Ten years ago, a man was seven times as old as his son. Two years hence, twice his age will be equal to five times the age of his son. What is the present age of the son?

### Concept & Formula The core concept is defining present ages in terms of a single variable based on past conditions, and then translating future conditions into a solvable algebraic equation. Let the son's age $10$ years ago be $x$. The father's age then was $7x$. Their present ages can be expressed by adding $10$ to both. ### Step-by-Step Solution * **Given:** * $10$ years ago: Man = $7 \times$ Son * $2$ years hence (future): $2 \times$ Man = $5 \times$ Son * **Calculation / Deduction:** * Let the son's age $10$ years ago be $x$. * Then, the man's age $10$ years ago was $7x$. * Their *present* ages are: * Son = $x + 10$ * Man = $7x + 10$ * Their ages $2$ years hence (from present) will be: * Son = $(x + 10) + 2 = x + 12$ * Man = $(7x + 10) + 2 = 7x + 12$ * The problem states that in this future time, twice the man's age equals five times the son's age: * $$2(7x + 12) = 5(x + 12)$$ * $$14x + 24 = 5x + 60$$ * Group the variables and constants: * $$14x - 5x = 60 - 24$$ * $$9x = 36$$ * $$x = 4$$ * We need the *present* age of the son, which is $x + 10$. * Present age = $4 + 10 = 14$ years. ### Exam Strategy & Shortcut Use Option Elimination with timeline matching. The son's present age minus $10$ years must yield a valid, workable past age. Testing option (c) $14$ years: If present son = $14$, past son ($10$ yrs ago) = $4$. Past man = $7 \times 4 = 28$. Present man = $28 + 10 = 38$. Future man ($+2$ yrs) = $40$. Future son ($+2$ yrs) = $16$. Check condition: $2 \times 40 = 80$, and $5 \times 16 = 80$. They match perfectly! ### Common Pitfall A major trap is successfully calculating $x = 4$ and assuming it is the final answer, or adding only $2$ instead of $10$ to find the present age. Always clearly label what time period your variable $x$ belongs to. ### Final Answer **Therefore, the correct answer is 14 years.**