**Best explanation to understand and solve the problems on trains**

This section is about the aptitude test based on PIPES AND CISTERNS. This is almost related to the concept of time and work, but with a slight difference in the terms where we use inlet and outlet here.

This article will definitely help you to clearly understand the topic and also to know about easy ways to solve pipes and cisterns aptitude problems.

Moreover, we have a collection of aptitude questions based on pipes and cisterns which you may use it to practice for your exams. Learn more about the topic and take up your exams confidently.

**What is called a pipe and cistern?**

Pipe is nothing but a tube that is used to convey water, gas, oil or other liquid substances.

Cisterns are nothing but the tank that stores water or other liquid substances.

A pipe is always connected to a tank or cistern. It is used to fill a tank or empty a tank. Accordingly it is called an inlet or an outlet.

**What is inlet?**

A pipe which is connected to a tank and is used to fill a tank with water is known as an inlet.

**What is outlet?**

A pipe which is connected to a tank and is used to empty a tank that contains water is known as an outlet.

**What is the concept of pipes and cisterns?**

Problems on pipes and cisterns are very similar to the problems on time and work.

In pipes and cistern aptitude, the amount of work done is nothing but the part of tank of filled or emptied with water or liquid. And, the time taken to do the work is the time taken to fill or empty a tank completely or to a given level.

**How to solve pipes and cisterns aptitude problems?**

Here are some important methods to solve pipes and cisterns problems easily.

1. An inlet is connected to a tank and it fills the tank in X hours. So the part of the tank filled in one hour can be given by 1/X

2. An outlet is connected to a tank and it empties the tank in Y hours. So the part of the tank emptied in one hour can be given by 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 can be given by

( 1 / X ) – ( 1 / Y )

Therefore, when both the pipes are open the time taken to fill the whole tank can be given by

= xy / ( y – x )

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 can be given by

( 1 / Y ) – ( 1 / X )

Therefore, when both the pipes are open the time taken to empty the full tank can be given by

= xy / ( x – Y )

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, then the net part of the tank filled in one hour can be given by

( 1 / X ) + ( 1 / Y )

Therefore, the time taken to fill the whole tank can be given by

= xy / ( y + x )

At the same time, 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 can be given by

( 1 / X ) + ( 1 / Y )

Therefore, the time taken to empty the full tank is given by

= xy / ( y + x )

5. Three inlets A, B, and C can fill a tank in X, Y and Z hours respectively. If all the inlets are opened together,then the time taken to fill the tank can be given by

= ( x + y + z ) / ( xy + yz + zx )

6. Two pipes can fill a tank in X and Y hours respectively and an outlet can empty the same tank in Z hours. If all the pipes are opened together, the part of the tank filled in one hour can be given by

(1/X + 1/Y – 1/Z)

Therefore time taken to fill the tank completely when all the pipes are working can be given by

= xyz / xy + yz + xz

7. A pipe can fill a tank in X hours but due to a small leak in the bottom, it can be filled only in Y hours. The time taken by the leak to empty the tank can be given by

= xy / y – x

8. An inlet A is X times faster than inlet B and takes Y minutes less than the inlet B. time taken to fill the tank when both the pipes are opened together can be given by

= xy / (x – y)2

And, A alone will fill the tank in ( y / x-1) minitues

And, B alone will fill the tank in ( xy / x-1 )minitues