If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone :

Answer: Option B

Let the original radius and height of the cone be r and h respectively
Then, Original volume = $$\frac{1}{3}\pi r^{2}h$$
New radius = $$\frac{r}{2}$$ and new hight = 2h
New volume :
$$\begin{array}{rl}& =\frac{1}{3}\times \pi \times {\left(\frac{r}{2}\right)}^2\times 3h\\ & =\frac{3}{4}\times \frac{1}{3}\pi r^{2}h\end{array}$$
∴ Decrease % :
$$\begin{array}{rl}& =\left(\frac{\frac{1}{4}\times \frac{1}{3}\pi r^{2}h}{\frac{1}{3}\pi r^{2}h}\times 100\right)\%\\ & =25\%\end{array}$$