Question

Age of father 10 years ago was three times the age of his son. After 10 years, father's age is twice that of his son. The ratio of their present ages is:

• A. 7:4
• B. 9:5
• C. 7:3
• D. 11:7

Hint:

Translate age problems into equations for past and future situations, then solve systematically.

Answer 1

The Correct Option is C

Step 1: Represent the present ages.
Let the present age of the father be \(F\) years and that of the son be \(S\) years.
Step 2: Use the first condition (10 years ago).
10 years ago, the father's age = \(F - 10\), son's age = \(S - 10\).
It is given that \(F - 10 = 3(S - 10)\).
So \(F - 10 = 3S - 30 \Rightarrow F = 3S - 20\).
Step 3: Use the second condition (10 years later).
After 10 years, father's age = \(F + 10\), son's age = \(S + 10\).
It is given that \(F + 10 = 2(S + 10)\).
So \(F + 10 = 2S + 20 \Rightarrow F = 2S + 10\).
Step 4: Solve the two equations.
From \(3S - 20 = 2S + 10\), we get \(S = 30\).
Then \(F = 2(30) + 10 = 70\).
Step 5: Find the ratio of present ages.
Ratio \(F:S = 70:30 = 7:3\).
Hence, the answer is \(\boxed{7:3}\).