NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2

Exercise 2.2 of Class 8 Maths Chapter 2 will help you understand how to work with equations that have numbers on either side of the equals sign. You will learn how to solve them using different operations like addition and subtraction.

This exercise is important because it continues from the previous exercise in this chapter on linear equations, and it teaches us how to solve slightly harder problems. The NCERT Solutions here adhere to the latest NCERT Syllabus and requirements for students in CBSE schools. 

Consistent practice of the NCERT Solutions provided in a step-by-step manner improves your conceptual understanding of the topic and also aids in other competitive exams like the olympiad.

1.0Download NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2: Free PDF

Exercise 2.2 teaches you how to solve equations when the variable or unknown is on both sides. The NCERT Solutions for Class 8 Maths Chapter 2 explain all the steps in a clear and simple manner. Click below to download the FREE PDF now:

2.0Key Concepts in Exercise 2.2 of Class 8 Maths Chapter 2

This exercise helps you understand how to solve equations when the unknown appears on both sides. Key concepts include:

• Solving equations with variables on both sides
• Using addition or subtraction to shift terms
• Simplifying both sides before solving
• Keeping the equation balanced while solving
• Checking answers by substituting the value

3.0NCERT Class 8 Maths Chapter 2: Other Exercises

4.0NCERT Class 8 Maths Chapter 2 Exercise 2.2: Detailed Solutions

1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?

Sol.

Let the required number be x.

According to the problem, (x − 1/2) × 1/2 = 1/8.

First, eliminate the multiplication by 1/2:

=> x − 1/2 = (1/8) ÷ (1/2)

=> x − 1/2 = (1/8) × 2

=> x − 1/2 = 2/8

=> x − 1/2 = 1/4

Now, isolate x by adding 1/2 to both sides:

=> x = 1/4 + 1/2

=> x = 1/4 + 2/4

=> x = (1+2)/4

=> x=3/4

Thus, the required number is 3/4.

2. 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?

Sol.

Let the breadth of the rectangular swimming pool be x metres.

Then, its length = (2x+2) metres (2 m more than twice its breadth).

Perimeter of a rectangle = 2 × (Length + Breadth)

Perimeter = 2 × [(2x+2) + x]

Perimeter = 2 × [3x+2]

Perimeter = 6x+4.

It is given that the perimeter of the swimming pool is 154 m.

So, 6x+4 = 154.

Subtract 4 from both sides:

=> 6x = 154−4

=> 6x = 150

Divide by 6:

=> x = 150/6

=> x=25.

Thus, the breadth = 25 metres.

And the length = (2 × 25 + 2) m = (50 + 2) m = 52 metres.

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 2/15 cm. What is the length of either of the remaining equal sides?

Sol.

Let each of the equal sides of an isosceles triangle be x cm.

The base is 4/3 cm.

Perimeter of a triangle = Sum of all its sides.

Perimeter = x + x + 4/3 cm

Perimeter = (2x + 4/3) cm.

It is given that the perimeter of the triangle is 4 2/15 cm.

Convert the mixed fraction to an improper fraction: 4 2/15 = (4 × 15 + 2) / 15 = (60+2)/15 = 62/15 cm.

So, 2x + 4/3 = 62/15.

To clear denominators, multiply both sides by the LCM of 3 and 15, which is 15:

=> 15 × (2x + 4/3) = 15 × (62/15)

=> (15 × 2x) + (15 × 4/3) = 62

=> 30x + (5 × 4) = 62

=> 30x + 20 = 62

Subtract 20 from both sides:

=> 30x = 62−20

=> 30x = 42

Divide by 30:

=> x = 42/30

Simplify the fraction:

=> x = 7/5.

Convert to mixed fraction: x = 1 2/5.

Thus, each equal side is 1 2/5 cm.

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Sol.

Let the smaller number be x.

Since one number exceeds the other by 15, the larger number will be (x+15).

The sum of the two numbers is 95.

So, x + (x+15) = 95.

=> 2x + 15 = 95

Subtract 15 from both sides:

=> 2x = 95−15

=> 2x = 80

Divide by 2:

=> x = 80/2

=> x=40. (This is the smaller number)

The larger number = x+15 = 40+15 = 55.

Hence, the numbers are 40 and 55.

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Sol.

Let the two numbers be 5x and 3x (representing the ratio 5:3).

Their difference is 18.

Since 5x is greater than 3x, the difference is 5x − 3x.

So, 5x − 3x = 18.

=> 2x = 18

Divide by 2:

=> x = 18/2

=> x=9.

Now, find the numbers:

First number = 5x = 5 × 9 = 45.

Second number = 3x = 3 × 9 = 27.

Hence, the numbers are 45 and 27.

6. Three consecutive integers add up to 51. What are these integers?

Sol.

Let the three consecutive integers be x, (x+1) and (x+2).

Their sum is 51.

So, x + (x+1) + (x+2) = 51.

=> 3x + 3 = 51

Subtract 3 from both sides:

=> 3x = 51−3

=> 3x = 48

Divide by 3:

=> x = 48/3

=> x=16.

Hence, the three consecutive integers are:

First integer = x = 16.

Second integer = x+1 = 16+1 = 17.

Third integer = x+2 = 16+2 = 18.

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Sol.

Let the three consecutive multiples of 8 be 8x, 8(x+1) and 8(x+2).

Their sum is 888.

So, 8x + 8(x+1) + 8(x+2) = 888.

=> 8x + 8x + 8 + 8x + 16 = 888

=> 24x + 24 = 888

Subtract 24 from both sides:

=> 24x = 888−24

=> 24x = 864

Divide by 24:

=> x = 864/24

=> x=36.

Hence, the three consecutive multiples of 8 are:

First multiple = 8x = 8 × 36 = 288.

Second multiple = 8(x+1) = 8 × (36+1) = 8 × 37 = 296.

Third multiple = 8(x+2) = 8 × (36+2) = 8 × 38 = 304.

8. 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.

Sol.

Let the three consecutive integers be x, (x+1) and (x+2).

When multiplied by 2, 3, and 4 respectively, and added up, the sum is 74.

So, 2x + 3(x+1) + 4(x+2) = 74.

=> 2x + 3x + 3 + 4x + 8 = 74

Combine like terms:

=> (2x+3x+4x) + (3+8) = 74

=> 9x + 11 = 74

Subtract 11 from both sides:

=> 9x = 74−11

=> 9x = 63

Divide by 9:

=> x = 63/9

=> x=7.

Hence, the three consecutive integers are:

First integer = x = 7.

Second integer = x+1 = 7+1 = 8.

Third integer = x+2 = 7+2 = 9.

9. 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?

Sol.

Let the present age of Rahul be 5x years.

Let the present age of Haroon be 7x years.

Four years later:

Rahul's age = (5x+4) years.

Haroon's age = (7x+4) years.

According to the given condition, the sum of their ages four years later will be 56 years.

So, (5x+4) + (7x+4) = 56.

=> 12x + 8 = 56

Subtract 8 from both sides:

=> 12x = 56−8

=> 12x = 48

Divide by 12:

=> x = 48/12

=> x=4.

Therefore, their present ages are:

Rahul's age = 5x = 5 × 4 = 20 years.

Haroon's age = 7x = 7 × 4 = 28 years.

10. The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Sol.

Let the number of girls be x.

The number of boys is 8 more than the number of girls, so the number of boys = (x+8).

The ratio of the number of boys to girls is 7:5.

So, (Number of boys) / (Number of girls) = 7/5.

(x+8) / x = 7/5.

Perform cross multiplication:

=> 5(x+8) = 7x

=> 5x + 40 = 7x

Subtract 5x from both sides:

=> 40 = 7x − 5x

=> 40 = 2x

Divide by 2:

=> x = 40/2

=> x=20. (This is the number of girls)

Number of boys = x+8 = 20+8 = 28.

Total class strength = Number of girls + Number of boys = 20 + 28 = 48 students.

11. Baichung's father is 26 years younger than Baichung's grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Sol.

Let Baichung's age be x years.

Baichung's father's age = (x+29) years (29 years older than Baichung).

Baichung's grandfather's age = (Baichung's father's age + 26) years (26 years older than father)

= (x+29) + 26 years

= (x+55) years.

The sum of the ages of all three is 135 years.

So, x + (x+29) + (x+55) = 135.

=> 3x + 84 = 135

Subtract 84 from both sides:

=> 3x = 135−84

=> 3x = 51

Divide by 3:

=> x = 51/3

=> x=17.

Therefore, their ages are:

Baichung's age = 17 years.

Baichung's father's age = 17+29 = 46 years.

Baichung's grandfather's age = 17+55 = 72 years.

12. Fifteen years from now, Ravi's age will be four times his present age. What is Ravi's present age?

Sol.

Let Ravi's present age be x years.

Fifteen years from now, Ravi's age will be (x+15) years.

It is given that after 15 years, Ravi's age will be four times his present age.

So, x+15 = 4x.

Subtract x from both sides:

=> 15 = 4x−x

=> 15 = 3x

Divide by 3:

=> x = 15/3

=> x=5.

Hence, Ravi's present age is 5 years.

13. A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get −7/12. What is the number?

Sol.

Let the required rational number be x.

According to the problem: x × (5/2) + 2/3 = −7/12.

=> 5x/2 + 2/3 = −7/12.

To eliminate the fractions, multiply all terms by the LCM of the denominators (2, 3, 12), which is 12:

=> 12 × (5x/2) + 12 × (2/3) = 12 × (−7/12)

=> (6 × 5x) + (4 × 2) = −7

=> 30x + 8 = −7

Subtract 8 from both sides:

=> 30x = −7−8

=> 30x = −15

Divide by 30:

=> x = −15/30

Simplify the fraction:

=> x = −1/2.

Hence, the required number is −1/2.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹100, ₹50 and ₹10 respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹4,00,000. How many notes of each denomination does she have?

Sol.

Let the number of ₹100 notes be 2x.

Let the number of ₹50 notes be 3x.

Let the number of ₹10 notes be 5x.

Value of 2x notes of ₹100 denomination = 2x × 100 = ₹200x.

Value of 3x notes of ₹50 denomination = 3x × 50 = ₹150x.

Value of 5x notes of ₹10 denomination = 5x × 10 = ₹50x.

The total cash with Lakshmi is ₹4,00,000.

So, 200x + 150x + 50x = 400000.

=> 400x = 400000

Divide by 400:

=> x = 400000/400

=> x=1000.

Thus, the number of notes of each denomination is:

Number of ₹100 notes = 2x = 2 × 1000 = 2000 notes.

Number of ₹50 notes = 3x = 3 × 1000 = 3000 notes.

Number of ₹10 notes = 5x = 5 × 1000 = 5000 notes.

15. I have a total of ₹300 in coins of denomination ₹1, ₹2 and ₹5. The number of ₹2 coins is 3 times the number of ₹5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Sol.

Let the number of ₹5 coins be x.

The number of ₹2 coins is 3 times the number of ₹5 coins, so number of ₹2 coins = 3x.

Total number of coins is 160.

Number of ₹1 coins = Total coins − (Number of ₹5 coins + Number of ₹2 coins)

= 160 − (x + 3x)

= 160 − 4x.

Value of ₹5 coins = x × 5 = ₹5x.

Value of ₹2 coins = 3x × 2 = ₹6x.

Value of ₹1 coins = (160−4x) × 1 = ₹(160−4x).

The total value of the money is ₹300.

So, 5x + 6x + (160−4x) = 300.

=> (5x+6x−4x) + 160 = 300

=> 7x + 160 = 300

Subtract 160 from both sides:

=> 7x = 300−160

=> 7x = 140

Divide by 7:

=> x = 140/7

=> x=20.

Thus, the number of coins of each denomination is:

Number of ₹5 coins = x = 20 coins.

Number of ₹2 coins = 3x = 3 × 20 = 60 coins.

Number of ₹1 coins = 160 − 4x = 160 − 4 × 20 = 160 − 80 = 80 coins.

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹100 and a participant who does not win gets a prize of ₹25. The total prize money distributed is ₹3000. Find the number of winners, if the total number of participants is 63.

Sol.

Let x be the number of winners.

Total number of participants is 63.

Number of participants who do not win = (63−x).

Prize money given to x winners = x × 100 = ₹100x.

Prize money given to (63−x) participants who do not win = (63−x) × 25 = ₹25(63−x).

The total prize money distributed is ₹3000.

So, 100x + 25(63−x) = 3000.

To simplify, divide the entire equation by 25:

=> (100x/25) + (25(63−x)/25) = 3000/25

=> 4x + (63−x) = 120

=> 3x + 63 = 120

Subtract 63 from both sides:

=> 3x = 120−63

=> 3x = 57

Divide by 3:

=> x = 57/3

=> x=19.

Therefore, the number of winners is 19.

5.0Key Features and Benefits of Class 8 Maths Chapter 2 Exercise 2.2

• This exercise covers equations with variables on both sides which is a part of the CBSE syllabus.
• The step by step questions allow students to grasp a clear understanding of each method used.
• All solutions are from the latest NCERT syllabus for Class 8 Maths.
• Will also build students' thinking and solving speed for competitions like the Olympiad.
• Regular practice reinforces student’s confidence in solving algebra and word problems.