Pipes And Cistern - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option B

Part filled by (A + B + C) in 3 minutes
$$\eqalign{ & = 3\left( {\frac{1}{{30}} + \frac{1}{{20}} + \frac{1}{{10}}} \right) \cr & = {3 \times \frac{{11}}{{60}}} = \frac{{11}}{{20}} \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,{\text{C}}\,{\text{in}}\,{\text{3}}\,{\text{minutes}} = \frac{3}{{10}} \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = {\frac{3}{{10}} \times \frac{{20}}{{11}}} = \frac{6}{{11}} \cr} $$

Answer: Option C

$$\eqalign{ & {\text{Net}}\,{\text{part}}\,{\text{filled}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = {\frac{1}{5} + \frac{1}{6} - \frac{1}{{12}}} = \frac{{17}}{{60}} \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\,\frac{{60}}{{17}}\,{\text{hours}} \cr & i.e.,\,3\frac{9}{{17}}\,{\text{hours}} \cr} $$

Answer: Option D

Work done by the leak in 1 hour
$$\eqalign{ & = {\frac{1}{2} - \frac{3}{7}} = \frac{1}{{14}} \cr} $$
∴ Leak will empty the tank in 14 hours

Answer: Option B

Let B be turned off after x minutes.
Then, Part filled by (A + B) in x min. + Part filled by A in (30 - x) min. = 1
$$\eqalign{ & \therefore x\left( {\frac{2}{{75}} + \frac{1}{{45}}} \right) + \left( {30 - x} \right).\frac{2}{{75}} = 1 \cr & \Rightarrow \frac{{11x}}{{225}} + \frac{{ {60 - 2x} }}{{75}} = 1 \cr & \Rightarrow 11x + 180 - 6x = 225 \cr & \Rightarrow x = 9 \cr} $$

Answer: Option C

Suppose, first pipe alone takes x hours to fill the tank.
Then, Second and third pipes will take (x - 5) and (x - 9) hours respectively to fill the tank.
$$\eqalign{ & \therefore \frac{1}{x} + \frac{1}{{ {x - 5} }} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \frac{{x - 5 + x}}{{x\left( {x - 5} \right)}} = \frac{1}{{ {x - 9} }} \cr & \Rightarrow \left( {2x - 5} \right)\left( {x - 9} \right) = x\left( {x - 5} \right) \cr & \Rightarrow {x^2} - 18x + 45 = 0 \cr & \Rightarrow \left( {x - 15} \right)\left( {x - 3} \right) = 0 \cr & \Rightarrow x = 15{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left[ {{\text{neglecting}}\,x = 3} \right] \cr} $$