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# Basic Concepts - Pipes and Cisterns - Quantitative Aptitude (Quant) - CAT

A pipe is connected to a tank or cistern. It is used to fill or empty the tank; accordingly, it is called an inlet or an outlet.

Inlet: A pipe which is connected to fill a tank is known as an inlet.
Outlet: A pipe which is connected to empty a tank is known as an outlet.

Problems on pipes and cisterns are similar to problems on time and work. In pipes and cistern problems, the amount of work done is the part of the tank of filled or emptied. And, the time taken to do a piece of work is the time take to fill or empty a tank completely or to a desired level.

Pipes and Cisterns Points to remember:
1) If an inlet connected to a tank fills it in X hours, part of the tank filled in one hour is = 1/X
2) If an outlet connected to a tank empties it in Y hours, part of the tank emptied in one hour is = 1/Y
3) An inlet can fill a tank in X hours and an outlet can empty the same tank in Y hours. If both the pipes are opened at the same time and Y > X, the net part of the tank filled in one hour is given by;
= (1/X – 1/Y)

Therefore, when both the pipes are open the time taken to fill the whole tank is given by;
= (XY/Y-X) Hours

If X is greater than Y, more water is flowing out of the tank than flowing into the tank. And, the net part of the tank emptied in one hour is given by;
= (1/Y – 1/X)

Therefore, when both the pipes are open the time taken to empty the full tank is given by;
= (YX/X-Y) Hours

4) An inlet can fill a tank in X hours and another inlet can fill the same tank in Y hours. If both the inlets are opened at the same time, the net part of the tank filled in one hour is given by;
= (1/X + 1/Y)

Therefore, the time taken to fill the whole tank is given by;
= (XY/Y+X) Hours

In a similar way, If an outlet can empty a tank in X hours and another outlet can empty the same tank in Y hours, the part of the tank emptied in one hour when both the pipes start working together is given by;
= (1/X + 1/Y)

Pipes and Cisterns Questions
Example 1:  A water tank has three taps A, B and C. A fills four buckets in 24 minutes, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together, a full tank is emptied in 2 hours. If a bucket can hold 5 liters of water, what is the capacity of the tank?
a.) 120 liters
b.) 240 liters
c.) 180 liters
d.) 60 liters
Solution:  A fills 4 buckets in 24 minutes. Thus, A fills 1 bucket in 24/4 = 6 minutes
Similarly, B fills 8 buckets in 1 hour. Thus B fills 1 bucket in 60/8 minutes
Similarly, C fills one bucket in 20/2 = 10 minutes
In 2 hours,
Number of buckets filled by A will be = 120/6 = 20 buckets
Number of buckets filled by B will be = 120/ (60/8) = (120 * 8) / 60  = 16 buckets
Number of buckets filled by C will be = 120 / 10 = 12 buckets
Total number of buckets filled = (20 + 16 + 12) = 48 buckets
Total amount of water coming out of the tank = capacity of the tank = 48 * 5 liters = 240 liters

Example 2:  There is a leak in the bottom of the tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admit 6 liters per hour and the tank is now emptied in 12 hours. What is the capacity of the tank?
a.) 8.8 liters
b.) 36 liters
c.) 144 liters
d.) Cannot be determined
Solution:  Since the leak can empty the tank in 8 hours,
In one hour, part of the tank emptied by the leak = 1/8
Also, after opening the tap, in one hour, part of the tank emptied = 1/12
Let the tap can fill the tank in x hours. Therefore, In one hour, part of the tank filled by the tap = 1/x
As per question, (1/x) – (1/8) = 1/12
Or x = 24
Since the tap admits 6 liters of water per hour, it will admit (6*24) =144 liters of water in 24 hours, which should be the capacity of the tank.

Example 3:  Three small pumps and one large pump are filling a tank. Each of the three small pump works at 2/3 of the rate of the large pump. If all four pumps work at the same time, they should fill the tank in what fraction of the time that it would have taken the large pump alone?
a.) 4/7
b.) 1/3
c.) 2/3
d.) ¾
Solution:  As per the question,
Capacity of three small pumps = Capacity of two large pumps
Also, if we want to express the capacity of three small pumps + one large pump in terms of large pump, we should add one large pump on both sides of the above equation
Adding one large pump on both sides of the above equation, we get
Three small pumps + one large pump = Three large pumps.
Thus, if all the four pumps are open together, they would fill the tank in 1/3 rd of the time large pump would have taken alone.

Example 4: A tank is fitted with 8 pipes, some of which that fill the tank and others that empty the tank. Each of the pipes that fills the tank fills it in 8 hours, while each of those that empty the tank empties it in 6 hours. If all the pipes are kept open when the tank is full, it will take 6 hours to drain the tank. How many of these are fill pipes?
a.) 2 fill pipes
b.) 4 fill pipes
c.) 6 fill pipes
d.) 5 fill pipes
Solution: Let the number of fill pipes be ‘n’
Therefore, there will be (8 – n) waste pipes.
Each of the fill pipes can fill the tank in 8 hours.
Therefore, each of the fill pipes will fill 1/8th of the tank in an hour.
Hence, n fill pipes will fill n/8th of the tank in an hour.
Similarly, each of the waste pipes will drain the full tank in 6 hours.
∴ each of the waste pipes will drain 1/6th of the tank in an hour.
(8 – n) waste pipes will drain (8-n)/6th of the tank in an hour.
Between the fill pipes and the waste pipes, they drain the tank in 6 hours.
That is, when all 8 of them are opened, 1/6th of the tank gets drained in an hour.
(Amount of water filled by fill pipes in 1 hour – Amount of water drained by waste pipes 1 hour) = (1/6th ) of the tank
Therefore,
(n/8) – ((8−n)/)6 = -1/6

Note: The right hand side has a negative sign because the tank gets drained.
Cross multiplying and solving the equations, 14n – 64 = -8
or 14n = 56 or n = 4
The correct answer is Choice (2).

Example 5:  Pipe A usually fills a tank in 2 hours. On account of a leak at the bottom of the tank, it takes pipe A 30 more minutes to fill the tank. How long will the leak take to empty a full tank if pipe A is shut?
a.) 2 hours 30 minutes
b.) 5 hours
c.) 4 hours
d.) 10 hours
Solution: Pipe A fills the tank normally in 2 hours.
Therefore, it will fill 1/2 of the tank in an hour.
Let the leak take x hours to empty a full tank when pipe A is shut.
Therefore, the leak will empty 1/x of the tank in an hour.
The net amount of water that gets filled in the tank in an hour when pipe A is open and when there is a leak =   (1/2 – 1/x) of the tank. —– (1)
Now, when there is a leak, the problem states that it takes two and a half hours to fill the tank. i.e. 5/2hours.
Therefore, in an hour, 2/5th of the tank gets filled. —– (2)
Equating (1) and (2), we get 1/2 – 1/x = 2/5
=> 1/x = 1/2 – 2/5 = 1/10
=> x = 10 hours.
The correct answer is Choice (4).

The document Basic Concepts - Pipes and Cisterns | Quantitative Aptitude (Quant) - CAT is a part of the CAT Course Quantitative Aptitude (Quant).
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## FAQs on Basic Concepts - Pipes and Cisterns - Quantitative Aptitude (Quant) - CAT

 1. What are pipes and cisterns? Ans. Pipes and cisterns are concepts used in mathematics and engineering to calculate the rate at which a pipe can fill or empty a cistern. It involves understanding the flow of water through pipes and the time it takes to fill or empty a container.
 2. How do you calculate the rate of filling a cistern using pipes? Ans. To calculate the rate of filling a cistern using pipes, you need to determine the individual rates at which each pipe can fill the cistern. Add up the rates of all the pipes that are filling the cistern to get the total rate of filling.
 3. Can multiple pipes fill a cistern faster than a single pipe? Ans. Yes, multiple pipes can fill a cistern faster than a single pipe. When multiple pipes are connected, their individual rates of filling add up to create a higher overall rate of filling. This is known as a parallel arrangement of pipes.
 4. How do you calculate the time taken to fill a cistern? Ans. To calculate the time taken to fill a cistern, divide the total volume of the cistern by the rate at which it is being filled. This will give you the time taken in hours.
 5. Can pipes and cisterns be used to solve real-life problems? Ans. Yes, pipes and cisterns concepts can be used to solve real-life problems related to water management, plumbing, and engineering. By understanding the flow rates and capacities of pipes and cisterns, one can optimize water usage, calculate filling or emptying times, and plan efficient systems.

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