Twelve years ago, the average age of a family having four members (father, mother, son and daughter) is (y-2) years. The ratio between the present ages of son and daughter is 3:5 respectively. Five years hence, the age of father is nine years more than double the age of daughter. If the present age of father is 64 years, then find out the age of mother 18 years hence.

Solution

The ratio between the present ages of son and daughter is 3:5 respectively.

Let’s assume the present ages of son and daughter are ‘3z‘ and ‘5z‘ respectively.

Five years hence, the age of father is nine years more than double the age of daughter.

So five years hence, the age of daughter = (5z+5)

Five years hence, the age of father = 2(5z+5)+9

= 10z+10+9

= 10z+19

Present age of father = 10z+19-5

= 10z+14

If the present age of father is 64 years

10z+14 = 64

10z = 64-14

10z = 50

z = 5

present age of son = 3z = 3x5 = 15 years

present age of daughter = 5z = 5x5 = 25 years

Twelve years ago, the average age of a family having four members (father, mother, son and daughter) is (y-2) years.

Twelve years ago, the age of father + Twelve years ago, the age of mother + Twelve years ago, the age of son + Twelve years ago, the age of daughter = 4(y-2)

(64-12)+Twelve years ago, the age of mother+(15-12)+(25-12) = 4(y-2)

52+Twelve years ago, the age of mother+3+13 = 4(y-2)

Here the information about two things which are the age of the mother and the value of ‘y’ are unknown. So we cannot determine the age of mother 18 years hence from the given information.