Question

Eighteen years ago, a man was three times as old as his son. Now, the man is twice as old as his son. The sum of the present ages of the man and his son is

• A. 72
• B. 100
• C. 105
• D. 108

Answer 1

Answer:

D. 108

Solution: Solution: Let the present age of the son be x. The present age of the man is twice as old as the son, so the present age of the man is 2x. Eighteen years ago, the man was three times as old as his son, so 2x - 18 = 3(x - 18) Simplifying this equation, we get: 2x - 18 = 3x - 54 ⇒ x = 36 So the present age of the son is 36. The present age of the man is 2x, or 2 × 36 = 72. The sum of their present ages is 36 + 72 = 108. Therefore, the sum of the present ages of the man and his son is 108.