Two-time comparisons, one solution: A father was 12 times as old as his son 20 years ago. Now he is exactly twice as old as his son. What are their present ages?

Introduction / Context:
Age problems with two different time references yield a solvable linear system. We align both statements to today's variables and solve simultaneously.

Given Data / Assumptions:

• Let present ages be F (father) and S (son).
• Now: F = 2S.
• Twenty years ago: F − 20 = 12(S − 20).

Concept / Approach:
Substitute the present-day relation F = 2S into the past relation to eliminate F, leaving a single equation in S.

Step-by-Step Solution:
1) 2S − 20 = 12(S − 20) = 12S − 240.2) Rearrange: 2S − 20 = 12S − 240 ⇒ 220 = 10S ⇒ S = 22.3) Then F = 2S = 44.

Verification / Alternative check:
Twenty years ago: F=24, S=2, and indeed 24 = 12 × 2. Now 44 = 2 × 22, consistent.

Why Other Options Are Wrong:
Other pairs do not satisfy both the “20 years ago” and “now” constraints simultaneously.

Common Pitfalls:
Placing 12 outside the parentheses incorrectly (e.g., 12S − 20) or forgetting to subtract 20 from both ages.

Final Answer:
Father 44 and son 22