Question

the diameter of a sphere is decreased by 25% by what percentage its volume decreases?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{orginal}\mathrm{diameter}\mathrm{of}\mathrm{sphere}=\mathrm{d} \\ \mathrm{Now},\mathrm{original}\mathrm{radius}=\mathrm{r}=\frac{\mathrm{d}}{2} \\ \mathrm{Now},\mathrm{original}\mathrm{volume}=\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^3=\frac{1}{6}{\mathrm{\pi d}}^3 \\ \mathrm{New}\mathrm{diameter}\mathrm{of}\mathrm{sphere},{\mathrm{d}}_1=\mathrm{d}-25\%\mathrm{of}\mathrm{d}=\mathrm{d}-\frac{\mathrm{d}}{4}=\frac{3\mathrm{d}}{4} \\ \mathrm{Now},\mathrm{new}\mathrm{radius}={\mathrm{r}}_1=\frac{{\mathrm{d}}_1}{2}=\frac{3\mathrm{d}}{8} \\ \mathrm{Now},\mathrm{new}\mathrm{volume}={\mathrm{V}}_1=\frac{4}{3}{{\mathrm{\pi r}}_1}^3=\frac{4}{3}\mathrm{\pi }{\left(\frac{3\mathrm{d}}{8}\right)}^3=\frac{9}{128}{\mathrm{\pi d}}^3 \\ \mathrm{decrease}\mathrm{in}\mathrm{volume}=\mathrm{V}-{\mathrm{V}}_1=\left[\frac{1}{6}-\frac{9}{128}\right]{\mathrm{\pi d}}^3=\frac{37}{384}{\mathrm{\pi d}}^3 \\ \mathrm{Decrease}\%=\left[\frac{\mathrm{decrease}}{\mathrm{original}\mathrm{volume}}\times 100\right]\%=\left[\frac{{\displaystyle \frac{37}{384}}{\mathrm{\pi d}}^3}{{\displaystyle \frac{1}{6}}{\mathrm{\pi d}}^3}\times 100\right]\%=42.1875\% \end{gathered}$$