Pipes And Cistern - Aptitude MCQ Questions and Solutions with Explanations

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

$$\eqalign{ & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,(A + B)\,{\text{in}}\,{\text{1}}\,{\text{minute}} \cr & = {\frac{1}{{60}} + \frac{1}{{40}}} = \frac{1}{{24}} \cr & {\text{Suppose}}\,{\text{the}}\,{\text{tank}}\,{\text{is}}\,{\text{filled}}\,{\text{in}}\,x\,{\text{minutes}} \cr & {\text{Then}},\,\frac{x}{2}\left( {\frac{1}{{24}} + \frac{1}{{40}}} \right) = 1 \cr & \Rightarrow \frac{x}{2} \times \frac{1}{{15}} = 1 \cr & \Rightarrow x = 30\,\text{minutes} \cr} $$

Time taken by one tap to fill half of the the tank = 3 hours
Part filled by the four taps in 1 hour
$$\eqalign{ & = {4 \times \frac{1}{6}} = \frac{2}{3} \cr & {\text{Remaining}}\,{\text{part}} = {1 - \frac{1}{2}} = \frac{1}{2} \cr & \therefore \frac{2}{3}:\frac{1}{2}::1:x \cr & \Rightarrow x = {\frac{1}{2} \times 1 \times \frac{3}{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4}\,hrs.\,\,i.e.,\,45\,\operatorname{mins} . \cr & {\text{So,}}\,{\text{total}}\,{\text{time}}\,{\text{taken}} = 3\,hrs.\,45\,mins. \cr} $$

$$\eqalign{ & \left( {{\text{A + B}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{15}}} = \frac{9}{{60}} = \frac{3}{{20}} \cr & \left( {{\text{A + C}}} \right){\text{'s 1 hour work}} \cr & {\text{ = }} {\frac{1}{{12}} + \frac{1}{{20}}} = \frac{8}{{60}} = \frac{2}{{15}} \cr & {\text{Part filled in 2 hrs}} \cr & {\text{ = }} {\frac{3}{{20}} + \frac{2}{{15}}} = \frac{{17}}{{60}} \cr & {\text{Part filled in 6 hrs}} \cr & {\text{ = }} {3 \times \frac{{17}}{{60}}} = \frac{{17}}{{20}} \cr & {\text{Remaining part}} \cr & {\text{ = }} {1 - \frac{{17}}{{20}}} = \frac{3}{{20}} \cr} $$
Now it is the turn of A and B and
$$\frac{3}{{20}}$$ part is filled by A and B in 1 hour
∴ Total time taken to fill tank
= (6 + 1) hrs
= 7 hrs

Part filled in 2 hours = $$\frac{2}{6}$$ = $$\frac{1}{3}$$
Remaining part = $$ {1 - \frac{1}{3}} $$  = $$\frac{2}{3}$$
∴ (A + B)'s 7 hour's work = $$\frac{2}{3}$$
(A + B)'s 1 hour's work = $$\frac{2}{{21}}$$
∴ C's 1 hour's work = {(A + B + C)'s 1 hour's work} - {(A + B's 1 hour's work}
$$\eqalign{ & = {\frac{1}{6} - \frac{2}{{21}}} \cr & = \frac{1}{{14}} \cr} $$
∴ C alone can fill the tank in 14 hours

We have the following table:

Period	Supply	Demand	Calculation	Excess Qty. in tank
0 - 3 hrs	30000	10000	30000 - 10000	20000
3 - 6 hrs	30000	10000	20000 + 30000 - 10000	40000
6 - 9 hrs	30000	45000	40000 + 30000 - 45000	25000
9 - 12 hrs	30000	25000	25000 + 30000 - 25000	30000
12 - 15 hrs	30000	40000	30000 + 30000 - 40000	20000
15 - 18 hrs	30000	15000	20000 + 30000 - 15000	35000
18 - 21 hrs	30000	60000	35000 + 30000 - 60000	5000
21 - 24 hrs	30000	35000	5000 + 30000 - 35000	0

The excess quantity in tank at any time does not exceed 40000 litres,
Which is the required minimum capacity to avoid wastage.

Net part filled in 1 hour
$$\eqalign{ & {\text{= }}\left( {\frac{1}{x} - \frac{1}{y}} \right) \cr & = \left( {\frac{{y - x}}{{xy}}} \right) \cr} $$
∴ The tank will be filled in
 $$ = \left( {\frac{{xy}}{{y - x}}} \right)\,\,{\text{hours}}$$

Let the total capacity of tank is 21 units
A's efficiency is 7 units/hr
A's efficiency after leakage 6 units/hr
leakage efficiency = 7 - 6 = 1 unit/hr
leakage will empty the filled tank
$$\frac{{{\text{Total Capacity}}}}{{{\text{Efficiency }}}} = \frac{{21}}{1}$$
= 21hours

Part of the cistern filled in 1 hour
$$ = \frac{1}{3} + \frac{1}{4} - \frac{1}{2}$$
(Cistern filled by 1^st pipe + Cistern filled by 2^nd pipe – Cistern emptied by 3^rd pipe )
$$\eqalign{ & \frac{{4 + 3 - 6}}{{12}} \cr & = \frac{1}{{12}} \cr} $$
Hence, the cistern will be filled in 12 hours.

If tank has 4x liters of total capacity and its holds 3x liters of water and if 30 liters of water is taken out, then tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x = 4 × 10 = 40 liters