Xavier Aptitude Test 2015 Paper

(d)At 5:00 PM, initial distance = AB = 1800 m Minutes hand height = CD = 200 $\sqrt{3}$ m, ∠CAB = 30° $\frac{CB}{AB}$ = tan 30° = $\frac{1}{\sqrt{3}}$ ⇒CB = 600 $\sqrt{3}$ Now, DB =$600\sqrt{3} - 200\sqrt{3} = 400\sqrt{3}$ At 5:10 PM: the minutes hand would move from CD to $C_1$D(i.e. 60°) as follows : $D$ and $D_1$ are at same height. And $\angle C_1 D D_1 = 30^{\circ}$ Now, $\frac{DD_1}{DC_1} = \cos 30^{\circ}$ $\Rightarrow \frac{DD_1}{DC} = \frac{\sqrt{3}}{2}$ $\Rightarrow \frac{DD_1}{200\sqrt{3}} = \frac{\sqrt{3}}{2}$ $\Rightarrow DD_1 = 300\ \text{m}$ Also $\frac{C_1 D_1}{D D_1} = \tan 30^{\circ}$ $\Rightarrow \frac{C_1 D_1}{300} = \frac{1}{\sqrt{3}}$ $\Rightarrow C_1 D_1 = 100\sqrt{3}\ \text{m}$At $5:10\ \text{PM},$ the person moves from A to $E$ , where$\angle C_1 E B_1 = 60^{\circ}.$ Now, $C_1 B_1 = B_1 D_1 + D_1 C_1 = BD + D_1 C_1$ $\Rightarrow C_1 B_1 = 400\sqrt{3} + 100\sqrt{3} = 500\sqrt{3}\ \text{m}$ Now, $\frac{E B_1}{C_1 B_1} = \cot 60^{\circ} \Rightarrow \frac{E B_1}{500\sqrt{3}} = \frac{1}{\sqrt{3}}$ $\Rightarrow E B_1 = 500\ \text{mts}$ The horizontal plane EBB $_{1}$ can be presented as below:Now, $DD_1 = BB_1 = 300\ \text{mts}$ . and $EB_1 = 500\ \text{m}$ Now, $EB = \sqrt{(500)^2 - (300)^2} = 400\ \text{m}$ In 10 minutes, distance travelled $= AE = AB - EB$$= 1800 - 400 = 1400\ \text{m}$ Speed$= \frac{1400 \times 60}{10 \times 1000} = 8.4$ km/hr