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?
Answer & Solution
Answer: Option
A
Solution:
$$\eqalign{
& {\text{Let}}\,{\text{Ronit's}}\,{\text{present}}\,{\text{age}}\,{\text{be}}\,x\,{\text{years}}. \cr
& {\text{Then,}}\,{\text{father's}}\,{\text{present}}\,{\text{age}}\, \cr
& = \left( {x + 3x} \right)\,{\text{years}} \cr
& = 4x\,{\text{years}} \cr
& \therefore \left( {4x + 8} \right) = \frac{5}{2}\left( {x + 8} \right) \cr
& \Rightarrow 8x + 16 = 5x + 40 \cr
& \Rightarrow 3x = 24 \cr
& \Rightarrow x = 8 \cr
& {\text{Hence,}}\,{\text{required}}\,{\text{times}} \cr
& = \frac{{ {4x + 16} }}{{ {x + 16} }} \cr
& = \frac{{48}}{{24}} \cr
& = 2 \cr} $$