These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4

Question 1.

Amina thinks of numbers and subtracts \(\frac{5}{2}\) from it. She multiplies the results by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Solution:

Let the number be x

By subtracting \(\frac{5}{2}\), we get x – \(\frac{5}{2}\)

According to the given question

\(8\left(x-\frac{5}{2}\right)=3 x\)

8x – \(\frac{8 \times 5}{2}\) = 3x

8x – 20 = 3x

By transposing 3x to L.H.S. and -20 to R.H.S., we get

8x – 3x = 20

5x = 20

Dividing both sides by 5, we get

\(\frac{5 \mathrm{x}}{5}=\frac{20}{5}\)

x = 4

∴ The required number is 4.

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 new number. What are the numbers?

Solution:

Let the number be x

The other positive number is 5x

on adding 21 to both numbers, we get (x + 21) and (5x + 21)

According to the question, we get

2(x + 21) = 5x + 21

2x + 42 = 5x + 21

Transposing 42 to R.H.S and 5x to L.H.S., we get

2x – 5x = 21 – 42

-3x = -21

Dividing both sides by -3, we get

\(\frac{-3 x}{-3}=\frac{-21}{-3}\)

x = 7

the other number 5x = 5 × 7 = 35

Thus, the required numbers are 7 and 35.

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 units digit be ‘x’.

the tens digits = 9 – x(sum of the digits is 9)

The original two digit number = 10(9 – x) + x

= 90 – 10x + x

= 90 – 9x

On interchanging the digits,

the new number = 10x + 9 – x = 9x + 9

According to the given question, we get

New number = Original number + 27

9x + 9 = 90 – 9x + 27

9x + 9 = 117 – 9x

Transposing 9 to R.H.S. and -9x to L.H.S., we get

9x + 9x = 117 – 9

18x = 108

Dividing both sides by 18, we get

\(\frac{18 \mathrm{x}}{18}=\frac{108}{18}\)

x = 6

∴ The original number = 90 – 9x = 90 – 54 = 36.

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 digit in the unit place be ‘x’

Then, the digit at tens place = 3x

The number = 10(3x) + x

= 30x + x

= 31x

On interchanging the digits,

The new number = 10x + 3x = 13x

According to the question

31x + 13x = 88

44x = 88

Dividing both sides by 44

\(\frac{44 x}{44}=\frac{88}{44}\)

x = 2

∴ The number = 31x = 31 × 2 = 62.

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 Shobos present age be ‘x’ years

Mother’s present age = 6x years

After 5 years

Shobos age = (x + 5) years

Shobos mothers age = (6x + 5) years

According to the question, we get

\(\frac{1}{3}\) (mothers present age) = Shobos age after 5 years

\(\frac{1}{3}\) × 6x = x + 5

2x = x + 5

Transposing x to LHS

2x – x = 5

x = 5

∴ Shobo’s present age = 5 years

Mothers present age = (6 × 5) = 30 years

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 of the rectangular plot be 11x metres

and the breadth of the rectangular plot be 4x metres

Perimeter of the plot = 2(l + b)

= 2(11x + 4x)

= 2 × 15x

= 30x

Cost of fencing = ₹ 75000

100 × 30x = 75000

3000x = 75000

Dividing both sides by 3000, we get

\(\frac{3000 \mathrm{x}}{3000}=\frac{75000}{3000}\)

x = 25

Length of the plot = 11 × 25 = 275 metres

Breadth of the plot = 4 × 25 = 100 metres

Question 7.

Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser 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 of ₹ 36,600. How much trouser material did he buy?

Solution:

Let the length of cloth for shirts be ‘3x’ metres

and the length of cloth for trousers be ‘2x’ metres

Cost of shirts cloth = 3x × 50 = ₹ 150x

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

S.P of shirts cloth at 12% profit

= ₹ \(\frac{112}{100}\) × 150x

= ₹ 168x

S.P. of trousers cloth at 10% profit

= \(\frac{110}{100}\) × 180x

= ₹ 198x

Total S.P = ₹ 36,600

168x + 198x = ₹ 36,600

366x = 36600

Dividing both sides by 366, we get

\(\frac{366 \mathrm{x}}{366}=\frac{36600}{366}\)

x = 100

Material bought for trousers (2 × 100) = 200 metres

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 total number of deer be ‘x’

Number of deer, grazing in the field = \(\frac{\mathrm{x}}{2}\)

Number of deer playing near by = \(\frac{3}{4}\left(x-\frac{x}{2}\right)=\frac{3 x}{4}-\frac{3 x}{8}=\frac{6 x-3 x}{8}=\frac{3 x}{8}\)

Number of deer drinking water = 9

\(\frac{x}{2}+\frac{3 x}{8}\) + 9 = x

Transposing 9 to RHS and x to LHS we get

\(\frac{x}{2}+\frac{3 x}{8}-x=-9\)

\(\frac{4 x+3 x-8 x}{8}=-9\)

\(\frac{-x}{8}=-9\)

Multiplying both sides by -8

\(\frac{-\mathrm{x}}{8}\) × (-8) = -9 × (-8)

x = 72

∴ Number of deer in herd = 72

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

Present age of grandfather = 10x years

According to the given question, we get

10x – x = 54

9x = 54

Dividing both sides by 9, we get

\(\frac{9 \mathrm{x}}{9}=\frac{54}{9}\)

x = 6

Present age of granddaughter = 6 years.

Present age of grandfather = 10 × 6 = 60 years.

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 son be ‘x’ years

Present age of Aman = 3x

Ten years ago Son’s age = (x – 10) years

Aman’s age = (3x – 10) years

According to the given question, we get

3x – 10 = 5(x – 10)

3x – 10 = 5x – 50

Transposing -10 to R.H.S. and 5x to L.H.S

3x – 5x = 10 – 50

-2x = -40

Dividing both sides by -2

\(\frac{-2 x}{2}=\frac{-40}{-2}\)

x = 20

∴ Sons present age = 20 years

Aman’s present age = (3 × 20) = 60 years.