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

Chapter 2 Linear Equations in One Variable Exercise 2.4

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?
Answer:

Let the number be x.
Given (x - 5/2) × 8 = 3x
= x - 5/2 = 3/8x
= x/1 - 3x/8 = 5/2
= 8x - 3x/8 = 5/2
= 5x/8 = 5/2
= x = 8/2
= x = 4
Verification
= (4 - 5/2) × 8 = 3 × 4
= (8 - 5/2) × 8 = 12
= 3/2 × 8 = 12
= 3 × 4 = 12
= 12 = 12
Therefore, the number is 12.

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?
Answer:

Let another number be x.
Given the number = 5x
Numbers are
x + 21
5x + 21
= x = 7
= 5x = 7 × 5 = 35
= x + 21 = 28
= 5x + 21 = 5 × 7 + 21 = 56
Therefore, the numbers are 28 and 56.

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?
Answer:

Let the units digit be x and tens digit be 9 - x.
Number
= 10(9 - x) + 1(x)
= 90 - 10x + x
= 90 - 9x
Number
= 10(x) + 1(9 - x)
= 10x + 9 - x
= 9x + 9
Given
= 9x - 9 - (90 - 9x) = 27
= 9x + 9 - 90 + 9x = 27
= 18x - 81 = 27
= 18x = 27 + 81
= 18x = 108
= x = 108/18
= x = 6
Numbers are

Therefore, the number is 36, if it is interchanged it becomes 63.

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 units digit be x and tens digit be 3x.

Number
= 10(3x) + 1(x)
= 30x + x
= 31x

Number
= 10(x) + 1(3x)
= 10x + 3x
= 13x
= 31x + 13x = 88
= 44x = 88
= x = 88/44
= x = 2


Therefore, the number is 26, if it is interchanged it becomes 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:

= x + 5 = 1/3 (6x)
= x + 5 = 2x
= 5 = 2x - x
= x = 5
Therefore, Shobo’s present age is 5 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 of narrow rectangular plot be 11x. Let the breadth of narrow rectangular plot be 4x.
Perimeter of rectangle = 2 × (l + b)
= 2(l + b) × 100 = 75000
= 2(11x + 4x) × 100 = 75000
= 2(15x) × 100 = 75000
= 3000x = 75000
= x = 75000/3000
= x = 25
Length = 11x = 25 × 11 = 275 m
Breadth = 4x = 25 × 4 = 100 m
Therefore, the length of the plot is 275 m while of the breadth of plot is 100 m.

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:


Shirt
= 75x + 9x
= 84x
Trouser
= 90x + 9x
= 99x
= 84x + 99x = 36000
= 183x = 36000
= x = 36000/183
= x = 200
Trouser material
= x = 200 m
Shirt material
= 3x/2 = 3 × 200/2 = 600/2 = 300 m
Therefore, he brought 200 metres of trouser material, brought 300 m of shirt material.

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:

Number of deer in herd = x
Grazing = x/2
Remaining = x - x/2 = x/2
Playing = 3/4 (x/2) = 3x/8
Drinking water = 9
= x/2 + 3x/8 + 9/1 = x
= 4x + 3x + 72/8 = x
= 7x + 72 = 8x
= 72 = 8x - 7x
= x = 72
So, number of deer
in the herd = x = 72
grazing = x/2 = 72/2 = 36
remaining = x/2 = 72/2 = 36
playing = 3x/8 = 216/8 = 27
drinking water = 9 = 9
Therefore, the total number of deer in the herd is 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 granddaughter’s age be x years.

= 10x - x = 54
= 9x = 54
= x = 54/9
= x = 6

Therefore, grandfather’s age is 60 years, while the granddaughter’s age is 6 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 son’s age be x years.
= 3x - 10 = 5(x - 10)
= 3x - 10 = 5x - 50
= -10 + 50 = 5x - 3x
= 40 = 2x
= x = 40/2
= x = 20

Therefore, Aman’s age is 60 years. His son’s age is 20 years.

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