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.

The Volume of the right circular cone of base radius 'r' and height 'h' is given by 'V' = $$\frac{1}{3} \pi r^2h$$

Given 'h' has been increased by 200%

$$\Rightarrow$$ New height h' = h[1 +$$\frac{200}{100}$$] = 3h

also,radius of the base is reduced by 50%

$$\Rightarrow$$ New base radius r' = r[1 - $$\frac{50}{100}$$] = $$\frac{r}{2}$$  

New Volume of the cone with new base radius r' and new height h' is given by V' = $$\frac{1}{3} \pi r'^2h'$$ = $$\frac{1}{3} \pi (\frac{r}{2})^2(3h) = \frac{3V}{4}$$.

Change in Volume = $$\frac{New Volume - Old Volume}{Old Volume}\times 100  = \frac{\frac{3V}{4} - V}{V}\times 100 = -25$$

Hence the new volume decreased by 25 %.