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?

The correct option is A: $2$ times

Let Ronit's present age be $x$ years. Then, father's present age $=(x+3x)$ years $=4x$ years. [His father is $3$ times more that ronit's age so, $x+3x$ will be his age now ]

After 8 years, Ronit,s Age $=x+8$

His father's age $=(4x+8)$

According to the question, we have

$\therefore (4x+8)=\frac{5}{2}(x+8)$
$\Rightarrow 8x+16=5x+40$
$\Rightarrow 3x=24$
$\Rightarrow x=8.$

Further $8$ years will means $16$ years from the current age.

After $16$ years from present, the ratio of their ages will be

$\frac{\text{His father's age}}{\text{Ronit' age}}=\frac{(4x+16)}{(x+16)}=\frac{4\times 8+16}{8+16}=\frac{48}{24}=2.$

$\therefore$ His father $=2\times$ Ronit age

Hence, his father will be $2$ times Ronit's present age after $16$ years from now.