Pipes and Cisterns questions with answers and concepts

Learn pipes and cisterns with concept notes and worked examples, then practise focused questions with explained answers. Build accuracy for placement and competitive aptitude tests before trying timed mode.

107 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Pipes and Cisterns problems deal with the filling and emptying of tanks or cisterns using pipes. These problems are analogous to Time and Work problems, where inlet pipes fill the tank (positive work) and outlet pipes empty it (negative work). Understanding rates of work and combined operations is essential for solving these efficiently.

Key formulas and rules

9 rules

If a pipe fills a tank in $x$ hours, its filling rate $= \frac{1}{x}$ per hour

If a pipe empties a tank in $y$ hours, its emptying rate $= \frac{1}{y}$ per hour

Net rate when both work together $= \frac{1}{x} - \frac{1}{y}$ per hour (when $y > x$ , tank fills; when $x > y$ , tank empties)

Time taken to fill or empty $= \frac{1}{Net Rate}$

Part of tank filled in $t$ hours $= t \times Filling Rate$

Part of tank emptied in $t$ hours $= t \times Emptying Rate$

If pipe A is open for $a$ hours and pipe B for $b$ hours: Part filled $= \frac{a}{x} \pm \frac{b}{y}$

For multiple pipes: Combined rate $= \frac{1}{x _{1}} \pm \frac{1}{x _{2}} \pm \frac{1}{x _{3}} \pm \dots$

Tank capacity $= LCM$ of individual times (useful for whole-number calculations)

Key concepts

Inlet and Outlet Pipes

An inlet pipe fills the tank and contributes positive work. An outlet pipe (or leak) empties the tank and contributes negative work. When calculating net work, inlet pipes are added and outlet pipes are subtracted. The sign convention is crucial - always verify whether the final answer represents filling or emptying.

Work Rate Concept

The fundamental principle is that work done = rate x time. For pipes, we consider the tank as 1 unit of work. A pipe's rate is the reciprocal of the time it takes to complete the work alone. For example, a pipe that fills in 5 hours works at 1/5 of the tank per hour. This rate remains constant unless stated otherwise.

Combined Operations

When multiple pipes operate simultaneously, their rates combine algebraically. If pipes A, B fill and C empties: Net Rate = 1/A + 1/B - 1/C. Positive result means filling; negative means emptying. The time to fill/empty = 1 / |Net Rate|. Always check the sign to determine the nature of work.

Pipes Operating in Sequence

Sometimes pipes operate one after another or for different durations. Calculate work done by each pipe separately and sum them. For example, if pipe A works for 2 hours and then pipe B for 3 hours: Work done = 2 x (1/A) + 3 x (1/B). The remaining work is 1 minus this sum.

Tank Capacity Method (LCM Approach)

To avoid fractions, assume tank capacity = LCM of all given times. Then calculate each pipe's hourly capacity as (LCM/time). This converts fractional rates to whole numbers, making calculations easier. For example, if pipes take 4, 6, 8 hours: Capacity = 24 units. Rates become 6, 4, 3 units/hour respectively.

Partial Operations and Leaks

When a pipe operates only partially or there's a leak, break the problem into phases. Calculate work done in each phase separately. For leak problems, treat the leak as an outlet pipe that may operate throughout or only when the tank is being filled. The key is identifying when each component is active.

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Pipe A can fill a tank in 6 hours and Pipe B can fill the same tank in 8 hours. If both pipes are opened simultaneously, how long will it take to fill the empty tank?

Solution

Step 1: Individual hourly filling rates.
- Rate of A: $\frac{1}{6} tank/hr$
- Rate of B: $\frac{1}{8} tank/hr$

Step 2: Combined net filling rate.

Net Rate = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24} tank/hr

Step 3: Compute filling time.

T = \frac{1}{\frac{7}{24}} = \frac{24}{7} = 3 \frac{3}{7} hours = 3 hours and (\frac{3}{7} \times 60) minutes \approx 3 h 26 m

Answer: $3 \frac{3}{7}$ hours (or approximately 3 hours 26 minutes).

Example 2

A pipe can normally fill a reservoir in 12 hours. Due to a leak at the bottom, it takes 15 hours to fill. In how many hours will the leak alone empty a completely full reservoir?

Solution

Step 1: Set up the net flow equation.

Filling Rate - Leak Rate = Effective Rate

\frac{1}{12} - \frac{1}{x} = \frac{1}{15}

Step 2: Solve for leak rate $\frac{1}{x}$ .

\frac{1}{x} = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}

x = 60 hours

Answer: 60 hours.

Example 3

Two inlets A and B can fill a cistern in 20 and 30 minutes respectively. Both pipes are opened together, but pipe B is shut off after 10 minutes. What is the total time required to fill the cistern?

Solution

Step 1: Joint filling during first 10 minutes.

Joint Rate = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} /min

Volume filled in 10 min = 10 \times \frac{1}{12} = \frac{5}{6} of tank

Step 2: Remaining volume completed by A alone.

Remaining Volume = 1 - \frac{5}{6} = \frac{1}{6} of tank

Time for A = \frac{\frac{1}{6}}{\frac{1}{20}} = \frac{20}{6} = \frac{10}{3} = 3 \frac{1}{3} minutes

Step 3: Total duration.

Total Time = 10 + 3 \frac{1}{3} = 13 \frac{1}{3} minutes = 13 minutes 20 seconds

Answer: $13 \frac{1}{3}$ minutes (or 13 minutes 20 seconds).

Example 4

Three pipes A, B, and C are connected to a tank. Inlets A and B can fill it in 4 and 6 hours respectively, while outlet C can empty it in 12 hours. If all three pipes are opened together, how long will it take to fill the tank?

Solution

Step 1: Combine filling and emptying rates.

Net Rate = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}

Net Rate = \frac{3 + 2 - 1}{12} = \frac{4}{12} = \frac{1}{3} tank/hr

Step 2: Calculate filling time.

T = \frac{1}{\frac{1}{3}} = 3 hours

Answer: 3 hours.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A pipe can fill a tank in 12 hours. How long will it take to fill 3/4 of the tank?

Answer: C. 9

Explanation

Work rate = 1/12 tank per hour.
Time needed = 3/4 / (1/12) = 3/4 x $12 = 9$ hours.
Answer = 9 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A pipe can fill a tank in 20 hours. How long will it take to fill 3/5 of the tank?

Answer: D. 12

Explanation

Work rate = 1/20 tank per hour.
Time needed = 3/5 / (1/20) = 3/5 x $20 = 12$ hours.
Answer = 12 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A pipe can fill a tank in 15 hours. How long will it take to fill 2/3 of the tank?

Answer: A. 10

Explanation

Work rate = 1/15 tank per hour.
Time needed = 2/3 / (1/15) = 2/3 x $15 = 10$ hours.
Answer = 10 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always identify inlet (filling) and outlet (emptying) pipes clearly and assign appropriate signs (+ for filling, - for emptying)
Use the LCM method when dealing with multiple pipes - it converts fractions to whole numbers and reduces calculation errors
When a leak is mentioned, treat it as an outlet pipe that works continuously while filling is happening
For pipes operating in sequence, calculate work done by each phase separately and track the remaining work
Check whether the answer should be in hours, minutes, or both - convert carefully using 1 hour = 60 minutes
If pipes are turned on/off at different times, break the problem into time intervals and sum the work done in each
For 'full tank' scenarios, remember that work completed = 1 (full tank). Partial fillings are represented as fractions
When comparing pipe efficiencies, the pipe with smaller time value is more efficient (works faster)
Verify your answer by working backwards - check if the calculated time with given rates actually fills/empties the tank

Ready to test your understanding?

Work through 107 questions with explanations after each answer.

Start practice

Next topic: Time and Distance

Pipes and Cisterns questions with answers and concepts

107 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

Practise this concept

Key formulas and rules

9 rules

If a pipe fills a tank in $x$ hours, its filling rate $= \frac{1}{x}$ per hour

If a pipe empties a tank in $y$ hours, its emptying rate $= \frac{1}{y}$ per hour

Net rate when both work together $= \frac{1}{x} - \frac{1}{y}$ per hour (when $y > x$ , tank fills; when $x > y$ , tank empties)

Time taken to fill or empty $= \frac{1}{Net Rate}$

Part of tank filled in $t$ hours $= t \times Filling Rate$

Part of tank emptied in $t$ hours $= t \times Emptying Rate$

If pipe A is open for $a$ hours and pipe B for $b$ hours: Part filled $= \frac{a}{x} \pm \frac{b}{y}$

For multiple pipes: Combined rate $= \frac{1}{x _{1}} \pm \frac{1}{x _{2}} \pm \frac{1}{x _{3}} \pm \dots$

Tank capacity $= LCM$ of individual times (useful for whole-number calculations)

Key concepts

Inlet and Outlet Pipes

Work Rate Concept

Combined Operations

Pipes Operating in Sequence

Tank Capacity Method (LCM Approach)

Partial Operations and Leaks

Check the idea while it is fresh.Open a focused exercise without losing this article.Practise this concept

Worked examples

Example 1

Pipe A can fill a tank in 6 hours and Pipe B can fill the same tank in 8 hours. If both pipes are opened simultaneously, how long will it take to fill the empty tank?

Solution

Step 1: Individual hourly filling rates.
- Rate of A: $\frac{1}{6} tank/hr$
- Rate of B: $\frac{1}{8} tank/hr$

Step 2: Combined net filling rate.

Net Rate = \frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24} tank/hr

Step 3: Compute filling time.

T = \frac{1}{\frac{7}{24}} = \frac{24}{7} = 3 \frac{3}{7} hours = 3 hours and (\frac{3}{7} \times 60) minutes \approx 3 h 26 m

Answer: $3 \frac{3}{7}$ hours (or approximately 3 hours 26 minutes).

Example 2

A pipe can normally fill a reservoir in 12 hours. Due to a leak at the bottom, it takes 15 hours to fill. In how many hours will the leak alone empty a completely full reservoir?

Solution

Step 1: Set up the net flow equation.

Filling Rate - Leak Rate = Effective Rate

\frac{1}{12} - \frac{1}{x} = \frac{1}{15}

Step 2: Solve for leak rate $\frac{1}{x}$ .

\frac{1}{x} = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}

x = 60 hours

Answer: 60 hours.

Example 3

Two inlets A and B can fill a cistern in 20 and 30 minutes respectively. Both pipes are opened together, but pipe B is shut off after 10 minutes. What is the total time required to fill the cistern?

Solution

Step 1: Joint filling during first 10 minutes.

Joint Rate = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} /min

Volume filled in 10 min = 10 \times \frac{1}{12} = \frac{5}{6} of tank

Step 2: Remaining volume completed by A alone.

Remaining Volume = 1 - \frac{5}{6} = \frac{1}{6} of tank

Time for A = \frac{\frac{1}{6}}{\frac{1}{20}} = \frac{20}{6} = \frac{10}{3} = 3 \frac{1}{3} minutes

Step 3: Total duration.

Total Time = 10 + 3 \frac{1}{3} = 13 \frac{1}{3} minutes = 13 minutes 20 seconds

Answer: $13 \frac{1}{3}$ minutes (or 13 minutes 20 seconds).

Example 4

Solution

Step 1: Combine filling and emptying rates.

Net Rate = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}

Net Rate = \frac{3 + 2 - 1}{12} = \frac{4}{12} = \frac{1}{3} tank/hr

Step 2: Calculate filling time.

T = \frac{1}{\frac{1}{3}} = 3 hours

Answer: 3 hours.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

A pipe can fill a tank in 12 hours. How long will it take to fill 3/4 of the tank?

Answer: C. 9

Explanation

Work rate = 1/12 tank per hour.
Time needed = 3/4 / (1/12) = 3/4 x $12 = 9$ hours.
Answer = 9 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 2

A pipe can fill a tank in 20 hours. How long will it take to fill 3/5 of the tank?

Answer: D. 12

Explanation

Work rate = 1/20 tank per hour.
Time needed = 3/5 / (1/20) = 3/5 x $20 = 12$ hours.
Answer = 12 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

A pipe can fill a tank in 15 hours. How long will it take to fill 2/3 of the tank?

Answer: A. 10

Explanation

Work rate = 1/15 tank per hour.
Time needed = 2/3 / (1/15) = 2/3 x $15 = 10$ hours.
Answer = 10 hours.

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Common mistakes and useful habits

Always identify inlet (filling) and outlet (emptying) pipes clearly and assign appropriate signs (+ for filling, - for emptying)
Use the LCM method when dealing with multiple pipes - it converts fractions to whole numbers and reduces calculation errors
When a leak is mentioned, treat it as an outlet pipe that works continuously while filling is happening
For pipes operating in sequence, calculate work done by each phase separately and track the remaining work
Check whether the answer should be in hours, minutes, or both - convert carefully using 1 hour = 60 minutes
If pipes are turned on/off at different times, break the problem into time intervals and sum the work done in each
For 'full tank' scenarios, remember that work completed = 1 (full tank). Partial fillings are represented as fractions
When comparing pipe efficiencies, the pipe with smaller time value is more efficient (works faster)
Verify your answer by working backwards - check if the calculated time with given rates actually fills/empties the tank

Ready to test your understanding?

Work through 107 questions with explanations after each answer.

Start practice

Next topic: Time and Distance

Key formulas and rules

Key concepts

Inlet and Outlet Pipes

Work Rate Concept

Combined Operations

Pipes Operating in Sequence

Tank Capacity Method (LCM Approach)

Partial Operations and Leaks

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Inlet and Outlet Pipes

Work Rate Concept

Combined Operations

Pipes Operating in Sequence

Tank Capacity Method (LCM Approach)

Partial Operations and Leaks

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Example 4

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.