NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable

NCERT
SOLUTIONS
CLASS - 8TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Class : 8th
Subject : Maths
Chapter : 2
Chapter Name : Linear Equations in One Variable

Exercise 2.1

Q1 Solve the following equations.
1.1 x – 2 = 7
1.2 y + 3 = 10
1.3 6 = z + 2
1.4 + x =
3 17

7 7

1.5 6 x=12
1.6 = 10
t

5

1.7 2x

3
= 18

1.8 1.6 =
y

1.5

1.9 7x − 9 = 16
1.10 14y − 8 = 13
1.11 17 + 6p = 9
1.12 + 1 =
x 7

3 15

Answer. 1.1
x–2=7
Transposing 2 to R.H.S, we obtain
x = 7+2 = 9
1.2
y + 3 = 10
Transposing 3 to R.H.S, we obtain
y = 10 - 3 = 7
1.3
6=z+2
Transposing 2 to L.H.S, we obtain
6-2=z
z=4
1.4
3 17
+ x =
7 7

Transposing 3

7
to R. H.S, we obtain

Page 1 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

17 3 14
x = − = = 2
7 7 7

1.5
6 x=12
Dividing both sides by 6, we obtain
6x 12
=
6 6

x = 2

1.6
t
= 10
5

Multiplying both sides by 5, we obtain
g
× 5 = 10 × 5
5

t = 50

1.7
2x
= 18
3

Multiplying both sides by 3/2 , we obtain

2x 3 3
× = 18 ×
3 2 2

x = 27

1.8
y
1.6 =
1.5

Multiplying both sides by 1.5, we obtain
y
1.6 × 1.5 = × 1.5
1.5

2.4 = y

1.9
7x − 9 = 16

Transposing 9 to R.H.S, we obtain

7x = 16 + 9

7x = 25

Dividing both sides by 7, we obtain

7x 25
=
7 7

25
x =
7

1.10
14y − 8 = 13

Transposing 8 to R.H.S, we obtain
14y = 13 + 8
14y = 21
Dividing both sides by 14, we obtain
14y 21
=
14 14

3
y =
2

1.11
17 + 6p = 9

Transposing 17 to R.H.S, we obtain

Page 2 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

6p = 9 - 17
6p = -8
Dividing both sides by 6, we obtain

6p 8
= −
6 6

4
p = −
3

1.12
x 7
+ 1 =
3 15

Transposing 1 to R.H.S, we obtain
x 7
= − 1
3 15

x 7−15
=
3 15

x 8
= −
3 15

Multiplying both sides by 3, we obtain

x 8
× 3 = − × 3
3 15

8
x = −
5

Page : 23 , Block Name : Exercise 2.1

Exercise 2.2

Q1 If you subtract 1

2
from a number and multiply the result by 1

2
, you get 1

2
. What is the
number?

Answer.
Let the number be x . According to the question,

1 1 1
(x − ) × =
2 2 8

On multiplying both sides by 2, we obtain

1 1 1
(x − ) × × 2 = × 2
2 2 8

1 1
x − =
2 4

On transposing 1/2 to R.H.S, we obtain

1 1
x = +
4 2
1 + 2 3
= =
4 4

Therefore, the number is 3

4

Page : 28 , Block Name : Exercise 2.2

Page 3 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Q2 The perimeter of a rectangular swimming pool is 154m. Its length is 2 m more than twice
its breadth. What are the length and the breadth of the pool?

Answer. Let the breadth be x m. The length will be (2x + 2) m.
Perimeter of swimming pool = 2(l + b) = 154 m
2(2x + 2 + x) = 154
2(3x + 2) = 154
Dividing both sides by 2, we obtain

2(3x+2) 154
=
2 2

3x + 2 = 77

On transposing 2 to R.H .S, we obtain
3x = 77 - 2
3x = 75
3x 75
=
3 3

x = 25

2x + 2 = 2 × 25 + 2 = 52

Hence, the breadth and length of the pool are 25 m and Q2 Which of the following numbers
would have digit 6 at unit place.
2 2 2
(i) 19 (ii) 24 (iii) 26

2 2
(iv) 36 (v) 34

52 m respectively

Page : 28 , Block Name : Exercise 2.2

Q3 The base of an isosceles triangle is 4

3
cm . The perimeter of the triangle is 4 2

15
cm . What is
the length of either of the remaining equal sides?

Answer. Let the length of equal sides be x cm.
Perimeter = x cm + x cm + Base = 4 cm 2

15
4 62
2x + =
3 15

On transposing 4

3
to R. H.S, we obtain
62 4
2x = −
15 3

62−4×5 62−20
2x = =
15 15

42
2x =
15

On dividing both sides by 2, we obtain

2x 42 1
= ×
2 15 2

7 2
x = = 1
5 5

Therefore, the length of equal sides is 1 2

5
cm

Page 4 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Page : 28 , Block Name : Exercise 2.2

Q4 Sum of two numbers is 95. If one exceeds the other by 15, nd the numbers.

Answer. Let one number be x. Therefore, the other number will be x + 15.
According to the question,
x + x + 15 = 95
2x + 15 = 95
On transposing 15 to R.H.S, we obtain
2x = 95 -15
2x = 80
On dividing both sides by 2, we obtain
2x 80
=
2 2

x = 40

x + 15 = 40 + 15 = 55

Hence, the numbers are 40 and 55.

Page : 28 , Block Name : Exercise 2.2

Q5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Answer. Let the common ratio between these numbers be x. Therefore, the numbers will be 5x
and 3x respectively.
Difference between these numbers = 18
5x - 3x = 18
2x = 18
Dividing both sides by 2,
2x 18
=
2 2

x = 9

First number = 5x = 5 × 9 = 45

Second number = 3x = 3 × 9 = 27

Page : 28 , Block Name : Exercise 2.2

Q6 Three consecutive integers add up to 51. What are these integers?

Answer. Let three consecutive integers be x, x + 1, and x + 2.
Sum of these numbers = x + x + 1 x + 2 = 51
3x + 3 = 51
On transposing both sides by 3, we obtain

Page 5 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

3x 48
=
3 3

x = 16

x + 1 = 17

x + 2 = 18

Hence, the consecutive integers are 16, 17, and 18.

Page : 28 , Block Name : Exercise 2.2

Q7 The sum of three consecutive multiples of 8 is 888. Find the multiples.

Answer. Let the three consecutive multiples of 8 be 8x, 8(x + 1) + 8 (x + 2).
Sum of these numbers = 8x + 8(x + 1) + 8 (x+2) = 88
8(x + x + 1 + x + 2) = 888
8(3x + 3 ) = 888
On Dividing both sides by 8, we obtain
8(3x+3) 888
=
8 8

3x + 3 = 111

On transposing 3 to R. H. S, we obtain

3x = 111 − 3

3x = 108

On dividing both sides by 3, we obtain

3x 108
=
3 3

x = 36

First multiple = 8x = 8 × 36 = 288

Second multiple = 8(x + 1) = 8 × (36 + 1) = 8 × 37 = 296

Third multiple = 8(x + 2) = 8 × (36 + 2) = 8 × 38 = 304

Hence, the required numbers are 288, 296, and 304 .

Page : 28 , Block Name : Exercise 2.2

Q8 Three consecutive integers are such that when they are taken in increasing order and
multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Answer. Let three consecutive be x, x+1 , x + 2 . according to the question,
2x + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
On transposing 11 to R.H.S, we obtain
9x = 74 -11
9x = 63
On dividing both sides by 9, we obtain

Page 6 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

9x 63
=
9 9

x = 7

x + 1 = 7 + 1 = 8

x + 2 = 7 + 2 = 9

Hence, the numbers are 7, 8, and 9.

Page : 28 , Block Name : Exercise 2.2

Q9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages
will be 56 years. What are their present ages?

Answer. Let common ratio between Rahul's age and Haroon's age be x.
Therefore, age of Rahul and Haroon will be 5x years and 7 x years respectively. After
4 years, the age of Rahul and Haroon will be (5x + 4) years and (7 x + 4) years respectively.
According to the given question, after 4 years, the sum of the ages of Rahul and Haroon is 56
years.
(5x + 4 + 7x + 4) = 56
12x + 8 = 56
12x = 48
On dividing both sides 12, we obtain
12x 48
=
12 12

x = 4

Rahul's age = 5x years = (5 × 4) years = 20 years

Haroon's age = 7x years = (7 × 4) years = 28 years

Page : 28 , Block Name : Exercise 2.2

Q10 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more
than the number of girls. What is the total class strength?

Answer. Let the common ratio between the number of boys and numbers of girls be x.
Number of boys = 7x
Number of girls = 5x
According to the given question,
Number of boys = Number of girls + 8
7x = 5x + 8
On transposing 5x to L.H.S, we obtain
7x - 5x = 8,br/>
2x = 8
On dividing both sides by 2, we obtain

Page 7 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

2x 8
=
2 2

x = 4

Number of boys = 7x = 7 × 4 = 28

Number of girls = 7x = 5 × 4 = 20

Hence, total class strength = 28 + 20 = 48 students

Page : 28 , Block Name : Exercise 2.2

Q11 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older
than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of
them?

Answer. Let Baichung's father's age be x years. Therefore, Baichung's age and Baichung's
grandfather's age will be (x 29) years and (x + 26) years respectively.
According to the given question, the sum of the ages of these 3 people is 135 years
x + x -29 + x + 26 = 135
3x - 3 = 135
On transposing 3 to R.H.S, we obtain
3x = 135 + 3
3x = 138
On dividing both sides bY 3, we obtain
3x 138
=
3 3

x = 46
Baichung's father's age = x years =46 years
Baichung's age =( x -29 ) years = (46 - 29) years = 17 years
Baichung's grandfather's age = (x + 26) years = (46 + 26) years =72 years

Page : 28 , Block Name : Exercise 2.2

Q12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s
present age?

Answer. Let Ravi's present age be x years.
Fifteen years later, Ravi's age = 4 x His present age
x + 15 = 4x
On transposing x to R.H.S, we obtain
15 = 4x - x
15 = 3x
On dividing both sides by 3,we obtain
15 3x
=
3 3

5 = x

Hence, Ravi's present age = 5 years

Page : 28 , Block Name : Exercise 2.2

Page 8 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Q13 A rational number is such that when you multiply it by 5

2
and add 2

3
to the product, you
get − 12
7
. What is the number?

Answer. Let the number be x.
According to the given question,
5 2 7
x + = −
2 3 12

2
On transposing to R.H.S, we obtain
3

5 7 2
x = − −
2 12 3

−7−(2×4)
5
x =
2 12
5 15
x = −
2 12

On multiplying both sides by 5, we obtain

15 2 1
x = − × = −
12 5 2

Hence, the rational number is 2 .

Page : 28 , Block Name : Exercise 2.2

Q14 Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100,Rs 50 and
Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with
Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Answer. Let the common ratio between the numbers of notes of different denominations be x.
Therefore, numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be2x, 3x, and
5x respectively.
Amount of Rs 100 notes = Rs (100 x 2x) = Rs 200x
Amount of Rs 50 notes = Rs (50 x 3x)= Rs 150x
Amount of Rs 10 notes = Rs (10 x 5x) = Rs 50x
It is given that total amount is Rs 400000.
200X + 150X + 50X = 400000
⇒ 400x = 400000

On dividing both sides by 400, we obtain
x = 1000
Number of Rs 100 notes = 2x = 2 x 1000 = 2000
Number of Rs 50 notes = 3x = 3 x 1000 = 3000
Number of Rs 10 notes = 5x = 5 x 1000 = 5000

Page : 28 , Block Name : Exercise 2.2

Q15 I have a total of Rs 300 in coins of denomination Rs 1, Rs 2 and Rs 5. The number of Rs 2
coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of
each denomination are with me?

Page 9 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Answer. Let the number of Rs 5 coins be x.
Number of Rs 2 coins = 3 x Number of Rs 5 coins = 3x
Number of Rs 1 coins = 160 - ( Number of coins of Rs 5 and of Rs 2)
= 160 - ( 3x + x) = 160 - 4x
Amount of Re 1 coins = Rs [1 x (160 - 4x)] = Rs (160 - 4x)
Amount of Rs 2 coins = Rs (2 x 3x) = Rs 6x
Amount of Rs 5 coins = Rs ( 5 x x) = Rs 5x
It is given that the total amount is Rs 300.
160 - 4x + 6x + 5x = 300
160 + 7x = 300
On transposing 160 to R.H.S, we obtain
7x = 300 - 160
7x = 140
On dividing both sides by 7, we obtain
7x 140
=
7 7

x = 20

Number of Re 1 coins = 160 − 4x = 160 − 4 × 20 = 160 − 80 = 80

Number of Rs 2 coins = 3x = 3 × 20 = 60

Number of Rs 5 coins = x = 20

Page : 28 , Block Name : Exercise 2.2

Q16 The organisers of an essay competition decide that a winner in the competition gets a
prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money
distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63.

Answer. Let the number of winners be x. Therefore, the number of participants who did not
win will be 63 — x.
Amount given to the winners = Rs (100 x x) = Rs 100x
Amount given to the participants who did not win = Rs [25(63 - x)] = Rs (1575 - 25x)
According to the given question,
100x + 1575 - 25x = 3000
On transposing 1575 to R.H.S, we obtain
75x = 3000 - 1575
75x = 1425
On dividing both sides by 75, we obtain
75x 1425
=
75 75

x = 19
Hence, number of winners = 19

Page : 28 , Block Name : Exercise 2.2

Exercise 2.3

Page 10 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Q1 Solve the following equations and check your results.
1.1 3x = 2x + 18
1.2 5t – 3 = 3t – 5
1.3 5x + 9 = 5 + 3x
1.4 4z + 3 = 6 + 2z
1.5 2x – 1 = 14 – x
1.6 8x + 4 = 3 (x – 1) + 7
1.7 x = (x + 10)
4

5

1.8 2x

3
+ 1 =
7x

15
+ 3

1.9 2y + 5

3
=
26

3
− y

1.10 3m = 5m − 8

5

Answer. 1.1
3x = 2x + 18
On transposing 2x to L.H.S, we obtain
3x - 2x = 18
x = 18
L.H.S = 3x = 3 x 18 = 54
R.H.S = 2x + 18 = 2 x 18 + 18 = 36 + 18 = 54
L.H.S. = R.H.S.
Hence, the result obtained above is correct.
1.2
5t – 3 = 3t – 5
On transposing 3t to L. H . S and − 3 to R.H.S, we obtain

5t − 3t = −5 − (−3)

2t = −2

On dividing both sides by 2, we obtain

t = −1

L.H.S = 5t − 3 = 5 × (−1) − 3 = −8

R.H.S = 3t − 5 = 3 × (−1) − 5 = −3 − 5 = −8

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

Hence, the result obtained above is correct.

1.3
5x + 9 = 5 + 3x
On transposing 3x to L.H.S and 9 to R.H.S, we obtain

Page 11 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

5x − 3x = 5 − 9

2x = −4

On dividing both sides by 2, we obtain

x = −2

L. H. S. = 5x + 9 = 5 × (−2) + 9 = −10 + 9 = −1

R. H.S. = 5 + 3x = 5 + 3 × (−2) = 5 − 6 = −1

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

Hence, the result obtained above is correct.
1.4
4z + 3 = 6 + 2z
On transposing 2z to L.H.S and 3 to R.H.S, we obtain
4z − 2z = 6 − 3

2z = 3

Dividing both sides by 2, we obtain

3
z =
2

3
L.H.S. = 4z + 3 = 4 × ( ) + 3 = 6 + 3 = 9
2

3
R.H.S. = 6 + 2z = 6 + 2 × ( ) = 6 + 3 = 9
2

L.H.S. = R.H.S.
Hence, the result obtained above is correct.
1.5
2x – 1 = 14 – x
Transposing x to L.H.S and 1 to R.H.S, we obtain
2x + x = 14 + 1
3x = 15
Dividings both sides by 3, we obtain
x=5
L.H.S = 2x -1 = 2 x (5) - 1 = 10 - 1 =9
R.H.S = 14 - x = 14 - 5 = 9
L.H.S. = L.H.S.
Hence, the result obtained above is correct.
1.6
8x + 4 = 3 (x – 1) + 7
8x + 4 = 3(x - 1) + 7
8x + 4 = 3x - 3 + 7
Transposing 3x to L.H.S and 4 to R.H.S, we obtain
8x - 3x = - 3 + 7 - 4
5x = -7 + 7
x=0
L.H.S = 8x + 4 = 8 x(0) + 4 = 4
R.H.S = 3(x-1 )+ 7 = 3 (0-1) + 7 = - 3 + 7 = 7
L.H.S. = R.H.S.
Hence, the result obtained above is correct.

Page 12 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

1.7
4
x = (x + 10)
5

Answer. Multiplying both sides by 5, we obtain
5x - 4x = 40
x = 40
L.H.S = x = 40
4

R.H.S= 5 5
(x+10)
=
4

5
(40 + 10) =
4

5
× 50 = 40

L.H.S. = R.H.S.
Hence, the result obtained above is correct.
1.8
2x 7x
+ 1 = + 3
3 15
7x
Transposing to L.H.S and 1 to R.H.S, we obtain
15

2x 7x
− = 3 − 1
3 15

5×2x−7x
= 2
15

3x
= 2
15

x
= 2
5

Multiplying both sides by 5, we obtain
2x 2 × 10 2 × 10 + 1 × 3 23
x = 10, + 1 + 1 = =
3 3 3 3

7x 7 × 10 7 × 2 14 14 + 3 × 3 23
+ 3 = + 3 = + 3 = + 3 = =
15 15 3 3 3 3

L.H.S. = R.H.S.
Hence, the result obtained above is correct.
1.9
5 26
2y + = − y
3 3
5
Transposing y to L.H.S and to R.H.S, we obtain
3

26 5
2y + y = −
3 3

21
3y = = 7
3

Dividing both sides by 3, we obtain

7
y =
3

5 7 5 14 5 19
L.H.S = 2y + = 2 × + = + =
3 3 3 3 3 3

26 26 7 19
R.H.S = − y = − =
3 3 3 3

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

Hence, the result obtained above is correct.

1.10
8
3m = 5m −
5

Page 13 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Transposing 5m to L.H.S, we obtain

8
3m − 5m = −
5

8
−2m = −
5

Dividing both sides by − 2, we obtain

4
m =
5

4 12
L.H.S =3m = 3 × =
5 5

8 4 8 12
R.H.S = 5m − = 5 × − =
5 5 5 5

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

Hence, the result obtained above is correct.

Page : 30 , Block Name : Exercise 2.3

Exercise 2.4

Q1 Amina thinks of a number and subtracts 5

2
from it. She multiplies the result by 8. The
result now obtained is 3 times the same number she thought of. What is the number?

Answer. Let the number be x. <>
According to the given question,
5
8 (x − )
2

8x − 20 = 3x

Transposing 3x x to L.H.S and -20 to R.H.S, we obtain

8x − 3x = 20

5x − 3x = 20

Dividing both sides by 5, we obtain

x = 4

Hence, the number is 4 .

Page : 31 , Block Name : Exercise 2.4

Q2 A positive number is 5 times another number. If 21 is added to both the numbers, then one
of the new numbers becomes twice the other new number. What are the
Numbers?

Answer. Let the numbers be x and 5x. According to the question,
21 + 5x = 2(x+ 21)
21+5x = 2x+42

Page 14 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Transposing 2x to L.H.S and 21 to R.H.S, we obtain
5x - 2x = 42 - 21
3x = 21
Dividing both sides by 3, we obtain
X=7
5x = 5 x 7 = 35
Hence, the numbers are 7 and 35 respectively.

Page : 31 , Block Name : Exercise 2.4

Q3 Sum of the digits of a two digit number is 9. When we interchange the digits it is found
that the resulting new number is greater than the original number by 27. What is the two-digit
number?

Answer. Let the digits at tens place and ones place be x and 9 - x respectively.
Therefore, new number after interchanging the digits = 10(9 - x) + x
= 90 - 10x + x
= 90 - 9x
According to the given question,
New number Original number + 27
90 - 9x = 9x + 9 + 27
90 - 9x = 9x + 36
Transposing 9x to R.H.S and 36 to L.H.S, we obtain
90 - 36 = 18x
54 = 18x
Dividing both sides by 18, we obtain
3 = x and 9 - x = 6
Hence, the digits at tens place and ones place of the numbers are 3 and 6 respectively.
Therefore, the two-digits number is 9x + 9 = 9 x 3 + 9 = 36

Page : 31 , Block Name : Exercise 2.4

Q4 One of the two digits of a two digit number is three times the other digit. If you
interchange the digit of this two-digit number and add the resulting number to the original
number, you get 88. What is the original number?

Answer. Let the digits at tens and ones place be x and 3x respectively.
Therefore, original number = I0x + 3x = 13x
On interchanging the digits, the digits at ones place and tens place will be x and 3x
respectively.
According to the given question,
Original number + New number = 88
13x + 31x = 88
44x = 88
Dividing both sides by 44, we obtain

Page 15 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

x=2
Therefore, original number = 13x = 13 x 2 = 26
By considering the tens place and ones place as 3x and x respectively, the two-digit number
obtained is 62.
Therefore, the two-digits number may be 26 or 62.

Page : 31 , Block Name : Exercise 2.4

Q5 Shobo's mother's present age is six times Shobo's present age. Shobo's age ve years from
now will be one third of this mother's present age. What are their present
ages?

Answer. Let Shobo's age be x years. Therefore, his mother's age will be 6x years.
According to the given question,
Shobo's mother's present age
After 5 years, Shobo'sage =
3

6x
x + 5 =
3

x + 5 =2x
Transposing x to R.H.S, we obtain
5 = 2x - x
5=x
6x = 6 x 5 = 30
Therefore, the present ages of Shobo and Shobo’s mother will be 5 years and 30 years
respectively.

Page : 31 , Block Name : Exercise 2.4

Q6 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and
breadth of the plot are in the ratio 11:4. At the rate Rs 100 per metre it will cost the village
panchayat Rs 75, 000 to fence the plot. What are the dimensions of the plot?

Answer. Let the common ratio between the length and breadth of the rectangular plot be x.
Hence, the length and breadth of the rectangular plot will be 1 lx m and 4x m
respectively.
Perimeter of the plot = 2(Length + Breadth) = [2(11x + 4x)]m = 30xm
It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75,
000.
:. 100 x Perimeter = 75000
100 x 30x = 75000
3000x = 75000
Dividing both sides by 3000, we obtain
x = 25
Length = 1 1x m = (11 x 25) m = 275 m
Breath = 4x m = (4 x 25) m = 100 m
Hence, the dimensions of the plot are 275 m and 100 m respectively.

Page 16 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Page : 31 , Block Name : Exercise 2.4

Q7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him
Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 2 meters of the
trouser material he buys 3 metres of the shirt material. He sells the materials at 12% and 10%
pro t respectively. His total sale is Rs 36660. How much trouser material did he buy?

Answer. Let 2x m of trouser material and 3x m of shirt material be bought by him.
Per metre selling price of trouser material = Rs(90 + 90×12

100
) = Rs 100.80

Per metre selling price of shirt material Rs(50 + 50×10

100
) = Rs 55

Given that, total amount Of selling = Rs 36660
100.80 x (2x) + 55 x (3x) = 36660
201.60x + 165x = 36660
366.60x = 36660
Dividing both sides by 366.60, we obtain
x = 100
Trouser material = 2x m = (2 x 100) m = 200 m

Page : 31 , Block Name : Exercise 2.4

Q8 Half Of a herd Of deer are grazing in the eld and three fourths Of the remaining are
playing nearby. The rest 9 are drinking water from the pond. Find the number Of deer in the
herd.

Answer. Let the number of deer be x.
Number of deer grazing in the eld = x

2

3
Number of deer playing nearby = × Number of remaining deer
4
3 x 3 x 3x
= × (x − ) = × =
4 2 4 2 8

Number of deer drinking water from the pond = 9

x 3x
x − ( + ) = 9
2 8

4x+3x
x − ( ) = 9
8

7x
x − = 9
8

x
= 9
8

Multiplying both sides by 8, we obtain
x = 72
Hence, the total number Of deer in the herd is 72.

Page : 32 , Block Name : Exercise 2.4

Page 17 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

Q9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her.
Find their present ages

Answer. Let the granddaughter's age be x years. Therefore, grandfather's age will be
I0x years.
According to the question,
Grandfather's age = Granddaughter's age + 54 years
l0x = x + 54
Transposing x to L.H.S, we obtain
l0x - x = 54
9x = 54
x=6
Granddaughter's age = x years = 6 years

Page : 32 , Block Name : Exercise 2.4

Q10 Aman’s age is three times his son’s age. Ten years ago he was ve times his son’s age.
Find their present ages.

Answer. Let Aman’s son’s age be x years. Therefore, Aman’s age will be 3x years. Ten years
ago, their age was (x − 10) years and (3x − 10) years respectively.
According to the question,
10 years ago, Aman’s age = 5 × Aman’s son’s age 10 years ago
3x − 10 = 5(x − 10)
3x − 10 = 5x − 50
Transposing 3x to R.H.S and 50 to L.H.S, we obtain
50 − 10 = 5x − 3x
40 = 2x
Dividing both sides by 2, we obtain
20 = x
Aman’s son’s age = x years = 20 years
Aman’s age = 3x years = (3 × 20) years = 60 years

Page : 32 , Block Name : Exercise 2.4

Exercise 2.5

Q1 Solve the following linear equations.
x 1 x 1 n 3n 5n 8x 17 5x
1.1 − = + 1.2 − + = 21 1.3 x + 7 − = −
2 5 3 4 2 4 6 3 6 2)

x−5 x−3 3t−2 2t+3 2 m−1 m−2
1.4 = 1.5 − = − t 1.6 m − = 1 −
3 5 4 3 3 2 3

Page 18 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

1.7 3(t − 3) = 5(2t + 1) 1.8 15(y − 4) − 2(y − 9) + 5(y + 6) = 0

1.9 3(5z − 7) − 2(9z − 11) = 4(8z − 13) − 17

1.10 0.25(4f − 3) = 0.05(10f − 9)

Answer. 1.1
x 1 x 1
− = +
2 5 3 4

L.C.M. of the denominators, 2, 3, 4, and 5, is 60 .

Multiplying both sides by 60, we obtain

x 1 x 1
60 ( − ) = 60 ( + )
2 5 3 4

⇒ 30x − 12 = 20x + 15( Opening the brackets)

⇒ 10x − 20x = 15 + 12

27
x =
10

1.2
n 3n 5n
− + = 21
2 4 6

L.C.M. of the denominators, 2, 4, and 6, is 12 .

Multiplying both sides by 12, we obtain

6n − 9n + 10n = 252

⇒ 7n = 252

252
⇒ n =
7

⇒ n = 36

1.3
8x 17 5x
x + 7 − = −
3 6 2

L.C.M. of the denominators, 2, 3, and 6, is 6 .

Multiplying both sides by 6, we obtain

6x + 42 − 16x = 17 − 15x

⇒ 6x − 16x + 15x = 17 − 42

⇒ 5x = −25

−25
⇒ x =
5

⇒ x = −5

1.4
x−5 x−3
=
3 5

L.C.M. of the denominators, 3 and 5, is 15 .

Multiplying both sides by 15, we obtain

5(x − 5) = 3(x − 3)

⇒ 5x − 25 = 3x − 9( Opening the brackets )

⇒ 2x − 3x = 25 − 9

16
⇒ x =
2

x=8

Page 19 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

1.5
3t−2 2t+3 2
− = − t
4 3 3

L.C.M. of the denominators, 3 and 4, is 12 .

Multiplying both sides by 12, we obtain

3(3t − 2) − 4(2t + 3) = 8 − 12t

⇒ 9t − 6 − 8t − 12 = 8 − 12t (Opening the brackets)

⇒ 13t − 8t + 12t = 8 + 6 + 12

⇒ 13t = 26

26
⇒ t =
13

⇒ t = 2

1.6
m−1 m−2
m − = 1 −
2 3

L.C.M. of the denominators, 2 and 3, is 6 .

Multiplying both sides by 6, we obtain

6m − 3(m − 1) = 6 − 2(m − 2)

⇒ 6m − 3m + 3 = 6 − 2m + 4 (Opening the brackets)

⇒ 6m − 3m + 2m = 6 + 4 − 3

⇒ 5m = 7

7
m =
5

1.7
3(t − 3) = 5(2t + 1)

⇒ 3t − 9 = 10t + 5( Opening the brackets )

⇒ −9 − 5 = 10t − 3t

⇒ −14 = 7t

−14
⇒ t =
7

⇒ t = −2

1.8
15(y − 4) − 2(y − 9) + 5(y + 6) = 0

⇒ 15y − 60 − 2y + 18 + 5y + 30 = 0 (Opening the brackets)

⇒ 18y − 12 = 0

⇒ 18y = 12

12 2
y = =
18 3

1.10
0.25(4f − 3) = 0.05(10f − 9)

1 1
(4f − 3) = (10f − 9)
4 20

Multiplying both sides by 20, we obtain

5(4f − 3) = 10f − 9

⇒ 20f − 15 = 10f − 9 (Opening the brackets)

Page 20 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

20f − 10f = −9 + 15

10f = 6

3
f = = 0.6
5

Page : 33 , Block Name : Exercise 2.5

Exercise 2.6

Q1 Solve the following equations.
8x−3 9x z 4
1.1 = 2 1.2 = 15 1.3 =
3x 7−6x z+15 9

3y+4 −2 7y+4 −4
1.4 = 1.5 =
2−6y 5 y+2 3

Answer. 1.1
8x−3
= 2
3x

On multiplying both sides by 3x, we obtain

8x- − 3 = 6x

⇒ 8x − 6x = 3

⇒ 2x = 3

3
x =
2

1.2
9x
= 15
7−6x

Onmultiplying both sides by 7 − 6x, we obtain

9x = 15(7 − 6x)

⇒ 9x = 105 − 90x

⇒ 9x + 90x = 105

⇒ 99x = 105

105 35
x = =
99 33

1.3
z 4
=
z+15 9

On multiplying both sides by 9(z + 15), we obtain

9z = 4(z + 15)

⇒ 9z = 4z + 60

⇒ 9z − 4z = 60

⇒ 5z = 60

⇒ z = 12

1.4

Page 21 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

3y+4 2
= −
2−6y 5

On multiplying both sides by 5(2 − 6y), we obtain

5(3y + 4) = −2(2 − 6y)

⇒ 15y + 20 = −4 + 12y

⇒ 15y − 12y = −4 + 12y

⇒ 3y = −24

⇒ y = −8

1.5
7y+4 4
= −
y+2 3

On multiplying both sides by 3(y + 2), we obtain

3(7y + 4) = −4(y + 2)

⇒ 21y + 12 = −4y − 8

⇒ 21y + 4y = −8 − 12

⇒ 25y = −20

4
y = −
5

Page : 35 , Block Name : Exercise 2.6

Q6 The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages
will be 3:4. Find their present ages.

Answer. Let the common ratio between their ages be x. Therefore, Hari's age and Harry's age
will be 5x years and 7x years respectively and four years later, their ages will be (5 + 4) years
and (7x + 4) years respectively.
According to the situation given in the question,
5x+4 3
=
7x+4 4

⇒ 4(5x + 4) = 3(7x + 4)

⇒ 20x + 16 = 21x + 12

⇒ 16 − 12 = 21x − 20x

⇒ 4 = x

Hari's age = 5x years = (5 x 4) years = 20 years
Harry's age = 7x years = (7 x 4) years = 28 years
Therefore, Hari's age and Harry's age are 20 years and 28 years respectively.

Page : 35 , Block Name : Exercise 2.6

Q7 The denominator of a rational number is greater than its numerator by 8. If the numerator
is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2 . Find
the rational number.

Answer. Let the numerator of the rational number be x. Therefore, its denominator will

Page 22 of 23 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-2 Maths

be x + 8.
x
The rational number will be . According to the question,
x+8

x+17 3
=
x+8−1 2

x+17 3
⇒ =
x+7 2

⇒ 2(x + 17) = 3(x + 7)

⇒ 2x + 34 = 3x + 21

⇒ 34 − 21 = 3x − 2x

⇒ 13 = x

Numerator of the rational number = x = 13

Denominator of the rational number = x + 8 = 13 + 8 = 21

Rational Number = 13

21

Page : 35 , Block Name : Exercise 2.6

Page 23 of 23 Aglasem Schools