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 :

Let the radius of a right circular cine be R cm and height be H cm
Volume of right circular cone $$ = \frac{1}{3}\pi {R^2}H{\text{ cu}}{\text{.cm}}$$
When height of right circular cone is increased by 200% and radius of the base is reduce by 50%
New volume :
$$\eqalign{ & {\text{ = }}\frac{1}{3}\pi {\left( {\frac{R}{2}} \right)^2}.3H \cr & = \frac{1}{3}\pi \frac{{{R^2}4}}{4}.3H \cr & = \frac{{\pi {R^2}H}}{4} \cr} $$
Difference :
$$\eqalign{ & = \pi {R^2}H\left( {\frac{1}{3} - \frac{1}{4}} \right) \cr & = \frac{1}{{12}}\pi {R^2}H \cr} $$
Decrease percentage :
$$\eqalign{ & = \frac{{\frac{1}{{12}}\pi {R^2}H}}{{\frac{1}{3}\pi {R^2}H}} \times 100 \cr & = 25\% \cr} $$