Question

a father is 7 times as old as his son two years ago the father was 13 times as old as his son how old are they now

Answer 1

Let the present age of the son be x years and two years ago his age would be ( x - 2 ) years .
So present age of father be 7x years and 2 years ago father's age would be ( 7x - 2 ) years



Present age of father = 7x years


Present age of son = x years




According to the question :


age of father 2 yeard ago = 13 × age of son years ago


( 7x - 2 ) years = 13( x - 2 ) years


7x - 2 = 13x - 26


26 - 2 = 13x - 7x


24 = 6x


4 = x




Therefore,
present age of son = x years = 4 years


present age of father = 7x year = 7( 4 ) years = 28 years


$\:$

Answer 2

Heya,

Let present age of son be x yrs.

present age of father be 7xyrs.

According to the question:-

• two years ago

age of father (7x-2)yrs

age of son = (x-2) yrs

7x-2 = 13 ( x-2)

7x-2 = 13x-26

7x-13x = -26+2

-6x = -24

x = -24 / -6

x= 4

Therefore present age of son is 4yrs
and present age of father 28yrs.ANSWER...

HOPE IT HELPS:-))