**NCERT
Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4**

NCERT Solutions for Class
8 Maths Chapter 2 Linear Equations in One
Variable Ex 2.4 are the part of NCERT Solutions for Class 8 Maths. Here you can find
the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4.

**Ex 2.4 Class 8 Maths Question 1.**

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?**Solution:****
**Let the required number be x.

Condition I: x – 5/2

Condition II: 8 × (x – 5/2)

Condition III: 8 × (x – 5/2) = 3x

⇒ 8x – 5/2 × 8 = 3x (Solving the bracket)

⇒ 8x – 20 = 3x

⇒ 8x – 3x = 20 (Transposing 3x to LHS and 20 to RHS)

⇒ 5x = 20

⇒ x = 20 ÷ 5 = 4 (Transposing 5 to RHS)

Thus, the required number is 4.

**Ex 2.4 Class 8 Maths Question 2.**

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 number. What are the numbers?**Solution:****
**Let the one positive number be x.

Then the other positive number = 5x

Condition I: x + 21 and 5x + 21

Condition II: 5x + 21 = 2(x + 21)

⇒ 5x + 21 = 2x + 42 (Solving the bracket)

⇒ 5x – 2x = 42 – 21 (Transposing 2x to LHS and 21 to RHS)

⇒ 3x = 21

⇒ x = 21 ÷ 3 = 7 (Transposing 3 to RHS)

Thus, the required numbers are 7 and 7 × 5 = 35.

**Ex 2.4 Class 8 Maths Question 3.**

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?**Solution:****
**Let the ones digit of the two-digit
number be x.

Then the ten’s digit of the two-digit number = 9 – x

Original number = 10(9 – x) + x

Condition I: 10x + (9 – x) (Interchanging the digits)

Condition II: New number = original number + 27

⇒ 10x + (9 – x) = 10(9 – x) + x + 27

⇒ 10x + 9 – x = 90 – 10x + x + 27 (Solving the brackets)

⇒ 9x + 9 = -9x + 117 (Transposing 9x to LHS and 9 to RHS)

⇒ 9x + 9x = 117 – 9

⇒ 18x = 108

⇒ x = 108 ÷ 18 (Transposing 18 to RHS)

⇒ x = 6

Therefore, the ones digit = 6

and the ten’s digit = 9 – 6 = 3

Thus, the required number = 3 × 10 + 6 = 30 + 6 = 36

**Ex 2.4 Class 8 Maths Question 4.**

One of the two digits of a two-digit number is three
times the other digit. If you interchange the digits of this two-digit number
and add the resulting number to the original number, you get 88. What is the
original number?**Solution:
**Let the ones digit of the two-digit
number be x.

Then the ten’s digit of the two-digit number = 3x

Original number = x + 3x × 10 = x + 30x = 31x

Condition I: 10x + 3x = 13x (Interchanging the digits)

Condition II: New number + original number = 88

13x + 31x = 88

⇒ 44x = 88

⇒ x = 88 ÷ 44 (Transposing 44 to RHS)

⇒ x = 2

Thus, the original number = 31x = 31 × 2 = 62

Hence, the required number = 62

**Ex
2.4 Class 8 Maths Question 5.**

Shobo’s mother’s present age is six times Shobo’s
present age. Shobo’s age five years from now will be one third of his mother’s
present age. What are their present ages?**Solution:
**Let Shobo’s present age be x years.

Then Shobo’s mother’s present age = 6x years.

After 5 years, Shobo’s age will be (x + 5) years.

According to the question, we have

x + 5 = 1/3 × 6x

⇒ x + 5 = 2x

⇒ 5 = 2x – x (Transposing x to RHS)

⇒ 5 = x

Hence, Shobo’s present age = 5 years

and Shobo’s mother’s present age = 6x = 6 × 5 = 30 years.

**Ex
2.4 Class 8 Maths Question 6.**

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 of ₹ 100 per metre, it will cost the village panchayat ₹
75000 to fence the plot. What are the dimensions of the plot?**Solution:
**Let the length and the breadth of the plot be 11x m
and 4x m, respectively.

Length of fence all around = perimeter of the rectangular plot

Perimeter of the plot = 75000/100 = 750 m

2(l + b) = 750

⇒ 2(11x + 4x) = 750

⇒ 2(15x) = 750

⇒ 30x = 750

⇒ x = 750 ÷ 30 = 25

Length = 11 × 25 m = 275 m

and breadth = 4 × 25 m = 100 m

**Ex 2.4 Class 8 Maths Question 7.**

Hasan buys two kinds of cloth materials for school
uniforms, shirt material that costs him ₹ 50 per metre and trousers material
that costs him ₹ 90 per metre. For every 3 metres of the shirt material, he
buys 2 metres of the trouser material. He sells the materials at 12% and 10%
profit respectively. His total sale is ₹ 36,600. How much trouser material did
he buy?**Solution:
**Ratio of shirt material bought to the trouser
material bought = 3 : 2

Let the length of the shirt material bought = 3x m

and the length of the trouser material bought = 2x m

Cost of shirt material = 50 × 3x = ₹ 150x

Cost of trouser material = 90 × 2x = ₹ 180x

168x + 198x = 36,600

⇒ 366x = 36,600

⇒ x = 36600 ÷ 366 = 100

Thus, the length of the trouser material bought = 2 × 100 = 200 m.

**Ex
2.4 Class 8 Maths Question 8.
**Half of a herd of deer are grazing in the field 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.

**Solution:
**Let the number of deer in the herd be x.

According to the question, we have

x/2 deer are grazing in the field.

⇒ 7x + 72 = 8x

⇒ 72 = 8x – 7x (Transposing 7x to RHS)

⇒ x = 72

Hence, the required number of deer is 72.

**Ex 2.4 Class 8 Maths Question 9.**

A grandfather is ten times older than his
granddaughter. He is also 54 years older than her. Find their present ages.**Solution:
**Let the present age of granddaughter be x years.

Then the present age of grandfather = 10x years.

According to the question, we have

10x – x = 54

⇒ 9x = 54

⇒ x = 54 ÷ 9 = 6 [Transposing 9 to RHS]

Hence, the present age of the granddaughter = 6 years

and the present age of grandfather = 6 × 10 = 60 years.

**Ex 2.4 Class 8 Maths Question 10.**

Aman’s age is three times his son’s
age. Ten years ago, he was five times his son’s age. Find their present ages.**Solution:****
**Let the present age of the son be x
years.

Then the present age of Aman = 3x years

10 years ago, the son’s age = (x – 10) years

10 years ago, the father’s age = (3x – 10) years

According to the question, we have

3x – 10 = 5(x – 10)

⇒ 3x – 10 = 5x – 50

⇒ 50 – 10 = 5x – 3x (Transposing 3x to RHS and 50 to LHS)

⇒ 40 = 2x

⇒ x = 40 ÷ 2 = 20

Hence, the son’s present age = 20 years.

and the present age of Aman = 20 × 3 = 60 years.

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