SSC CGL 20 Aug 2021 Shift 1 Paper

Let speed of $\mathit{A}$ is $\mathit{x}\mathit{k}\mathit{m}∕\mathit{h}$.
Speed of $\mathit{B}$ is $\mathit{y}\mathit{k}\mathit{m}∕\mathit{h}$.
We know that, Speed
$\because$ Distance $=40\mathit{k}\mathit{m}$
Time difference $=2\mathit{h}$
According to the question,
$\frac{40}{\mathit{x}}-\frac{40}{\mathit{y}}=2$ . . . (i)
A speed is double and distance is $80\mathit{k}\mathit{m}$
Then, $\frac{80}{2\mathit{x}}-\frac{80}{\mathit{y}}=\frac{3}{2}$ . . . (ii)
Subtracting Eq. (i) by Eq. (ii),
$\Rightarrow \frac{40}{\mathit{x}}-\frac{40}{\mathit{y}}-\frac{40}{\mathit{x}}+\frac{80}{\mathit{y}}$
$=2-\frac{3}{2}=\frac{\left(4-3\right)}{2}$
$\Rightarrow \frac{40}{\mathit{y}}=\frac{1}{2}$


$80\mathit{k}\mathit{m}∕\mathit{h}.$