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1. Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?

A. 2 times

B. $2 \frac{1}{2} times$

C. $2 \frac{3}{4} times$

D. 3 times

Solution:

\begin{aligned} Let Ronit's present age be x years . \\ Then, father's present age \\ = (x + 3 x) years \\ = 4 x years \\ ∴ (4 x + 8) = \frac{5}{2} (x + 8) \\ \Rightarrow 8 x + 16 = 5 x + 40 \\ \Rightarrow 3 x = 24 \\ \Rightarrow x = 8 \\ Hence, required times \\ = \frac{4 x + 16}{x + 16} \\ = \frac{48}{24} \\ = 2 \end{aligned}

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