A geometry problem by nishant kumar

Let the radius of the original sphere be $r$ then the radius of the new sphere is $2r$ . The volume of a sphere is given by $v=\dfrac{4}{3} \pi r^3$ where $r$ is the radius. So the volume of the original sphere is $\dfrac{4}{3}\pi r^3$ and the volume of the new sphere is $\dfrac{4}{3} \pi (2r)^3 = \dfrac{32}{3}\pi r^3$ .

$\%~increased = \dfrac{new~volume-original~volume}{original~volume} \times 100\%=\dfrac{\dfrac{32}{3}\pi r^3-\dfrac{4}{3}\pi r^3}{\dfrac{4}{3}\pi r^3}\times 100\%=\dfrac{\dfrac{32}{3}-\dfrac{4}{3}}{\dfrac{4}{3}}\times 100\%=\dfrac{\dfrac{28}{3}}{\dfrac{4}{3}}\times 100\%=\dfrac{28}{3}\times \dfrac{3}{4}\times 100\%=\boxed{700\%}$