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

Volume of cone = $$\ \frac{\ 1}{3}\pi\ r^2h$$ 
Let initial height be 'h' and radius be 'r' of the cone.
So, Initial volume =  $$\ \frac{\ 1}{3}\pi\ r^2h$$
New height H = h(1+200%) = h(1+$$\ \frac{\ 200}{100}$$) = 3h
New radius R = r(1-50%) = r(1-0.5) = $$\ \frac{\ r}{2}$$
New volume = $$\ \ \frac{\ 1}{3}\pi\ \times\ \left(\frac{r}{2}\right)^2\times\ 3h$$ = $$\ \frac{\ \pi\ r^2h}{4}$$
Change in volume = $$\frac{\left(\frac{\ \pi\ r^2h}{4}-\frac{\ \pi\ r^2h}{3}\right)}{\ \frac{\ \pi\ r^2h}{3}}\times\ 100$$ = -25%
So, volume decreased by 25%.
Option B is the answer.