Pipes and Cisterns problems deal with a tank or cistern being filled or emptied by one or more pipes. A pipe may act as an inlet, which fills the tank, or an outlet, which empties the tank.
These questions are based mainly on the relationship between work, time and rate.
Rate = Work / Time
Time = Work / Rate
If a pipe fills a tank completely in x hours, then the part of the tank filled in one hour is:
Rate = 1/8
Therefore, the pipe fills 1/8 of the tank per hour.
If an outlet pipe empties a full tank in x hours, then the amount emptied in one hour is:
The important difference is that an outlet has a negative rate when combined with an inlet.
Emptying rate = 1/12 tank per hour.
If Pipe A fills a tank in x hours and Pipe B fills it in y hours, their combined rate is the sum of their individual rates.
Time together = xy/(x+y)
Combined rate
= 1/10 + 1/15
= 3/30 + 2/30
= 5/30
= 1/6
Time = 6 hours
Combined rate = 1/12 + 1/15 + 1/20
= 5/60 + 4/60 + 3/60
= 12/60
= 1/5
Time = 5 hours
Inlet pipe → Positive rate
Outlet pipe → Negative rate
Always add filling rates and subtract emptying rates.
When one pipe fills the tank and another pipe empties it, subtract the outlet rate from the inlet rate.
Net rate = 1/6 − 1/12
= 2/12 − 1/12
= 1/12
Time = 12 hours
Net rate = 1/10 + 1/15 − 1/30
= 3/30 + 2/30 − 1/30
= 4/30
= 2/15
Time = 15/2
= 7.5 hours
A leak behaves like an outlet pipe. Therefore, its rate must be subtracted from the filling pipe's rate.
Net rate = 1/8 − 1/24
= 3/24 − 1/24
= 2/24
= 1/12
Time = 12 hours
If a pipe fills a tank in x hours but, because of a leak, the tank takes y hours, then the leak's rate is:
Leak rate = 1/10 − 1/15
= 3/30 − 2/30
= 1/30
Therefore, the leak alone empties the tank in 30 hours.
If a pipe fills the tank in x hours, then the fraction filled in t hours is:
Work = 4/10
= 2/5
Therefore, 40% of the tank is filled.
Remaining = 1 − 3/8
= 5/8
The LCM method is one of the easiest methods for solving pipes and cisterns questions involving different filling times.
Take the total capacity of the tank as the LCM of the given times. Then calculate each pipe's hourly rate in units.
LCM of 8 and 12 = 24
Assume tank capacity = 24 units.
A's rate = 24/8 = 3 units/hour
B's rate = 24/12 = 2 units/hour
Together = 3 + 2 = 5 units/hour
Time = 24/5
= 4.8 hours
Sometimes pipes are opened one after another, for example Pipe A for one hour, Pipe B for the next hour, and so on.
Calculate the work completed during one complete cycle and then determine how many cycles are required.
A's rate = 1/6
B's rate = 1/12
Work in 2 hours = 1/6 + 1/12
= 1/4
In 8 hours:
4 × 1/4 = 1
Tank fills in 8 hours.
If one pipe is opened first and another is opened later, calculate the work done by the first pipe before the second pipe is opened. Then calculate the remaining work using the combined rate.
A's work in 2 hours = 2/10 = 1/5
Remaining = 4/5
Combined rate:
= 1/10 + 1/15
= 1/6
Time for remaining work:
= (4/5) ÷ (1/6)
= 24/5
= 4.8 hours
Total time = 2 + 4.8
= 6.8 hours
If a pipe fills a tank at a known rate, tank capacity can be found from the quantity of water delivered in a given time.
Capacity supplied = 50 × 40
= 2000 litres
If two pipes have rates in the ratio a : b, their work done in the same time will also be in the ratio a : b.
3 : 5
| Situation | Operation |
|---|---|
| Inlet + Inlet | Add rates |
| Inlet + Outlet | Subtract outlet rate |
| Inlet + Leak | Subtract leak rate |
| Multiple Inlets | Add all filling rates |
| Multiple Outlets | Subtract all emptying rates |
| Partial Tank | Use remaining fraction |
| Alternate Pipes | Calculate cycle work |
| Concept | Formula |
|---|---|
| Filling Rate | 1/x |
| Emptying Rate | 1/x |
| Two Inlets | 1/x + 1/y |
| Inlet + Outlet | 1/x − 1/y |
| Two Inlets + Outlet | 1/x + 1/y − 1/z |
| Three Inlets | 1/x + 1/y + 1/z |
| Net Rate | Total Filling Rate − Total Emptying Rate |
| Work Done | Rate × Time |
| Remaining Work | Total Work − Completed Work |
| Leak Rate | Original Rate − Actual Rate |
| Time | Work ÷ Rate |
Rate = 1/12 + 1/18
= 3/36 + 2/36
= 5/36
Time = 36/5 = 7.2 hours
Net rate = 1/8 − 1/24
= 2/24
= 1/12
Time = 12 hours
Combined rate = 1/15 + 1/20
= 4/60 + 3/60
= 7/60
Work in 3 hours = 3 × 7/60
= 7/20
- A pipe fills a tank in 10 hours. What fraction of the tank does it fill in one hour?
- A pipe fills a tank in 12 hours and another in 18 hours. How long will they take together?
- A pipe fills a tank in 8 hours and an outlet empties it in 24 hours. Find the time required when both are open.
- Two pipes fill a tank in 15 and 20 hours respectively. Find their combined filling time.
- A pipe fills a tank in 10 hours. How much of the tank will it fill in 4 hours?
- A pipe fills a tank in 6 hours and another pipe fills it in 12 hours. If they work alternately, starting with the first pipe, find the time required to fill the tank.
- A pipe fills a tank in 10 hours. Due to a leak, it takes 15 hours. Find the time in which the leak alone can empty the tank.
- Two pipes can fill a tank in 12 and 15 hours, while a third pipe can empty it in 20 hours. Find the time when all three are opened.
- A pipe supplies 40 litres per minute. How much water will it supply in 45 minutes?
- A pipe fills 3/5 of a tank in 6 hours. Find the time required to fill the complete tank at the same rate.
2. 7.2 hours
3. 12 hours
4. 60/7 hours = 84/7 hours
5. 2/5 of the tank
6. 8 hours
7. 30 hours
8. 10 hours
9. 1800 litres
10. 10 hours
1. Identify each pipe as an inlet or outlet.
2. Convert completion time into hourly rate.
3. Give inlet rates a positive sign.
4. Give outlet/leak rates a negative sign.
5. Add all rates to obtain the net rate.
6. Time = Required Work ÷ Net Rate.
7. For partial filling, calculate the fraction already filled.
8. For alternate operation, calculate one complete cycle.
9. Check whether the answer is logically reasonable.
