Question

a father says to his son 7 year ago i was 7 times as old youwere after 3 year i will be 3 times as old you will be find there present age

Answer 1

Let s = son’s age now
Then 7 years ago, son’s age was s - 7 
Since father’s age, 7 years ago was 7 times his son’s, then father’s age then was 7(s – 7) = 7s - 49
In 3 years time, son’s and father’s ages would be their ages 7 years ago, plus 10 years
Therefore, their ages, 3 years from now, or 10 years from 7 years ago will be s – 7 + 10, or s + 3 (son’s), and 7s – 49 + 10, or 7s – 39 (father’s). 
Now, since in 3 year’s time, the father’s age will be twice that of his son’s, then we have: 2(s + 3) = 7s – 39 
2s + 6 = 7s – 39
-5s = - 45
s = 9 
Therefore, son is now 9 years-old. 
Now, since 7 years ago, father’s age was 7 times his son’s, then 7 years ago, his son was (9 – 7) = 2, which means that the father was 7(2), or 14 years-old then. 
If father was 14 years old 7 years ago, that makes him 21 years-old now. 

Answer 2

Let the present age of the father be f

Let the present age of the son be s

Given,  $f-7=7(s-7)$ .....(1)

$f+3=3(s+3)$ .....(2)

Solving Eqn (1) and (2)

$7+7(s-7)=3(s+3)-3\\ 7+7s-49=3s+9-3\\ 4s=48\\ s=12 \\\therefore f =42$

The present age of the father is 42 years and that of his son is 12 years. #SPJ3