Pipes and Cistern - Solved Questions and Answers

Last Updated : 10 Jul, 2025

Pipes and Cisterns deal with calculating the time taken to fill or empty a tank (cistern) using inlet pipes and outlet pipes.
A pipe represents an inlet (filling) or outlet (draining) connected to a cistern (or tank/reservoir) is a container that holds liquid.

Pipes and Cisterns questions and answers are provided below for you to learn and practice.

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 C 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 C 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) T
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.

Question 8: 9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps needed to empty the same reservoir working 6 hours a day in 16 days.

Solution:

Apply formula P₁D₁H₁ /W₁= P₂D₂H₂/W₂
9 x 20 x 8 = P2 x 16 x 6
P₂= 15 pumps required

Question 9: A leak can empty a completely filled tank in 10 hours. If a tap is opened in completely filled tank which admits 4 liters of water per minute, then leak takes 15 hours to empty the tank. How many litres of water does the tank hold?

Solution:

Take LCM (10, 15) = 30
Let leak pipe is A and A’s efficiency = 30/10 = 3
here combine efficiency = 2
So efficiency of the leakage (Pipe A) and another Pipe (Pipe B) which is filling the tank is= 30/15 = 2
Pipe A is emptying at 3 units/hour and Pipe B is filling using then emptying rate down to 2 units/hour.
So, filling pipe efficiency is 3 - 2 = 1 unit/ hour.
Pipe B will fill tank in 30/1=30 hours
Filling rate = 4 litre/minute
It will fill 4 x 60 = 240 litre/hour.
Total capacity= 240 x 30 = 7200 litres

Question 10: A, B and C can fill a tank in 6 hours together. After working for 2 hours, C is closed and A and B fill it in 7 hours more. Find the time taken by C alone to fill the tank?

Solution:

Let the total capacity is 42 units.
(A + B + C) per hour work = 42/6 = 7 units.
They all worked for 2 hour.
Total water filled = 7 x 2 = 14 units.
Remaining capacity= 42 - 14 =28 units.
(A+B)’s efficiency= 28/7 = 4 units/hr
C’s efficiency = 7 - 4 =3 units/hr
C alone can fill the tank in 42/3 = 14 hours

Question 11: A tap drives at a rate of one drop/second and 800 drops = 100 ml. The number of litres water wasted in 30 days of a month is ?
Solution:

1 sec = 1 drop
Number of seconds in 30 days= 30_days x 24_hrs x 60_min x 60_sec
Number of milli-liters wasted = (100 x 30_days x 24_hrs x 60_min x 60_sec)/800 = 324000 ml
Number of liter= 324000/1000 =324 liters water wasted.

Question 12: Two pipes A and B independently can fill a tank in 20 hours and 25 hours. Both are opened together for 5 hours after which the second pipe is turned off. What is the time taken by first pipe alone to fill the remaining portion of the tank?

Solution:

Total unit water = LCM(20, 25) = 100 unit
A’s efficiency = 100/20 = 5 unit/hour
B’s efficiency =100/25 = 4 unit/hour
After 5 hour the water filled by A and B together = 5 x 9 =45 unit
Remaining unit = 100 - 45 = 55 unit
Time taken by A alone = 55/5 = 11 hours