Management Aptitude Test 2014 Paper

(b) Let the speed of faster swimmer be x km/h.
Then, speed of slower swimmer = $\frac{75}{100}\times \mathit{x}=\frac{3}{4}\mathit{x}$ km/h
Now, distance covered by both in 2 h.
Faster swimmer = 2x km
and slower swimmer = $\frac{3}{4}\mathit{x}\times 2=\frac{3}{2}\mathit{x}$ km
After, 2 h the distance between them is 100 km.

From the figure,
${\left(\mathrm{OE}\right)}^2+{\left(\mathrm{OS}\right)}^2={\left(\mathrm{ES}\right)}^2$
⇒ ${\left(\frac{3}{2}\mathit{x}\right)}^2+{\left(2\mathit{x}\right)}^2={\left(100\right)}^2$
[Here, S = South direction, E = East direction, O = Origin]
⇒ $\frac{9}{4}{\mathit{x}}^2+4{\mathit{x}}^2=10000$
⇒ $\frac{9{\mathit{x}}^2+16{\mathit{x}}^2}{4}=10000$
⇒ $\frac{25{\mathit{x}}^2}{4}=10000$
⇒ ${\mathit{x}}^2=\frac{10000\times 4}{25}$
∴ x = $\sqrt{\frac{10000\times 4}{25}}$ = $\frac{100\times 2}{5}$
= 40 km/h
∴ Speed of faster swimmer = 40 km/h
Direction of faster swimmer can also be taken towards East.