Question

a man is five times as old as his daughter. After 25 years he will be twice as old as his son
at that time. find their present ages​

Answer 1

Answer:

Let father age be x, his son age be y.

Before 2 years,

⇒(x−2)=5(y−2)−−−−−−−(1)

After 2 years

⇒(x+2)=8+3(y+2)−−−−−−−−(2)

From (1) x−5y+8=0

From (2)

⇒x+2−3y−6−8=0

⇒x−3y−12=0

Solving them

⇒−2y+20=0

⇒y=10

When y=10,x=42

So the present age of the father is 42 and the son age is 10.