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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) $$2\frac{1}{2}\,{\text{times}}$$
c) $$2\frac{3}{4}\,{\text{times}}$$
d) 3 times

Answer: a
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{Ronit’s}}\,{\text{present}}\,{\text{age}}\,{\text{be}}\,x\,{\text{years}}. \cr & {\text{Father’s}}\,{\text{present}}\,{\text{age}}\, = \left( {x + 3x} \right)\,{\text{years}} \cr & = 4x\,{\text{years}} \cr & \left( {4x + 8} \right) = \frac{5}{2}\left( {x + 8} \right) \cr & 8x + 16 = 5x + 40 \cr & 3x = 24 \cr & \Rightarrow x = 8 \cr & {\text{Required}}\,{\text{times}} \cr & = \frac{{ {4x + 16} }}{{ {x + 16} }} \cr & = \frac{{48}}{{24}} \cr & = 2 \cr} $$

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