Pipe and Cistern Tips and Tricks for Government Exams

Theory

Pipes and cisterns problems are almost the same as those of Time and work problems. Thus, if a pipe fills a tank in 6 hrs, then the pipe fills 1/6th of the tank in 1 hr. the only difference with Pipes and Cisterns problems is that there are outlets as well as inlets. Thus, there are agents (the outlets) which perform negative work too.

Formula

• If x hours are required to fill up a tank, then part filled in 1 hr =1/x
• If y hours are required to empty the tank, then part emptied in 1 hour = 1/y
• If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (y-x)}, provided y>x
• If a pipe can fill a tank in x hours and can empty the same tank in y hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr = {(xy) / (x-y)}, provided x>y
• Net work done = (Sum of work done by Inlets) – (Sum of work done by Outlets)
• One inlet can fill the tank in x hr and the other inlet can fill the same tank in y hrs, if both the inlets are opened at the same time, the time taken to fill the whole tank = {(xy) / (y+x)}
• If two pipes take x and y hours respectively to fill a tank of water and a third pipe is opened which takes z hours to empty the tank, then the time taken to fill the tank = {1 / (1/x)+(1/y)+(1/z)} and the net part of the tank filled in 1 hr = (1/x)+(1/y)-(1/z)

Tips

• One must be familiar with terms like inlet, outlet, leak, filling a tank, emptying a tank and the formulas related to the same. Only then can a candidate answer these questions without getting confused
• If you are unable to crack the question, then ensure that you do not spend too much time on a single question 
• Memorize the formulas and practise as much as possible to understand the concept better

Solved Examples