# Ncert Math Solutions for CBSE Class 8 | Linear Equations in One Variable | Exercise 2.2 Solved

## NCERT Math Solutions for CBSE Class 8

### Class 8 Mathematics Solutions of NCERT Maths Exercise 2.2

#### Linear Equations in One Variable

(page 28)
View solutions for Question No.1 - 8

Q 9: The ages for Rahul and Haroon are in the ration of 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Solution:
Let the ratios are in x
Then Rahul’s present age = 5x years
Haroon’s present age = 7x years
Four years later Rahul’s age will be = (5x + 4) years
Four years later Haroon’s age will be = (7x + 4) years
As per the given condition,
5x + 4 + 7x + 4 = 56
or, x = 4
Putting the value of x above we get,
Rahul’s present age = 5 x 4 = 20 years
Haroon’s present age = 7 x 4 = 28 years
Q 10: The number of boys and girls in a class are in the ratio of 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Solution: (taking hint from Q.9 try to solve this question yourself.)
Q 11: Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung’s. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Solution:
Let Baichung’s age = x years.
Baichung’s father’s age = x + 29 years
Baichung’s grandfather’s age = x + 29 + 26 = x +55 years
Given that sum of the ages of all three persons = 135 years.
or,  x + x + 29 + x + 55 = 135
or, 3x + 84 = 135
or, x = 17
Hence, Baichung’s age = 17 years.
Baichung’s father’s age = 17 + 29 = 46 years
Baichung’s grandfather’s age = 17 + 29 + 26 = 72 years
Q 12: Fifteen years from now Ravi’s age will be 4 times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age = x years
15 years later Ravi’s age will be x + 15 years
According to the given condition,
4x = x + 15
or, x = 5 years
Hence, Ravi’s present age = 5 years.
Q 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 number be x.
Q 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?
Solution:
Let the currency notes are in the ratios of x.
(a) Number of Rs 100 notes = 2x and
its value = Rs 100 x 2x = Rs 200x.
(b) Number of Rs 50 notes = 3x and
its value = Rs 50 x 3x = Rs 150x.
(c) Number of Rs 10 notes = 5x and
its value = Rs 10 x 5x = Rs 50x.
Given the total value of all notes = 4,00,000
or, 200x + 150x + 50x = 400000
or, x = 1000
Therefore,
Number of Rs 100 notes = 2 x 1000 = 2000
Number of Rs 50 notes = 3 x 1000 = 3000
Number of Rs 10 notes = 5 x 1000 = 5000.
Q 15: I have a total of Rs 300 in coins denomination Rs 1, Rs 2 and Rs 5. The number of Rs 2 coins is three times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Solution:
Let the number of Rs 1 coins be x and its value then Rs x,
Number of Rs 5 coins be y and its value then Rs 5y,
Number of Rs 2 coins be 3y and its value then Rs 6y
(since it is given that the number of Rs 2 coins is three times the number of Rs 5 coins).
Now, x + y + 3y = 160 (given that total number of coins = 160) . . . (i)
x + 5y + 6y = 300 (given that total value Rs 300) . . . . (ii)
Solving equations (i) and (ii) we obtain,
x = 80 and y = 20
Therefore, number of Rs 1 coins = 80
Number of Rs 5 coins = 20
Number of Rs 2 coins = 60
Q 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 3000. Find the number of winners, if the total number of participants is 63.
Solution:
The total number of participants = 63.
Let the number of winner be x.
Winners get Rs 100x.
Number of participants who does not win = 63 – x.
They get Rs (63 – x) x 25.
Now as per the given conditions,
100x +  (63 – x) x 25 = 3000
or, 75x + 1575 = 3000
or, x = Number of winners = 19.
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