Quantitative Aptitude Pipe and Cistern - General Knowledge Today

Pipe and Cistern

Quantitative Aptitude Questions and Answers section on “Pipe and Cistern” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.

Two pipes can fill a cistern in 3 hours and 4 hours respectively and a waste pipe can empty it in 2 hours. If all the three pipes are kept open, then the cistern will be filled in :
[A]10 hours
[B]12 hours
[C]5 hours
[D]8 hours

12 hours
Part of the cistern filled in 1 hour = $\frac{1}{3}+\frac{1}{4}-\frac{1}{2}$
(Cistern filled by 1st pipe + Cistern filled by 2nd pipe – Cistern emptied by 3rd pipe )
$=>\frac{4+3-6}{12}=\frac{1}{12}$
Hence, the cistern will be filled in 12 hours.
So option [B] is the right answer.

A tap can fill a cistern in 8 hours and another tap can empty it in 16 hours. If both the taps are open, the time (in hours) taken to fill the tank will be:
[A]24 hours
[B]16 hours
[C]10 hours
[D]8 hours

16 hours
Part of the cistern filled in 1 hour = 1 $\div$ 8
Part of the cistern emptied in 1 hour = 1 $\div$ 16
When both the taps are opened simultaneously, part of cistern fill in 1 hour = $\frac{1}{8}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16}$
Hence, the cisatern wil be filled in 16 hours.
So option [B] is the right answer.

A cistern can be filled with water by a pipe in 5 hours and it can be emptied by a second pipe in 4 hours. If both the pipes are opned when the cistern is full, the time in which it will be emptied is :
[A]18 hours
[B]9 hours
[C]20.5 hours
[D]20 hours

20 hours
According to the question,
Cistern filled in 1 hour = 1/5 part
Cistern emptied in 1 hour = 1/4 part
When both pipes are opened simultaneously;
Cistern emptied in 1 hour = $\frac{1}{4}-\frac{1}{5}=\frac{5-4}{20}=\frac{1}{20}$
∴The time in which it will be emptied = 20 hours.
Therefore option [D] is correct.

Two pipes can fill a tank in 15 hours and 20 hours resoectively, while the third can empty it in 30 hours. If all the pipes are opened simultaneously, the empty tank will be filled in :
[A]12 hours
[B]10 hours
[C]15 hours
[D]15.5 hours

12 hours
Part of the tank filled in 1 hour when all three pipes are opened simultaneously = $\frac{1}{15}+\frac{1}{20}-\frac{1}{30}$
$=\frac{4+3-2}{60}=\frac{5}{60}=\frac{1}{12}$
Hence, the tank will be filled in 12 hours.
So option [A] is the right answer.

Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened together, the time taken to fill the tank is:
[A]15 minutes
[B]25 minutes
[C]50 minutes
[D]12 minutes

12 minutes
Part of the tank filled by both pipes in one minute = $\frac{1}{20}+\frac{1}{30}$
Required time = $\frac{1}{\frac{1}{20}+\frac{1}{30}}=\frac{20\times 30}{50}=\frac{60}{5}$
∴ Required time = 12 Minutes
Hence option [D] is the right answer.

A tap can empty a tank in one hour. A second tap can empty it in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?
[A]45 minutes
[B]20 minutes
[C]40 minutes
[D]30 minutes

20 minutes
1 hour = 60 minutes
Rate of emptying the tank by the two taps are $\frac{1}{60}$ and $\frac{1}{30}$ of the tank per minute respectively.
Rate of emptying the tank when both operate simultaneously =
$= \frac{1}{60} + \frac{1}{30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20}$
of the tank per minute.
∴ Time taken by the two taps together to empty the tank = 20 minutes.
Hence option [B] is the right answer.

Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 minutes. In how much time the third pipe alone can empty the cistern?
[A]100 minutes
[B]120 minutes
[C]90 minutes
[D]110 minutes

100 minutes
Let the third pipe empty the cistern in x minutes.
Part of cistren filled in 1 minute when all three pipes are opened simultaneously
$= \frac{1}{60} + \frac{1}{75} - \frac{1}{x}$
According to the question,
$= \frac{1}{60} + \frac{1}{75} - \frac{1}{x} = \frac{1}{50}$
$=> \frac{1}{x} = \frac{1}{60} + \frac{1}{75} - \frac{1}{50}$
$=>\frac{5+4-6}{300}= \frac{3}{300}$
$=> \frac{1}{x} = \frac{3}{300}$
$\therefore x = \frac{300}{3} = 100 minutes$
Hence option [A] is the right answer.

A pipe can fill a tank in ‘x’ hours and another pipe can empty it in ‘y’ hours (y > x). If both the pipes are open, in how many hours will the tank be filled?
[A](y – x) hours
[B](x – y) hours
[C] $\frac{xy}{y-x} hours$
[D] $\frac{xy}{x-y} hours$

$\frac{xy}{y-x} hours$
Part of the tank filled in 1 hour = $\frac{1}{x}$
Part of the tank emptyied in 1 hour = $\frac{1}{y}$
Part of the tank filled in 1 hour when both are opened = $\frac{1}{x}-\frac{1}{y}= \frac{y-x}{xy}$
∴ Tank will be filled in $\frac{xy}{y-x}$ hours
Hence option [C] is the right answer.

Two pipes can fill a cistern separately in 10 hours and 15 hours. They can together fill the cistern in
[A]9 hours
[B]7 hours
[C]6 hours
[D]8 hours

6 hours
Part of the cistern filled by both pipes in 1 hour = $\frac{1}{10}+\frac{1}{15} = \frac{3+2}{30} = \frac{1}{6}$
∴ The cistern will be filled in 6 hours, hence option [C] is the right answer.

10.

Three taps A, B and C together can fill an empty cistern in 10 minutes. The tap A alone can fill it in 30 minutes and the tap B alone in 40 minutes. How long will the tap C alone take to fill it?
[A]40 minutes
[B]32 minutes
[C]16 minutes
[D]24 minutes

24 minutes
Part of the cistern filled by taps A, B and C in 1 minutes = $\frac{1}{10}$
Part of the cistern filled by taps A and B in 1 minutes = $\frac{1}{30}+\frac{1}{40}=\frac{4+3}{120}=\frac{7}{120}$
∴ Part of the cistern filled by tap C in 1 minute =
$= \frac{1}{10}-\frac{7}{120} = \frac{12-7}{120} = \frac{5}{120} = \frac{1}{24}$
∴ Tap C will fill the cistern in 24 minutes.
Hence option [D] is the right answer.

Pipe and Cistern

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