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### Download PDF of NCERT Solutions for class 8 Maths Chapter 2- Linear Equations in One Variable Exercise 2.4

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### Access other exercise solutions of Class 8 Maths Chapter 2- Linear Equations in One Variable

Exercise 2.1 Solutions 12 Questions (12 Short Answer Questions)

Exercise 2.2 Solutions 16 Questions (6 Long Answer Questions, 10 Short Answer Questions)

Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions)

Exercise 2.5 Solutions 10 Questions (1 Long Answer Questions, 9 Short Answer Questions)

Exercise 2.6 Solutions 7 Questions (1 Long Answer Questions, 6 Short Answer Questions)

### Access Answers of Maths NCERT Class 8 Chapter 2- Linear Equations in One Variable Exercise 2.4 Page Number 31

**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 number be x,

According to the question,

(x â€“ 5/2) Ã— 8 = 3x

â‡’ 8x â€“ 40/2 = 3x

â‡’ 8x â€“ 3x = 40/2

â‡’ 5x = 20

â‡’ x = 4

Thus, the number is 4.

**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 one of the positive number be x then other number will be 5x. According to the question,

5x + 21 = 2(x + 21)

â‡’ 5x + 21 = 2x + 42

â‡’ 5x â€“ 2x = 42 â€“ 21

â‡’ 3x = 21

â‡’ x = 7

One number = x = 7

Other number = 5x = 5Ã—7 = 35 The two numbers are 7 and 35.

**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 digit at tens place be x then digit at ones place will be (9-x).

Original two-digit number = 10x + (9-x)

After interchanging the digits, the new number = 10(9-x) + x

According to the question,

10x + (9-x) + 27 = 10(9-x) + x

â‡’ 10x + 9 â€“ x + 27 = 90 â€“ 10x + x

â‡’ 9x + 36 = 90 â€“ 9x

â‡’ 9x + 9x = 90 â€“ 36

â‡’ 18x = 54

â‡’ x = 3

Original number = 10x + (9-x) = (10Ã—3) + (9-3) = 30 + 6 = 36

Thus, the number is 36.

**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 at tens place be x then digit at ones place will be 3x.

Original two-digit number = 10x + 3x

After interchanging the digits, the new number = 30x + x

According to the question,

(30x + x) + (10x + 3x) = 88

â‡’ 31x + 13x = 88

â‡’ 44x = 88

â‡’ x = 2

Original number = 10x + 3x = 13x = 13Ã—2 = 26

**5. Shoboâ€™s motherâ€™s present age is six times Shoboâ€™s present age. Shoboâ€™s age five years from now will be one third of his motherâ€™s present age. What are their present ages?**

Solution:

Let the present age of Shobo be x then age of her mother will be 6x.

Shoboâ€™s age after 5 years = x + 5

According to the question,

(x + 5) = (1/3) Ã— 6x

â‡’ x + 5 = 2x

â‡’ 2x â€“ x = 5

â‡’ x = 5

Present age of Shobo = x = 5 years

Present age of Shoboâ€™s mother = 6x = 30 years.

**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 â‚¹100 per metre it will cost the village panchayat â‚¹75000 to fence the plot. What are the dimensions of the plot?**

Solution:

Let the length of the rectangular plot be 11x and breadth be 4x.

Rate of fencing per metre = â‚¹100

Total cost of fencing = â‚¹75000

Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 2Ã—15x = 30x

Total amount of fencing = (30x Ã— 100)

According to the question,

(30x Ã— 100) = 75000

â‡’ 3000x = 75000

â‡’ x = 75000/3000

â‡’ x = 25

Length of the plot = 11x = 11 Ã— 25 = 275m

Breadth of the plot = 4 Ã— 25 = 100m.

**7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him â‚¹50 per metre and trouser material that costs him â‚¹90 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 â‚¹36,600. How much trouser material did he buy?**

Solution:

Let 2x m of trouser material and 3x m of shirt material be bought by him

Selling price of shirt material per meter = â‚¹ 50 + 50 Ã—(12/100) = â‚¹ 56

Selling price of trouser material per meter = â‚¹ 90 + 90 Ã— (10/100) = â‚¹ 99

Total amount of sale = â‚¹36,600

According to the question,

(2x Ã— 99) + (3x Ã— 56) = 36600

â‡’ 198x + 168x = 36600

â‡’ 366x = 36600

â‡’ x = 36600/366

â‡’ x = 100

Total trouser material he bought = 2x = 2 Ã— 100 = 200 m.

**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.

Deer grazing in the field = x/2

Deer playing nearby = x/2 Ã— Â¾ = 3x/8

Deer drinking water = 9

According to the question,

x/2 + 3x/8 + 9 = x

(4x + 3x)/8 + 9 = x

â‡’ 7x/8 + 9 = x

â‡’ x â€“ 7x/8 = 9

â‡’ (8x â€“ 7x)/8 = 9

â‡’ x = 9 Ã— 8

â‡’ x = 72

**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 age of granddaughter be x and grandfather be 10x.

Also, he is 54 years older than her.

According to the question, 10x = x + 54

â‡’ 10x â€“ x = 54

â‡’ 9x = 54

â‡’ x = 6

Age of grandfather = 10x = 10Ã—6 = 60 years.

Age of granddaughter = x = 6 years.

**10. Amanâ€™s age is three times his sonâ€™s age. Ten years ago he was five times his sonâ€™s age. Find their present ages.**

Solution:

Let the age of Amanâ€™s son be x then age of Aman will be 3x.

According to the question,

5(x â€“ 10) = 3x â€“ 10

â‡’ 5x â€“ 50 = 3x â€“ 10

â‡’ 5x â€“ 3x = -10 + 50

â‡’ 2x = 40

â‡’ x = 20

Amanâ€™s son age = x = 20 years

Aman age = 3x = 3Ã—20 = 60 years

The Exercise 2.4 of NCERT Solutions for Class 8 Maths Chapter 2- Linear Equations in One Variable is based on the application of solving equations having the variables on both sides. By solving this exercise, the students will get an idea of how the questions related to the application of solving equations having the variables on both sides should be solved.