A right circular cone of largest volumeis cut out from a solid wooden hemisphere. The remaining material is what percentage of the volume ofthe original hemisphere?

The volume of hemisphere =  $$\dfrac{2}{3} \pi r^3 $$

volume of cone = $$ \dfrac{1}{3} \pi r^2 h $$

where  r = h 

volume of cone = $$\dfrac{1}{3} \pi r^2 r - \dfrac{1}{3} \pi r^3 $$

so volume of remaining = $$ \dfrac{2}{3} \pi r^3 - \dfrac{1}{3} \pi r^3 $$

$$\Rightarrow \dfrac{1}{3} \pi r^3 $$ 

% of remaining material =$$ \dfrac{\dfrac{1}{3}\pi r^3} {\dfrac{2}{3} \pi r^3} \times 100 $$

$$\Rightarrow \dfrac{1}{2} \times 100 $$

$$\Rightarrow 50 Ans %