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 can score good marks easily as it merely requires thorough practice. By referring to the ML Aggarwal solutions, students can cross-check their answers while solving 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 syllabus. ML Aggarwal Solutions, available in PDFs, enhance the problem-solving skills of 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
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 + 
(ii) – 4 / 9 + 
Solution:
(i) Given
5 / 11 +
This can be written as,
5 / 11 + 39 / 9
Taking the 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 numbers,
45 / 99 + 429 / 99 = (45 + 429) / 99
= 474 / 99
=
Dividing the numerator and denominator by 3,
= 4 (78 ÷ 3) / (99 ÷ 3)
We get,
=
(ii) Given
– 4 / 9 +
This can be written as,
– 4 / 9 + 38 / 13
Taking the 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 numbers,
– 52 / 117 + 342 / 117 = (- 52 + 342) / 117
We get,
= 290 / 117
=
3. Verify the 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 numbers,
= – 4 / 3 + 3 / 7
Taking the L.C.M., we get,
= (- 28 + 9) / 21
= – 19 / 21
And
3 / 7 + (- 4 / 3)
Again, taking the 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 the L.C.M., we get,
= (6 + 5) / 15
= 11 / 15
And 1 / 3 + 2 / 5
Again, taking the 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 numbers,
= 9 / 11 + 2 / 13
Taking the L.C.M., we get,
= (117 + 22) / 143
We get,
= 139 / 143
And 2 / 13 + 9 / 11
Again, taking the 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 the 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 the L.C.M., we get,
= (12 + 70) / 63
= 82 / 63
=
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 (Existence 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 the 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) 
from – 3 / 7
(ii) – 4 / 9 from 
(iii) 
from 
Solution:
(i)
from – 3 / 7
= – 3 / 7 – (13 / 5)
Taking the L.C.M., we get,
= (- 15 – 91) / 35
= – 106 / 35
=
Hence, the subtraction of
from – 3 / 7 is
(ii) – 4 / 9 from
This can be written as,
– 4 / 9 from 29 / 8
= 29 / 8 – (- 4 / 9)
= 29 / 8 + 4 / 9
Taking the L.C.M., we get,
= (261 + 32) / 72
= 293 / 72
=
(iii)
from
This can be written as,
= – 16 / 5 from – 43 / 9
= – 43 / 9 – (- 16 / 5)
= – 43 / 9 + 16 / 5
Taking the L.C.M., we get,
= (- 215 + 144) / 45
We get,
= – 71 / 45
=
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 numbers is – 2 / 7
Hence, the other number is calculated as follows:
Other number = 3 / 5 – (- 2 / 7)
= 3 / 5 + 2 / 7
Taking the 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 the 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 
to get
?
Solution:
The required number can be calculated as follows:
(
) – (
)
This can be written as,
(- 23 / 5) + (7 / 2)
On further calculation, we get
= (- 46 + 35) / 10
= – 11 / 10
=
Therefore, the required number is
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 the L.C.M., we get,
= (133 + 284) / 84
= 417 / 84
=
=
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 the L.C.M., we get,
= (- 20 – 14) / 35
= – 34 / 35
And
y – x = 2 / 5 – (- 4 / 7)
= 2 / 5 + 4 / 7
Taking the 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 the 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 the 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 the subtraction of rational numbers.
(v) Associative property does not hold for the subtraction of rational numbers.
(vi) 0 is the identity element for the 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)
×
(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)
×
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
=
(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 the commutative property of multiplication for the following pairs of rational numbers:
(i) 4 / 5 and – 7 / 8
(ii) 
and 
(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)
and
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 also name the property:
(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 the 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 the 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 =
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 terms, 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 the 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 the 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 the 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. Is 4 / 5 the multiplicative inverse of 
? 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 the 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
=
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) 
÷ (- 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)
÷ (- 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
=
(iii) – 8 / 9 ÷ – 3 / 5
= – 8 / 9 × 5 / – 3
= – 40 / – 27
= {- 40 × (- 1)} / {- 27 × (- 1)}
We get,
= 40 / 27
=
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 the division is not true.
(iv) The given statement is true.
(v) The given statement is true.
(vi) The given statement is false.
Correct: It is a rational number.
3. The product of two rational numbers is – 11 / 12. If one of them is 
, find the other.
Solution:
Given
Product of two rational numbers = – 11 / 12
One of the numbers =
= 22 / 9
The other number is calculated as given 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 given 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
=
Therefore, the required number is
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,
The 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 given 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
=
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)
(iii) – 9 / 7
(iv) – 2 / – 5
Solution:
(i) 11 / 4 =
The given rational number on the number line is as shown below:
(ii) 
The given rational number on the number line is as shown below:
(iii) – 9 / 7 =
The given rational number on the number line is as shown below:
(iv) – 2 / – 5 = – 2 × (- 1) / – 5 × (- 1)
We get,
= 2 / 5
The given rational number on the number line is shown as below:
2. Write the rational numbers for each point labelled with a letter:
Solution:
(i) The rational numbers for each point labelled with a letter are as follows:
A = 3 / 7
B = 7 / 7 = 1
C = 8 / 7 =
D = 12 / 7 =
E = 13 / 7 =
(ii) The rational numbers for each point labelled 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 shown 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 shown 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
The 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 a 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 is 20 kg of fruits. If 
kg of these fruits are oranges and 
kg of these are apples, and the rest are grapes, find the mass of the grapes in the bag.
Solution:
Given
Total fruits in a bag = 20 kg
Oranges =
kg i.e 43 / 6 kg
Apples =
kg 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
=
kg
Therefore, the mass of the grapes in the bag is
kg
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
The population of adult males = (1 / 2) of 6,63,432
= 3,31,716
The population of adult females = (1 / 3) of 6,63,432
= 2,21,144
Remaining population can be calculated as given below:
Remaining population = 6,63,432 – (3,31,716 + 2,21,144)
= 6,63,432 – 5,52,860
We get,
= 1,10,572
Therefore, the number of children in a city is 1,10,572.
3. In an election of the housing society, there are 30 voters. Each of them gives a 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 persons 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 the total votes = 1 / 3 of 30
= (1 / 3) × 30
= 10
The remaining votes can be calculated as given 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 
on food and Rs 
on petrol, how much did he save on that day?
Solution:
Given
A person’s earnings in a day = Rs 100
Money spent on food = Rs
= Rs 100 / 7
Money spent on petrol = Rs
= 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
Hence, the person saved Rs 
on 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 the examination.
Solution:
Given
The number of students who appeared in the exam = 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, the number of boys who failed = 270 – 180
= 90
Hence, 90 boys failed in the examination.
6. A car is moving at the speed of 
km / h. Find how much distance it will cover in 9 / 10 hrs.
Solution:
Given
Speed of a car =
km / h = 122 / 3 km / h
Distance covered in 9 / 10 hours can be calculated as follows:
Distance = (122 / 3) × (9 / 10)
= 366 / 10
We get,
= 36.6 km
=
km
Therefore, the distance covered by the car in 9 / 10 hours is
km
7. Find the area of a square lawn whose one side is
m long.
Solution:
Given
One side of a square lawn =
m = 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
=
sq. m
Therefore, the area of a square lawn is
sq. m
8. Perimeter of a rectangle is 
m. If the length is 
m, find its breadth.
Solution:
Given
The perimeter of a rectangle =
m
= 108 / 7 m
So,
Length + Breadth = (108 / 7) ÷ 2
= (108 / 7) × (1 / 2)
We get,
= 54 / 7 m
Given length =
= 30 / 7 m
Hence, the breadth of a rectangle can be calculated as,
Breadth = (54 / 7) – (30 / 7)
= 24 / 7
=
m
Therefore, the breadth of a rectangle is
m
9. Rahul had a rope of 
m long. He cut off a 
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 =
m
Length of one piece of rope after cut off =
m
The remaining length of a rope can be calculated as shown below:
=
–
We get,
=
m
= 876 / 5 m
This length divided into three equal parts
So, the length of each part can be calculated as follows:
Length of each part = (876 / 5) ÷ 3
= (876 / 5) × (1 / 3)
We get,
= 292 / 5 m
=
m
Therefore, the length of each part of a rope is
m
10. If 
litre of petrol costs Rs 
, then find the cost of 4 litres of petrol.
Solution:
Given
Cost of
litre = 7 / 2 litre of petrol = Rs
= Rs 2163 / 8
Hence, the cost of one litre can be calculated as shown below:
Cost of one litre = Rs (2163 × 2) / (8 × 7)
The cost of 4 litres of petrol can be calculated as shown below:
Cost of 4 litre = Rs (2163 × 2 ×4) / (8 × 7)
We get,
= Rs 309
Therefore, the cost of 4 litres of petrol is Rs 309.
11. Ramesh earns Rs 40,000 per month. He spends 3 / 8 of his 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
The 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 the rest of the bill was shared equally by C, D and E. What fractions of the bill were 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
The 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 the total number of students of a school come by car, while 1 / 4 of the 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 the 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
The number of students who come by walking can be calculated as shown below:
Number of students who come by walking = 1 / 3 of 7 / 20
= 7 / 60
Now, 7 / 60 of the 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’s contribution = 3 / 8 of his mother’s contribution
Younger son’s contribution = 1 / 2 of his mother’s share
Let the mother’s 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, the 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, the 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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