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?

12 times
2 $2 \frac{1}{2} times$ $2\frac{1}{2}{\text{times}}$
3 $2 \frac{3}{4} times$ $2\frac{3}{4}{\text{times}}$
43 times

Answer:- 1

Explanation:-

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 ratio \\ = \frac{(4 x + 16)}{(x + 16)} = \frac{48}{24} = 2 \end{aligned}

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