The pipe and cistern type problems are quite common in competitive exams like CAT, Bank PO etc. I have included some easy way of solving these kind of problems.

Some general rules are as follows:

1. If a pipe attached to reservior or tank can fill in X hours, in 1 hours the tank will filled 1/x

2. If an outlet attached to a tank is opened to empty a tank takes Y hours to flush out till end, in 1 hour the tank wull 1/y part empty

3. If the tank is attached with a pipe that fills it full in X hours and another pipe that empties it full in Y hours then on opening both the pipes togrther the part emptied or full can be determined using Part = (1/x) – (1/y) when Y > X and (1/y) – (1/x) when X>Y

4. If there are 2 pipes ( X, Y)simultaneously filling (emptying) the tank then the part emptied/filled in = (1/x ) + (1/y)

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is there some other trick for these problems??? this is already know.. pls mail some other tricks to lilly.lilly91@yahoo.co.in