Pipes and Cisterns

Inlet Pipes

The questions on the pipes and cisterns are more like the work-play questions. In this section, you will have to answer questions on the rate at which a pipe of some available dimensions will fill a container. Here we will see questions on the inlet pipes. Suppose one pipe fills a tank in 1 hour and another one fills it in two hours. Then if we open both of these pipes, when will we have a full tank? Let us see all the concepts, tips and tricks that you need to solve the section. In this section, we will focus on the inlet pipes.

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Inlet Pipes

Let us see some basic terminology that we will encounter in this section.

Inlet Pipe:  A pipe used to fill up a tank, or a cistern or a reservoir is the ‘Inlet’ pipe. The inlet pipe adds to the volume of the water. The work done (volume added) is thus additive in nature. In order to differentiate it with the Outlet pipes that decrease the volume, the rate of work done by the inlet pipes is positive.

Important Formulae

Let us say that an inlet pipe takes ‘x’ hours to fill up a tank of some volume. Then portion or the part of the tank that the inlet pipe fills in 1 hr = 1/x units or parts.

Now that we know that, we can move on to another important concept. Let us say that there are two inlet pipes and three outlet pipes. Then how would we calculate the rate at which a certain volume of the tank will be filled up? The answer is below:

Net Work Done = (Sum of work done by inlet pipes) – (Sum of work done by outlet pipes)

Some Tips and Tricks

If two pipes take ‘x’ & ‘y’ hr respectively to fill a tank or a reservoir or any container, and there is a third pipe that takes ‘z’ hr to empty the same container, tank or reservoir, and all of them are opened together then:

1. The net part filled in 1hr = (1/x) + (1/y) – (1/z)

Hence, The time it takes to fill the tank = [ 1/(1/x) + (1/y) – (1/z)]

2. The leakage case: Consider a pipe fills the tank in ‘x’ hrs. If there is a leakage in the bottom, the tank is filled in ‘y’ hrs. If the tank is full, the time is taken by the leak to empty the tank =[1/(1/x) – (1/y)] hr

3. Suppose that pipe ‘A’ fills the tank as fast as the other pipe ‘B’. If pipe ‘B’ (slower) & pipe ‘A’ (faster) take ‘x’ min & ‘x/n’ min respectively to fill up an empty tank together, then Part of the tank filled in 1 hr = (n + 1)/x.

Solved Example

Example 1: There are two pipes A and B of some width. The two pipes A and B can fill a tank in 36 hours and 45 hours respectively. However, if both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Answer: Let us first calculate the part filled by A in 1 hour. Hence, we may write:

The part filled by A in 1 hour = 1/36; Part filled by B in 1 hour = 1/45

Part filled by (A + B) in 1 hour = [(1/36) + (1/45)] = 9/180 = 1/20.

Hence, both the pipes together will fill the tank in 20 hours.

Example 2: If two pipes function simultaneously, the reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir?

Answer: Let the reservoir be filled by the first pipe in x hours. Then, the second pipe will fill it in (x + 10) hours. Therefore, we can write:

(1/x) + {1/(x+10)} = 1/12. This means that we may write:

(x + 10 + x)/x(x+10) = 1/12 or x2 -14x -120 = 0.

Therefore, neglecting the negative values of ‘x’, we have: x = 20.

So the second pipe will take (20+10) hrs or 30 hrs to fill the reservoir.

More Example

Example 3: Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it takes 32 minutes more to fill the cistern. When the cistern is full, in what time will the leak empty it?

Answer: Work done by the two pipes in 1 hour = [(1/14) + (1/16)] = 15/112

Therefore, the time taken by these pipes to fill the tank = 112/15 hours = 7 hours 28 minutes.

Due to leakage, time taken  = 7 hours 28 minutes + 32 min = 8 hours.

therefore, work done by the two pipes + leak in one hour = 1/8

And work done by the leak in one hour = [15/112 – 1/8] = 1/112

Thus the leak will empty the full cistern in 112 hours.

Practice Problems

Q 1: Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

A) 10 h                        B) 12 h                        C) 14 h                     D) 16 h

Ans: C) 14 h

Q 2: Two pipes A and B can fill a cistern in 37(1/2) minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after:

A) 5 min                           B) 9 min                           C) 10 min                                    D) 15 min

Ans:   B) 9 min

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