## NCERT Solution for Class 8 Maths Chapter – 2 Exercise 2.2

Q1. If you subtract 1/4 from a number and multiply the result by 1/2 you get 1/8 what is the number ?

Solution:-

Let the number be x ,

A.T.Q

(x – 1/2) x 1/2 = 1/8

X/2 – 1/4 = 1/8

X/2  = 1/8 + 1/4

X/2  = ( 1 + 2)/8

X/2 = 3/8

X = (3/8) x 2

X = 3/4

Q2. The perimeter of a rectangular swimming pool is 154m. Its length is 2m more than twice its breadth . What are the length and the breadth of the pool?

Solution:-

Perimeter of the swimming pool =

2( l + b )

= 2 ( 2x + 2 + x ) = 154 m

= 2( 3x + 2) = 154 m

= 3x + 2 = 154/2

= 3x = 77 – 2

= 3x = 75

= x = 25m

Therefore , breadth  = x = 25m

Length = 2x + 2

= (2 x 25) + 2

= 50 + 2

= 52 m

Q3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is  62/15 cm. What is the length of the either of the remaining equal sides?

Solution:-

Perimeter of triangle = 62/15

4/3 + x + x = 62/15

2x = ( 62/15 – 4/3 ) cm

2x = ( 62 – 20) / 15 cm

= 2x = 42/15

= x = (42/30() x ( 1/2)

= x = 42/30

= x = 7/5 cm.

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

Q4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution:-

X + x + 15 = 95

2x + 15 = 95

2x = 95 – 15

2x = 80

X = 40

First number = x = 40

Other number = x + 15 = 40

40 + 15 = 55

Q5. Two  numbers are in the ratio 5:3 If they differ by 18, what are the numbers?

Solution:-

5x – 3x = 18

2x = 18

X = 18/2

X = 9

Thus,

The number are 5x = 5 x 9 = 45

And 3x = 3 x 9 = 27.

Q6. Three consecutive integers add up to 51. What are these integers?

Solution:-

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

3x + 3 = 51

3x = 48

X = 48/3

X = 16

Thus, the integers are

X = 16

X + 1 = 17

X + 2 = 8

Q7. The sum of three consecutive multiplies of 8 is 888. Find the multiplies.

Solution:-

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

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

8(3x + 3) = 888

3x + 3 = 888/8

3x + 3 = 111

3x = 111 – 3

3x = 108

X = 108/3

X = 36

Thus, the three consecutive multiplies of 8 are:

8x = 8 x 36 = 288

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

8( x + 2) = 8 x ( 36 + 2) = 8 x 38 = 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:-

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

2x + 3x + 3 + 4x + 8 + 74

9x + 11 = 74

9x = 74 – 11

9x = 63

X = 7

Thus, the numbers are:

X = 7

X + 1 = 7

X + 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:-

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

5x + 4 + 7x + 4 = 56

12x + 8 = 56

12x = 56 – 8

12x = 48

X = 48/12

X = 4

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

And, present age of haroon = 7x = 7 x 4 = 28

Q10. 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:-

7x = 5x + 8

7x – 5x = 8

2x = 8

X = 8/2

X = 4

Therefore, Number of boys = 7 x 4 = 28

And, Numbers of girls = 5 x 4 = 20

Total number of students = 20 + 28 = 48

Q11. Baichung father is 26 years younger then baichung 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:-

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

=  3x + 26 – 29 = 135

3x – 3 = 135

3x = 135 + 3

3x = 138

X = 138/3

X = 46

Age of baichung father = x = 46

Age of baichung grandfather = ( x + 26) = 46 + 26 = 72

Age of baichung = ( x-29) = 46 – 29 = 17

Q12. Fifteen years from now Ravi age will be four times his present age. What is Ravi present age?

Solution:-

X + 15 = 4x

4x – x = 15

3x = 15

X = 5

Therefore, Present age of ravi = 5 years.

Singular Plural English & Hindi Grammar

Q13. 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:-

X x (5/2) + 2/3 = -7/12

5x/2 + 2/3 = -7/12

5x/2 = -7/12 – 2/3

5x/2 = -5/4

x = (-5/4) x (2/5)

X = -10/20

X = -1/2

Q14. 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 400000. How many notes of each denomination does she have?

Solution:-

200x + 150x + 50x = 400000

400x = 400000

X = 400000/400

X = 1000

Number of Rs 100 notes = 2x = 2000

Number of Rs 50 notes = 3x = 3000

Number of Rs 10 notes = 5x = 5000

Q15. I have a total of Rs 300 in coins of  denomination Rs 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 ?

Solution:-

5x + 6x +(160 – 4x) = 300

11x + 160 -4x = 300

7x = 140

X = 140/7

X = 20

Number of Rs 5 coins = x = 20

Number of Rs 2 coins = 3x= 60

Number of Rs 1 coins = (160 – 4x) = `160 – 80 = 80

Q16. The organisers of an essay competition decise that a winner in the competition gets a prize of Rs 100 and a participants who does not win gets a prize of Rs 25 The total prize money distributed is Rs 3000 Find the number of winner, if the total numbers of participants is 63.

Solution:-

100x + 1575 – 25x = 3000

75x = 3000 – 1575

75x = 1425

X = 1425/75

= 19

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NCERT Solution for Class 8 Maths Chapter – 2 Exercise 2.2   Q1. If you subtract 1/4 from a number and…