NCERT Solutions for Class 8 Maths Exercise 2.2 of Chapter 2Â Linear Equations in One Variable is a set of all the answers to the questions listed under the exercise. NCERT Solutions cover each concept thoroughly using simple language. It is the prerogative of Class 8 students to practice the concepts learned as much as possible. Without practising the CBSE Class 8 NCERT solutions, students may not be exam ready, which is very essential. Therefore, the NCERT Solutions act as a great resource for scoring well in the exam.

### Download PDF of NCERT Solutions for class 8 Maths Chapter 2- Linear Equations in One Variable Exercise 2.2

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### Access other exercise solutions of class 8 Maths Chapter 2- Linear Equations in One Variable

Exercise 2.1 Solutions 12 Questions (12 Short Answer Questions)

Exercise 2.3 Solutions 10 Questions (3 Long Answer Questions, 7 Short Answer Questions)

Exercise 2.4 Solutions 10 Questions (4 Long Answer Questions, 6 Short Answer Questions)

Exercise 2.5 Solutions 10 Questions (1 Long Answer Questions, 9 Short Answer Questions)

Exercise 2.6 Solutions 7 Questions (1 Long Answer Questions, 6 Short Answer Questions)

### Access Answers of Maths NCERT class 8 Chapter 2- Linear Equations in One Variable 2.2 Page Number 28

**1. If you subtract Â½ from a number and multiply the result by Â½, you get 1/8 what is the number?**

Solution:

Let the number be x.

According to the question,

(x â€“ 1/2) Ã— Â½ = 1/8

x/2 â€“ Â¼ = 1/8

x/2 = 1/8 + Â¼

x/2 = 1/8 + 2/8

x/2 = (1+ 2)/8

x/2 = 3/8

x = (3/8) Ã— 2

x = Â¾

**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?**

Solution:

Given that,

Perimeter of rectangular swimming pool = 154 m Let the breadth of rectangle be = x

According to the question,

Length of the rectangle = 2x + 2 We know that,

Perimeter = 2(length + breadth)

â‡’ 2(2x + 2 + x) = 154 m

â‡’ 2(3x + 2) = 154

â‡’ 3x +2 = 154/2

â‡’ 3x = 77 â€“ 2

â‡’ 3x = 75

â‡’ x = 75/3

â‡’ x = 25 m

Therefore, Breadth = x = 25 cm

Length = 2x + 2

= (2 Ã— 25) + 2

= 50 + 2

= 52 m

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

Solution:

Base of isosceles triangle = 4/3 cm

Perimeter of triangle =

image cm = 62/15

Let the length of equal sides of triangle be x.

According to the question,

4/3 + x + x = 62/15 cm

â‡’ 2x = (62/15 â€“ 4/3) cm

â‡’ 2x = (62 â€“ 20)/15 cm

â‡’ 2x = 42/15 cm

â‡’ x = (42/30) Ã— (Â½)

â‡’ x = 42/30 cm

â‡’ x = 7/5 cm

The length of either of the remaining equal sides are 7/5 cm.

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

Solution:

Let one of the numbers be= x.

Then, the other number becomes x + 15 According to the question,

x + x + 15 = 95

â‡’ 2x + 15 = 95

â‡’ 2x = 95 â€“ 15

â‡’ 2x = 80

â‡’ x = 80/2

â‡’ x = 40

First number = x = 40

And, other number = x + 15 = 40 + 15 = 55

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

Solution:

Let the two numbers be 5x and 3x. According to the question,

5x â€“ 3x = 18

â‡’ 2x = 18

â‡’ x = 18/2

â‡’ x = 19

Thus,

The numbers are 5x = 5 Ã— 9 = 45

And 3x = 3 Ã— 9 = 27.

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

Solution:

Let the three consecutive integers be x, x+1 and x+2. According to the question,

x + (x+1) + (x+2) = 51

â‡’ 3x + 3 = 51

â‡’ 3x = 51 â€“ 3

â‡’ 3x = 48

â‡’ x = 48/3

â‡’ x = 16

Thus, the integers are

x = 16

x + 1 = 17

x + 2 = 18

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

Solution:

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

8x + 8(x+1) + 8(x+2) = 888

â‡’ 8 (x + x+1 + x+2) = 888 (Taking 8 as common)

â‡’ 8 (3x + 3) = 888

â‡’ 3x + 3 = 888/8

â‡’ 3x + 3 = 111

â‡’ 3x = 111 â€“ 3

â‡’ 3x = 108

â‡’ x = 108/3

â‡’ x = 36

Thus, the three consecutive multiples of 8 are:

8x = 8 Ã— 36 = 288

8(x + 1) = 8 Ã— (36 + 1) = 8 Ã— 37 = 296

8(x + 2) = 8 Ã— (36 + 2) = 8 Ã— 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.**

Solution:

Let the three consecutive integers are x, x+1 and x+2. According to the question,

2x + 3(x+1) + 4(x+2) = 74

â‡’ 2x + 3x +3 + 4x + 8 = 74

â‡’ 9x + 11 = 74

â‡’ 9x = 74 â€“ 11

â‡’ 9x = 63

â‡’ x = 63/9

â‡’ x = 7

Thus, the numbers are:

x = 7

x + 1 = 8

x + 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?**

Solution:

Let the ages of Rahul and Haroon be 5x and 7x. Four years later,

The ages of Rahul and Haroon will be (5x + 4) and (7x + 4) respectively. According to the question,

(5x + 4) + (7x + 4) = 56

â‡’ 5x + 4 + 7x + 4 = 56

â‡’ 12x + 8 = 56

â‡’ 12x = 56 â€“ 8

â‡’ 12x = 48

â‡’ x = 48/12

â‡’ x = 4

Therefore, Present age of Rahul = 5x = 5Ã—4 = 20

And, present age of Haroon = 7x = 7Ã—4 = 28

**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?**

Solution:

Let the number of boys be 7x and girls be 5x.

According to the question,

7x = 5x + 8

â‡’ 7x â€“ 5x = 8

â‡’ 2x = 8

â‡’ x = 8/2

â‡’ x = 4

Therefore, Number of boys = 7Ã—4 = 28

And, Number of girls = 5Ã—4 = 20

Total number of students = 20+28 = 48

**11. Baichungâ€™s father is 26 years younger than Baichungâ€™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?**

Solution:

Let the age of Baichungâ€™s father be x.

Then, the age of Baichungâ€™s grandfather = (x+26)

and, Age of Baichung = (x-29) According to the question,

x + (x+26) + (x-29) = 135

â‡’ 3x + 26 â€“ 29 = 135

â‡’ 3x â€“ 3 = 135

â‡’ 3x = 135 + 3

â‡’ 3x = 138

â‡’ x = 138/3

â‡’ x = 46

Age of Baichungâ€™s father = x = 46

Age of Baichungâ€™s grandfather = (x+26) = 46 + 26 = 72

Age of Baichung = (x-29) = 46 â€“ 29 = 17

**12. Fifteen years from now Raviâ€™s age will be four times his present age. What is Raviâ€™s present age?**

Solution:

Let the present age of Ravi be x.

Fifteen years later, Ravi age will be x+15 years. According to the question,

x + 15 = 4x

â‡’ 4x â€“ x = 15

â‡’ 3x = 15

â‡’ x = 15/3

â‡’ x = 5

Therefore, Present age of Ravi = 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?**

Solution:

Let the rational be x.

According to the question,

x Ã— (5/2) + 2/3 = -7/12

â‡’ 5x/2 + 2/3 = -7/12

â‡’ 5x/2 = -7/12 â€“ 2/3

â‡’ 5x/2 = (-7- 8)/12

â‡’ 5x/2 = -15/12

â‡’ 5x/2 = -5/4

â‡’ x = (-5/4) Ã— (2/5)

â‡’ x = â€“ 10/20

â‡’ x = -Â½

Therefore, the rational number is -Â½.

**14. Lakshmi is a cashier in a bank. She has currency notes of denominations â‚¹100, â‚¹50 and â‚¹10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is â‚¹4,00,000. How many notes of each denomination does she have?**

Solution:

Let the numbers of notes of â‚¹100, â‚¹50 and â‚¹10 be 2x, 3x and 5x respectively.

Value of â‚¹100 = 2x Ã— 100 = 200x

Value of â‚¹50 = 3x Ã— 50 = 150x

Value of â‚¹10 = 5x Ã— 10 = 50x According to the question,

200x + 150x + 50x = 4,00,000

â‡’ 400x = 4,00,000

â‡’ x = 400000/400

â‡’ x = 1000

Numbers of â‚¹100 notes = 2x = 2000

Numbers of â‚¹50 notes = 3x = 3000

Numbers of â‚¹10 notes = 5x = 5000

**15. I have a total of â‚¹300 in coins of denomination â‚¹1, â‚¹2 and â‚¹5. The number of â‚¹2 coins is 3 times the number of â‚¹5 coins. The total number of coins is 160. How many coins of each denomination are with me?**

Solution:

Let the number of â‚¹5 coins be x.

Then,

number â‚¹2 coins = 3x

and, number of â‚¹1 coins = (160 â€“ 4x) Now,

Value of â‚¹5 coins = x Ã— 5 = 5x

Value of â‚¹2 coins = 3x Ã— 2 = 6x

Value of â‚¹1 coins = (160 â€“ 4x) Ã— 1 = (160 â€“ 4x)

According to the question,

5x + 6x + (160 â€“ 4x) = 300

â‡’ 11x + 160 â€“ 4x = 300

â‡’ 7x = 140

â‡’ x = 140/7

â‡’ x = 20

Number of â‚¹5 coins = x = 20

Number of â‚¹2 coins = 3x = 60

Number of â‚¹1 coins = (160 â€“ 4x) = 160 â€“ 80 = 80

**16. The organisers of an essay competition decide that a winner in the competition gets a prize of â‚¹100 and a participant who does not win gets a prize of â‚¹25. The total prize money distributed is â‚¹3,000. Find the number of winners, if the total number of participants is 63.**

Solution:

Let the numbers of winner be x.

Then, the number of participants who didnâ€™t win = 63 â€“ x

Total money given to the winner = x Ã— 100 = 100x

Total money given to participant who didnâ€™t win = 25Ã—(63-x)

According to the question,

100x + 25Ã—(63-x) = 3,000

â‡’ 100x + 1575 â€“ 25x = 3,000

â‡’ 75x = 3,000 â€“ 1575

â‡’ 75x = 1425

â‡’ x = 1425/75

â‡’ x = 19

Therefore, the numbers of winners are 19.

The Exercise 2.2 of NCERT Solutions for Class 8 Maths Chapter 2- Linear Equations in One Variable is based on some Applications of Linear Equations In One Variable. These will help the students in understanding the context and situations in which the concepts they learned in the chapter, Linear Equations in One Variable, can be applied.

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