In this section we will learn how to solve aptitude problems based on Pipes and Cisterns. The introduction gives a list of formulae that can be used to solve aptitude question and answers. The exercises are divided into sections based on two pipe and three pipe problems. The following kinds of problems are addressed in the exercises:
- Pipes that fill a tank
- Combination of pipes that can fill or empty a tank
Introduction to Pipes and Cisterns
Problems on pipes and cisterns are similar to the problems based on time and work. In pipes and cisterns, we mainly come across problems that are concerned with filling or emptying a tank. The following points will be helpful in solving problems on pipes and cisterns
- Inlet pipe : A pipe that fills a tank or cistern is the inlet pipe.
- Outlet pipe : A pipe that empties a tank or cistern is the outlet pipe.
- If a pipe takes $x$ hours to fill a tank, then the portion of the tank filled in 1 hour is $\frac1x$
- If a pipe takes $x$ hours to empty a tank, then the portion of the tank emptied in 1 hour is $\frac1x$
- If a pipe takes $x$ hours to fill a tank and another pipe takes $y$ hours ($y$>$x$) to empty a full tank, then the portion of the tank filled in 1 hour when both the pipes are open is $\frac1x$ – $\frac1y$
- If a pipe takes $x$ hours to fill a tank and another pipe takes $y$ ($x$>$y$) hours to empty a full tank, then the portion of the tank emptied in 1 hour when both the pipes are open is $\frac1y$ – $\frac1x$
Problems
Pipes and Cisterns - Exercise - Blog
15 questions
- Two pipes can fill a tank in 4 hours and 6 hours respectively. Find the time taken by both the pipes to fill the tank.
- 2.4 hours
- 1.2 hours
- 1.8 hours
- 3 hours
Show the answer
Answer: 2.4 hours
Portion of the tank filled by first pipe in 1 hour = $\frac14$
Portion of the tank filled by second pipe in 1 hour = $\frac16$
Portion of the tank filled by both the pipes together in 1 hour = $\frac14$ + $\frac16$
= $\frac{5}{12}$
Hence if $\frac{5}{12}$ of the tank is filled in 1 hour, the entire tank will be full in $\frac{12}{5}$ = 2.4 hours or 2 hours 24 minutes.
- Pipe A can fill a cistern in 3 hours while pipe B can empty the cistern in 5 hours. If both the pipes are open together find the total time taken to fill the tank.
- 5 hours and 20 minutes
- 6 hours and 10 minutes
- 7 hours and 30 minutes
- 4 hours and 40 minutes
Show the answer
Answer: 7 hours and 30 minutes
In 1 hour pipe A can fill $\frac{1}{3}$ of the cistern, while pipe B can empty $\frac{1}{5}$ of the cistern.
If both the pipes are opened together, then the amount of cistern filled in 1 hour = $\frac{1}{3}$ - $\frac{1}{5}$
= $\frac{2}{15}$
Time taken to fill the tank when both the pipes are open = $\frac{15}{2}$ = 7 hours and 30 minutes
- A pipe normally fills a tank in 30 minutes but due to a leak in the tank it takes 40 minutes to fill the tank. In how much time does the full tank empty due to the leak.
- 1 hours
- 1.5 hours
- 2 hours
- 2.5 hours
Show the answer
Answer: 2 hours
Difference in time taken per minute to fill the tank with out leak and with leak gives the time taken per minute by the leak to empty the tank.
i.e,
Normal time taken to fill the tank per minute = $\frac{1}{30}$
Time taken to fill the tank per minute with leak = $\frac{1}{40}$
Time taken to empty the tank by the leak per minute = $\frac{1}{30}$ - $\frac{1}{40}$
= $\frac{1}{120}$
$\frac{1}{120}$ of the tank is emptied by the leak in 1 minute. Hence a full tank is emptied by the leak in $\frac{120}{1}$ = 120 minutes or 2 hours
- Pipes A and B can fill a tank in 9 hours and 6 hours respectively. If both the pipes are opened together and after 2 hours pipe B is turned off. Find the total time in which the tank will be full.
- 5 hours
- 6 hours
- 4 hours
- 3 hours
Show the answer
Answer: 6 hours
In 1 hour pipe A can fill $\frac{1}{9}$ of the tank
In 1 hour pipe B can fill $\frac{1}{6}$ of the tank
If both the pipes are open together, amount of tank filled in 1 hour = $\frac{1}{9}$ + $\frac{1}{6}$
= $\frac{5}{18}$
Since pipe B is closed after 2 hours, amount of tank filled in 2 hours =$\frac{5}{18}$ x 2
= $\frac{5}{9}$
Remaining part of tank = 1 - $\frac{5}{9}$ = $\frac{4}{9}$
Now,
$\frac{1}{9}$ of the tank will be filled by pipe A in 1 hour,
hence, $\frac{4}{9}$ of the tank will be filled by pipe A in $\dfrac{\frac{4}{9}} {\frac{1}{9}}$ = 4 hours
Total time taken to fill the tank = 2 + 4 = 6 hours.
- Two pipes can fill a tank in 6 hours and 8 hours respectively. If they are turned on alternatively for 1 hour each, find the time in which the tank is full. (Assume pipe A is opened first)
- 3 hours and 30 minutes
- 4 hours and 45 minutes
- 5 hours and 30 minutes
- 6 hours and 45 minutes
Show the answer
Answer: 6 hours and 45 minutes
In 1 hour pipe A fills $\frac{1}{6}$ of the tank
In 1 hour pipe B fills $\frac{1}{8}$ of the tank
Since pipes A and B are turned on alternatively starting with A,
In 2 hours pipes A and B can fill $\frac{1}{6}$ + $\frac{1}{8}$ = $\frac{7}{24}$
Now, we can see that after 6 hours A and B can alternatively fill $\frac{7}{24}$ x 3 = $\frac{7}{8}$ part of the tank.
Remaining portion of the tank => 1 - $\frac{7}{8}$ = $\frac{1}{8}$
$\frac{1}{8}$ of the tank is filled by A in$\dfrac{\frac{1}{8}}{\frac{1}{6}}$ = $\frac{3}{4}$ of an hour or 45 minutes.
Hence the total time taken to fill the tank = 6 hours + 45 minutes = 6 hours and 45 minutes.
- Three pipes can fill a tank in 30, 40 and 50 minutes respectively. Find the time taken by all the three pipes to fill the tank together.
- 11.5 hours
- 12.7 hours
- 13.3 hours
- 10.2 hours
Show the answer
Answer: 12.7 hours
Since the three pipes fill the tank in 30, 40 and 50 minutes respectively, part of tank filled by them in 1 minute is $\frac{1}{30}$, $\frac{1}{40}$ and $\frac{1}{50}$ respectively.
Hence, part of the tank filled by all the three pipes together in 1 minute = $\frac{1}{30}$ + $\frac{1}{40}$ + $\frac{1}{50}$
= $\frac{47}{600}$
Hence the time taken by all the three pipes together to fill the tank = $\frac{600}{47}$
= 12.76 hours or 12 hours 46 minutes approx.
- Two pipes P and Q can fill a tank in 3 and 4 hours respectively while a pipe R can empty the tank in 6 hours. If all the 3 pipes are opened together, find the time in which the tank is full.
- 2 hours and 24 minutes
- 1 hours and 30 minutes
- 4 hours and 24 minutes
- 3 hours and 30 minutes
Show the answer
Answer: 2 hours and 24 minutes
In 1 hour pipes P and Q can fill $\frac{1}{3}$ and $\frac{1}{4}$ portions of the tank respectively, while pipe R can empty $\frac{1}{6}$ of the tank.
Portion of the tank filled when all the three pipes are opened together is
$\frac{1}{3}$ + $\frac{1}{4}$ - $\frac{1}{6}$ = $\frac{5}{12}$
Hence, when all the three pipes are open together, the tank will be full in $\frac{12}{5}$ = 2 hours and 24 minutes
- Two pipes take 3 hours and 4 hours respectively to fill a tank. But due to a leak they take 5 hours to fill the tank. Find the time taken by the leak to empty a full tank.
- 1$\frac{11}{20}$ hours
- 2$\frac{14}{23}$ hours
- 2$\frac{7}{9}$ hours
- 4$\frac{1}{2}$ hours
Show the answer
Answer: 2$\frac{14}{23}$ hours
In 1 hour portion of the tank filled by the two pipes is $\frac{1}{3}$ and $\frac{1}{4}$.
In 1 hour portion of the tank filled by both the pipes along with the leak is $\frac{1}{5}$
Now, amount of leak in 1 hour => $\frac{1}{3}$ + $\frac{1}{4}$ - $\frac{1}{5}$
=> $\frac{7}{12}$ - $\frac{1}{5}$ = $\frac{23}{60}$
Hence the time taken by the leak to empty a full tank => $\frac{60}{23}$ = 2$\frac{14}{23}$ hours
- Pipes P, Q and R can fill a tank in 3, 4 and 5 hours respectively. If all the pipes are opened together and after 30 minutes pipes Q and R are turned off, find the total time in which the tank is full.
- 2 $\frac{13}{40}$ hours
- 1 $\frac{2}{3}$ hours
- 2 $\frac{10}{11}$ hours
- 1 $\frac{13}{40}$ hours
Show the answer
Answer: 2 $\frac{13}{40}$ hours
In 1 hour, pipes P, Q and R fill $\frac{1}{3}$, $\frac{1}{4}$, and $\frac{1}{5}$ portion of the tank respectively.
Portion of the tank filled by all the 3 pipes together in 1 hour=> $\frac{1}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ = $\frac{47}{60}$
Portion of the tank filled by the 3 pipes in 30 minutes => $\frac{47}{60}$ x $\frac{1}{2}$ = $\frac{47}{120}$
Remaining portion of the tank => 1 - $\frac{47}{120}$ = $\frac{73}{120}$
Since pipe P fills $\frac{1}{3}$ of the tank in 1 hour,
$\frac{73}{120}$ portion of the tank is filled in $\dfrac{\frac{73}{120}}{\frac{1}{3}}$ = $\frac{73}{40}$ hours
Total time in which the tank is full = $\frac{73}{40}$ + $\frac{1}{2}$
=> $\frac{93}{40}$ = 2 $\frac{13}{40}$ hours
- Pipes A, B and C can fill a cistern in 20, 30 and 40 hours respectively. If pipes B and C are turned on alternatively for 1 hour (assume B is opened first) and A is turned on the whole time. Find the total time in which the tank is full.
- 10 hours and 10 minutes
- 12 hours and 12 minutes
- 10 hours and 36 minutes
- 12 hours and 36 minutes
Show the answer
Answer: 12 hours and 36 minutes
Portion of the tank filled by A and B in 1 hour => $\frac{1}{20}$ + $\frac{1}{30}$ = $\frac{5}{60}$ = $\frac{1}{12}$
Portion of the tank filled by A and C in 1 hour => $\frac{1}{20}$ + $\frac{1}{40}$ = $\frac{3}{40}$
Portion of the tank filled in 2 hours =>$\frac{1}{12}$ + $\frac{3}{40}$ = $\frac{10+9}{120}$ = $\frac{19}{120}$
In 12 hours portion of the tank filled => $\frac{19}{120}$ x 6 = $\frac{19}{20}$
Remaining portion of the tank => 1 - $\frac{19}{20}$ = $\frac{1}{20}$
In the next 1 hour, portion of the tank filled by A and B => $\frac{\frac{1}{20}}{\frac{1}{12}}$ = $\frac{3}{5}$ hours or 36 minutes
Hence the total time in which the tank is full is 12 hours + 3/5 hours => 12 hours and 36 minutes
- Two pipes together can fill a tank in 4 hours. If one pipe takes double the time of the other pipe to fill the tank separately, find the time in which the faster pipe fills the tank.
- 2 hours
- 4 hours
- 6 hours
- 8 hours
Show the answer
Answer: 6 hours
Let the faster pipe fill the tank in $x$ hours, then the other pipe fills the tank in 2$x$ hours.
Time taken by both the pipes together to fill the tank in 1 hour = $\frac{1}{4}$
=> $\frac{1}{x}$ + $\frac{1}{2x}$ = $\frac{1}{4}$
=> $\frac{3}{2x}$ = $\frac{1}{4}$
=> 2$x$ = 12
=> $x$ = 6 hours.
Hence, the faster pipe fills the tank in 6 hours.
- Two pipes P and Q can fill a tank in 30 and 45 minutes respectively. If both the pipes were open for few minutes after which Q was closed and the tank was full in 25 minutes, find the time for which pipe Q was open.
- 5 minutes
- 7.5 minutes
- 9 minutes
- 11 minutes
Show the answer
Answer: 7.5 minutes
Let Q be open for $x$ minutes, then portion of the tank filled by P and Q together in $x$ minutes =>
($\frac{1}{30}$ + $\frac{1}{45}$) x $x$ _________ 1
Since, the tank was full in 25 minutes, portion of the tank filled by P in 25-$x$ minutes is
$\frac{1}{30}$(25-$x$) _________ 2
For the tank to be full
($\frac{1}{30}$ + $\frac{1}{45}$) x $x$ + $\frac{1}{30}$(25-$x$) = 1
=>$\frac{x}{45}$ + $\frac{25}{30}$ = 1
=>$\frac{x}{45}$ = $\frac{5}{30}$
=> $x = \frac{45}{6}$
x = 7.5 minutes
- Three pipes open 16 hours a day can empty a tank in 4 days. How many hours a day must 4 pipes be open to empty the tank in 2 days.
- 8 hours
- 6 hours
- 10 hours
- 12 hours
Show the answer
Answer: 6 hours
If 3 pipes open 16 hours/day can empty the tank in 4 days, then the total number of hours required by 3 pipes is 16 x 4 = 64 hours.
If 3 pipes together need 64 hours to fill the tank, then 4 pipes can fill the tank in$\frac{64 x 3}{4}$ = 48 hours.
Now, to fill the tank in 2 days, the four pipes need $\frac{48}{4x2}$ = 6 hours per day
- Three pipes A, B and C attached to a tank can fill it completely in 6 hours. After opening together for 2 hours, C is closed. A and B fill the remaining part in 7 hours. The number of hours required by C to fill the tank is?
- 14 hours
- 12 hours
- 10 hours
- 16 hours
Show the answer
Answer: 14 hours
Let A, B and C fill the tank in $x$, $y$ and $z$ hours respectively, then in 1 hour A, B and C fills $\frac{1}{x}$, $\frac{1}{y}$ and $\frac{1}{z}$ of the tank .
Since A, B and C together can fill the tank in 6 hours. In 1 hour,
$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = $\frac{1}{6}$ ________ 1
Given, C is closed after 2 hours. Portion of the tank filled by 3 pipes in 2 hours is $\frac{1}{6}$ x 2 => $\frac{1}{3}$
i.e, in 2 hours,
$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = $\frac{1}{3}$
Given remaining portion of the tank is filled by A and in 7 hours.
Remaining portion of the tank => 1 - $\frac{1}{3}$ = $\frac{2}{3}$
Hence,
In 7 hours portion of the tank filled by A and B is $\frac{2}{3}$
=> In 1 hour portion of tank filled by A and B is $\frac{\frac{2}{3}}{7}$ = $\frac{2}{21}$
i.e $\frac{1}{x}$ + $\frac{1}{y}$ = $\frac{2}{21}$ ________ 2
Substituting 2 in 1
$\frac{2}{21}$ + $\frac{1}{z}$ = $\frac{1}{6}$
$\frac{1}{z}$ = $\frac{1}{6}$ - $\frac{2}{21}$
$\frac{1}{z}$ = $\frac{9}{126}$
$z$ = 14 hrs
- A cistern is filled by 3 pipes P, Q and R with uniform flow. The second pipe Q takes 3232 times the time taken by P to fill the tank, while R takes twice the time taken by Q to fill the tank. If all the three pipes can fill the tank in 5 hours, find the time required by pipe A alone to fill the tank.
- 6 hours
- 8 hours
- 10 hours
- 12 hours
Show the answer
Answer: 10 hours
Let the time taken by P to fill the tank be $x$ hours, then Q takes $\frac{3}{2}x$ hours and R takes 2 x $\frac{3}{2}x$ = $3x$ hours to fill the tank.
Given all three pipes together can fill the tank in 5 hours, hence
$\frac{1}{x}$ + $\dfrac{1}{\frac{3}{2}x}$ + $\frac{1}{3x}$ = $\frac{1}{5}$
$\frac{6}{3x}$ = $\frac{1}{5}$
$x$ = 10 hours.
Hence pipe A alone can fill the tank in 10 hours.
