CAT 2019 Shift 2 Paper

To get the minimum number of cylinders, the volume of each of the cylinder must be HCF of $405,783,$ and 351
$\Rightarrow \mathit{H}\mathit{C}\mathit{F}\left(405,783,351\right)=27$
Therefore, number of cylinders of iron $=\frac{405}{27}=15$
and, number of cylinders of aluminum $=\frac{783}{27}=29$
and, number of cylinders of copper $=\frac{351}{27}=13$
Hence, the total number of a cylinders $=15+29+13=57$
Also, volume of each cylinder $=27\mathit{c}\mathit{c}$
$\Rightarrow \mathit{\pi }{\mathit{r}}^2\mathit{h}=27$
$\Rightarrow \mathit{\pi }\times 3^2\times \mathit{h}=27$
$\Rightarrow \mathit{h}=\frac{3}{\mathit{\pi }}$
And total surface area of each cylinder $=2\mathit{\pi }\mathit{r}\left(\mathit{r}+\mathit{h}\right)$
$=2\mathit{\pi }\times 3\left(3+\frac{3}{\mathit{\pi }}\right)=18\left(\mathit{\pi }+1\right)$
Hence, total surface area of 57 cylinders $=57\times 18\left(\mathit{\pi }+1\right)$
$=1026\left(\mathit{\pi }+1\right)$