NCERT MATHS SOLUTION CLASS 8 EXERCISE 2.4 | LINEAR EQUATIONS IN ONE VARIABLE

Find here step-by-step NCERT Maths Solution Class 8 Exercise 2.4 | Applications of Linear Equations | Chapter 2 Linear Equations in One Variable from CBSE NCERT textbook.

Before moving to maths solution class 8 exercise 2.4 , it is advised to go through the following topics:

• Linear Equations in One Variable

NCERT Maths Solution Class 8 Exercise 2.4

Problems in Exercise 2.4 require the knowledge of Applications of Linear Equations in different problems on numbers, ages, perimeters, combination of currency notes, etc.

First we assign a variable to the unknown. Next we construct an equation based on the conditions given in the problem. Then we solve the equation for the variable. Hence we find the solution of the problem.

Class 8 Maths
EXERCISE 2.4
Applications of Linear Equations

QUESTION 1

Aminathinks of a number and subtracts \frac{2}{5} 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,

A/Q

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

\implies 8x - \frac{16}{5} = 3x

\implies 8x - 3x = \frac{16}{5}

\implies 5x = \frac{16}{5}

\implies x = \frac{16}{5\times 5}

\implies x = \frac{16}{25}

Therefore, the number Amina thinks of = x = \frac{16}{25}

QUESTION 2

Apositive 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,

A/Q,

5x + 21 = 2(x + 21)

\implies 5x + 21 = 2x + 42

\implies 5x - 2x = 42 - 21

\implies 3x = 21

\implies x = \frac{21}{3}

\implies x = 7

Therefore, the numbers are x = 7 and 5x = 5 × 7 = 35.

QUESTION 3

Sumof 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

Sum of the digits of a two-digit number = 9,

Let, the digit at unit place of the original number be x,

So, the digit at tens place of the original number be (9-x),

Then, original number be {10(9 – x) + x}
= (90 – 9x)

When digits are interchanged the number will become {10(x) + (9 – x)}
= (9x + 9)

A/Q,

9x + 9 = (90 - 9x) + 27

\implies 9x + 9 = 90 - 9x + 27

\implies 9x + 9 = 117 - 9x

\implies 9x + 9x = 117 - 9

\implies 18x = 108

\implies x = \frac{108}{18}

\implies x = 6

Therefore, the two digit number = 10(9 – x) + x

= 10(9 – 6) + 6 = 10×3 + 6

= 30 + 6 = 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?

ANSWER

Let, the digit at unit place of the original number be x,

So, the digit at tens place of the original number be 3x,

Then, original number be {10(3x) + x}
= 30x + x = 31x

When digits are interchanged the number will be {10(x) + 3x}
= 10x + 3x = 13x

A/Q,

31x + 13x = 88

\implies 44x = 88

\implies x = \frac{88}{44}

\implies x = 2

Therefore, the two digit 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?

ANSWER

Let, the present age of Shobo be x years,
and the present age of Shobo’s mother be 6x years,

Five years from now,

The age of Shobo = (x + 5) years,

A/Q,

(x + 5) = \frac{1}{3}\times 6x

\implies x + 5 = 2x

\implies x - 2x = -5

\implies -x = -5

\implies x = 5

Therefore, the present age of Shobo = x years = 5 years,

and the present age of Shobo’s mother be 6x years = 6 × 5 years
= 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 ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

ANSWER

Let, the length (l) of the plot be 11x m,
and the breadth (b) of the plot be 4x m,

Then, Perimeter of the plot = 2(l+b)
= 2(11x + 4x) m = (2 × 15x) m = 30x m,

A/Q,

30x \times 100 = 75000

\implies 3000x = 75000

\implies x = \frac{75000}{3000}

\implies x = 25

Therefore, the length (l) of the plot = 11x m
= 11 × 25 m = 275m,

and the breadth (b) of the plot = 4x m
= 4 × 25 m = 100m.

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 meters 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?

ANSWER

Let, the shirt material be 3x m,
So, the trouser material be 2x m,

Cost of shirt material per meter = ₹ 50
So, the cost price of total shirt material, CP_{shirt} = ₹ (50 × 3x)
= ₹ 150x

Profit percent on shirt material = 12%

=> Profit on shirt = ₹\frac{12}{100}\times 150x
= ₹ 18x

Then,SP_{shirt} = ₹ 150x + 18x
= ₹ 168x

Similarly,

Cost of trouser material per meter = ₹ 90
So, the cost price of total trouser material, CP_{trouser}= ₹ (90 × 2x)
= ₹ 180x

Profit percent on trouser material = 10%

=> Profit on trouser = ₹ \frac{10}{100}\times 180x
= ₹ 18x

Then,SP_{trouser} = ₹ 180x + 18x
= ₹ 198x

A/Q,

168x + 198x = 36,600

\implies 366x = 36,600

\implies x =\frac{36,600}{366}

\implies x = 100

Therefore, trouser material bought by Hasan = 2x m = 200m.

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.

ANSWER

Let, the number of deers in the herd be x,

Then, the number of deers grazing in the field be \frac{1}{2}x,

and the number of deers playing nearby be \frac{3}{4}\frac{1}{2}x = \frac{3}{8}

The number of deers drinking water = 9,

A/Q,

\implies x - 9 = \frac{1}{2}x + \frac{3}{8}x

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

\implies x - 9 = \frac{7}{8}x

\implies 8(x - 9) = 7x

\implies 8x - 72 = 7x

\implies 8x - 7x = 72

\implies x = 72

Therefore, the number of deers in the herd = x = 72.

QUESTION 9

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

ANSWER

Let, the present age of granddaughter be x years,

So, the present of grandfather be 10x years,

A/Q,

10x = x + 54

\implies 10x - x = 54

\implies 9x = 54

\implies x = \frac{54}{9}

\implies x = 6

Therefore, the present age of granddaughter = x years = 6years,

and the present of grandfather = 10x years = 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.

ANSWER

Let, the present age of Aman’s son be x years,

So, the present age of Aman be 3x years,

Then, ten years ago,

The age of Aman’s son = (x – 10) years,
The age of Aman = (3x – 10) years,

A/Q,

3x - 10 = 5(x - 10)

\implies 3x - 10 = 5x - 50

\implies 3x - 5x = - 50 + 10

\implies - 2x = - 40

\implies x = \frac{40}{2}

\implies x = 20

Therefore, the present age of Aman’s son = x years
= 20 years,

and the present age of Aman = 3x years
= 60 years.


We hope NCERT Maths Solution Class 8 Exercise 2.4 has helped you understand how to solve word problems using linear equations.

Please write in the comment section for any error or any solution related queries from the exercise.

Check NCERT Solution of other exercises from Class 8 Maths Chapter 2 Linear Equations in One Variable by clicking the links given below.


NCERT MATHS SOLUTION | CLASS 8 | CHAPTER 2 | LINEAR EQUATIONS IN ONE VARIABLE

Exercise 2.1 | Exercise 2.2 | Exercise 2.3

Exercise 2.4 | Exercise 2.5 | Exercise 2.6

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