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1. 18 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. 54 years

B. 72 years

C. 105 years

D. 108 years

Solution:

Let the son's age 18 years ago be x years,
Then man's age 18 years ago = 3x years

\begin{aligned} (3 x + 18) = 2 (x + 18) \\ \Rightarrow 3 x + 18 = 2 x + 36 \\ \Rightarrow x = 18 \end{aligned}

Sum of their present ages

\begin{aligned} \Rightarrow (3 x + 18 + x + 18) years \\ \Rightarrow (4 x + 36) years \\ \Rightarrow (4 \times 18 + 36) years \\ \Rightarrow 108 years \end{aligned}

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