Pipes and Cistern Concept Problems Shortcuts and Solutions

Pipes and cistern problems are asked in many competitive examinations including SSC, IBPS etc. solving these problems requires more time if the candidate does not know the shortcut method to solve this type of problem. It is more important to clear basic concept in these problems. So here is a shortcut method to solve pipes and cistern problems with basic concept.

The below given 2 problems with solution will let you understand the concept easily.

Suppose you are given a problem like the one given below.

The shortcut method to solve this type of pipes and cisterns problems is that you have to put them in ratio to 1 and add the values of pipes which are filling the tank and subtract the value which is emptying the tank. The process is shown below.

Tank will be filled in 1 hours= (1/10)+(1/15)-(1/20)= (6+4-3)/60

(LCM of 10,15,20 is 60 by shortcut method)

Tank filled in 1 hour = 7/60

So full tank will be filled in =60/7 hours Ans.

The above question clears the basic concept of pipes and cistern. The basic point is you have to add the pipe’s value filling the tank and subtract the pipe’s value emptying it.

Now let us take another example.

Two pipes A and B can fill a cistern in 36 hours and 45 hours respectively and a third pipe C can empty it in 30 hours. Both A and B pipes are kept opened for 7 hours and then third pipe C is also opened. Determine the time required to fill the tank.

Solution: now in this question,

Tank filled by 2 pipes filling together in 7 hour= 7[(1/36)+(1/45)]= 7/20

Remaining part= 1-(7/20)= 13/20

Tank filled by all the 3 pipes together in 1 hour= (1/36)+(1/45)-(1/30)= 1/60

Now all 3 pipes working together can fill 1/60 tank in 1 hour.

So 13/20 tank will be filled in = (13/20)*60= 39 hours

Total time to fill the full tank= 39+7= 46 hours Ans.