Question

a father is twice as old as his son 20 years before the fatjer was four times as old as his son find their age.​

Answer 1

Answer:

Let his son's present age = x years  

So his father's present age = 2x years

Before 20 years his son's age=(x-20) years

Before 20 years father's age=(2x-20) years

4(x-20)=(2x-20)

 

4x-80=2x-20

4x-2x=80-20

2x=60

x=30

His son's present age = 30

Father's present age=60

Step-by-step explanation:

Answer 2

Answer:

60

Step-by-step explanation:

let his son's present age= x years

So his father's present age= 2x years

Before 20 years his son's age = (x -20) years

Before 20 years father's age = (2x- 20) years

4( x- 20)=( 2x- 20)

4x- 80= 2x-20

4x- 2x= 80- 20

2x = 60

X= 2/60

x = 30

His son's present age= 30

His father's present age= 60