Question

A father is twice as old as his son.8 year ago their ages were in the ratio8:3 find the present age?​

Answer 1

if age of son is x years

father's age is 2x years

8 years ago age of son was x-8

age of father was 2x-8

their ages were in ratio 8:3

2x-8/x-8=8/3

(2x-8)(3)=(x-8)(8)

6x-24=8x-64

64-24=2

2x=40.

x=20

age of son is is 20 years

age of father is 40 years

Answer 2

Let the sons age be x.

The fathers age be 2x.

8 years ago:-

Father's age be 2x-8

Son's age be x-8

According to the question

  x-8    =8

 2x-8     3

3(x-8)=8(2x-8)

3x-24=16x-64

3x-16x=-64+24

-13x=-40

x=-13/-40