152 2 4, 8, 12 Alternative Method
4 8 2 2, 4, 6
1 1 + 2 3 – 1152
= 5 + 19 – 17 1, 2, 3 4 8
4 8 12
LCM = 2×2×1×2×3=24 = 5 + 19 – 1172
4 8
L.C.M. of 4, 8 and 12 is 24

= 5 × 6 + 19 × 3 – 17 × 2 = 5 × 6 + 19 × 3 – 17 × 2
4 6 8 3 12 2 24

= 3240 + 57 – 34 = 30 + 57 – 34
24 24 24

= 30 + 57 – 34 = 2543
24
= 2254
= 87 – 34 = 5234 = 2 5
24 24

Exercise 14.2

1. Add:

(a) 2 + 1 2 (b) 1 3 + 1 + 3
3 3 4 4 4

(c) 6 1 + 5 1 + 8 1 (d) 1 +2 5 + 5
3 3 3 6 6 6

2. Subtract:

(a) 130 – 1 (b) 5 2 – 2 1 (c) 3 5 – 2 7 (d) 5 1 – 3 3
10 3 3 9 9 4 4

3. Add together:

(a) 15 + 2 + 4 (b) 2 + 4 + 1
3 5 5 5 4

(c) 11 + 9 + 3 (d) 1 + 3 + 2 3
12 8 4 2 4 8

4. Subtract the following:

(a) 1 – 1 (b) 4 – 8 (c) 4 1 – 7 2
5 6 3 5 5 3

(d) 12 1 – 10 7 (e) 5 – 3 1
3 9 6

5. Simplify:

(a) 3 – 7 + 1 (b) 1 1 + 5 – 2 1
8 6 4 3 6 9

(c) 5 3 – 2 1 + 4172 (d) 4 3 + 2 3 – 5 1 (e) 4 + 2 3 – 13
4 6 8 4 8 5 10

Oasis School Mathematics – 7 187

6. (a) What must be added to 3 to get 4 ?
5 7

(b) What must be added to -7 to get 12 ?
5 7

(c) What must be subtracted from 7 to get 3 ?
9 4

(d) What must be subtracted from 2 to get -4 ?
3 5



Answer

1. (a) 2 1 (b) 2 3 (c) 20 (d) 3 5 2. (a) 1 (b) 3 1
3 4 6 5 3

(c) 97 (d) 1 1 3. (a) 132 (b) 1 9 (c) 2 19 (d) 3 5
2 20 24 8

4. (a) 1 (b) –4 (c) -3175 (d) 1 5 (e) 1 5
30 15 9 6

5. (a) – 13 (b) 1 (c) 8 1 (d) 2 (e) 5130
24 18 6

6. (a) - 1 (b) 3 4 (c) 1 (d) 1 7
35 35 36 15

14.3 Verbal Problems on Fractions

There are many situations in the real life where subtraction of fraction is used. Lets
discuss it with some examples.

Worked Out Examples

Example: 1

In a village 5 part of the people are male, find what part of the people are female?
11
Solution:
Part of male = 151
5
Then, the part of female = 1 – 11

= 11–5
11

= 6
11

∴ 151 part of the people are female.

188 Oasis School Mathematics – 7

Example: 2

A completed 2 part of a work and B completed 3 part of the work, find who completed
5 7
more work and by how much?

Solution: 2
5
Work completed by A =

Work completed by B = 3
7

Now, L.C.M. of 5 and 7 is 35

So, 2 = 2 × 7 = 14
5 5 7 35

3 = 3 × 5 = 15
7 7 5 35

Since, 1355 > 1345 , 3 > 2
7 5

Hence, 'B' completed more work.

Again, 3 – 2 = 3 × 5–2 × 7
7 5 35

= 15–14
35

= 1
35

∴ B completed 1 part of work more than A.
35

Example: 3

A boy travelled 9 1 km by cycle, 7 1 km by bus and 3 2 km by walking. How much
2 4 3

did he travel?

Solution: Distance travelled by cycle = 9 1 km = 129 km
2

Distance travelled by bus =7 1 km = 249 km
4

Distance travelled by walking = 3 2 km = 131 km
3

( )Total distance travelled = 9 1 + 7 1 +3 2 km
2 4 3

= 129 + 249 + 131

= 19 × 6 × 29 × 3 × 11 × 4
12

= 114+87+44 km
12

= 245 km
12

= 20152 km

Oasis School Mathematics – 7 189

Example: 4

A man gave 1 of his property to his elder son, 1 to the younger and 1 to his daughter.
4 3 5

How much property is left with him?

Solution:

Let total property be Re. 1.

Property given to his elder son = 1
4

Property given to his younger son = 1
3

Property given to his daughter = 1
5

Total property given to his children = 1 + 1 + 1
4 3 5

= 15 + 20 + 12
60

= 6407

Property left with him = 1– 4607 = 60–47 = 6103 .
60

Example: 5

In a school, 5 of the students play cricket, 1 play table tennis and 1 play football.
12 3 4

Find in which game more students participate.

Solution:

Students playing cricket = 152 2 12, 3, 4

Students playing table tennis = 1 2 6, 3, 2
3 3 3, 3, 1

Students playing football = 1 1, 1, 1
4

In order to find in which game more students participated, we first change the unlike
fractions into like fractions

Now, L.C.M. of 12, 3 and 4 = 12

So, 152 = 5×1 = 152, 1 = 1×4 = 142, 1 = 1 × 3 = 3
12×1 3 3×4 4 4 × 3 12

Obviously, 3 < 4 < 5 ⇒ 5 > 1 > 1
12 12 12 12 3 4

This shows that more students participate in cricket.

190 Oasis School Mathematics – 7

Exercise 14.3

1. (a) If 163 part of students in a class are girls, find the part of boy students.

(b) If 1 part of girls are taller, find the part of shorter girls.
3

2. (a) A and B bought a cake. A ate 2 of the cake and B ate 130 of it,
5

(i) who ate more? (ii) how much more?

(b) Rohan completed 1 part of a work and Sohan completed 2 part of the work.
3 5

Find who completed more work and by how much?

3. (a) There are 3 persons whose weight is 35 2 kg. 42 1 kg. and 45 4 kg. respectively.
3 3 5
What is their total weight?

(b) Ram travelled 25 km by bus, 8 1 km on cycle and 2 3 km on foot. How far did
2 4
he go?

4. (a) Two persons together weight 85 2 kg. If weight of one person is 45 1 kg, what
5 3
is the weight of the second person?

(b) A man gave 2 of his property to his son, 1 to his daughter and the remaining
5 3
to his wife. Find the share of his wife.

(c) The distance between two places is 10 km. Ram travelled first 5 2 km by bus
3
and the remaining distance on foot. Find the distance travelled on foot.



Answer

1. (a) 7 (b) 2 2. (a) (i) A (ii) 1 (b) Sohan by 1
13 3 10 15

3. (a) 123 4 kg (b) 36 1 km. 4. (a) 40 1 kg (b) 4 (c) 4 1 km.
5 4 15 15 3

14.4 Multiplication of Fractions

To multiply one fraction by another, multiply the numerators for the numerator
of the product, and multiply the denominators for its denominator and reduce the
fraction obtained after multiplication in lowest term.

Remember !

When fractions are multiplied,

Product of fractions = Product of numerators
Product of denominators

Oasis School Mathematics – 7 191

e.g. 31 1 1 1 1 1 or 1
2 3 2 6 3 2
1
× = × = 2 = 1
6

Reciprocal of a Fraction

Two fractions are called reciprocal of one another, if their product is one.

For examples; 4 is reciprocal of 1 and 4 is the reciprocal of 5 .
4 5 4

Division of Fraction

We know that division by fraction is the inverse of multiplication. So, to divide one
fraction by another fraction, multiply the dividend by the reciprocal of the divisor.

Example:

10 ÷ 1 = 10 × 4 [Dividing a whole number by a fraction.]
4 1
= 40

3 ÷ 2 = 3 × 1 [ Dividing a fraction by a whole number.]
4 4 2

= 3
8

3 ÷ 1 = 3 × 2 = 6 [Dividing a fraction by a fraction.]
7 2 7 1 7

When any number or fraction is divided by a fraction, we multiply the dividend by
the reciprocal of the divisor.

Multiplication of rational number follows the following properties

Closure property: If a and b are any two rational numbers. Then ab is also a rational number.

Example: 1 × 1 = 115 is a rational number.
3 5

Commutative property: If a and b are any two rational numbers,

then, a × b = b × a

Example: 1 × 1 = 1
3 5 15

Also, 1 × 1 = 1
3 5 15
1 1 1 1
∴ 3 × 5 = 5 × 3

Associative property: Let a, b and c are any three rational numbers,

then (a × b) × c = a × (b × c)

Example: ( 1 × 1 ) × 1 = 210 × 1 = 1
4 5 2 2 40

Also, 1 × ( 1 × 1 ) = 1 × 110 = 410
4 5 2 4

192 Oasis School Mathematics – 7

∴ ( 1 × 1 ) × 1 = 1 × ( 1 × 1 ) .
4 5 2 4 5 2

Multiplicative identity: Let a be any rational number,

then we have, a × 1 = 1 × a = a.

i.e. 1 is called multiplicative identity.

Example: 1 × 1 = 1
3 3

Also, 1 × 1 = 1
3 3

∴ 1 × 1 = 1 × 1 = 1
3 3 3

Multiplicative inverse: Let a be any rational number other than zero. Then, there

exists a rational number 1 such that: a × 1 = 1.
a a

Example: 2 × 3 = 1.
3 2

Hence, 3 is multiplicative inverse of 2 .
2 3

Distributive property over addition:

Let a, b and c are any three rational numbers. Then, a(b + c) = ab + ac.

Example: = 1 ( 1 + 2 ) = 1 × 1151 = 1310
2 3 5 2

Also, 1 × 1 + 1 × 2 = 1 + 120 = 11
2 3 2 5 6 30

Hence, 1 ( 1 + 2 ) = 1 × 1 + 1 × 2
2 3 5 2 3 2 5

Simplification of Expressions Involving Fractions

To simplify expressions involving fractions, we remember the 'BODMAS' rule
where order of the letters of this word are as follows:

Steps:
Remove bar — , brackets ( ), { }, [ ] in order by simplifying
all the operations within it or by simplifying all the operations
within it. (B)
Perform the operation involving 'of' (O)
Perform the operation involving division (D)
Perform the operation involving multiplication (M)
Perform the operation involving addition (A)
Perform the operation involving subtraction (S)

Oasis School Mathematics – 7 193

Worked Out Examples

Example: 1 Example: 2

Find the product of the following fractions: Find the product of:

(a) 9 × 4 (b) 3 3 × 5 × 2110
12 18 7 8

(a) Solution: (b) Solution:

9 × 4 3 3 × 5 × 2110
12 18 7 8
3 13
19×41 24×5×21
= 312×182 = 17×8×10 2

= 1×1 = 3×1×3
3×2 1×1×2

= 1 = 9
6 2

Example: 3 = 4 1
2

Evaluate: 3
7
(a) 6 ÷ 2 (b) 5 2 ÷9 (c) 18
3 5
35

(a) Solution: (b) Solution: (c) Solution:

6÷ 2 5 2 ÷9 3
3 5
3 7
= 6 × 2 1 = 27 ÷ 9 18
3 = 5 1
=3× 27 3 91 35
5 ×
=9 = 3 ÷ 18
7 35
= 3×1 = 3
5×1 5 1 3 × 35 5
7 18 6
=
1

= 1×5 = 5
1×6 6

Example: 4 Example: 5

Find the value of 4 of Rs. 1000. If 3 of a number is 81, find the number.
5 Sol5ution:

Solution: Here, 3 of a number = 81.
5
4 of Rs. 1000 = 4 × Rs. 1000 3
5 5 or, 5 × a number = 81.

= Rs. 800 or, A number = 81 ÷ 3
5
5
= 81 × 3 = 27 × 5 = 135

194 Oasis School Mathematics – 7