Question

A mother was twice as old as her son 18 years ago. Thirty-two years ago, the father was in

the age equal to the sum of the ages of the mother and son. The father is six years older

than the mother. Find their present ages.​

Answer 1

Given:

Mother was twice as old as her son 18 years ago.

The father is 6 years older than the mother.

32 years ago, the father's age is the sum of the mother and son.

To Find:

Their present age

Solution:

Define the son's age now:

Let the son's age be x

18 years ago:

Son's age = x - 18

Mother is twice the age of the son:

Mother's age = 2(x - 18) = 2x - 36

Find the mother's age now:

Mother's age = 2x - 36 + 18

Mother's age = 2x - 18

Find the father's age now:

Mother's age = 2x - 18

Father is 6 years older than mother:

Father's age = 2x - 18 + 6

Father's age = 2x - 12

Also, given that 32 years ago, the father's age is the sum of the son and the mother:

(2x - 12) - 32 =  [ (2x - 18) - 32 ] + (x - 32)

2x - 12 - 32 = 2x - 18 - 32 + x - 32

2x - 44 = 3x - 82

x = 38

Find their present age:

Son = x = 38 years old

Mother = 2x - 18 = 2(38) - 18 = 58 years old

Father = 2x - 12 = 2(38) - 12 = 64 years old

Answer: Son = 38 years old, Mother = 58 years old, Father = 64 years old

Answer 2

Answer:

Son is 38 years old, mother is 58 years old, father is 64 years old.