# 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: Option A
Let Ronit’s present age be x yeras. Then, Father’s present age $\begin{array}{l}=(x+3x)\text{ years }\\ =4x\text{ years }\\ \therefore(4x+8)=\frac{5}{2}(x+8)\\ \Rightarrow8x+16=5x+40\\ \Rightarrow3x=24\\ \Rightarrow x=8\\ \text{Hence, required times}\\ \begin{array}{l}=\frac{4x+16}{x+16}\\ =\frac{48}{24}\\ =2\end{array}\end{array}$