Pipes And Cisterns

Concept 1. If an inlet pipe can fill a cistern in ‘x’ hours, then


Concept 2. If an outlet pipe can empty a cistern in ‘y’ hours, then


Concept 3. Net work done in 1 hours = (filling work in 1 hour) – (Empty work in 1 hour) If W is ve, then cistern is emptied.


Concept 4.

Concept 5. If more than one inlet pipe or more than one out let pipes are fitted, then



Time taken to fill or Empty = 1/part filled or emptied in 1 hour

Concept 6. If one inlet pipe can fill in t1 hours and one outlet pipe can empties it in t2 hours, then part of cistern filled or emptied in 1 hours = (1/t1 – 1/t2)


Note : Chain rule will work here same as Time and Work problem.

Problem on Leakage:

Two fill pipes can respectively fill a cistern, say, in ‘x’ hours and ‘y’ hours respectively, but due to leakit takes ‘P’ hours extra to fill the cistern. Now both pipes are closed and the fill cistern can be emptied through the leak in ‘T’ hours,


If there is only one fill pipe, then above relation can be reduces to Empty time by leak, T