Management Aptitude Test Sep 2020 Paper

Here, length of cistern $=4.5\mathit{m}$
Breadth of cistern $=3\mathit{m}$
Volume \/capacity of cistern $=50{\mathit{m}}^3$
$\therefore$ Height of cistern $=\frac{50}{4.5\times 3}=\frac{100}{27}\mathit{m}$
Area to be lined with sheet lead
$=4.5\times 3+2\times \frac{100}{27}\left(4.5+3\right)$
$=13.5+2\times \frac{100}{27}\times 7.5$
$=13.5+\frac{1500}{27}=\frac{1864.5}{27}{\mathit{m}}^2$
$\therefore$ Weight of lead required
$=\frac{1864.5}{27}\times 27=1864.5\mathit{k}\mathit{g}\approx 1864.62\mathit{k}\mathit{g}$