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1. A cistern, open at the top, is to be lined with sheet of lead which weights 27 kg/m². The cistern is 4.5 m long and 3 m wide and holds 50 m³. The weight of lead required is :

A. 1660.5 kg

B. 1764.5 kg

C. 1860.5 kg

D. 1864.5 kg

Solution:

Let the depth of the cistern be h metresThen,

\begin{aligned} 4.5 \times 3 \times h = 50 \\ \Rightarrow h = \frac{50}{13.5} \\ \Rightarrow h = \frac{100}{27} \end{aligned}

Area of sheet required :

\begin{aligned} = l b + 2 (b h + l h) \\ = l b + 2 h (l + b) \\ = [4.5 \times 3 + 2 \times \frac{100}{27} \times (4.5 + 3)] m^{2} \\ = (13.5 + \frac{200}{27} \times 7.5) m^{2} \\ = (\frac{27}{2} + \frac{500}{9}) m^{2} \\ = \frac{1243}{18} m^{2} \end{aligned}

∴ Weight of lead :

\begin{aligned} = (27 \times \frac{1243}{18}) k g \\ = (\frac{3729}{2}) k g \\ = 1864.5 k g \end{aligned}

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