Question

The sum of the present ages of a father and his son is 70 years. Ten years ago, the father’s age was thrice the age of the son. Now the son’s age will be:

• A. 22.5 years
• B. 18 years
• C. 15 years
• D. 20.5 years

Hint:

Always remember to subtract the time from both parties when going into the past. Common mistake: Subtracting 10 only from the father.

Answer 1

The Correct Option is A

Step 1: Set up the system of equations.
Let son's present age = $s$ years and father's present age = $f$ years; from the given information: $f + s = 70$ ... (i).
Step 2: Use the second condition to form another equation.
Ten years ago, father's age = 3 times son's age: $f - 10 = 3(s - 10)$, expanding: $f = 3s - 20$ ... (ii).
Step 3: Solve and verify.
Substituting (ii) into (i): $(3s - 20) + s = 70$, so $4s = 90$, giving $s = 22.5$ years; father's age = $70 - 22.5 = 47.5$ years; verify: 10 years ago son = 12.5, father = 37.5 = $3 \times 12.5$. Correct!
\[ \boxed{22.5 \text{ years}} \]