2. Linear Equations in One Variable-Exercise 2.2 Mathematics Class 8 In English
Last Updated : 06 March 2026
Get updated solved mathematics for class 8 chapter 2. Linear Equations in One Variable Topic Exercise 2.2 with clear explanations free study material in English medium.
2. Linear Equations in One Variable-Exercise 2.2 Mathematics Class 8 In English
Last Updated : 06 March 2026
2. Linear Equations in One Variable-Exercise 2.2 Mathematics Class 8 In English
2. Linear Equations in One Variable
Exercise 2.2
Exercise 2.2
Q1. If you subtract from a number and multiply the result by , you get . What is the number?
Solution:
Let the number be x
so, the equation will be
Q2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Solution:
Let the breadth of the swimming pool x.
So, the equation will be 
so, the length of pool 2x + 2
= 2×25 + 2
= 50 + 2
= 52
and the breadth of pool x = 25
Q3. The base of an isosceles triangle cm the perimeter of a triangle is 4 cm what is the length of either remaining equal sides?
Solution:
Let the length of one remaining equal sides x.So, equation wil 
Q4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the second number is x.
So, the equation will be x + x + 15 = 95
So, the first number will be x + 15
= 40 + 15
= 55
And second number: x = 40
Q5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Solution:
Let,
Both numbers will be 5x and 3x
So, the equation will be 5x - 3x = 18
⇒2x = 18
so, the first number will be 5x
= 5 × 9
= 45
and the second number will be = 3 × 9
= 27
Q6. Three consecutive integers add up to 51. What are these integers?
Solution:
Let, all numbers x, x+1, x+2 respectively.
So, the equation will be x+x+1+x+2=51
⇒ 3x +3=51
⇒ 3x=51-3
So, the first number will be x = 16
, second number will be x + 1
= 16 + 1
= 17
and the third number will be x + 2 = 18
Q7. The sum of three consecutive multiples of 8 is 888. Find the multiples?
Solution:
Let, three consecutive multiples of 8 is x, x+8, and x+16 respectively
So, the equations will be : x + x + 8 + x + 16 = 888
⇒ 3x + 24 = 888
⇒ 3x = 888 - 24
So, the first multiple x = 288
Second multiple x + 8 = 288+8 = 296
Third multiple x + 16 = 288 + 16 = 304
Q8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Solution:
Let, all numbers be x, x+1, x+2
So, the equation will be: 2(x) + 3(x+1) +4(x+2) =74
⇒ 2x+ 3x+3+ 4x+8 = 74
⇒ 9x+11=74
⇒ 9x=74-11
So, first number is x = 7
Second number is x + 1
= 7 + 1 = 8
Third number is x + 2
= 7 + 2
= 9
Q9. The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Let, the present ages of Rahul and Haroon be 5x and 7x respectively
So, the equation will be 5x + 4 + 7x + 4 = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 - 8
So, the present age of Rahul: 5x = 5 × 4 = 20
The present age of Haroon: 7x = 7 × 4 = 28
Q10. 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 x
⇒5(x - 10) = 3x - 10
⇒ 5x – 50 = 3x - 10
⇒5x – 3x = -10 + 50
⇒2x = 40
The present age of Aman: 3x = 3 × 20 = 60
The present age of his son: x = 20
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NCERT Solutions for Class 8 mathematics Chapter 2. Linear Equations in One Variable Topic Exercise 2.2
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2. Linear Equations in One Variable-Exercise 2.2 Mathematics Class 8 In English
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