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 – CAT 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

Answer Choice: B

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.

Correct Answer: 24 minutes

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.

Answer Choice b

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) = (1/6^th ) 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

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).

Type 1: Pipes and Cistern (Tank)

In order to solve questions falling into the first category, imagine each of the pipes as a human doing some work. (Again, we are using a Work and Time reference here). The time taken by a single pipe to fill up the tank by itself, is the time taken by that human to do the entire work. Find out each pipe’s (or human’s) rate of doing work, with respect to each other. For instance, if Pipe A can fill the tank in 8 hours, and Pipe B can fill it in 4 hours, you know that the rate of B is twice that of A.

Once, you have the rates of doing work, all you have to do is add them up to find out the overall rate of filling the tank by all the pipes together. Remember, rate is always inversely proportional to the time taken, so the rates you add up should be reciprocals of the time taken by each pipe.

An easy tip that helps, is to assign any one pipe’s rate as ‘x’, and calculate the others’ rate in terms of ‘x’.

E.g. 1. Pipe A fills a tank in 3½ hours whereas Pipe B fills the same tank in 10½ hours.

Solution:

Since Pipe B takes more time, let us take ‘x’ as the rate at which B fills the tank.
Since time taken by B is 3 times the time taken by A, the rate of A will be 3 times faster than the rate of B. So the rate of A is ‘3x’.

Another advantage of using the rates, is that rates can be added up, not time. Which means that if two pipes take 4 and 8 hours respectively to fill up a tank, you can add the reciprocals of 4 and 8 to get the net rate. What you cannot do, is add 4 and 8 up, directly. That would result in a wrong answer.

The logic is simple. If you add 4 and 8 up, to find the time taken by both together to fill up the tank, you will get 12. But think, if one pipe takes 4 hours and the other takes 8 hours to fill up the tank, they couldn’t possibly take 12 hours together!
So let us use the rates in the following example:

E.g. 2. Pipes P, Q, R and S can fill a certain tank in 3, 5, 6 and 10 hours, respectively. How much time will it take for the tank to be filled if all pipes are opened together?

Solution:

Let ‘x’ be the rate of Pipe S, which is the slowest. So the rates of all the pipes are as follows. Make sure to make as few calculations as possible – this is the trick to cracking your exam! Try the following method:

So the total efficiency of the four pipes is: x + 2x + 3.33x + 1.66x ≈ 8x = 8 × (1/10)
Thus the total time taken for the four pipes working together to fill the tank would be:
10/8 hours = 1.25 hours = 1 hour 15 mins.

Shorten Your Calculations

By choosing the pipe taking the maximum time first and basing your calculations on that, you effectively bring down the number of reciprocal calculations you make. Of course, you need to pick and choose which pipe’s efficiency to calculate next. Like in the example, after S, we chose Q because its time taken was a direct multiple of the time taken by S. This lets us calculate its reciprocal easily. Next, we were left with Pipes P and R, both of whose times are not multiples of the time taken by S. However, we all know that 10 is 3.33 times 3. So we use this to first find the efficiency of P. Next we know that 6 is two times 3. So clearly the efficiency of R is half of P.

Type 2: Speed of Filling/Emptying

For the second category, take each pipe filling the tank as positive, and each hole, drain, outlet or leakage emptying the tank as negative. When you add these negative and positive rates together (being very careful with the signs), you get an equation giving you the net rate by which the tank is being filled up. If this figure is negative, that means the tank is being emptied, rather than filled up.

Once you have the net rate of the tank emptying or filling up, all you have to do is factor in the size of the tank, to find out the time it takes to empty or fill up. The size of the tank can be obtained from other information provided in the question. The question will either tell you how long the tank takes to empty through one hole, or it will give you the capacity of the tank. Either way, once you’re at this point of the solving process, it should be easy to get to the end. Simple math will get you there!

E.g. 3. A pipe fills a tank in 3 hours, whereas an outlet can empty the filled tank completely in 12 hours. In what time will the tank be filled, if both the pipe and the outlet are opened simultaneously?

Solution:

Let ‘x’ be the efficiency of the outlet. However since it is emptying the tank, we always use a negative sign with it.
And efficiency of pipe = (12/3)x = 4x
∴ Total Efficiency = 4x – x = 3x
Since x = 1/12, we can say that the tank will be filled in time = 12/3 = 4 Hours.