NIFT PG 2008 Question Paper with solutions

Let length and breadth of the room be$\frac{3\mathit{x}}{2}$m and x m respectively.,Then $\frac{3\mathit{x}}{2}\times \mathit{x}=\frac{14400}{150}$
$\Rightarrow {\mathit{x}}^2=\frac{96\times 2}{3}=64$
$\therefore \mathit{x}=8$
Hence length $=\frac{3\mathit{x}}{2}\mathit{m}$
$=\frac{3\times 8}{2}=12\mathit{m}$
breadth = x m = 8 m
Let height of the room = h m ; then
$\Rightarrow 2\left(12+8\right)\times \mathit{h}=\frac{625}{5}$
$\mathit{h}=\frac{125}{2\times 20}=\frac{25}{8}\mathit{m}$
$=3\frac{1}{8}\mathit{m}$
Hence, length, breadth and heights are 12m,8m and $3\frac{1}{8}$m.