RRB NTPC Mains or Stage 2 Model Paper 10 with solutions

(d) Let the length, breadth and height of the cuboid be x, 2x and 3x, respectively.Therefore, volume $=\mathit{x}\times 2\mathit{x}\times 3\mathit{x}=6{\mathit{x}}^3$New length, breadth and height = 2x, 6x and 9x, respectively.
New volume $=108{\mathit{x}}^3$
Thus, increase in volume $=\left(108\mathit{–}6\right){\mathit{x}}^3=102{\mathit{x}}^3$
$\frac{\mathit{I}\mathit{n}\mathit{c}\mathit{r}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{i}\mathit{n}\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}}{\mathit{O}\mathit{r}\mathit{i}\mathit{g}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}}=\frac{102{\mathit{x}}^3}{6{\mathit{x}}^3}=17$