Some candidates find Pipes And Cisterns problem difficult to solve, so we here are trying to help those candidates in solving such problems by telling them some quick tips to follow to get the answer in very less time. So that they can use these fundas for solving such Pipes And Cisterns problem in the exam very quickly without facing any kind of difficulty.

The method of solving Pipes And Cisterns problem is identical to problems on time and work. They can be solved using the same concept and formulae. Such as for example-

1) If pipe A can fill a tank in ‘x’ hours, then part filled by the pipe A in 1 hour = 1/x

2) And if a pipe B can empty a full tank in ‘y’ hours, then part emptied in 1 hour = 1/y

3) If pipe A can fill a tank in ‘x’ hours and empty in ‘y’ hours (where x>y), then on opening both the pipes, the net part filled in 1 hour = 1/x – 1/y

4) If pipe A can fill a tank in ‘x’ hours and empty in ‘y’ hours (where y>x), then on opening both the pipes, the net part filled in 1 hour = 1/y – 1/x

Example question:

Ques. 1) Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in how much time?

Ans. Net part filled in 1 hour: (1/5 + 1/6 – 1/12) =17/60

Hence,The tank will be full in = 60/17 hours = 3.53 hours.

Ques. 2) A pump can fill a tank in 3 minutes. Because of a leak in the tank, it took 7/2 minutes to fill the tank. If the tank is full, how much tie will the leak take to empty the tank?

Ans. Work done ue to the leak in 1 minute = {1/3 – 2/7} = 1/21
Hence,The tank will be empty due to leakage in 21 minutes.

If you are still facing any confusion in such type of questions, comment below.
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