Symbiosis National Aptitude Test SNAP 2015 Paper

Let r be the radius of cylindrical block, then height will be 2r.
Volume of block = $\mathit{\pi }\left({\mathit{r}}^2\right)\left(2\mathit{r}\right)$ = $2\mathit{\pi }{\mathit{r}}^3$
A sphere of maximum possible volume is carved out whose radius will be r.
∴ Volume of sphere = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$
∴ Volume of utilised wood = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$
and volume of wasted wood = $2\mathit{\pi }{\mathit{r}}^3-\frac{4}{3}\mathit{\pi }{\mathit{r}}^3$
= $\frac{6\mathit{\pi }{\mathit{r}}^3-4\mathit{\pi }{\mathit{r}}^3}{3}$
= $\frac{2\mathit{\pi }{\mathit{r}}^3}{3}$
∴ Required ratio = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^3:\frac{2}{3}\mathit{\pi }{\mathit{r}}^3$
= 2 : 1