Class 8 NCERT Solutions – Chapter 2 Linear Equations in One Variable – Exercise 2.4

Last Updated : 03 Apr, 2024

This exercise has been deleted as per the new NCERT Syllabus

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 number be x,

According to the question,

(x â€“ 5/2) Ã— 8 = 3x

Solving for x,

â‡’ 8x â€“ 40/2 = 3x

â‡’ 8x â€“ 3x = 40/2 (Bring like terms together)

â‡’ 5x = 20 (Divide both sides by 5)

â‡’ x = 4

Thus, the number is 4.

Question 2. A positive number is 5 times another number. If 21 is added to both numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution:

According to the question,

Let one of the positive number be x then other number will be 5x.

Therefore,

5x + 21 = 2(x + 21)

Solving for x,

â‡’ 5x + 21 = 2x + 42

â‡’ 5x â€“ 2x = 42 â€“ 21 (Bring like terms together)

â‡’ 3x = 21 (Dividing both sides by 3)

â‡’ x = 7

1st number = x = 7

2nd number = 5x = 5Ã—7 = 35

Question 3. The 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

Solving for x,

â‡’ 10x + 9 – x + 27 = 90 – 10x + x

â‡’ 9x + 36 = 90 â€“ 9x

â‡’ 9x + 9x = 90 â€“ 36 (Bring like terms together)

â‡’ 18x = 54 ( Divide both sides by 18)

â‡’ x = 3

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

Thus, the number is 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 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

Solving for x,

â‡’ 31x + 13x = 88

â‡’ 44x = 88 (Divide both sides by 44)

â‡’ x = 2

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

Question 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

Solving for x,

â‡’ x + 5 = 2x

â‡’ 2x â€“ x = 5 (Bring like terms together)

â‡’ x = 5

Present age of Shobo = x = 5 years

Present age of Shoboâ€™s mother = 6x = 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 â‚¹100 per meter 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 meter = â‚¹100

Total cost of fencing = â‚¹75000

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

Question 7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him â‚¹50 per meter and trouser material that costs him â‚¹90 per meter. For every 3 meters of the shirt material he buys 2 meters 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.

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.

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

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 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 = 10 Ã— 6 = 60 years.

Age of granddaughter = x = 6 years.

Question 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 = 3 Ã— 20 = 60 years

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