Question

Surface area of a sphere A is $300\%$ more that the surface area of another sphere B. If the volume of sphere is $p\%$ less than the volume of sphere A. Find the value of P.

Answer 1

Let the radius of sphere $A=r^1$ and radius of sphere $B=r^2$.
Surface Area of sphere $A=\frac{4}{3}\pi \times r^1{r}^2$ and SA of sphere $B=\frac{4}{3}\times \pi \times r{2}^2.$
Ratio of SA of A to B = $(\frac{r^1}{r^2}{)}^2=4.\Rightarrow \frac{r^1}{r^2}=2.$
Volume of A = x
Volume of B = $(1-\frac{p}{100})x$
$\frac{Va}{Vb}=\frac{1(1-p}{100})=(\frac{r^1}{r^2}{)}^3=(2{)}^3=8.$
$\Rightarrow 1-\frac{p}{100}=\frac{1}{8}$
$\Rightarrow 1-\frac{1}{8}=\frac{p}{100}$
$\frac{7}{8}=\frac{p}{100}orp=\frac{700}{8}=87.5\%$