Indian Institute of Foreign Trade 2009 Solved Paper

Let the initial position of the warship be A and initial and final positions of the submarine be P and Q respectively.
AP $=50$km
Ratio of speed of warship to submarine $=2:1$
$\Rightarrow \mathit{A}\mathit{B}=2\mathit{P}\mathit{Q}$---------------Eqn(I)
In ΔAPN
$\Rightarrow \sin 30=\frac{\mathit{A}\mathit{N}}{\mathit{A}\mathit{P}}$
$\Rightarrow \frac{1}{2}=\frac{\mathit{A}\mathit{N}}{50}$
$\Rightarrow \mathit{A}\mathit{N}=\frac{50}{2}=25$km
$\Rightarrow \mathit{B}\mathit{M}=\mathit{A}\mathit{N}=50\mathit{k}\mathit{m}$
Similarly $\mathit{P}\mathit{N}=25\sqrt{3}=\mathit{A}\mathit{R}$
∴ In ΔBMQ
$\Rightarrow \sin 60=\frac{\mathit{B}\mathit{M}}{\mathit{B}\mathit{Q}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{25}{\mathit{B}\mathit{Q}}$
$\Rightarrow \mathit{B}\mathit{Q}=\frac{50}{\sqrt{3}}$km
Similarly MQ$=\frac{25}{\sqrt{3}}$
Let RB = PM = x and using Eqn (I), we get :
$\Rightarrow \left(\mathit{A}\mathit{R}+\mathit{R}\mathit{B}\right)=2\left(\mathit{P}\mathit{M}+\mathit{M}\mathit{Q}\right)$
$\Rightarrow \left(25\sqrt{3}+\mathit{x}\right)=2\left(\mathit{x}+\frac{25}{\sqrt{3}}\right)$
$\Rightarrow \left(25\sqrt{3}+\mathit{x}\right)=2\mathit{x}+\frac{50}{\sqrt{3}}$
$\Rightarrow 2\mathit{x}-\mathit{x}=25\sqrt{3}-\frac{50}{\sqrt{3}}$
$\Rightarrow \mathit{x}=\frac{25}{\sqrt{3}}$
$\therefore \mathit{A}\mathit{B}=25\sqrt{3}+\frac{25}{\sqrt{3}}$
$\Rightarrow \mathit{A}\mathit{B}=\frac{100}{\sqrt{3}}$
∴ Speed of warship $=2\times \frac{100}{\sqrt{3}}=\frac{200}{\sqrt{3}}$km/h