SSC CGL 17 Aug 2021 Shift 1 Paper

Let speed of $\mathit{A}$ be $\mathit{x}$ and $\mathit{B}$ be $\mathit{y}\mathit{k}\mathit{m}∕\mathit{h}$.
According to the question,
$\frac{160}{\mathit{x}}-\frac{160}{\mathit{y}}=8\mathit{h}$ . . . (i)

If speed of $\mathit{A}$ is double

From Eq. (i) - Eq. (ii),
$\Rightarrow \frac{160}{\mathit{x}}-\frac{160}{\mathit{y}}-\frac{160}{2\mathit{x}}+\frac{160}{\mathit{y}}=8-3$
$\Rightarrow \frac{320-160}{2\mathit{x}}=5$
$\Rightarrow \frac{160}{2\mathit{x}}=5\Rightarrow \mathit{x}=16\mathit{k}\mathit{m}∕\mathit{h}$
Value of $\mathit{x}$ put in Eq. (i),
$\Rightarrow \frac{160}{16}-\frac{160}{\mathit{y}}=8$
$\Rightarrow \frac{160}{\mathit{y}}=10-8=2$
$\mathit{y}=\frac{160}{2}=80\mathit{k}\mathit{m}∕\mathit{h}$
$\therefore$ Speed of $\mathit{B}$ is $80\mathit{k}\mathit{m}∕\mathit{h}$.