Pipe and Cistern question condition, formulas and solution

Pipe and cistern aptitude condition and formula
Already we have covered three tutorials on Pipe and Cistern math tricks including basic concept, shortcut formula and example question with the easy solution. The previous tutorial on pipe and cistern was the tricky quicker solution. As per the request of some candidates such as my student and friends, we are going to discuss some advanced level math question and solution. Earlier we have recommended you that learn the shortcut math formula by giving enough time. Otherwise, you could not use all shortcut formula in the competitive examination hall. Because some examiner is asking some question about the difficult problem. Then the basic concept helps you to take the challenge in time. One more thing that you may know the practice makes the man perfect.
However, we are going to discuss some most important aptitude question on the math chapter Pipe and Cistern. If you did not visit the previous tutorial, please visit to get a clear concept.
Example question of pipe and cistern problem by quick solution math formula.


Pipe and cistern aptitude condition and shortcuts formula

When a chamber is connected by A&B both inlet pipes, And they can fill the tank X and Y hour respectively. If both pipes started filling the tank, then
i)                   When the pipe A should be closed to fill the tank in T hour?

ii)                When the pipe B should be closed to fill the tank in T hour?
To solve this kind of question applies this formula.
Pipe X should be closed after:-

Math problems and solution of Pipe and Cistern

Sample Question:-  A chamber is connected by two pipes X and Y. They can fill the tank in 40 and 50 minutes respectively. Both pipes opened to filling tank, When should be closed the first tan to complete filling the tank in 30 minutes?
Solution by math tricks of pipe and cistern:-
Directly apply the first formula. Here the first pipe X and its value are 40 minutes, the 2nd pipe is y, and its value is 50 minutes, and T=30 is the time value to complete the fill chamber. Let’s look at below how easily, we are getting the answer!
The formula is X*(1-T/Y)

Prove of this shortcut formula:
We need to prove, whether the shortcut formula is working correctly or not! As per the question and answer, both pipes are working 16 minutes together, and the second pipe Y is working (30-16)=14 minutes alone. So we have to find out first how many parts they have filled out in 16 minutes by both pipes. Then the remaining portion must be filled by Y in 14 minutes.
So let’s see X and Y can complete work in one days= (1/40+1/50) = 9/200

They can fill the cistern in 16 days (9/200) *16 =18/25 parts of the cistern. The work will remain (1-18/25) =7/25 parts. If the Y  works for 14 days and complete the work 7/25, then we will satisfy with this formula.
So let’s see Y can do the work in one day 1/50 parts.
Y can do the work in 14 days (1/50)*14 = 7/25 days.
Friends it is proved that the formula is working correctly. So we can use this method in our upcoming competitive exam for saving time.

Aptitude question and solution for the 2nd formula:-  I hope that you understood this shortcut formula very well. Yet we are giving you another example of the 2nd method. Where 2nd pipe will stop after a specific time and the complete work will have done in particular time.
Question:- A tank is connected by two inlet pipe, and they can fill cisterns in 60 and 80 minutes. Both pipes are opened at 6.00 A.M when the second pipe should close to fill the cistern on 6.40 A.M?
Solution:- only we will use the second formula. Let’s assume the first pipe is X, and its time value is 60 minutes. Second pipe Y and its time value are 80 minutes. We have to complete in 6.00 A.M to 6.40 A.M means in 40 minutes.
The formula is Y*(1- T/X) hour.
Now input value 60*(1-40/80)
=60*40/80 =60* ½= 30 minutes
The second pipe should be close at 6.00A.M + 30 minute= 6.30 A.M to Complete filling the tank on 6.40 A.M.
Please note that this formula also is used in time and works math chapter, Where two people will work, and one of them will leave and rest work will do by another worker.


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