Placements | QA | Pipes and Cisterns

Pipes and Cisterns are somewhat similar to the concepts of Work and Wages.

  • The problems of pipes and cisterns usually have two kinds of pipes, Inlet pipe and Outlet pipe / Leak. Inlet pipe is the pipe that fills the tank/reservoir/cistern and Outlet pipe / Leak is the one that empties it.
  • If a pipe can fill a tank in ‘n’ hours, then in 1 hour, it will fill ‘1 / n’ parts. For example, if a pipe takes 6 hours to fill a tank completely, say of 12 liters, then in 1 hour, it will fill 1 / 6 th of the tank, i.e., 2 liters.
  • If a pipe can empty a tank in ‘n’ hours, then in 1 hour, it will empty ‘1 / n’ parts. For example, if a pipe takes 6 hours to empty a tank completely, say of 18 liters, then in 1 hour, it will empty 1 / 6 th of the tank, i.e., 3 liters.
  • If we have a number of pipes such that some fill the tank and some empty it, and we open all of them together, then in one hour, part of the tank filled/emptied = ∑ (1 / mi) – ∑ (1 / nj), where ‘mi‘ is the time taken by inlet pipe ‘i’ to fill the tank completely if only it were open and ‘nj‘ is the time taken by outlet pipe ‘j’ to empty the tank completely if only it were open. If the sign of this equation is positive, the tank would be filled and if the sign is negative, the tank would be emptied.

This topic is really simple if you are able to solve the problems in Work and Wages.

Sample Problems

Question 1 : Two pipes A and B can fill a tank separately in 12 and 16 hours respectively. If both of them are opened together when the tank is initially empty, how much time will it take to completely fill the tank?
Solution : Part of tank filled by pipe A in one hour working alone = 1 / 12
Part of tank filled by pipe B in one hour working alone = 1 / 16
=> Part of tank filled by pipe A and pipe B in one hour working together = (1 / 12) + (1 / 16) = 7 / 48
Therefore, time taken to completely fill the tank if both A and B work together = 48 / 7 hours
 
Another Method
Let the capacity of tank be LCM (12, 16) = 48 units
=> Efficiency of pipe A = 48 / 12 = 4 units / hour
=> Efficiency of pipe B = 48 / 16 = 3 units / hour
=> Combined efficiency of pipes A and B = 7 units / hour
Therefore, time taken to completely fill the tank = 48 / 7 hours
 
Question 2 : Three pipes A, B and C are connected to a tank. Out of the three, A and B are the inlet pipes and C is the outlet pipe. If opened separately, A fills the tank in 10 hours, B fills the tank in 12 hours and C empties the tank in 30 hours. If all three are opened simultaneously, how much time does it take to fill / empty the tank ?
Solution : Part of tank filled by pipe A in one hour working alone = 1 / 10
Part of tank filled by pipe B in one hour working alone = 1 / 12
Part of tank emptied by pipe B in one hour working alone = 1 / 30
=> Part of tank filled by pipes A,B and C in one hour working together = (1 / 10) + (1 / 12) – (1 / 30) = 3 / 20
Therefore, time taken to completely fill the tank if both A and B work together = 20 / 3 hours = 6 hours 40 minutes
 
Another Method
Let the capacity of tank be LCM (10, 12, 30) = 60 units
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 12 = 5 units / hour
=> Efficiency of pipe C = – 60 / 30 = – 2 units / hour (Here, ‘-‘ represents outlet pipe)
=> Combined efficiency of pipes A, B and C = 6 + 5 – 2 = 9 units / hour
Therefore, time taken to completely fill the tank = 60 / 9 = 6 hours 40 minutes
 
Question 3 : Three pipes A, B and C are connected to a tank. Out of the three, A is the inlet pipe and B and C are the outlet pipes. If opened separately, A fills the tank in 10 hours, B empties the tank in 12 hours and C empties the tank in 30 hours. If all three are opened simultaneously, how much time does it take to fill / empty the tank ?
Solution : Part of tank filled by pipe A in one hour working alone = 1 / 10
Part of tank emptied by pipe B in one hour working alone = 1 / 12
Part of tank emptied by pipe B in one hour working alone = 1 / 30
=> Part of tank filled by pipes A, B and C in one hour working together = (1 / 10) – (1 / 12) – (1 / 30) = -1 / 60
Therefore, time taken to completely empty the tank if all pipes are opened simultaneously = 1 / 60 hours = 60 hours
 
Another Method
Let the capacity of tank be LCM (10, 12, 30) = 60 units
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = – 60 / 12 = – 5 units / hour (Here, ‘-‘ represents outlet pipe)
=> Efficiency of pipe C = – 60 / 30 = – 2 units / hour (Here, ‘-‘ represents outlet pipe)
=> Combined efficiency of pipes A, B and C = 6 – 5 – 2 = – 1 units / hour (Here, ‘-‘ represents outlet pipe)
Therefore, time taken to completely empty the tank = 60 / (1) = 60 hours
 
Question 4 : A cistern has two pipes. Both working together can fill the cistern in 12 minutes. First pipe is 10 minutes faster than the second pipe. How much time would it take to fill the cistern if only second pipe is used ?
Solution : Let the time taken by first pipe working alone be ‘t’ minutes.
=> Time taken by second pipe working alone = t + 10 minutes.
Part of tank filled by pipe A in one hour working alone = 1 / t
Part of tank filled by pipe B in one hour working alone = 1 / (t + 10)
=> Part of tank filled by pipe A and B in one hour working together = (1 / t) + (1 / t+10) = (2t + 10) / [t x (t + 10)]
But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together.
=> (2t + 10) / [t x (t + 10)] = 1 / 12
=> t x (t + 10) / (2t + 10) = 12
=> t2 + 10t = 24t + 120
=> t2 – 14t – 120 = 0
=> (t – 20) (t + 6) = 0
=> t = 20 minutes (Time cannot be negative)
Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes
 
Another Method
Let the time taken by first pipe working alone be ‘t’ minutes.
=> Time taken by second pipe working alone = t + 10 minutes.
Let the capacity of cistern be t x (t + 10) units.
=> Efficiency of first pipe = t x (t + 10) / t = (t + 10) units / minute
=> Efficiency of second pipe = t x (t + 10) / (t + 10) = t units / minute
=> Combined efficiency of pipes = (2t + 10) units / minute
=> Time taken to fill the cistern completely = t x (t + 10) / (2t + 10)
But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together.
=> t x (t + 10) / (2t + 10) = 12
=> t2 + 10t = 24t + 120
=> t2 – 14t – 120 = 0
=> (t – 20) (t + 6) = 0
=> t = 20 minutes (Time cannot be negative)
Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes
 
Question 5 : Three pipes A, B and C are connected to a tank. Out of the three, A and B are the inlet pipes and C is the outlet pipe. If opened separately, A fills the tank in 10 hours and B fills the tank in 30 hours. If all three are opened simultaneously, it takes 30 minutes extra than if only A and B are opened. How much time does it take to empty the tank if only C is opened?
Solution : Let the capacity of tank be LCM (10, 30) = 30 units
=> Efficiency of pipe A = 30 / 10 = 3 units / hour
=> Efficiency of pipe B = 30 / 30 = 1 units / hour
=> Combined efficiency of pipes A and B = 4 units/hour
Therefore, time taken to completely fill the tank if only A and B are opened = 30 / 4 = 7 hours 30 minutes
=> Time taken to completely fill the tank if all pipes are opened = 7 hours 30 minutes + 30 minutes = 8 hours
=> Combined efficiency of all pipes = 30 / 8 = 3.75 units / hour
Now, efficiency of pipe C = Combined efficiency of all three pipes – Combined efficiency of pipes A and B
Therefore, efficiency of pipe C = 4 – 3.75 = 0.25 units / hour
Thus, time taken to empty the tank if only C is opened = 30 / 0.25 = 120 hours
 
Question 6 : Time required by two pipes A and B working separately to fill a tank is 36 seconds and 45 seconds respectively. Another pipe C can empty the tank in 30 seconds. Initially, A and B are opened and after 7 seconds, C is also opened. In how much more time the tank would be completely filled ?
Solution : Let the capacity of the tank be LCM (36, 45, 30) = 180 units
=> Efficiency of pipe A = 180 / 36 = 5 units / second
=> Efficiency of pipe B = 180 / 45 = 4 units / second
=> Efficiency of pipe C = – 180 / 30 = – 6 units / second
Now, for the first 7 seconds, A and B were open.
=> Combined efficiency of A and B = 5 + 4 = 9 units / second
=> Part of the tank filled in 7 seconds = 7 x 9 = 63 units
=> Part of tank empty = 180 – 63 = 117 units
Now, all pipes are opened.
=> Combined efficiency of all pipes = 5 + 4 – 6 = 3 units / second
Therefore, more time required = 117 / 3 = 39 seconds
 
Question 7 : Two pipes A and B can fill a tank in 20 hours and 30 hours respectively. If both the pipes are opened simultaneously, find after how much time should pipe B be closed so that the tank is full in 18 hours?
Solution : Let the capacity of the tank be LCM (20, 30) = 60 units
=> Efficiency of pipe A = 60 / 20 = 3 units / hour
=> Efficiency of pipe B = 60 / 30 = 2 units / hour
=> Combined efficiency of pipes A and B = 5 units / hour
Let both A and B be opened for ‘n’ hours and then, B be closed and only A be opened for the remaining ’18 – n’ hours.
=> 5n + 3 x (18 – n) = 60
=> 2n + 54 = 60
=> 2n = 6
=> n = 3
Therefore, B should be closed after 3 hours.
 

Quiz on Pipes and Cisterns

 
This article has been contributed by Nishant Arora
 
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