A father is twice as old as his son. If 30 years ago the age of the father was…

Concept

Linear age problems are solved by translating each verbal relationship into an algebraic equation in the unknown present ages, then solving the system. A relationship 'A is k times B' becomes A = k·B, and a past condition 't years ago' shifts every age by subtracting t.

Application

1. Let the son's present age be s and the father's present age be f.

2. 'Father is twice as old as his son' gives f = 2s.

3. '30 years ago' subtracts 30 from each age, and 'father was 5 times the son' gives f − 30 = 5(s − 30).

4. Substitute f = 2s: 2s − 30 = 5s − 150.

5. Collect terms: 150 − 30 = 5s − 2s, so 120 = 3s, giving s = 40.

6. Therefore f = 2s = 2 × 40 = 80.

The father's present age is 80 years.

Cross-check

Son is 40 and father is 80, so the father is twice the son today. 30 years ago the father was 50 and the son was 10, and 50 = 5 × 10, so both conditions hold.