Question

5 years ago A man was 7 times as old as his son 5 years has the father will be 3 times as old as his son find their present ages

Answer 1

Let the present age of the son be x.
Then the present age of his father will be 7x.


Given that 5 years ago a man was 7 times as old as his son.
The son's age is x + 5.
The father's age will be 7x + 5.


5 years hence :
The age of son be x + 5 + 5 = x + 10.
The age of father is 7x + 5 + 5 = 7x + 10.



Given that father age will be 3 times as old as his son:

7x + 10 = 3(x+10)

7x + 10 = 3x + 30

4x = 20

x = 5.

Then the father's age will be 7x = 7 * 5 = 35.

Present age of son be x + 5 = 10.
Present age of father will be 7x + 5 = 40.


Hope this helps! :)

Answer 2

let the present age of son be Y

& the man age be x

A.T.Q
x-5=7(y-5)
x-5=7y-35
x-7y-5+35=0
x-7y+30=0.......I

x+5=3(Y+5)
x+5=3y+15
x-3y+5-15=0
x-3y-10=0........ii
sub ii from I
x-7y+30-(x-3y-10)=0
x-7y+30-x+3y+10=0
-4y+40=0
-4y=-40
Y=10.....
putting the value of Y in eq ii
x-3×10-10=0
x-30-10=0
x-40=0
x=40




The present age of the son is 10
& the present age of man is 40.....
Hope this helps you.......!!!!!