Father–son age relation across time: A father is presently 34 years older than his son. Twelve years ago, the father's age was 18 times the age of his son. What is the son's present age (in years)?

Introduction / Context:
Age problems mixing “difference now” and “multiple in the past” are classic linear-equation setups. A fixed difference today remains fixed at any time shift, while multiplicative relationships apply at the specified time reference.

Given Data / Assumptions:

• If son's present age is s, father's present age is s + 34.
• Twelve years ago: father's age = 18 × (son's age).

Concept / Approach:
Translate the past condition using present variables shifted by 12 years: (s + 34 − 12) = 18 (s − 12). Solve the resulting linear equation for s.

Step-by-Step Solution:
1) Past condition: s + 22 = 18(s − 12).2) Expand: s + 22 = 18s − 216.3) Rearrange: 22 + 216 = 18s − s ⇒ 238 = 17s.4) Solve: s = 238 / 17 = 14.5) Therefore, son's present age is 14 years.

Verification / Alternative check:
12 years ago, son was 2; father was 36. Indeed, 36 = 18 × 2. The numbers are consistent.

Why Other Options Are Wrong:

• 12/16/18/20 years do not satisfy the “18 times” condition when verified with the 34-year present difference.

Common Pitfalls:
Forgetting to subtract 12 from both ages or mishandling the difference (which remains 34 at any time) often leads to wrong results.

Final Answer:
14 years