Pipes and Cisterns

Pipes and Cisterns is an important quantitative aptitude topic based on work-rate concepts.

Quantitative Aptitude Pipes and Cisterns Competitive Exams

Pipes and Cisterns is an important quantitative aptitude topic based on work-rate concepts. It deals with pipes filling or emptying a tank, combined work of multiple pipes, inlet and outlet pipes, leakage, and time taken to fill or empty a cistern.

What are Pipes and Cisterns?

A cistern is a tank or container that can be filled or emptied by pipes. A pipe that fills the cistern is called an inlet pipe, and a pipe that empties the cistern is called an outlet pipe.

Pipes and cisterns questions are very similar to time and work problems. The main idea is to calculate the work done by each pipe in one unit of time.

Quick idea: If a pipe fills a tank in \(x\) hours, then its one-hour work is \(\frac{1}{x}\) of the tank.

Term	Meaning	Example
Inlet Pipe	Pipe that fills the tank	Fills tank in 6 hours
Outlet Pipe	Pipe that empties the tank	Empties tank in 8 hours
Cistern	Tank or container	Water tank
Rate of Work	Part of tank filled or emptied per hour	\(\frac{1}{6}\) tank per hour
Net Rate	Combined effect of inlet and outlet	Filling rate minus emptying rate

“Pipes and cisterns are time-and-work problems with filling and emptying signs.”

Aptitude Tip

Key points

Inlet pipe adds water.
Outlet pipe removes water.
Filling rate is positive.
Emptying rate is negative.
Combined rate is found by adding rates algebraically.
Time taken equals total work divided by net rate.

inlet outlet rate tank

Visual Understanding

These diagrams show the basic meaning of inlet pipe, outlet pipe, and combined pipe action.

Inlet Pipe

\[ \text{Rate} = \frac{1}{x} \]

If an inlet fills the tank in \(x\) hours, its rate is \(\frac{1}{x}\) tank per hour.

Outlet Pipe

\[ \text{Rate} = -\frac{1}{y} \]

If an outlet empties the tank in \(y\) hours, its rate is treated as negative.

Inlet and Outlet Together

\[ \text{Net Rate} = \frac{1}{x}-\frac{1}{y} \]

When inlet and outlet work together, subtract the outlet rate from inlet rate.

Important Formulas and Rules

Filling Rate

\[ \text{One-hour work}=\frac{1}{x} \]

If a pipe fills the tank in \(x\) hours.

Emptying Rate

\[ \text{One-hour work}=-\frac{1}{y} \]

If a pipe empties the tank in \(y\) hours.

Two Inlets Together

\[ \text{Rate}=\frac{1}{a}+\frac{1}{b} \]

Add rates if both pipes fill the tank.

Inlet and Outlet

\[ \text{Rate}=\frac{1}{a}-\frac{1}{b} \]

Subtract outlet rate from inlet rate.

Time Taken

\[ \text{Time}=\frac{1}{\text{Net Rate}} \]

When total tank work is considered as \(1\).

Work Done

\[ \text{Work}=\text{Rate}\times\text{Time} \]

Useful when pipe is opened only for some time.

Remaining Work

\[ \text{Remaining}=1-\text{Work Done} \]

Used in partial filling problems.

Capacity Method

\[ \text{Rate}=\frac{\text{Capacity}}{\text{Time}} \]

Useful when tank capacity is given in litres.

Rule: Filling pipes are positive, emptying pipes are negative. Always combine their rates with proper signs.

Common Types of Pipes and Cisterns Questions

Single Pipe

One pipe fills or empties the tank.

Simple rate calculation
One-hour work
Partial filling
Basic time concept

Two Inlets

Two pipes fill the tank together.

Add both rates
Faster filling
Find total time
LCM method useful

Inlet and Outlet

One pipe fills and another empties.

Subtract outlet rate
Net filling rate
Leakage problems
Careful sign use

Alternate Opening

Pipes are opened at different times.

Partial work
Sequential filling
Remaining tank
Step-by-step solution

Exam approach: First decide whether each pipe is an inlet or outlet. Then calculate their one-hour work.

Method Bank

One Pipe Fills

Pipe fills tank in 6 hours:

\[ \text{Rate}=\frac{1}{6} \]

One Pipe Empties

Pipe empties tank in 8 hours:

\[ \text{Rate}=-\frac{1}{8} \]

Two Fill Together

Pipes fill in 6 and 12 hours:

\[ \frac{1}{6}+\frac{1}{12}=\frac{1}{4} \]

Fill and Leak

Fill in 6 hours, leak empties in 12 hours:

\[ \frac{1}{6}-\frac{1}{12}=\frac{1}{12} \]

Tip: Treat the full tank as \(1\). Then each pipe’s work is a fraction of the tank per hour.

LCM Capacity Method

If pipes take 6 hours and 12 hours, assume tank capacity \(12\) units. Their rates become \(2\) units/hour and \(1\) unit/hour.

\[ \text{Combined Rate}=2+1=3\text{ units/hour} \]

\[ \text{Time}=\frac{12}{3}=4\text{ hours} \]

Step-by-Step Solving Method

Step	Action	Example
Step 1	Identify whether each pipe fills or empties.	A fills, B empties.
Step 2	Write one-hour work of each pipe.	A: \(\frac{1}{6}\), B: \(-\frac{1}{12}\)
Step 3	Add rates with signs.	\(\frac{1}{6}-\frac{1}{12}\)
Step 4	Find net rate.	\(\frac{1}{12}\) tank per hour
Step 5	Find time taken.	\(\text{Time}=12\) hours

Important: If the net rate is positive, the tank fills. If the net rate is negative, the tank empties.

Solved Examples

Question	Method	Answer
Pipe A fills a tank in 6 hours and Pipe B fills it in 12 hours. How long will they take together?	\[ \text{Combined Rate}=\frac{1}{6}+\frac{1}{12} \] \[ \text{Combined Rate}=\frac{2+1}{12}=\frac{3}{12}=\frac{1}{4} \] \[ \text{Time}=4\text{ hours} \]	4 hours
Pipe A fills a tank in 8 hours and Pipe B fills it in 24 hours. Find the time taken together.	\[ \text{Rate}=\frac{1}{8}+\frac{1}{24} \] \[ \text{Rate}=\frac{3+1}{24}=\frac{4}{24}=\frac{1}{6} \]	6 hours
A pipe fills a tank in 6 hours. A leak empties it in 12 hours. If both are open, how long will the tank take to fill?	\[ \text{Net Rate}=\frac{1}{6}-\frac{1}{12} \] \[ \text{Net Rate}=\frac{2-1}{12}=\frac{1}{12} \]	12 hours
Pipe A fills a tank in 10 hours and Pipe B empties it in 15 hours. If both are open, how long will the tank take to fill?	\[ \text{Net Rate}=\frac{1}{10}-\frac{1}{15} \] \[ \text{Net Rate}=\frac{3-2}{30}=\frac{1}{30} \]	30 hours
A pipe fills a tank in 5 hours. How much of the tank will it fill in 2 hours?	\[ \text{One-hour work}=\frac{1}{5} \] \[ \text{Work in 2 hours}=2\times\frac{1}{5}=\frac{2}{5} \]	\(\frac{2}{5}\) of the tank
A pipe fills \(\frac{3}{4}\) of a tank in 6 hours. How long will it take to fill the full tank?	If \(\frac{3}{4}\) tank takes \(6\) hours: \[ \text{Full tank time}=6\times\frac{4}{3} \] \[ =8\text{ hours} \]	8 hours
Two pipes can fill a tank in 20 hours and 30 hours respectively. An outlet can empty it in 60 hours. Find time if all are opened together.	\[ \text{Net Rate}=\frac{1}{20}+\frac{1}{30}-\frac{1}{60} \] \[ \text{Net Rate}=\frac{3+2-1}{60}=\frac{4}{60}=\frac{1}{15} \]	15 hours
A tank has capacity 600 litres. A pipe fills it in 10 hours. How many litres does it fill per hour?	\[ \text{Rate}=\frac{600}{10}=60 \]	60 litres/hour

Note: In most exam questions, taking the full tank as \(1\) makes calculation easier.

Common Traps and Shortcuts

Common Traps

Adding outlet rate instead of subtracting it.
Forgetting that emptying work is negative.
Confusing time taken with rate of work.
Using direct average of times instead of adding rates.
Ignoring partial opening of pipes.
Not checking whether tank is filling or emptying overall.

Useful Shortcuts

Full tank can be treated as \(1\).
Pipe filling in \(x\) hours has rate \(\frac{1}{x}\).
Outlet emptying in \(y\) hours has rate \(-\frac{1}{y}\).
For multiple pipes, add rates with signs.
Use LCM method to avoid fractions.
Time is reciprocal of net rate.

Exam approach: Never average pipe times directly. Always convert time into one-hour work first.

Practice

A) Multiple Choice Questions

Pipe A fills a tank in 10 hours and Pipe B fills it in 15 hours. How long will they take together?
4 hours 5 hours 6 hours 8 hours
A pipe fills a tank in 12 hours. How much tank will it fill in 3 hours?
\(\frac{1}{2}\) \(\frac{1}{3}\) \(\frac{1}{4}\) \(\frac{1}{6}\)
A pipe fills a tank in 8 hours and an outlet empties it in 16 hours. If both are open, time taken to fill is:
8 hours 12 hours 16 hours 24 hours
Pipe A fills a tank in 20 hours and Pipe B fills it in 30 hours. Their combined rate is:
\(\frac{1}{10}\) \(\frac{1}{12}\) \(\frac{1}{15}\) \(\frac{1}{20}\)
A tank of 900 litres is filled in 15 hours. Rate of filling is:
45 litres/hour 50 litres/hour 60 litres/hour 75 litres/hour

B) Solve the Higher-Order Problems

Pipe A fills a tank in 12 hours and Pipe B fills it in 18 hours. Find the time taken together. (Hint: Add their one-hour work.)
A pipe fills a tank in 9 hours, but a leak empties it in 18 hours. Find the time taken to fill the tank when both are active. (Hint: Subtract leak rate.)
Two pipes fill a tank in 15 hours and 20 hours. An outlet empties it in 30 hours. Find the net filling time if all are opened together. (Hint: Add inlet rates and subtract outlet rate.)
A pipe fills \(\frac{2}{3}\) of a tank in 8 hours. Find time to fill the full tank. (Hint: Scale up from partial work.)
A tank has capacity 1200 litres. Pipe A fills it at 80 litres/hour and Pipe B fills it at 120 litres/hour. Find time taken together. (Hint: Add litre-per-hour rates.)

C) Match the Concept with the Correct Meaning

Concept	Correct Meaning
Inlet Pipe	Pipe that fills the tank
Outlet Pipe	Pipe that empties the tank
One-hour Work	Part of tank filled or emptied in one hour
Net Rate	Combined effect of all pipes
Leakage	Acts like an outlet pipe
Cistern	Tank or container

Aptitude Reminder

Pipes and cisterns problems are solved by converting time into rate. Inlet rates are positive, outlet or leakage rates are negative. Add all rates algebraically and take the reciprocal of the net rate to find time.

Task: Create five questions using single pipe, two inlets, inlet with outlet, partial filling, and tank capacity in litres.

Show Suggested Answers

Multiple Choice

6 hours

\[ \frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6} \]
\(\frac{1}{4}\)

\[ 3\times\frac{1}{12}=\frac{3}{12}=\frac{1}{4} \]
16 hours

\[ \frac{1}{8}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16} \]
\(\frac{1}{12}\)

\[ \frac{1}{20}+\frac{1}{30}=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12} \]
60 litres/hour

\[ \frac{900}{15}=60 \]

Higher-Order Problems

Pipe A: \(12\) hours, Pipe B: \(18\) hours.
\[ \frac{1}{12}+\frac{1}{18}=\frac{3+2}{36}=\frac{5}{36} \]

\[ \text{Time}=\frac{36}{5}=7.2 \]
Answer = 7.2 hours.
Filling pipe: \(9\) hours, leak: \(18\) hours.
\[ \frac{1}{9}-\frac{1}{18}=\frac{2-1}{18}=\frac{1}{18} \]
Answer = 18 hours.
Two inlets and one outlet:
\[ \frac{1}{15}+\frac{1}{20}-\frac{1}{30} \]

\[ \frac{4+3-2}{60}=\frac{5}{60}=\frac{1}{12} \]
Answer = 12 hours.
\(\frac{2}{3}\) tank is filled in \(8\) hours.
\[ \text{Full tank time}=8\times\frac{3}{2}=12 \]
Answer = 12 hours.
Capacity \(=1200\) litres. Combined rate:
\[ 80+120=200\text{ litres/hour} \]

\[ \text{Time}=\frac{1200}{200}=6 \]
Answer = 6 hours.

Concept Matching

Inlet Pipe → Pipe that fills the tank
Outlet Pipe → Pipe that empties the tank
One-hour Work → Part of tank filled or emptied in one hour
Net Rate → Combined effect of all pipes
Leakage → Acts like an outlet pipe
Cistern → Tank or container

Clue Explanation

The most important step is converting time into one-hour work. After that, add inlet rates and subtract outlet or leakage rates.

Exam tips

Full tank can be treated as \(1\).
Inlet pipe rate is positive.
Outlet pipe rate is negative.
Never average the given pipe times.
Use LCM method to avoid fractions.
Check whether the net result is filling or emptying.

Practice MCQs

Learning Modules