SSC CHSL 20 Dec 2015 Shift 1 Paper

Given ${\mathit{r}}_1=1$ cm and ${\mathit{r}}_2=6$ cm
Volume of 1st sphere = $\frac{4}{3}\mathit{\pi }{{\mathit{r}}^3}_1=\frac{4}{3}\mathit{\pi }{1}^3=\frac{4}{3}\mathit{\pi }$
Volume of 2nd sphere = $\frac{4}{3}\mathit{\pi }{{\mathit{r}}^3}_2=\frac{4}{3}\mathit{\pi }{\left(6\right)}^3$
Combined volume of two spheres = $\frac{4}{3}\mathit{\pi }\left(217\right)$
Let outer radius of hollow sphere = x then inner radius = x-1
Volume of the hollow sphere is given by,
$\frac{4}{3}\mathit{\pi }\left({\mathit{x}}^3-{\left(\mathit{x}-1\right)}^3\right)=\frac{4}{3}\mathit{\pi }\left(217\right)$
${\mathit{x}}^3-{\left(\mathit{x}-1\right)}^3=217$
${\mathit{x}}^3-{\mathit{x}}^3-1-3{\mathit{x}}^2+3\mathit{x}=217$
$3{\mathit{x}}^2-3\mathit{x}-216=0$
x = 9 or -8 (which cannot be a solution)
Hence, option C is the correct answer.