Problems on Ages — “A father’s present age is one more than five times his son’s present age. After 3 years, the father’s age will be two less than four times the son’s age at that time. Find the father’s current age.”

Introduction / Context:
This is a two-equation ages problem using simple linear relationships now and in the future. We model the father’s and son’s present ages with variables and translate the text into equations, then solve systematically.

Given Data / Assumptions:

• Let the son’s present age = s, father’s present age = f.
• Now: f = 5s + 1.
• After 3 years: father will be “two less than four times” the son ⇒ f + 3 = 4(s + 3) − 2.
• Ages are whole numbers and positive.

Concept / Approach:
Convert both sentences into linear equations in s and f, then equate/eliminate to find s first, followed by f. Cross-check both conditions to confirm correctness.

Step-by-Step Solution:

Verification / Alternative check:
In 3 years: father 34, son 9. Four times son is 36; “two less” is 34, which matches the father’s 34. Condition satisfied.

Why Other Options Are Wrong:
29/30/33/40 do not satisfy both the “5s + 1” relation and the “four-times minus two after 3 years” condition simultaneously.

Common Pitfalls:
Misreading “two less than four times” as “four times less two” but applied to the present instead of the future, or forgetting to add 3 years to both ages in the future statement.

Final Answer:
31 years