Given:
Father is 30 years older than the son.
After 8 years , he will be 4 times as old as his son.
Find:
their present ages
Solution:
Let the son's age be x and age of father be y years
⟹ Father's age = Son's age + 30
⟹ y = x + 30 -- equation (1)
Also,
Son's age after 8 years = (x + 8) years.
Father's age after 8 years = (y + 8) years.
According to the above condition,
⟹ y + 8 = 4(x + 8)
⟹ y + 8 = 4x + 32
Substitute the value of y from equation (1).
⟹ x + 30 + 8 - 32 = 4x
⟹ 6 = 4x - x
⟹ 6 = 3x
⟹ 6/3 = x
⟹ 2 = x
Substitute the value x in equation (1).
⟹ y = 2 + 30
⟹ y = 32
Therefore,
Son's present age = x = 2 years.
Father's present age = y = 30 years.
I hope it will help you.
Regards.