Pipes and Cisterns Practice Test 2 with solutions

Let time taken by Pipe A alone = x
Time taken by Pipe B alone = 2x
Part of pool emptied by A and B together
= $\frac{1}{\mathit{x}}+\frac{1}{2\mathit{x}}$
= $\frac{1}{\mathit{x}}\left(1+\frac{1}{2}\right)$
= $\frac{3}{2}\times \frac{1}{\mathit{x}}$
So, time taken by A and B together = $\frac{2\mathit{x}}{3}$
Time taken by C = 2 × $\frac{2\mathit{x}}{3}$ = $\frac{4\mathit{x}}{3}$
Part of pool emptied by A, B and C together
= $\frac{1}{\mathit{x}}+\frac{1}{2\mathit{x}}+\frac{3}{4\mathit{x}}$
= $\frac{1}{\mathit{x}}\left[1+\frac{1}{2}+\frac{3}{4}\right]$
= $\frac{1}{\mathit{x}}\left[\frac{4+2+3}{4}\right]$ = $\frac{9}{4\mathit{x}}$
Time taken by all together = $\frac{4\mathit{x}}{9}$
$\frac{4}{9}\mathit{x}$ = 4 hours
$\frac{4}{9}\mathit{x}$ = 4
x = $\frac{9}{4}\times 4$
x = 9 hours
So, time taken by pipe A alone is 9 hours