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 next  8 years, how many times would he be of Ronit’s age? 

$$\begin{gathered} \text{ Let Ronit's present age be }x\text{ years. Then, father's present age } \\ =(x+3x)\text{ years }=4x\text{ years.} \\ \text{[His father is }3\text{ times more that ronit's age so, }x+3x\text{ will be his age now ] } \\ \text{ After }8\text{ years, Ronit,s Age }=x+8 \\ \text{ His father's age }=(4x+8) \\ \therefore (4x+8)=\frac{5}{2}(x+8) \\ \Rightarrow 8x+16=5x+40 \\ \Rightarrow 3x=24 \\ x=8 \\ \text{ Further }8\text{ years will means }16\text{ years from the current age. } \\ \text{ After }16\text{ years from present, the ratio of their ages will be } \\ \frac{\text{ His father's age }}{\text{ Ronit' age }}=\frac{(4x+16)}{(x+16)}=\frac{4\times 8+16}{8+16}=\frac{48}{24}=2 \\ \therefore \text{ His father }=2\times \text{ Ronit age } \end{gathered}$$