Question

Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages, by solving step by step.

Answer 1

Answer:

Let present age of father=x,and that of the son =y

So according to question

(x-5)=7(y-5)--------------1

(x+5)=3(y+5)------------2

So from 1

x-5=7y-35

x-7y=-30------------3

From 2

x+5=3y+15

x-3y=10--------------4

So substracting 3 and 4

x-7y=-30

-

x-3y=10

---------------

-4y=-40

y=10

So putting value of y in 4

x-3(10)=10

x=30+10

x=40

So x=40,y=10

HENCE THE PRESENT AGE OF FATHER IS 40 AND PRESENT AGE OF HIS SON IS 10YEARS

Hope it helps ☺️