Question

Seven year back the age of a father was 5 times the age of

his son. It will be twice after 14 years. Find their ages at
present.

Answer 1

Explanation:

x = age of father

y = age of son

x - 7 = 5(y - 7) = 5y - 35

x = 5y - 28

x + 14 = 2(y + 14) = 2y + 28

we use the identity of the first equation in the second equation :

5y - 28 + 14 = 2y + 28

3y = 28 + 28 - 14 = 42

y = 14

x = 5y - 28 = 5×14 - 28 = 70 - 28 = 42

the father is currently 42 years old.

the son is currently 14 years old.