Pipes And Cistern - Aptitude MCQ Questions and Solutions with Explanations

Net volume filled in 1 minute
= (x - y) liters
∴ The tank will be filled in
= $$\frac{{\text{T}}}{{\left( {x - y} \right)}}$$  minutes

Net part filled in 1 hour
$$\eqalign{ & {\text{ = }}\frac{1}{3} - \left( {\frac{1}{7} + \frac{1}{{10}}} \right) \cr & = \frac{1}{3} - \frac{{17}}{{70}} \cr & = \frac{{19}}{{210}} \cr} $$
∴ The tank will be filled in $$\frac{{210}}{{19}}$$  hours i.e.
$$\eqalign{ & {\text{= 11}}\frac{1}{{19}}{\text{ hours}} \cr & \cong 11\,{\text{hours }} \cr} $$

$$\eqalign{ & {\text{Net part filled in 1 hour}} \cr & = \frac{2}{3} - \left( {\frac{2}{9} + \frac{1}{3}} \right) \cr & = \left( {\frac{2}{3} - \frac{5}{9}} \right) \cr & = \frac{1}{9} \cr & \therefore \,\frac{1}{9}\,:\,\frac{1}{2}\,::\,1\,:\,x \cr & {\text{or}}\,\,\,x = \left( {\frac{1}{2} \times 9} \right) = 4\frac{1}{2}{\text{ hours}} \cr & {\text{So, the tank will be emptied in}} \cr & {\text{ = }}4\frac{1}{2}{\text{ hours}} \cr} $$

Work done by the C pipe in 1 minute
$$\eqalign{ & = \frac{1}{{15}} - \left( {\frac{1}{{20}} + \frac{1}{{24}}} \right) \cr & = \left( {\frac{1}{{15}} - \frac{{11}}{{120}}} \right) \cr & = - \frac{1}{{40}}\,\left[ { - {\text{ve}}\,{\text{means}}\,{\text{emptying}}} \right] \cr} $$
∴ Volume of $$\frac{1}{{40}}$$ part = 3 gallons.
Volume of whole = (3 × 40) gallons = 120 gallons.

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{(Total Capacity)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{120}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline { \downarrow \,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow } \cr & {\text{efficiency 30}}\,\,\,\,\,\,15\,\,\,\,\,\,\,\,10\,\,\, - 12 \cr & {\text{hours}} \to \,\,\mathop 4\limits_{\left( {\text{P}} \right)}^ \downarrow \,\,\,\,\,\,\mathop 8\limits_{\left( {\text{Q}} \right)}^ \downarrow \,\,\,\,\,\,\,\mathop {12}\limits_{\left( {\text{R}} \right)}^ \downarrow \,\,\,\,\,\,\,\,\,\,\,\mathop {10}\limits_{\left( {\text{S}} \right)}^ \downarrow \cr} $$
In order to fill the cistern in less time.
So, efficiency of filling should be more
now, check all options
(A) → Q efficiency 15 units/hr
(B) → (P + R - S) efficiency
          = 30 + 10 - 12 = 28 units/hr
(C) → (P - S) efficiency
          = 30 - 12 = 18 units/hr
(D) → (P + Q - S) efficiency
          = 30 + 15 - 12 = 33 units/hr
Option (D) is answer
Since efficiency of option (D) is highest

Let the total capacity of the tank is 30 units.
The efficiency of Leakage(Pipe A) will be $$\frac{30}{10}$$  = 3
And the efficiency of the leakage (Pipe A) and another Pipe (B) which is filling the tank will be $$\frac{30}{15}$$  = 2
Pipe A is emptying at 3 units/hr and when filling pipe B started then the emptying rate will come down to 2 units/hr.
∴ Filling Pipe B efficiency is 3 - 2 = 1unit/hr
Pipe B will be fill the tank in $$\frac{30}{1}$$  = 30 hrs
Filling rate of Pipe B per minute is 4 litter
∴ Total Capacity of tank will be = (4 × 60) × 30 = 7200 litters

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

Let the slower pipe alone fill the tank in x minutes
Then, Faster pipe alone will fill it in $$\frac{x}{3}$$ minutes
$$\eqalign{ & \therefore \frac{1}{x} + \frac{3}{x} = \frac{1}{{36}} \cr & \Rightarrow \frac{4}{x} = \frac{1}{{36}} \cr & \Rightarrow x = 144 \cr} $$
So slower pipe alone will fill the tank in 144 min.

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

Capacity of the tank
= (12 × 13.5) litres
= 162 litres
Capacity of each bucket = 9 litres
Number of buckets needed
$$\eqalign{ & {\text{= }}\left( {\frac{{162}}{9}} \right) \cr & = 18 \cr} $$