Management Aptitude Test Sep 2020 Paper

Let the speed of faster swimmer be $\times \mathit{k}\mathit{m}∕\mathit{h}$.
Then, speed of slower swimmer $=\frac{75}{100}\times \mathit{x}$
$=\frac{3\mathit{x}}{4}\mathit{k}\mathit{m}∕\mathit{h}$
Now, distance covered by both in $2\mathit{h}$.
Faster swimmer $=2\mathit{x}\mathit{k}\mathit{m}$
and slower swimmer $=\frac{3\mathit{x}}{4}\times 2$
$=\frac{3\mathit{x}}{2}\mathit{k}\mathit{m}$
After, $2\mathit{h}$ the distance between them is $100\mathit{k}\mathit{m}$.From the figure,
${\left(\mathit{O}\mathit{E}\right)}^2+{\left(\mathit{O}\mathit{S}\right)}^2={\left(\mathit{E}\mathit{S}\right)}^2$
$\Rightarrow {\left(\frac{3\mathit{x}}{2}\right)}^2+{\left(2\mathit{x}\right)}^2={\left(100\right)}^2$
[Here, $\mathit{S}=$ South direction, $\mathit{E}=$ East direction, $\mathit{O}=$ Origin]
$\Rightarrow \frac{9{\mathit{x}}^2}{4}+4{\mathit{x}}^2=10000$
$\Rightarrow \frac{9{\mathit{x}}^2+16{\mathit{x}}^2}{4}=10000$
$\Rightarrow \frac{25{\mathit{x}}^2}{4}=10000$
$\Rightarrow {\mathit{x}}^2=\frac{10000\times 4}{25}$
$\therefore \mathit{x}=\sqrt{\frac{10000\times 4}{25}}=\frac{100\times 2}{5}$
$=40\mathit{k}\mathit{m}∕\mathit{h}$
$\therefore$ Speed of faster swimmer $=40\mathit{k}\mathit{m}∕\mathit{h}$
Note Direction of faster swimmer can also be taken towards East.