Management Aptitude Test 2014 Paper

(b) Let the speed of faster swimmer be x km/h. Then, speed of slower swimmer = $\frac{75}{100}\times x = \frac{3}{4}x$ km/h Now, distance covered by both in 2 h. Faster swimmer = 2x km and slower swimmer = $\frac{3}{4}x \times 2 = \frac{3}{2}x$ km After, 2 h the distance between them is 100 km. From the figure, $(OE)^2 + (OS)^2 = (ES)^2$⇒ $\left(\frac{3}{2}x\right)^2 + (2x)^2 = (100)^2$[Here, S = South direction, E = East direction, O = Origin] ⇒ $\frac{9}{4}x^2 + 4x^2 = 10000$⇒ $\frac{9x^2 + 16x^2}{4} = 10000$⇒ $\frac{25x^2}{4} = 10000$⇒ $x^2 = \frac{10000 \times 4}{25}$∴ 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.