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
Also Read:- बिहार राशन कार्ड सूची 2021