SSC CPO SI and ASI Model Paper 2

(3) Number of small cubes
$=\frac{\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{o}\mathit{f}\mathit{l}\mathit{a}\mathit{r}\mathit{g}\mathit{e}\mathit{r}\mathit{c}\mathit{u}\mathit{b}\mathit{e}}{\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{o}\mathit{f}\mathit{s}\mathit{m}\mathit{a}\mathit{l}\mathit{l}\mathit{e}\mathit{r}\mathit{c}\mathit{u}\mathit{b}\mathit{e}}$ $=\frac{8^3}{2^3}=64$
Surface area of the larger cube =$6{\left(\mathit{e}\mathit{d}\mathit{g}\mathit{e}\right)}^2=6{\left(8\right)}^2=384\mathit{c}{\mathit{m}}^2$
Surface area of each smaller cube$=6{\left(2\right)}^2=24\mathit{c}{\mathit{m}}^2$
Total surface area of all the smallercubes
= Number of smaller cubes × Surfacearea of each smaller cube
$=64\times 24=1536\mathit{c}{\mathit{m}}^2$
Required ratio = 384 : 1536 = 1 : 4