# NCERT Solutions for Class 8 Maths Chapter 2 (Ex 2.2) Linear Equations in One Variable

## NCERT Solutions for Class 8 Chapter 2 Linear Equations in One Variable -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 2 – Linear Equations in One Variable solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 2 – Linear Equations in One Variable Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.

Maths Revision Notes for Class 8

 Chapter Name Linear Equations in One Variable Chapter Chapter 2 Exercise Exercise 2.2 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions

# NCERT SOLVED

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

Ans. Let the number be

According to the question,

[Multiplying both sides by 2]

[Adding  to both sides ]

Hence, the required number is

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 breadth?

Ans. Let the breadth of the pool be  m.

Then, the length of the pool =  m

Perimeter =

154 =

[Dividing both sides by 2]

[Subtracting 2 from both sides]

[Dividing both sides by 3]

m

Hence, length of the pool =

= 50 + 2 = 52 m

And,breadth of the pool = 25 m.

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

Ans. Let each of equal sides of an isosceles triangle be  cm.

Perimeter of a triangle = Sum of all three sides

[Subtracting  from both the sides]

[Dividing both sides by 2]

cm

Hence, each equal side of an isosceles triangle is  cm.

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

Ans. Sum of two number = 95

Let the first number be

then another number be .

According to the question,

[Subtracting 15 from both sides]

[Dividing both sides by 2]

Hence, the first number = 40

And another number = 40 + 15 = 55.

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

Ans. Let the two numbers be  and

According to question,

[Dividing both sides by 2]

Hence, first number =  = 45 and second number =  = 27.

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

Ans. Let the three consecutive integers be  and

According to the question,

[Subtracting 3 from both sides]

[Dividing both sides by 3]

Hence, first integer = 16,

second integer = 16 + 1 = 17 and

third integer = 16 + 2 = 18.

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

Ans. Let the three consecutive multiples of 8 be  and

According to question,

[Subtracting 24 from both sides]

[Dividing both sides by 3]

Hence, first multiple of 8 = 288,

second multiple of 8 = 288 + 8 = 296 and

third multiple of 8 = 288 + 16 = 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.

Ans. Let the three consecutive integers be  and

According to the question,

[Subtracting 11 from both sides]

[Dividing both sides by 9]

Hence first integer = 7, second integer

= 7 + 1 = 8 and third integer = 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?

Ans. Let the present ages of Rahul and Haroon be  years and  years respectively.

According to condition,

[Subtracting 8 from both sides]

[Dividing both sides by 12]

Hence, present age of Rahul =  = 20 years and

present age of Haroon =  = 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?

Ans. Let the number of girls be

Then, the number of boys =

According to the question,

[Transposing  to L.H.S. and 40 to R.H.S.]

[Dividing both sides by ]

Hence the number of girls = 20 and number of boys = 20 + 8 = 28.

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?

Ans. Let Baichung’s age be  years,

then Baichung’s father’s age =  years

and Baichung’s grandfather’s age =  years.

According to condition,

[Subtracting 84 from both sides]

[Dividing both sides by 3]

years

Hence, Baichung’s age = 17 years,

Baichung’s father’s age = 17 + 29 = 46 years

And Baichung’s grandfather’s age = 17 + 29 + 26 = 72 years.

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

Ans. Let Ravi’s present age be  years.

After fifteen years, Ravi’s age =  years.

Fifteen years from now, Ravi’s age =  years.

According to question,

[Transposing  to L.H.S.]

[Dividing both sides by 3]

years

Hence, Ravi’s present age be 5 years.

13. A rational number is such that when you multiply it by  and add to the product, you get  What is the number?

Ans. Let the rational number be

According to the question,

[Subtracting from both sides]

[Dividing both sides by 60]

Hence, the rational number is

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

Ans. Let number of notes be  and

According to question,

[Dividing both sides by 400]

Hence, number of denominations of  Rs. 100 notes =  = 2,000

Number of denominations of Rs.  50 notes = = 3,000

Number of denominations of Rs. 10 notes =  = 5000

Therefore, required denominations of notes of Rs.  100, Rs.  50 and Rs. 10 are 2000, 3000 and 5000 respectively.

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

Ans. Total sum of money = Rs.  300

Let the number of Rs. 5 coins be

number of Rs.  2 coins be  and

number of Re. 1 coins be

According to question,

[Subtracting 160 from both sides]

[Dividing both sides by 7]

Hence, the number of coins of Rs.  5 denomination = 20

Number of coins of Rs.  2 denomination = = 60

Number of coins of Rs. 1 denomination = = 160 – 80 = 80

16. The organizers of an essay competition decide that a winner in the competition gets a prize of Rs. 100 and a participant who does not win, gets a prize of Rs. 25. The total prize money distributed is Rs. 3,000. Find the number of winners, if the total number of participants is 63.

Ans. Total sum of money =  Rs. 3000

Let the number of winners of  Rs. 100 be

And those who are not winners =

According to the question,

75x + 1575 – 1575 = 3000 – 1575  [Subtracting 1575 from both sides]

75x = 1425

75x75=14257575×75=142575   [Dividing both sides by 75]

Hence, the number of winner is 19.