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

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 – 5/2) × 8 = 3x
⇒ 8x – 40/2 = 3x
⇒ 8x – 3x = 40/2
⇒ 5x = 20
⇒ 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)
⇒ 5x + 21 = 2x + 42
⇒ 5x – 2x = 42 – 21
⇒ 3x = 21
⇒ x = 7
One number = x = 7
Other number = 5x = 5 × 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
⇒ 10x + 9 – x + 27 = 90 – 10x + x
⇒ 9x + 36 = 90 – 9x
⇒ 9x + 9x = 90 – 36
⇒ 18x = 54
⇒ x = 3
Original number = 10x + (9 – x) = (10 × 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
⇒ 31x + 13x = 88
⇒ 44x = 88
⇒ x = 2
Original number = 10x + 3x = 13x = 13 × 2 = 26

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 the present age of Shobo be x then age of her mother will be 6x.
Shobo’s age after 5 years = x + 5
According to the question,
(x + 5) = (1/3) × 6x
⇒ x + 5 = 2x
⇒ 2x – x = 5
⇒ x = 5
Present age of Shobo = x = 5 years
Present age of Shobo’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 ₹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 and breadth be 4x.
Rate of fencing per metre = ₹100
Total cost of fencing = ₹75000
Perimeter of the plot = 2(l + b) = 2(11x + 4x) = 2×15x = 30x
Total amount of fencing = (30x × 100)
According to the question,
(30x × 100) = 75000
⇒ 3000x = 75000
⇒ x = 75000/3000
⇒ x = 25
Length of the plot = 11x = 11 × 25 = 275m
Breadth of the plot = 4 × 25 = 100m.

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?

Solution:
Let 2x m of trouser material and 3x m of shirt material be bought by him
Selling price of shirt material per meter = ₹ 50 + 50 ×(12/100) = ₹ 56
Selling price of trouser material per meter = ₹ 90 + 90 × (10/100) = ₹ 99
Total amount of sale = ₹36,600
According to the question,
(2x × 99) + (3x × 56) = 36600
⇒ 198x + 168x = 36600
⇒ 366x = 36600
⇒ x = 36600/366
⇒ x = 100
Total trouser material he bought = 2x = 2 × 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 × 3/4 = 3x/8
Deer drinking water = 9
According to the question,
x/2 + 3x/8 + 9 = x
(4x + 3x)/8 + 9 = x
7x/8 + 9 = x
9 = x – 7x/8
9 = (8x – 7x)/8
9 × 8 = x
 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
⇒ 10x – x = 54
⇒ 9x = 54
⇒ x = 6
Age of grandfather = 10x = 10 × 6 = 60 years.
Age of granddaughter = x = 6 years.

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 age of Aman’s son be x then age of Aman will be 3x.
According to the question,
5(x – 10) = 3x – 10
⇒ 5x – 50 = 3x – 10
⇒ 5x – 3x = -10 + 50
⇒ 2x = 40
⇒ x = 20
Aman’s son age = x = 20 years
Aman age = 3x = 3 × 20 = 60 years