SSC CGL Tier II 1 Dec 2016 Paper 1

(3) AB = height of hill = h metre
Let speed of vehicle be v metre/minute.
Time taken to reach B from D = t minutes
CD = 10v metre
BD = vt metre
In ΔABC,
tan 45° = $\frac{\mathrm{AB}}{\mathrm{BC}}$
$\Rightarrow 1=\frac{\mathit{h}}{\mathit{B}\mathit{C}}$
⇒ BC = h
= (10v + vt) metre ....(i)
In ΔABD,
tan 60° = $\frac{\mathrm{AB}}{\mathrm{BD}}$
$\Rightarrow \sqrt{3}=\frac{\mathit{h}}{\mathit{v}\mathit{t}}$
$\Rightarrow \mathit{h}=\sqrt{3}\mathit{v}\mathit{t}$
$\Rightarrow 10\mathit{v}+\mathit{v}\mathit{t}=\sqrt{3}\mathit{v}\mathit{t}$
$\Rightarrow 10=\sqrt{3}\mathit{t}-\mathit{t}$
$\Rightarrow 10=\mathit{t}\left(\sqrt{3}-1\right)$
$\Rightarrow \mathit{t}=\frac{10}{\sqrt{3}-1}$
$=\frac{10\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{10\left(\sqrt{3}+1\right)}{2}$
= 5 (1.732 + 1) = 5 × 2.732
= 13.66 minutes
= 13 minutes 40 seconds