ML Aggarwal Solutions for Class 8 Maths Chapter 1: Rational Numbers

ML Aggarwal Solutions for Class 8 Maths Chapter 1 Rational Numbers are curated by experienced teachers in a simple language, to boost students’ exam preparation. In a subject like Mathematics, students score good marks easily as it merely requires thorough practice. Referring to ML Aggarwal solutions, students can cross check their answers while revising the exercise problems. Here, the students can find ML Aggarwal Solutions for Class 8 Maths Chapter 1 Rational Numbers PDF, from the links given below

Chapter 1 mainly deals with the study of Rational Numbers, in accordance with the current ICSE board. ML Aggarwal Solutions available in PDF format, enhance problem solving skills among students. It also helps them in improving their time management skills, which are important from the exam perspective.

ML Aggarwal Solutions for Class 8 Maths Chapter 1: Rational Numbers Download PDF

 

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Access ML Aggarwal Solutions for Class 8 Maths Chapter 1: Rational Numbers

Exercise 1.1

1. Add the following:

(i) 4 / 7 and 5 / 7

(ii) 7 / – 13 and 4 / – 13

Solution:

(i) Given

4 / 7 and 5 / 7

Adding both the numbers

4 / 7 + 5 / 7 = (4 + 5) / 7

We get,

= 9 / 7

∴ The addition of 4 / 7 and 5 / 7 is 9 / 7

(ii) Given

7 / – 13 and 4 / – 13

Consider

7 / – 13 = {7 × (-1)} / {- 13 × (-1)}

= – 7 / 13

Also,

4 / – 13 = {4 × (-1)} / {- 13 × (-1)}

= – 4 / 13

Now,

Adding both the numbers

(7 / – 13) + (4 / – 13) = (- 7 – 4) / 13

We get,

= – 11 / 13

2. Simplify:

(i) 5 / 11 + ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 1

(ii) – 4 / 9 + ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 2

Solution:

(i) Given

5 / 11 +
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 3

This can be written as,

5 / 11 + 39 / 9

Taking L.C.M we get,

5 / 11 = (5 × 9) / (11 × 9)

We get,

= 45 / 99

39 / 9 = (39 × 11) / (9 × 11)

We get,

= 429 / 99

Now,

Adding both the numbers,

45 / 99 + 429 / 99 = (45 + 429) / 99

= 474 / 99

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 4

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 5

Dividing numerator and denominator by 3,

= 4 (78 ÷ 3) / (99 ÷ 3)

We get,

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 6

(ii) Given

– 4 / 9 +
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 7

This can be written as,

– 4 / 9 + 38 / 13

Taking L.C.M we get,

– 4 / 9 = (-4 × 13) / (9 × 13)

We get,

= – 52 / 117

38 / 13 = (38 × 9) / (13 × 9)

We get,

= 342 / 117

Now,

Adding both the numbers,

– 52 / 117 + 342 / 117 = (- 52 + 342) / 117

We get,

= 290 / 117

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 8

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 9

3. Verify commutative property of addition for the following pairs of rational numbers.

(i) – 4 / 3 and 3 / 7

(ii) – 2 / – 5 and 1 / 3

(iii) 9 / 11 and 2 / 13

Solution:

(i) – 4 / 3 and 3 / 7

Adding both the numbers,

= – 4 / 3 + 3 / 7

Taking L.C.M. we get,

= (- 28 + 9) / 21

= – 19 / 21

And

3 / 7 + (- 4 / 3)

Again taking L.C.M. we get,

= (9 – 28) / 21

= – 19 / 21

Therefore,

– 4 / 3 + 3 / 7 = 3 / 7 + (- 4 / 3)

(ii) – 2 / – 5 and 1 / 3

Consider,

– 2 / – 5 = { – 2 × (- 1)} / {- 5 × (- 1)}

= 2 / 5

Now,

2 / 5 + 1 / 3

Taking L.C.M. we get,

= (6 + 5) / 15

= 11 / 15

And 1 / 3 + 2 / 5

Again taking L.C.M. we get,

= (5 + 6) / 15

= 11 / 15

Therefore,

2 / 5 + 1 / 3 = 1 / 3 + 2 / 5

(iii) 9 / 11 and 2 / 13

Adding both the numbers,

= 9 / 11 + 2 / 13

Taking L.C.M. we get,

= (117 + 22) / 143

We get,

= 139 / 143

And 2 / 13 + 9 / 11

Again taking L.C.M. we get,

= (22 + 117) / 143

We get,

= 139 / 143

Therefore,

9 / 11 + 2 / 13 = 2 / 13 + 9 / 11

4. Find the additive inverse of the following rational numbers:

(i) 2 / – 3

(ii) – 7 / – 12

Solution:

(i) Given

2 / – 3

Additive inverse of

2 / – 3 = – (2 / – 3)

We get,

= 2 / 3

(ii) Given

– 7 / -12

Additive inverse of

– 7 / – 12 = – (- 7 / – 12)

We get,

= – 7 / 12

5. Verify that – (- x) = x for

(i) x = 10 / 13

(ii) x = – 15 / 17

Solution:

(i) x = 10 / 13

– x = – 10 / 13

– (- x) = – (- 10 / 13)

= 10 / 13

Hence, – (- x) = x

(ii) x = – 15 / 17

– x = 15 / 17

– (- x) = – (15 / 17)

= – 15 / 17

Hence, – (- x) = x

6. Using appropriate properties of addition, find the following:

(i) 4 / 5 + 11 / 7 + (-7 / 5) + (- 2 / 7)

(ii) 3 / 7 + 4 / 9 + (- 5 / 21) + (2 / 3)

Solution:

(i) 4 / 5 + 11 / 7 + (- 7 / 5) + (- 2 / 7)

= 4 / 5 + (- 7 / 5) + 11 / 7 + (- 2 / 7)

= {4 + (- 7)} / 5 + {11 + (- 2)}/ 7

= (4 – 7) / 5 + (11 – 2) / 7

On further calculation, we get,

= – 3 / 5 + 9 / 7

Now, taking L.C.M. we get,

= (- 21 + 45) / 35

= 24 / 35

(ii) 3 / 7 + 4 / 9 + (- 5 / 21) + 2 / 3

= 3 / 7 + (- 5 / 21) + 4 / 9 + 2 / 3

On simplifying, we get,

= {9 + (-5)} / 21 + (4 + 6) / 9

= 4 / 21 + 10 / 9

Taking L.C.M. we get,

= (12 + 70) / 63

= 82 / 63

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 10

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 11

7. Fill in the blanks:

(i) (- 4 / 9) + (2 / 7) is a ……….. number

(ii) (43 / 89) + (- 51 / 47) = …….. + (43 / 89)

(iii) 2 / 7 + …… = 2 / 7 = 0 + ………

(iv) 4 / 11 + {(- 7 / 12) + 9 / 10} = {(4 / 11) + (- 7 / 12)} + …..

(v) 5 / 9 + …… = 0 = (- 5 / 9) + ………

Solution:

(i) (- 4 / 9) + (2 / 7) is a rational number

(ii) (43 / 89) + (- 51 / 47) = (- 51 / 47) + (43 / 89) (Commutative property)

(iii) 2 / 7 + 0 = 2 / 7 = 0 + 2 / 7 (Commutative property)

(iv) 4 / 11 + {(- 7 / 12) + 9 / 10} = {(4 / 11) + (- 7 / 12)} + 9 / 10 (Associative property)

(v) 5 / 9 + (- 5 / 9) = 0 = (- 5 / 9) + 5 / 9 (Existance of zero property)

8. If a = – 11 / 27, b = 4 / 9 and c = – 5 / 18, then verify that a + (b + c) = (a + b) + c

Solution:

Given

a = – 11 / 27, b = 4 / 9 and c = – 5 / 18

a + (b + c) = (a + b) + c

Consider,

L.H.S. = a + (b + c)

= – 11 / 27 + {4 / 9 + (- 5 / 18)}

= – 11 / 27 + (4 / 9 – 5 / 18)

On simplification, we get

= – 11 / 27 + (8 – 5) / 18

= – 11 / 27 + 3 / 18

Taking L.C.M. we get,

= (- 22 + 9) / 54

= – 13 / 54

R.H.S. = (a + b) + c

= (- 11 / 27 + 4 / 9) + (- 5 / 18)

On further calculation, we get

= {(- 11 + 12) / 27} + (- 5 / 18)

= (1 / 27) + (- 5 / 18)

= (2 – 15) / 54

= – 13 / 54

Hence,

L.H.S. = R.H.S.

Exercise 1.2

1. Subtract:

(i) ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 12 from – 3 / 7

(ii) – 4 / 9 from ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 13

(iii) ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 14 from ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 15

Solution:

(i)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 16from – 3 / 7

= – 3 / 7 – (13 / 5)

Taking L.C.M. we get,

= (- 15 – 91) / 35

= – 106 / 35

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 17

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 18

Hence, the subtraction of
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 19from – 3 / 7 is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 20

(ii) – 4 / 9 from
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 21

This can be written as,

– 4 / 9 from 29 / 8

= 29 / 8 – (- 4 / 9)

= 29 / 8 + 4 / 9

Taking L.C.M. we get,

= (261 + 32) / 72

= 293 / 72

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 22

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 23

(iii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 24from
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 25

This can be written as,

= – 16 / 5 from – 43 / 9

= – 43 / 9 – (- 16 / 5)

= – 43 / 9 + 16 / 5

Taking L.C.M. we get,

= (- 215 + 144) / 45

We get,

= – 71 / 45

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 26

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 27

2. Sum of two rational numbers is 3 / 5. If one of them is – 2 / 7, find the other.

Solution:

Given

Sum of two rational numbers is 3 / 5

One of the number is – 2 / 7

Hence, the other number is calculated as follows:

Other number = 3 / 5 – (- 2 / 7)

= 3 / 5 + 2 / 7

Taking L.C.M. we get,

= (21 + 10) / 35

= 31 / 35

Therefore, the other number is 31 / 35

3. What rational number should be added to – 5 / 11 to get – 7 / 8?

Solution:

Given

According to the statement,

Sum of two numbers = – 7 / 8

One number = – 5 / 11

Hence, the other number is calculated as below:

Other number = – 7 / 8 – (- 5 / 11)

= – 7 / 8 + 5 / 11

Taking L.C.M. we get,

= (- 77 + 40) / 88

= – 37 / 88

Therefore, the other number is – 37 / 88

4. What rational number should be subtracted from ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 28 to getML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 29 ?

Solution:

The required number can be calculated as follows:

(
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 30) – (
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 31)

This can be written as,

(- 23 / 5) + (7 / 2)

On further calculation, we get

= (- 46 + 35) / 10

= – 11 / 10

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 32

Therefore, the required number is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 33

5. Subtract the sum of – 5 / 7 and – 8 / 3 from the sum of 5 / 2 and – 11 / 12.

Solution:

Sum of – 5 / 7 and – 8 / 3 can be calculated as,

– 5 / 7 and – 8 / 3 = (- 5 / 7) + (- 8 / 3)

On further calculation, we get

= (- 15 – 56) / 21

= – 71 / 21

Now,

Sum of 5 / 2 and – 11 / 12 can be calculated as,

5 / 2 + (- 11 / 12) = 5 / 2 – 11 / 12

On simplification, we get,

= (30 – 11) / 12

= 19 / 12

Now,

19 / 12 – (- 71 / 21)

= 19 / 12 + 71 / 21

Taking L.C.M. we get,

= (133 + 284) / 84

= 417 / 84

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 34

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 35

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 36

6. If x = – 4 / 7 and y = 2 / 5, then verify that x – y ≠ y – x

Solution:

Given

x = – 4 / 7 and y = 2 / 5

Now,

x – y = – 4 / 7 – (2 / 5)

= – 4 / 7 – 2 / 5

Taking L.C.M. we get,

= (- 20 – 14) / 35

= – 34 / 35

And

y – x = 2 / 5 – (- 4 / 7)

= 2 / 5 + 4 / 7

Taking L.C.M. we get,

= (14 + 20) / 35

= 34 / 35

Therefore, x – y ≠ y – x

7. If x = 4 / 9, y = – 7 / 12 and z = – 2 / 3, then verify that x – (y – z) ≠ (x – y) – z

Solution:

Given

x = 4 / 9, y = – 7 / 12, z = – 2 / 3

x – (y – z) ≠ (x – y) – z

L.H.S. = x – (y – z)

= 4 / 9 – {- 7 / 12 – (- 2 / 3)}

= 4 / 9 – (- 7 / 12 + 2 / 3)

On further calculation, we get

= 4 / 9 – {(- 7 + 8) / 12}

= 4 / 9 – (1 / 12)

= 4 / 9 – 1 / 12

Taking L.C.M. we get,

= (16 – 3) / 36

= 13 / 36

Now,

R.H.S = (x – y) – z

= {4 / 9 – (- 7 / 12)} – (- 7 / 12)

= (4 / 9 + 7 / 12) + 7 / 12

On further calculation, we get

= {(16 + 21) / 36} + 7 / 12

= 37 / 36 + 7 / 12

Again taking L.C.M. we get,

= (37 + 21) / 36

= 58 / 36

Therefore, x – (y – z) ≠ (x – y) – z

8. Which of the following statement is true / false?

(i) 2 / 3 – 4 / 5 is not a rational number.

(ii) – 5 / 7 is the additive inverse of 5 / 7.

(iii) 0 is the additive inverse of its own.

(iv) Commutative property holds for subtraction of rational numbers.

(v) Associative property does not hold for subtraction of rational numbers.

(vi) 0 is the identity element for subtraction of rational numbers.

Solution:

(i) 2 / 3 – 4 / 5

Taking L.C.M

= (10 – 12) / 15

= – 2 / 15

Is a rational number

Hence, the given statement is false

(ii) The given statement is true

(iii) The given statement is true

(iv) Let us take,

5 / 4 – 3 / 4 = 2 / 4

We know that,

3 / 4 – 5 / 4 = – 2 / 4

2 / 4 ≠ – 2 / 4

Therefore, the given statement is false

(v) The given statement is true

(vi) Let us take,

7 / 8 – 0 = 7 / 8

But 0 – 7 / 8 = – 7 / 8

7 / 8 ≠ – 7 / 8

Therefore, the given statement is false

Exercise 1.3

1. Multiply and express the result in the lowest form:

(i) 6 / – 7 × 14 / 30

(ii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 37 ×
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 38

(iii) 25 / – 9 × – 3 / 10

Solution:

(i) 6 / – 7 × 14 / 30

= (6 × 14) / (- 7 × 30)

We get,

= 84 / – 210

= (84 ÷ 42) / (- 210 ÷ 42)

∵ HCF of 84, 210 = 42

= 2 / – 5

= {2 × (- 1)} / {- 5 × (- 1)}

= – 2 / 5

(ii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 39×
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 40

This can be written as,

= 20 / 3 × 9 / 7

= (20 × 9) / (3 × 7)

= 180 / 21

= (180 ÷ 3) / (21 ÷ 3)

∵ HCF of 180, 21 = 3

We get,

= 60 / 7

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 41

(iii) 25 / – 9 × – 3 / 10

= {25 × (- 3)} / {(- 9) × 10}

= – 75 / – 90

= {- 75 ÷ (- 15)} / {- 90 ÷ (- 15)}

∵ HCF of 75, 90 = 15

We get,

= 5 / 6

2. Verify commutative property of multiplication for the following pairs of rational numbers:

(i) 4 / 5 and – 7 / 8

(ii) ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 42 and ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 43

(iii) – 7 / – 20 and 5 / – 14

Solution:

(i) 4 / 5 and – 7 / 8

Now,

4 / 5 × – 7 / 8

= {4 × (- 7)} / 5 × 8

We get,

= – 28 / 40

and

– 7 / 8 × 4 / 5

= (- 7 × 4) / (8 × 5)

We get,

= – 28 / 40

Therefore, 4 / 5 × (- 7 / 8) = – 7 / 8 × 4 / 5

(ii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 44and
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 45

This can be written as,

40 / 3 and 9 / 8

Now,

40 / 3 × 9 / 8

= (40 × 9) / (3 × 8)

= 360 / 24

We get,

= 15

and

9 / 8 × 40 / 3

= (9 × 40) / (8 × 3)

= 360 / 24

We get,

= 15

Therefore, 40 / 3 × 9 / 8 = 9 / 8 × 40 / 3

(iii) – 7 / – 20 and 5 / – 14

– 7 / – 20 = {- 7 × (- 1)} / {- 20 × (- 1)}

= 7 / 20

Now, 7 / 20 and 5 / – 14

7 / 20 × 5 / – 14

= (7 × 5) / 20 × (- 14)

= 35 / – 280

and

5 / – 14 × 7 / 20

= (5 × 7) / (- 14 × 20)

= 35 / – 280

Therefore, 7 / 20 × 5 / – 14 = 5 / – 14 × 7 / 20

3. Verify the following and name the property also:

(i) 3 / 5 × (- 4 / 7 × – 8 / 9) = (3 / 5 × – 4 / 7) × – 8 / 9

(ii) 5 / 9 × (- 3 / 2 + 7 / 5) = 5 / 9 × – 3 / 2 + 5 / 9 × 7 / 5

Solution:

(i) 3 / 5 × (- 4 / 7 × – 8 / 9) = (3 / 5 × – 4 / 7) × – 8 / 9

L.H.S. = 3 / 5 × (- 4 / 7 × – 8 / 9)

= 3 / 5 × (- 4 × – 8) / 7 × 9

= 3 / 5 × 32 / 63

= (3 × 32) / (5 × 63)

We get,

= 96 / 315

R.H.S. = (3 / 5 × – 4 / 7) × – 8 / 9

= – 12 / 35 × – 8 / 9

= {- 12 × (- 8)}/ (35 × 9)

We get,

= 96 / 315

Hence, 3 / 5 × (- 4 / 7 × – 8 / 9) = (3 / 5 × – 4 / 7) × – 8 / 9

The name of the property is Associative property of multiplication

(ii) 5 / 9 × (- 3 / 2 + 7 / 5) = 5 / 9 × – 3 / 2 + 5 / 9 × 7 / 5

L.H.S = 5 / 9 × (- 3 / 2 + 7 / 5)

= 5 / 9 × {(- 15 + 14) / 10}

We get,

= 5 / 9 × (- 1 / 10)

= – 5 / 90

= (- 5 ÷ 5) / (90 ÷ 5)

We get,

= – 1 / 18

R.H.S. = 5 / 9 × (- 3 / 2) + 5 / 9 × 7 / 5

On further calculation, we get,

= – 15 / 18 + 35 / 45

Taking L.C.M. we get,

= (- 75 + 70) / 90

= – 5 / 90

= (- 5 ÷ 5) / (90 ÷ 5)

= – 1 / 18

Hence, L.H.S. = R.H.S.

4. Find the multiplication inverse of the following:

(i) 12

(ii) 2 / 3

(iii) – 4 / 7

(iv) – 3 / 8 × (- 7 / 13)

Solution:

(i) The multiplication inverse of 12 is 1 / 12

(ii) The multiplication inverse of 2 / 3 is 3 / 2

(iii) The multiplication inverse of – 4 / 7 is 7 / – 4

(iv) – 3 / 8 × (- 7 / 13) = 21 / 104

The multiplication inverse of 21 / 104 is 104 / 21 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 46

5. Using the appropriate properties of operations of rational numbers, evaluate the following:

(i) 2 / 5 × – 3 / 7 – 1 / 14 – 3 / 7 × 3 / 5

(ii) 8 / 9 × 4 / 5 + 5 / 6 – 9 / 5 × 8 / 9

(iii) – 3 / 7 × 14 / 15 × 7 / 12 × (- 30 / 35)

Solution:

(i) 2 / 5 × – 3 / 7 – 1 / 14 – 3 / 7 × 3 / 5

= 2 / 5 × – 3 / 7 – 3 / 7 × 3 / 5 – 1 / 14

Taking common term, we get

= – 3 / 7 (2 / 5 + 3 / 5) – 1 / 14

= – 3 / 7 × (2 + 3) / 5 – 1 / 14

= – 3 / 7 × 1 – 1 / 14

= – 3 / 7 – 1 / 14

Taking L.C.M. we get,

= (- 6 – 1) / 14

= – 7 / 14

= (- 7 ÷ 7) / (14 ÷ 7)

We get,

= – 1 / 2

(ii) 8 / 9 × 4 / 5 + 5 / 6 – 9 / 5 × 8 / 9

= 8 / 9 × 4 / 5 – 9 / 5 × 8 / 9 + 5 / 6

Taking common terms, we get,

= 8 / 9 (4 / 5 – 9 / 5) + 5 / 6

= 8 / 9 {(4 – 9) / 5} + 5 / 6

= 8 / 9 × – 5 / 5 + 5 / 6

= 8 / 9 × (- 1) + 5 / 6

On further calculation, we get

= – 8 / 9 + 5 / 6

Taking L.C.M. we get,

= (- 16 + 15) / 18

= – 1 / 18

(iii) – 3 / 7 × 14 / 15 × 7 / 12 × (- 30 / 35)

= (- 3 / 7 × 14 / 15) × (7 / 12 × – 30 / 35)

On further calculation, we get

= – 2 / 5 × – 1 / 2

We get,

= 1 / 5

6. If p = – 8 / 27, q = 3 / 4 and r = – 12 / 15, then verify that

(i) p × (q × r) = (p × q) × r

(ii) p × (q – r) = p × q – p × r

Solution:

Given

p = – 8 / 27, q = 3 / 4 and r = – 12 / 15

(i) p × (q × r) = (p × q) × r

L.H.S. = p × (q × r)

= – 8 / 27 × (3 / 4 × – 12 / 15)

= – 8 / 27 × – 3 / 5

On further calculation, we get,

= {(- 8) × (- 3)} / (27 × 5)

= 24 / (27 × 5)

We get,

= 8 / 45

Now,

R.H.S. = (p × q) × r

= (- 8 / 27 × 3 / 4) × – 12 / 15

= – 2 / 9 × – 12 / 15

We get,

= 8 / 45

Therefore, L.H.S. = R.H.S.

(ii) p × (q – r) = p × q – p × r

L.H.S. = p × (q – r)

= – 8 / 27 × {(3 / 4) – (- 12 / 5)}

Taking L.C.M. we get,

= – 8 / 27 × {(45 + 48) / 60}

= – 8 / 27 × 93 / 60

We get,

= – 62 / 135

R.H.S. = p × q – p × r

= – 8 / 27 × 3 / 4 – (8 / 27 × – 12 / 15)

= – 2 / 9 – 32 / 135

= (- 30 – 32) / 135

We get,

= – 62 / 135

Therefore, L.H.S. = R.H.S.

7. Fill in the following blanks:

(i) 2 / 3 × – 4 / 5 is a …… number.

(ii) 54 / 81 × – 63 / 108 = ………. × 54 / 81

(iii) 4 / 5 × 1 = …… = 1 × ……

(iv) 5 / – 12 × …… = 1 = – 12 / 5 × ……

(v) 3 / 7 × (- 2 / 8 × …..) = (3 / 7 × – 2 / 8) × 5 / 9

(vi) – 8 / 9 × {4 / 13 + 5 / 17} = – 8 / 9 × 4 / 13 + ………

(vii) – 6 / 13 × {8 / 9 – 4 / 7} = – 6 / 13 × ……. – (- 6 / 13) × (4 / 7)

(viii) 16 / 23 × ……… = 0

(ix) The reciprocal of 0 is ………..

(x) The numbers …… and …… are their own reciprocals.

(xi) If y be the reciprocal of x, then the reciprocal of y2 in terms of x will be ……

(xii) The product of a non-zero rational number and its reciprocal is ……….

(xiii) The reciprocal of a negative rational number is ……….

Solution:

(i) 2 / 3 × – 4 / 5 is a rational number.

(ii) 54 / 81 × – 63 / 108 = ……. × 54 / 81

54 / 81 × – 63 / 108 = – 63 / 108 × 54 / 81

(iii) 4 / 5 × 1 = …… = 1 × ……

4 / 5 × 1 = 4 / 5 = 1 × 4 / 5

(iv) 5 / – 12 × …… = 1 = – 12 / 5 × ……

5 / – 12 × – 12 / 5 = 1 = – 12 / 5 × 5 / – 12

(v) 3 / 7 × (- 2 / 8 × ….) = (3 / 7 × – 2 / 8) × 5 / 9

3 / 7 × (- 2 / 8 × 5 / 9) = (3 / 7 × – 2 / 8) × 5 / 9

(vi) – 8 / 9 × (4 / 13 + 5 / 17) = – 8 / 9 × 4 / 13 + ………

– 8 / 9 × (4 / 13 + 5 / 17) = – 8 / 9 × 4 / 13 + – 8 / 9 × 5 / 17

(vii) – 6 / 13 × (8 / 9 – 4 / 7) = – 6 / 13 × ….- (- 6 / 13) × (4 / 7)

– 6 / 13 × (8 / 9 – 4 / 7) = – 6 / 13 × 8 / 9 – (- 6 / 13) × (4 / 7)

(viii) 16 / 23 × …. = 0

16 / 23 × 0 = 0

(ix) The reciprocal of 0 is not defined

(x) The numbers 1 and – 1 are their own reciprocals

(xi) If y be the reciprocal of x, then the reciprocal of y2 in terms of x will be x2

(xii) The product of a non-zero rational number and its reciprocal is 1

(xiii) The reciprocal of a negative rational number is a negative rational number

8. If 4 / 5 the multiplicative inverse of ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 47? Why or why not?

Solution:

No, the multiplicative inverse of 4 / 5 is not – 5 / 4

The multiplicative inverse of 4 / 5 is 5 / 4

9. Using distributivity, find

(i) {7 / 5 × (- 3 / 12)} + {7 / 5 + 5 / 12}

(ii) {9 / 16 × 4 / 12} + {9 / 16 × (- 3 / 9)}

Solution:

(i) {7 / 5 × (3 / 12)} + {7 / 5 + 5 / 12}

Taking common factor, we get

= 7 / 5 × (- 3 / 12 + 5 / 12)

= 7 / 5 × {(- 3 + 5) / 12}

= 7 / 5 × 2 / 12

We get,

= 7 / 30

(ii) {9 / 16 × 4 / 12} + {9 / 16 × (- 3 / 9)}

Taking common factor, we get

= 9 / 16 × {4 / 12 + (- 3 / 9)}

= 9 / 16 × (1 / 3 – 1 / 3)

We get,

= 9 / 16 × 0

= 0

10. Find the sum of additive inverse and multiplication inverse of 9.

Solution:

The additive inverse of 9 is – 9

The multiplicative inverse of 9 is 1 / 9

Hence,

– 9 + 1 / 9 = (- 81 + 1) / 9

We get,

= – 80 / 9

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 48

11. Find the product of additive inverse and multiplicative inverse of – 3 / 7

Solution:

The additive inverse of – 3 / 7 is 3 / 7

The multiplicative inverse of – 3 / 7 is – 7 / 3

Therefore,

3 / 7 × (- 7 / 3) = – 1

Exercise 1.4

1. Find the value of the following:

(i) – 3 / 7 ÷ 4

(ii) ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 49 ÷ (- 4 / 9)

(iii) – 8 / 9 ÷ – 3 / 5

Solution:

(i) – 3 / 7 ÷ 4

= – 3 / 7 × 1 / 4

We get,

= – 3 / 28

Hence, the value of – 3 / 7 ÷ 4 = – 3 / 28

(ii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 50÷ (- 4 / 9)

This can be written as,

= 37 / 8 ÷ (- 4 / 9)

= 37 / 8 × 9 / – 4

We get,

= 333 / – 32

= {333 × (- 1)} / {- 32 × (- 1)}

= – 333 / 32

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 51

(iii) – 8 / 9 ÷ – 3 / 5

= – 8 / 9 × 5 / – 3

= – 40 / – 27

= {- 40 × (- 1)} / {- 27 × (- 1)}

We get,

= 40 / 27

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 52

2. State whether the following statements are true or false:

(i) – 9 / 13 ÷ 2 / 7 is a rational number.

(ii) 4 / 13 ÷ 11 / 12 = 11 / 12 ÷ 4 / 13

(iii) – 3 / 4 ÷ (5 / 9 ÷ – 4 / 11) = (- 3 / 4 ÷ 5 / 9) ÷ – 4 / 11

(iv) 13 / 14 ÷ – 5 / 7 ≠ – 5 / 7 ÷ 13 / 14

(v) (- 7 ÷ 4 / 5) ÷ – 9 / 10 ≠ – 7 ÷ (4 / 5 ÷ – 9 / 10)

(vi) – 7 / 24 ÷ 6 / 11 is not a rational number.

Solution:

(i) The given statement is true

(ii) The given statement is false

Correct: Commutative property is not true for the division

(iii) The given statement is false

Correct: Associative in division is not true

(iv) The given statement is true

(v) The given statement is true

(vi) The given statement is flase

Correct: It is a rational number

3. The product of two rational numbers is – 11 / 12. If one of them is ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 53, find the other.

Solution:

Given

Product of two rational numbers = – 11 / 12

One of the number =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 54= 22 / 9

The other number is calculated as below

– 11 / 12 ÷ 22 / 9

= – 11 / 12 × 9 / 22

We get,

= – 3 / 8

Therefore, the other number is – 3 / 8

4. By what rational number should – 7 / 12 be multiplied to get the product as 5 / 14?

Solution:

Given

Product = 5 / 14

The required number can be calculated as below

5 / 14 ÷ – 7 / 12

= 5 / 14 × 12 / – 7

We get,

= 30 / – 49

= {30 × (- 1)} / {- 49 × (- 1)}

= – 30 / 49

Hence, the required number is – 30 / 49

5. By what rational number should – 3 is divided to get – 9 / 13?

Solution:

The required number can be calculated as follows:

– 3 ÷ – 9 / 13

= – 3 × 13 / – 9

We get,

= – 13 / – 3

= {- 13 × (- 1)} / {- 3 × (- 1)}

= 13 / 3

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 55

Therefore, the required number is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 56

6. Divide the sum of – 13 / 8 and 5 / 12 by their difference.

Solution:

Given

Sum of – 13 / 8 and 5 / 12 is calculated as,

= – 13 / 8 + 5 / 12

On further calculation, we get

= (- 39 + 10) / 24

We get,

= – 29 / 24

Now,

Difference of – 13 / 8 and 5 / 12 is calculated as,

= – 13 / 8 – 5 / 12

We get,

= (- 39 – 10) / 24

= – 49 / 24

Now,

– 29 / 24 ÷ – 49 / 24

= – 29 / 24 × 24 / – 49

= – 29 / – 49

= {- 29 × (- 1)} / {- 49 × (- 1)}

We get,

= 29 / 49

7. Divide the sum of 8 / 3 and 4 / 7 by the product of – 3 / 7 and 14 / 9.

Solution:

Sum of 8 / 3 and 4 / 7 is calculated as below

8 / 3 + 4 / 7 = (56 + 12) / 21

We get,

= 68 / 21

Product of – 3 / 7 and 14 / 9 is calculated as follows:

– 3 / 7 × 14 / 9 = – 2 / 3

Hence,

68 / 21 ÷ – 2 / 3 = 68 / 21 × 3 / – 2

We get,

= 34 / – 7

= {34 × (- 1)} / {- 7 × (- 1)}

= – 34 / 7

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 57

8. If p = – 3 / 2, q = 4 / 5 and r = – 7 / 12, then verify that (p ÷ q) ÷ r ≠ p ÷ (q ÷ r)

Solution:

Given

p = – 3 / 2, q = 4 / 5 and r = – 7 / 12

(p ÷ q) ÷ r ≠ p ÷ (q ÷ r)

LHS = (p ÷ q) ÷ r

= (- 3 / 2 ÷ 4 / 5) ÷ (- 7 / 12)

= (- 3 / 2 × 5 / 4) ÷ (- 7 / 12)

= – 15 / 8 ÷ – 7 / 12

= – 15 / 8 × 12 / – 7

We get,

= – 45 / – 14

= {- 45 × (- 1)} / {- 14 × (- 1)}

= 45 / 14

Now,

RHS = p ÷ (q ÷ r)

= – 3 / 2 ÷ (4 / 5) ÷ (- 7 / 12)

= – 3 / 2 ÷ (4 / 5× 12 / – 7)

We get,

= – 3 / 2 ÷ 48 / – 35

= – 3 / 2 × – 35 / 48

We get,

= 35 / 32

Therefore, LHS ≠ RHS

 

Exercise 1. 5

1. Represent the following rational numbers on the number line.

(i) 11 /4

(ii)
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 58

(iii) – 9 / 7

(iv) – 2 / – 5

Solution:

(i) 11 / 4 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 59

The given rational number on the number line is shown as below:

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 60

(ii) ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 61

The given rational number on the number line is shown as below

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 62

(iii) – 9 / 7 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 63

The given rational number on the number line is shown as below

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 64

(iv) – 2 / – 5 = – 2 × (- 1) / – 5 × (- 1)

We get,

= 2 / 5

The given rational number on the number line is shown as below

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 65

2. Write the rational numbers for each point labeled with a letter:

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 66

ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 67

Solution:

(i) The rational numbers for each point labeled with a letter are as follows:

A = 3 / 7

B = 7 / 7 = 1

C = 8 / 7 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 68

D = 12 / 7 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 69

E = 13 / 7 =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 70

(ii) The rational numbers for each point labeled with a letter are as follows:

P = – 3 / 8

Q = – 4 / 8 or – 1 / 2

R = – 7 / 8

S = – 11 / 8

T = – 12 / 8 or – 3 / 2

3. Find twenty rational numbers between – 3 / 7 and 2 / 3

Solution:

Twenty rational numbers between – 3 / 7 and 2 / 3 can be calculated as follows:

We know that,

LCM of 7, 3 = 21

Hence,

– 3 / 7 = (- 3 × 3) / (7 × 3)

We get,

= – 9 / 21

2 / 3 = (2 × 7) / (3 × 7)

We get,

= 14 / 21

Now, twenty rational numbers between – 9 / 21 and 14 / 21 are,

– 8 / 21, – 7 / 21, – 6 / 21, – 5 / 21, – 4 / 21, – 3 / 21, – 2 / 21, – 1 / 21, 0, 1 / 21, 2 / 21, 3 / 21, 4 / 21, 5 / 21, 6 / 21, 7 / 21, 8 / 21, 9 / 21, 10 / 21, 11 / 21, 12 / 21 and 13 / 21

4. Find six rational numbers between – 1 / 2 and 5 / 4

Solution:

Six rational numbers between – 1 / 2 and 5 / 4 can be calculated as below

LCM of 2, 4 = 4

– 1 / 2 = (- 1 × 2) / (2 × 2)

We get,

= – 2 / 4

Now, six rational numbers between – 1 / 2 and 5 / 4 are as follows:

– 1 / 4, 0, 1 / 4, 2 / 4, 3 / 4 and 4 / 4

5. Find three rational numbers between – 2 and – 1

Solution:

Three rational numbers between – 2 and – 1 can be calculated as below:

First rational number = 1 / 2 (- 1 – 2)

We get,

= – 3 / 2

Second rational number – 2 and – 3 / 2

= 1 / 2 {- 2 – (3 / 2)}

= 1 / 2 (- 7 / 2)

We get,

= – 7 / 4

Third rational number between – 3 / 2 and – 1

= 1 / 2 {(- 3 / 2) – 1}

= 1 / 2 (- 5 / 2)

= 1 / 2 × – 5 / 2

We get,

= – 5 / 4

Therefore, three rational numbers are – 7 / 4, – 3 / 2, – 5 / 4

6. Write ten rational numbers which are greater than 0.

Solution:

Ten rational numbers which are greater than 0

There can be the finite number of a rational number greater than 1.

Here, we shall take only 10 rational numbers.

The numbers are as follows:

(1 / 2), 1, (3 / 2), 2, (5 / 2), 3, (7 / 2), 4, (9 / 2), 5 etc.

7. Write five rational numbers which are smaller than – 4

Solution:

Five rational numbers which are smaller than – 4

These can be finite number of rational numbers smaller than – 4

Here, we shall take only 5 rational numbers.

The numbers are as follows:

(- 9 / 2), – 5, (- 11 / 2), – 6, (- 13 / 2), etc.

8. Identify the rational number which is different from the other three. Explain your reasoning

(- 5 / 11), (- 1 / 2), (- 4 / 9), (- 7 / 3)

Solution:

Given four rational number are,

(- 5 / 11), (- 1 / 2), (- 4 / 9), (- 7 / 3)

Among the given numbers,

– 7 / 3 is different from the other three numbers.

Because in – 7 / 3 its denominator is less than its numerator

In other numbers, denominators are greater than their numerators respectively.

Exercise 1.6

1. In a bag, there are 20 kg of fruits. If ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 71 kg of these fruits be oranges and ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 72 kg of these are apples and rest are grapes. Find the mass of the grapes in the bag.

Solution:

Given

Total fruits in a bag = 20 kg

Oranges =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 73kg i.e 43 / 6 kg

Apples =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 74kg i.e 26 / 3 kg

Remaining fruits in a bag = 20 – {(43 / 6) + (26 / 3)} kg

= 20 – {(43 + 52) / 6}

On further calculation, we get

= 20 – (95 / 6)

= (120 – 95) / 6

= 25 / 6

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 75kg

Therefore, the mass of the grapes in the bag is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 76kg

2. The population of a city is 6, 63,432. If 1 / 2 of the population are adult males and 1 / 3 of the population are adult females, then find the number of children in the city.

Solution:

Given

Population of a city = 6, 63,432

Population of adult males = (1 / 2) of 6,63,432

= 3,31,716

Population of adult females = (1 / 3) of 6,63,432

= 2,21,144

Remaining population can be calculated as below

Remaining population = 6,63,432 – (3,31,716 + 2,21,144)

= 6,63,432 – 5,52,860

We get,

= 1,10,572

Therefore, number of children in a city are 1,10,572

3. In an election of housing society, there are 30 voters. Each of them gives the vote. Three persons X, Y and Z are standing for the post of Secretary. If Mr X got 2 / 5 of the total votes and Mr Z got 1 / 3 of the total votes, then find the number of votes which Mr Y got.

Solution:

Given

Number of votes = 30

Number of person for election = X, Y, Z

X got (2 / 5) of total votes = (2 / 5) of 30

= (2 / 5) × 30

= 12

Z got 1 / 3 of total votes = 1 / 3 of 30

= (1 / 3) × 30

= 10

Remaining votes can be calculated as below

= 30 – (12 + 10)

= 30 – 22

We get,

= 8

Therefore, Mr Y got 8 votes

4. A person earns Rs 100 in a day. If he spent Rs ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 77 on food and Rs ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 78 on petrol. How much did he save on that day?

Solution:

Given

A person’s earning in a day = Rs 100

Money spent on food = Rs
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 79= Rs 100 / 7

Money spent on petrol = Rs
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 80= Rs 92 / 3

The savings of a person is calculated as follows:

Savings = Rs 100 – {(100 / 7 + 92 / 3)}

= Rs 100 – {(300 + 644) / 21}

On further calculation, we get

= Rs 100 – (944 / 21)

= (2100 – 944) / 21

= Rs 1156 / 21

= Rs
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 81

Hence, a person saved Rs
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 82on that day.

5. In an examination, 400 students appeared. If 2 / 3 of the boys and all 130 girls passed in the examination, then find how many boys failed in an examination?

Solution:

Given

Number of students appeared exams = 400

(2 / 3) of total boys and all 130 girls passed in the examination

Hence,

Number of total boys = 400 – 130

= 270

Number of boys passed = (2 / 3) of 270

= (2 / 3) × 270

= 180

So, number of boys failed = 270 – 180

= 90

Hence, 90 boys failed in an examination.

6. A car is moving at the speed of ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 83 km / h. Find how much distance will it cover in 9 / 10 hrs?

Solution:

Given

Speed of a car =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 84km / h = 122 / 3 km / h

Distance covered in 9 / 10 hour can be calculated as follows:

Distance = (122 / 3) × (9 / 10)

= 366 / 10

We get,

= 36.6 km

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 85km

Therefore, the distance covered by the car in 9 / 10 hours is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 86km

7. Find the area of a square lawn whose one side is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 87 m long.

Solution:

Given

One side of a square lawn =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 88m = 52 / 9 m

The area of a square lawn can be calculated as follows:

Area = (side)2

= (52 / 9)2

We get,

= 2704 / 81 sq. m

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 89sq. m

Therefore, the area of a square lawn is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 90sq. m

8. Perimeter of a rectangle is ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 91 m. If the length is ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 92 m, find its breadth.

Solution:

Given

Perimeter of a rectangle =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 93m

= 108 / 7 m

So,

Length + Breadth = (108 / 7) ÷ 2

= (108 / 7) × (1 / 2)

We get,

= 54 / 7 m

Given length =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 94

= 30 / 7 m

Hence, breadth of a rectangle can be calculated as,

Breadth = (54 / 7) – (30 / 7)

= 24 / 7

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 95m

Therefore, the breadth of a rectangle is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 96m

9. Rahul had a rope of ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 97 m long. He cut off a ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 98 m long piece, then he divided the rest of the rope into 3 parts of equal length. Find the length of each part.

Solution:

Given

Length of a rope =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 99m

Length of one piece of rope after cut off =
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 100m

Remaining length of a rope can be calculated as below

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 101
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 102

We get,

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 103m

= 876 / 5 m

This length divided into three equal parts

So, length of each part can be calcualted as follows:

Length of each part = (876 / 5) ÷ 3

= (876 / 5) × (1 / 3)

We get,

= 292 / 5 m

=
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 104m

Therefore, the length of each part of a rope is
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 105m

10. If ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 106 litre of petrol costs Rs ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 107, then find the cost of 4 litre of petrol.

Solution:

Given

Cost of
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 108litre = 7 / 2 litre of petrol = Rs
ML Aggarwal Solutions Mathematics Class 8 Chapter 1 - 109

= Rs 2163 / 8

Hence, the cost of one litre can be calculated as below:

Cost of one litre = Rs (2163 × 2) / (8 × 7)

The cost of 4 litre of petrol can be calculated as below

Cost of 4 litre = Rs (2163 × 2 ×4) / (8 × 7)

We get,

= Rs 309

Therefore, the cost of 4 litre of petrol is Rs 309

11. Ramesh earns Rs 40,000 per month. He spends 3 / 8 of the income on food, 1 / 5 of the remaining on LIC premium and then 1 / 2 of the remaining on other expenses. Find how much money is left with him?

Solution:

Ramesh earnings per month = Rs 40,000

Expenditure on food = (3 / 8) of Rs 40, 000

= Rs 15,000

Remaining amount = 40,000 – 15,000

= Rs 25,000

Expenditure on LIC premium = (1 / 5) of Rs 25,000

= Rs 5000

Remaining amount = Rs 25000 – Rs 5000

= Rs 20,000

Expenditure on other expenses = (1 / 2) of Rs 20,000

= Rs 10,000

Remaining amount left = Rs 20,000 – Rs 10,000

= Rs 10,000

Therefore, the remaining amount left with Ramesh is Rs 10,000

12. A, B, C, D and E went to a restaurant for dinner. A paid 1 / 2 of the bill, B paid 1 / 5 of the bill and rest of the bill was shared equally by C, D and E. What fractions of the bill was paid by each?

Solution:

Let us consider the total bill of the restaurant = 1

Bill paid by A = 1 / 2

Bill paid by B = 1 / 5

Remaining bill can be calculated as below:

Remaining bill = 1 – {(1 / 2) + (1 / 5)}

= 1 – {(5 + 2) / 10}

= 1 – (7 / 10)

We get,

= 3 / 10

Shares of the three persons = (3 / 10) ÷ 3

= (3 / 10) × (1 / 3)

= 1 / 10

Therefore, each paid (1 / 10) of the bill.

13. 2 / 5 of total number of students of a school come by car while 1 / 4 of students come by bus to school. All the other students walk to school of which 1 / 3 walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in the school?

Solution:

Let total number of students be 1

Students who come by car = 2 / 5

Students who come by bus = 1 / 4

Students who come by walking = 1 / 3 of remaining

Rest students = 1 – (2 / 5 + 1 / 4)

= 1 – (8 + 5) / 20

= 1 – (13 / 20)

We get,

= 7 / 20

Number of students who come by walking can be calculated as below

Number of students who come by walking = 1 / 3 of 7 / 20

= 7 / 60

Now, 7 / 60 of total students = 224

Total students = (224 × 60) / 7

= 32 × 60

= 1920

Hence, 1920 students study in the school

14. A mother and her two sons got a room constructed for Rs 60,000. The elder son contributes 3 / 8 of his mother’s contribution while the younger son contributes 1 / 2 of his mother’s share. How much do the three contribute individually?

Solution:

The cost of a room = Rs 60,000

Elder son contribution = 3 / 8 of his mother’s contribution

Younger son contribution = 1 / 2 of his mother’s share

Let the mother contribution be 1

Elder son’s contribution = 3 / 8

Younger son’s contribution = 1 / 2

Now,

Ratios in their share = 1: (3 / 8): (1 / 2)

= 8: 3: 4

Sum of ratios = 8 + 3 + 4

= 15

Therefore,

Mother’s share = (60000 × 8) / 15

= Rs 32000

Elder son’s share = (60000 × 3) / 15

= Rs 12000

Younger son’s share = (60000 × 4) / 15

= Rs 16000

15. In a class of 56 students, the number of boys is 2 / 5 th of the number of girls. Find the number of boys and girls.

Solution:

Total number of students in a class = 56

Let the number of girls be 1

Then number of boys will be = 2 / 5 of 1

= 2 / 5

Ratios in girls and boys = 1: (2 / 5)

= 5: 2

Number of girls = {56 / (5 + 2)} × 5

= (56 / 7) × 5

= 40

And number of boys = (56 / 7) × 2

= 16

Therefore, number of boys = 16 and number of girls = 40

16. A man donated 1 / 10 of his money to a school, 1 / 6 th of the remaining to a church and the remaining money he distributed equally among his three children. If each child gets Rs 50000, how much money did the man originally have?

Solution:

Let the money of a man be 1

Money donated to a school = 1 / 10

Remaining money = 1 – (1 / 10)

= 9 / 10

Money donated to a church = 1 / 6 of 9 / 10

= 3 / 20

Hence, remaining money = (9 / 10) – (3 / 20)

= (18 – 3) / 20

= 15 / 20

A man divides equally to his three children

Hence,

Share of each child = (15 / 20) ÷ 3

= (15 / 20) × (1 / 3)

= 1 / 4

Here, each child gets Rs 50000

Therefore, his total money = Rs 50000 × (4 / 1)

= Rs 200000

17. If 1 / 4 of a number is added to 1 / 3 of that number, the result is 15 greater than half of that number. Find the number.

Solution:

Let us consider the number as x

Then as per the condition,

(1 / 4) x + (1 / 3) x – (1 / 2) x = 15

(3x + 4x – 6x) / 12 = 15

(1 / 12) x of a number = 15

x = 15 × 12 / 1

x = 180

Therefore, the required number is 180

18. A student was asked to multiply a given number by 4 / 5. By mistake, he divided the given number by 4 / 5. His answer was 36 more than the correct answer. What was the given number?

Solution:

Let the given number be x

According to the condition,

x × 4 / 5 = (4 / 5) x

But by mistake a student divides the given number

Then,

x ÷ 4 / 5 = x × 5 / 4

= (5 / 4) x

Hence,

(5 / 4) x – (4 / 5) x = 36

(25x – 16x) / 20 = 36

9x / 20 = 36

9x = 36 × 20

x =(36 × 20) / 9

We get,

x = 80

Therefore, the given number is 80

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