Management Aptitude Test Feb 2012 Paper

(2 ) Let the distance travelled by the ship be x m.
Now , in ΔACD

tan 60° = $\frac{100}{\mathit{C}\mathit{D}}$ ⇒ CD = $\frac{100}{\sqrt{3}}$
And in ΔABD
tan 30° = $\frac{100}{\mathit{B}\mathit{D}}$ ⇒ BD = 100$\sqrt{3}$
⇒ BC + CD = 100$\sqrt{3}$
⇒ x + $\frac{100}{\sqrt{3}}$ = 100$\sqrt{3}$
⇒ x = 100$\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)$
⇒ x = 100 x $\frac{2}{\sqrt{3}}$ = $\frac{200}{\sqrt{3}}=\frac{200}{1.732}$
= 115.47 m = 115. 5 m