Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later… · ABC26GN3706
Subject: General Aptitude · Chapter: Time and Distance · Exam: SNAP 2008
Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was 17 km, and another 15 minutes later, one sea trawler was 10.5 km farther from the origin than the other. Find the speed of each sea trawler.
(a)16 km/hr, 30 km/hr
(b)18 km/hr, 24 km/hr
(c)20 km/hr, 22 km/hr
(d)18 km/hr, 36 km/hr
Answer
Answer (as printed): A
Explanation
Suppose the two trawlers start from a point $O$ and move in the directions $O A$ and $O B$ respectively. Let the speeds of the two sea trawlers be $x \mathrm{km} / \mathrm{hr}$ and y km/hr. respectively. Then, $$\Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{4}=289 \Rightarrow x^{2}+y^{2}=1156$$ And, $\left(x \times \frac{3}{4}\right)-\left(y \times \frac{3}{4}\right)=10.5 \Rightarrow x-y=10.5 \times \frac{4}{3}=14$ Now, $(x+y)^{2}+(x-y)^{2}=2\left(x^{2}+y^{2}\right)$ $$\begin{aligned} & \Rightarrow(x+y)^{2}=2 \times 1156-(14)^{2}=2312-196=2116 \\ & \Rightarrow x+y=\sqrt{2116}=46 \end{aligned}$$ Adding (ii) and (iii), we get : $2 x=60$ or $x=30$. Putting $x=30$ in (ii), we get : $y=16$. Hence, the speeds of the two sea-trawlers are 30 km/hr and 16 km/hr.