The present age of a man is 4 years more than twice the present age of his son. After 6 years, the sum of their ages will be 58 y…

Question

The present age of a man is 4 years more than twice the present age of his son. After 6 years, the sum of their ages will be 58 years. What is the present age of the son?

1. A. 10 years
2. B. 11 years
3. C. 14 years (Correct answer)
4. D. 9 years

Correct Answer

Option C — 14 years

Detailed Solution & Explanation

The correct answer is 14 years.

Shortcut Trick

• Test the options against the second condition — with four choices this is often faster than forming equations.
• If the son is 14, the man is $2(14) + 4 = 32$. After 6 years they are 20 and 38, summing to 58 ✓.

Alternate Method

• Let the son's present age be s. Then the man's age is $2s + 4$.
• After 6 years: son $= s + 6$, man $= 2s + 10$.
• Given their sum is 58:
  • $(s + 6) + (2s + 10) = 58$
  • $3s + 16 = 58$
  • $3s = 42 \Rightarrow s = \textbf{14}$
• The man is therefore 32 now, and in 6 years they will be 20 and 38.

Additional Information

• The reliable habit in age problems: add the same number of years to *every* person. Here two people gain 6 years each, so the combined sum rises by 12, not 6 — the single most common error in this question type.
• Working backwards, the sum today is $58 - 12 = 46$, which also gives $3s + 4 = 46 \Rightarrow s = 14$.
• The difference between two people's ages never changes, so any statement about their difference applies at every point in time.