# Solution

Explanation :-

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Then, father’s present age = 3x + x = 4x
After 8 years, father’s age = (5/2) times of Sunils’ age
After further 8 years,
(4x + 8) = (5/2)(x + 8)
8x + 16 = 5x + 40
3x = 40−16
x = 8
After further 8 years,
Sunil’s age = (x + 8) + 8 = 8 + 8 + 8 = 24
Father’s age = 4x + 8 = (4*8) + 8 = 48
Father’s age/Sunil’s age = 48/24 = 2

Explanation :-

Let the present age of P and Q be 3x and 4x respectively.
Ten years ago, P was half of Q in age=> 2(3x – 10) = (4x – 10)
=> 6x – 20 = 4x – 10
=> 2x = 10
=> x = 5

Explanation :-

Let the number  be x and y
Then 6y – x = 71 and 7x + y = 62
By Solving these equations
x = 7,  y = 13

Explanation :-

a : b = 2 : 3
b : c = 5 : 8 = (5∗3/5) : (8∗3/5) = 3 : 24/5
a : b : c = 2 : 3 : 24/ 5 = 10:15:24
b = 98∗15/49 = 30

Explanation :-

Boys : Girls = 7 : 8
20% Increase in Boys = 20% of 7 = 1.4
10% Increase in Girls = 10% of 8 = 0.8
New Ratio => (7+1.4) : (8 + 0.8) = 8.4 : 8.8
84/10 : 88/10 => 21:22

Explanation :-

Total students = 40 × 15 = 600
So, 18 x 16 + (12 + 14) = 456
Left student is 10 and their average is  = 600 – 456 = 144 ÷ 10 = 14.4

Explanation :-

Let Arun’s weight by x kg.
According to Arun, 65 < x < 72
According to Arun’s brother, 60 < x < 70.
According to Arun’s mother, x <= 68
The values satisfying all the above conditions are 66, 67 and 68.

Explanation :-

Total marks = 87*8 = 696
previous marks of two subjects = 98 + 108 = 206
marks after revaluation = 138 + 132 = 270
Increased marks  =  270 – 206 = 64
Average after revaluation = (696 + 64)/ 8 = 95

Explanation :-

Total Marks of Rahul, Manish and Suresh = 63*3 = 189
Ajay’s score = 63 + 30 = 93, Therefore
Rahul’s Score = 93 -15 = 78
Manish’s score = 78 – 10 = 68
Suresh’s score = 189 – (78 + 68) = 43
The sum of Manish’s and Suresh’s score = 43 + 68 = 111

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