SSC CHSL 10 Jan 2017 Shift 1 Paper

Let the external radius, the internal radius and the height of the hollow cylinder be R, r and h respectively.
=> R = 7 cm , h = 22 cm and r = 7 - 1 = 6 cm
Total surface area of hollow cylinder = $\left(2\mathit{\pi }\mathit{R}\mathit{h}\right)+\left(2\mathit{\pi }\mathit{r}\mathit{h}\right)+2\mathit{\pi }\left({\mathit{R}}^2-\mathit{r}\right)$
= $\left(2\times \frac{22}{7}\right)[\left(7\times 22\right)$ $+\left(6\times 22\right)+\left(7^2-6^2\right)]$
= $\left(\frac{44}{7}\right)\times \left(154+132+13\right)$
= $\frac{44}{7}\times 299=1879.428$ $\approx 1879.43\mathit{c}{\mathit{m}}^2$