Pipes and Cisterns Problems: Concepts and Worked Examples
Pipes and cisterns from first principles: fill rates, empty rates, LCM shortcut, and 8 worked examples covering every question type in campus placement aptitude tests.
Pipes and cisterns problems are time-and-work algebra with one sign change: filling pipes add to the combined rate, emptying pipes subtract from it.
That single rule handles every variant in the topic. The sections below derive the formula from first principles, show the LCM shortcut, and work through eight exam-style examples with verified solutions.
From Time-and-Work to Pipes: The Rate Foundation
In time-and-work problems, a worker who finishes a job in T days works at 1/T of the job per day. Two workers together complete 1/T1 + 1/T2 of the job per day.
Pipes follow the same logic, with one addition: some pipes remove water.
- A pipe that fills a tank in Tf hours adds at rate 1/Tf of the tank per hour.
- A pipe that drains the same tank in Td hours removes at rate 1/Td per hour.
- With both open, net rate = 1/Tf minus 1/Td per hour.
- Time to fill the tank = 1 / net rate, provided the net rate is positive.
If the drain pipe is faster than the fill pipe, net rate is negative. The tank doesn’t fill; it empties. Always check the sign first.
Key Formulas
| Situation | Net Rate (per hour) | Time to Fill |
|---|---|---|
| Single fill pipe, time T | 1/T | T |
| Two fill pipes, times T1 and T2 | 1/T1 + 1/T2 | T1 x T2 / (T1 + T2) |
| Fill pipe Tf, drain pipe Td | 1/Tf - 1/Td | Tf x Td / (Td - Tf) |
| n fill pipes, m drain pipes | Sum of 1/Tfi minus Sum of 1/Tdi | 1 / net rate |
The fill-plus-drain formula assumes Td > Tf (drain is slower than fill). If Td < Tf the net rate is negative and the tank empties.
The LCM Method: A Faster Path
Fraction arithmetic mid-exam slows most students. The LCM method converts every rate to a whole number.
Set total capacity = LCM of all individual pipe times. Each pipe’s rate becomes capacity divided by its individual time (a whole number). Add fill rates, subtract drain rates, then divide capacity by net rate.
This is not an approximation. Picking LCM as capacity selects a unit system where every rate is an integer, and the ratio capacity / net rate gives a numerically identical result to the fraction method.
Worked Example: LCM in Action
A pipe fills a tank in 6 hours; a second pipe drains it in 8 hours. Both are open together. How long to fill?
- LCM(6, 8) = 24. Set capacity = 24 units.
- Fill rate = 24 / 6 = 4 units per hour.
- Drain rate = 24 / 8 = 3 units per hour.
- Net rate = 4 - 3 = 1 unit per hour.
- Time = 24 / 1 = 24 hours.
Cross-check with fractions: 1/6 - 1/8 = 4/24 - 3/24 = 1/24 per hour. Time = 24 hours. Same answer.
Eight Worked Examples
Example 1: Single-Pipe Fraction
A pipe fills a tank in 5 hours. What fraction of the tank is filled in 2 hours?
- Fill rate = 1/5 of the tank per hour.
- Work done in 2 hours = 2 x (1/5) = 2/5.
- Answer: 2/5 of the tank is filled.
Example 2: Two Fill Pipes Together
Two pipes fill a tank in 4 hours and 6 hours respectively. How long to fill the tank with both open?
- Pipe A rate = 1/4 per hour.
- Pipe B rate = 1/6 per hour.
- Combined rate = 1/4 + 1/6 = 3/12 + 2/12 = 5/12 per hour.
- Time = 12/5 = 2.4 hours = 2 hours 24 minutes.
Example 3: Fill Pipe Plus Leak
A pipe fills a tank in 3 hours. A leak drains it in 5 hours. How long to fill the tank?
- Fill rate = 1/3 per hour, drain rate = 1/5 per hour.
- Net rate = 1/3 - 1/5 = 5/15 - 3/15 = 2/15 per hour.
- Time = 15/2 = 7.5 hours.
Example 4: Three Pipes (Two Fill, One Drain)
Pipe A fills in 4 hours, Pipe B fills in 6 hours, Pipe C drains in 12 hours. All three are open together. How long to fill the tank?
- LCM(4, 6, 12) = 12. Set capacity = 12 units.
- Pipe A rate = 12/4 = 3 units/hour.
- Pipe B rate = 12/6 = 2 units/hour.
- Pipe C rate = 12/12 = 1 unit/hour (drain, so subtract).
- Net rate = 3 + 2 - 1 = 4 units/hour.
- Time = 12/4 = 3 hours.
Example 5: One Pipe Opened Later
Pipe A fills a tank in 10 hours. Pipe B fills it in 15 hours. Pipe A runs alone for 4 hours, then Pipe B is also opened. How long does the complete fill take?
- Pipe A alone in 4 hours: 4 x (1/10) = 2/5 of the tank filled.
- Remaining fraction: 1 - 2/5 = 3/5.
- Combined rate (both open) = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 per hour.
- Time for remaining 3/5 at rate 1/6: (3/5) / (1/6) = (3/5) x 6 = 18/5 = 3.6 hours.
- Total time = 4 + 3.6 = 7.6 hours (7 hours 36 minutes).
Example 6: Finding the Unknown Pipe’s Time
Two pipes together fill a tank in 12 hours. Pipe A alone fills it in 20 hours. How long does Pipe B alone take?
- Combined rate = 1/12, Pipe A rate = 1/20.
- Pipe B rate = 1/12 - 1/20 = 5/60 - 3/60 = 2/60 = 1/30 per hour.
- Pipe B alone fills the tank in 30 hours.
Example 7: Detecting a Leak from Fill Time
A tank fills in 8 hours without a leak. With a leak active, it takes 10 hours. How long does the leak take to drain a full tank on its own?
- Fill rate (no leak) = 1/8 per hour.
- Net fill rate (with leak) = 1/10 per hour.
- Leak rate = 1/8 - 1/10 = 5/40 - 4/40 = 1/40 per hour.
- The leak drains a full tank in 40 hours.
Example 8: Net-Negative Rate (Tank Drains)
A tank is one-third full. Pipe A fills at rate 1/9 per hour. Pipe B drains at 1/6 per hour. Both are open. What happens to the water level?
- Net rate = 1/9 - 1/6 = 2/18 - 3/18 = -1/18 per hour.
- The rate is negative: the tank is draining, not filling.
- Time to fully drain from the current 1/3 level: (1/3) / (1/18) = (1/3) x 18 = 6 hours.
- The tank empties completely in 6 hours.
This is a deliberate exam trap. The correct first step is always to check the sign of the net rate before computing fill time.
Three Errors That Cost Marks
Most mistakes on this topic trace back to three specific points:
- Wrong sign for the drain pipe. The drain rate always subtracts. A pipe that empties in 8 hours contributes -1/8 to the net rate, never +1/8.
- LCM of rates instead of times. Take the LCM of the individual times (4 hours, 6 hours, 12 hours), not the rates. Then divide capacity by each time to get the per-hour rate.
- Decimal hours to minutes. 2.4 hours is not 2 hours 40 minutes. The decimal portion is a fraction of an hour: 0.4 x 60 = 24 minutes. So 2.4 hours = 2 hours 24 minutes.
Where This Topic Appears in Placement Tests
Pipes and cisterns questions appear in the quantitative sections of most campus aptitude tests. TCS NQT’s numerical ability block draws from this topic alongside time-and-work and speed-distance; the official TCS NQT page lists quantitative ability as one of four test sections. IndiaBix’s pipes and cisterns practice set offers a graded problem bank if you want to test speed and accuracy after working through the examples above.
Company-specific tests at campus placement evaluation tests and analytical-reasoning-heavy assessments such as the Mu Sigma aptitude test also include pipes-style problems. For the parent topic these problems extend, Time and Work aptitude questions covers the rate method, LCM shortcut, and efficiency ratios in full.
Working out that Pipe B alone takes 30 hours (Example 6 above) uses the same reciprocal-rate decomposition that appears in capacity analysis for multi-stage data pipelines. If that kind of systematic analysis interests you beyond placement prep, TinkerLLM (₹499) is a browser-based environment where you can apply the same thinking to real LLM workflows.
Primary sources
Frequently asked questions
What is the difference between pipes and cisterns and time-and-work problems?
The algebra is identical. Both use rate = 1/T. The only difference is that cistern problems include emptying pipes, which subtract from the combined rate instead of adding. In time-and-work, all contributors add to the rate.
How does the LCM method work in pipes and cisterns?
Set total capacity equal to the LCM of all individual pipe times. Each pipe's rate becomes capacity divided by its individual time, giving a whole number. Add fill rates, subtract drain rates, then divide capacity by net rate for the answer.
When is a tank never filled even with both fill and drain pipes open?
When the drain pipe is faster than the fill pipe, the net rate is negative. For example, fill in 4 hours and drain in 3 hours gives net rate = 1/4 minus 1/3 = negative 1/12 per hour. The tank drains rather than fills.
How do I solve a problem with three pipes (two filling, one draining)?
Use the LCM of all three times as the tank capacity. Each pipe's rate = capacity divided by its individual time. Net rate = fill-rate-1 plus fill-rate-2 minus drain-rate. Time to fill = capacity divided by net rate.
How do I handle a pipe that opens later than the others?
Calculate how much the first pipe fills during its solo period. Subtract that fraction from 1 to get remaining work. Divide remaining fraction by the combined rate of all open pipes. Total time equals the solo period plus the additional time.
If a tank fills in 8 hours normally but in 10 hours with a leak, how long does the leak take to drain the full tank?
Leak rate = 1/8 minus 1/10 = 5/40 minus 4/40 = 1/40 per hour. The leak alone drains a full tank in 40 hours.
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