Management Aptitude Test Dec 2013 Paper

(b) Here, length of cistern = 4.5 m
Breadth of cistern = 3 m
Volume / capacity of cistern = 50 ${\mathit{m}}^3$
∴ Height of cistern = $\frac{50}{45\times 3}=\frac{100}{27}$ m
Area to be lined with sheet lead
= 4.5 x 3 + 2 x $\frac{100}{27}$(4.5 + 3)
= 13.5 + 2 x $\frac{100}{27}$ x 7.5
= 13.5 + $\frac{1500}{27}$
= $\frac{1864.5}{27}$ ${\mathit{m}}^2$
∴ Weight of lead required
= $\frac{1864.5}{27}$ x 27
= 1864.5 kg ≈ 1864.62 kg