Pipes and cisterns aptitude problems concept short tricks

Pipes and cisterns aptitude problems concept short tricks

Introduction-Pipes and cisterns aptitude problems

Problems based on pipes and cistern are quite similar to that of time and work.So,it is very necessary to understand the concept of time and work before starting this chapter.In day today life, we see various examples of pipes and cisterns.Ex- The water tank is filled by using electric pumps and the stored water is used through pipes.

Basic Definitions:-

Inlet – The pipe through which a tank,cistern or reservoir is filled is callet inlet.

Outlet- The pipe through which the tank is emptied is called outlet.

Pipes and cisterns aptitude problems concept short tricks :-

Let a pipe A fills a tank in x hours.
In 1 hour, it fills (1/x) part of the tank.

Similarly let a pipe B empties it in y hours.
In 1 hour, it empties (1/y) part of the tank.

This concept will be more clear by taking a simple example.

Intext Example-
If a home water tank is filled by a pipe in 2 hours. What part of water tank is filled in 1 hour?

Ans- The part of water tank filled in 1 hour = (1/2)
It means half of water tank is filled in 1 hour.

Basic Situations:-

Case-1.A tank can be filled/emptied by more than 1 pipe.
Case-2.Some pipes can fill the tank while other pipes can empty.

Case 1

When more than one pipes are filling the tank,it is very similar to more than one person doing a work.

Example
Let a pipe A can fill the tank in 5 hours and pipe B can fill in 10 hours.What will be the time to fill the tank if both are opened simultaneously?

Answer-The part of tank filled by A in 1 hour=(1/5)
The part of tank filled by B in 1 hour=(1/10)
The part of tank filled by A+B in 1 hour=(1/5) + (1/10) = (3/10)
Time taken to fill the tank when both opened = (reciprocal of part filled in 1 hour)=(10/3) hours

Note-In the case of emptying tank the same procedure is followed.

Case2

Some pipes can fill the tank while other pipes can empty.

Example
Let A can fill the tank in 5 hours and pipe B can empty in 10 hours. What will be the time to fill the tank if both are opened simultaneously?

Answer-The part of tank filled by A in 1 hour=(1/5)
The part of tank emptied by B in 1 hour=(1/10)
The part of tank filled by A+B in 1 hour=(1/5) – (1/10) = (1/10)
Time taken to fill the tank when both opened = (reciprocal of part filled in 1 hour)=(10) hours

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