The fundamental ideas of "Time and Work" are applied in the same way to the pipe and cistern problems. On the other hand, the query may also refer to outlet pipes in addition to intake pipes.
The pipe that either fills or drains the cistern is referred to as the inlet pipe or the outlet pipe, respectively.
Since the intake pipes are assumed to be performing positive work, and the outlet pipes are assumed to be performing negative work, the concept of work and time can be applied here.
Let us look at a few different types of questions to get a feel for the process, strategies, and quick cuts that can help us solve them.
Shortcuts and Tricks to solve Pipes and Cisterns
- Time required by the Leaking pipe to drain the cistern empty.
If it takes an inlet pipe X minutes to fill a cistern, but it takes an extra Y minutes because of a leak, then the leak can empty a full cistern in Y minutes.
Here if X is greater than Y then, it will take more time to fill the cistern/tank
If X is less than Y then, the tank will get empty faster or it will never get filled up.
The Time required to fill the cistern = \(\frac{X^2+ XY}{Y}\) min
- Finding the capacity of the cistern
A leak in a tank can make it empty in X hours. When a pipe that lets in Y liters of water per hour is turned on, the tank is now empty in Z hours.
Capacity of the Cistern = \(\frac{X.Y.Z}{Z-X}\) litres
- Finding the work done by a pipe when two pipes are opened for different durations
If two pipes A and B are opened for different time durations.
Assuming the work done by A & B is 100% or 1
A’s amount of work + B’s amount of work = 1
\(\frac{A's \ Opening \ Time}{Time \ in \ which\ A\ alone\ fills\ the\ tank}+ \frac{B's \ Opening \ Time}{Time \ in \ which\ B\ alone\ fills\ the\ tank}\) = 1
- The rate of flow in the pipe is directly proportional to the diameter
Pipes and Cisterns: An Important Formula
Here are a few important numbers that will help you answer questions about pipes and cisterns faster and better:
- If it takes x hours to fill a tank, then the amount filled in 1 hour is equal to \(\frac{1}{x}\).
- If it takes y hours to empty the tank, then the amount that was emptied in one hour is \(\frac{1}{y}\).
- If a pipe can fill a tank in x hours and empty the same tank in y hours, then it has a flow rate of y. When both lines are opened at the same time, the net amount of water that goes into the tank in one hour is \(\frac{xy}{y-x}\), as long as y is greater than x.
- If a pipe can fill a tank in x hours and empty the same tank in y hours, then it has a flow rate of y. When both lines are opened at the same time, the net amount of water that goes into the tank in one hour is \(\frac{xy}{x-y}\), as long as x is greater than y.
- Net work done = (Total amount of work done by inputs – total amount of work done by outputs).
- If one inlet can fill the tank in x hours and the other inlet can fill the same tank in y hours, and both inlets are opened at the same time, the time it takes to fill the whole tank is \(\frac{x+y}{y+x}\).
- If two pipes take x and y hours to fill a tank with water, and a third pipe takes z hours to empty the tank, then the time it takes to fill the tank = \(\frac{1}{\frac{1}{x}+ \frac{1}{y}+ \frac{1}{z}}\), and the amount of the tank that was filled in 1 hour = \(\frac{1}{x} +\frac{1}{y}-\frac{1}{z}\)
Previous Year Questions
Ques 1: Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is: (CAT 2021 Slot 2)
- 140
- 120
- 144
- 264
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Ans: (A)
Let the filling and emptying capacity of A and B be 'a' and 'b'
units/hour respectively.
Case 1: A is opened at 2 pm and B at 3 pm
Total work done till 10 pm = 8a - 7b
Case 2: A is opened at 2 pm and B at 4 pm
Total work done till 6 pm = 4a - 2b
As work done is same for A and B
8a -7b = 4a - 2b
4a = 5b
Now time taken by A alone to fill the tank
= \(\frac{Total\ work }{a} = \frac{4a-2b}{b}= \frac{1}{x}= \frac{5b-2b}{5b/4}= \frac{12}{5}\) hours = 144 minutes.
Ques 2: A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open? (CAT 2018 Slot 2)
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Ans: Suppose the inlet pipes of type A fill in water at the rate 'a' units per
minute and the inlet pipes of type B fill in water at the rate 'b' units per minute.
Therefore we have the following
30(10a + 45b) = 60(8a + 18b)
300a + 1350b = 480a + 1080b
180a = 270b
a = 1.5b
Total capacity of the tank
300a + 1350b = 300(1.5b) + 1350b = 1800b
If 7 inlet pipes of type A and 27 inlet pipes of type B are opened, the
volume of water filled in every minute
7a + 27b = 7(1.5b) + 27b = 37.5b
Therefore the number of minutes taken to fill the tank = \(\frac{1800}{37.5}\) = 48
Ques 3: A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank get completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on? (CAT 2018 Slot 1)
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Ans: Let a filling pipe fills the tank at 'a' liters per hour and a draining pipe drains at 'b' liters per hour.
6(6a - 5b) = 60(5a - 6b)
.: 6a - 5b = 50a - 60b
:. 44a = 55b
:. a = 5b/4
Let the tank gets filled completely in 'm' hours when one draining pipe
and two filling pipes are on.
m(2a – b) = 6(6a – 5b)
m (2.5/4 b – b)= = 6(6. 5/4 b – 5b)
m(6b/4) = 6 (10b/4)
m = 10
Ques 4: A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full? (CAT 2017 Slot 2)
- 20
- 30
- 40
- 45
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Ans: (A) The Inlet pipe will fill \(\frac{1}{8}\)th of the tank in an hour.
The outlet pipe will empty the tank in 10 hours.
Outlet pipe will empty \(\frac{1}{10}\)th of the tank.
\(\frac{1}{8}- \frac{1}{B}= \frac{1}{10}\)
10(B-8) =8B
B= 40
Outlet pipe will take 40 hours to empty the tank fully
So, to half empty the tank, it will take 20 hours.
Ques5: A chemical plant has four tanks (A, B, C, and D), each containing 1000 litres of a chemical. The chemical is being pumped from one tank to another as follows:
From A to B @ 20 litres/minute
From C to A @ 90 litres/minute
From A to D @ 10 litres/minute
From C to D @ 50 litres/minute
From B to C @ 100 litres/minute
From D to B @ 110 litres/minute
Which tank gets emptied first and how long does it take (in minutes) to get empty after pumping starts? (CAT 2005)
- A, 16.66
- C, 20
- D, 20
- D. 25
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Ans: (C)
The change in the amount of chemical in each tank after every minute is as follows:
A: -20 - 10 + 90 = 60
B: -100 + 110 + 20 = 30
C: -50 - 90 + 100 = -40
D: -110 + 10 + 50 = -50
Since tank D loses the maximum amount of chemical in a minute, it will be emptied first
Let n minutes be the time taken by tank D to get empty.
1000 - 50n = 0
n = 20 minutes
Ques 6: Three small pumps and a large pump are filling a tank. Each small pump works at (2/3)rd the rate of the large pump. If all four work at the same time, they should fill the tank in what fraction of the time it would have taken the large pump alone? (CAT 2003)
- 4/7
- 1/3
- 2/3
- ¾
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Ans: (B)
Small pump = \(\frac{2}{3}\)rd the rate of large pump 3
⇒ 3 small pump = 2 large pump
Hence, 3 small pump +1 large pump = 3 large pump.
Hence, required fraction of time = \(\frac{1}{3}\)
Ques 7: Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindrical tank holds 500 L more than the conical tank. After 200 L of fuel has been pumped out from each tank, the cylindrical tank contains twice the amount of fuel in the conical tank. How many litres of fuel did the cylindrical tank have when it was full ? (CAT 2000)
- 700L
- 1000L
- 1100L
- 1200L
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Ans: (D)
Let the conical tank hold x L of fuel, then cylindrical tank will hold = (x + 500) L
Given, (x + 300) = 2(x − 200)
⇒ x = 700 L
Hence, cylindrical tank will hold (700 + 500) = 1200 L of fuel.
Ques 8: There is a leak in the bottom of the tank. This leak can empty a full tank in 8 h. When the tank is full, a tap is opened into the tank which intakes water at rate of 6 L/h and the tank is now emptied in 12 h. What is the capacity of the tank? (CAT 1994)
- 28.8 L
- 36 L
- 144 L
- Cannot be determined
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Ans: (C)
Working efficiency of leak per hour = (1/8)
Working efficiency of leak with tap = 1/12
∴Working efficiency of tap = 1/8 – 1/12 = 1/24 tank
∴Tap can fill the tank in 24 h.
Hence capacity of tank =24×6=144L.
Ques 9: A water tank has three taps: A, B and C. A fills 4 buckets in 24 min, B fills 8 buckets in 1 h and C fills 2 buckets in 20 min. If all the taps are opened together, a full tank is emptied in 2 h. If a bucket contains 5 L water, what is the capacity of the tank ? (CAT 1994)
- 120 L
- 240 L
- 180 L
- 60 L
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Ans: (B)
Tap A fills 20 L water in 24 min.
Tap B fills 40 L water in 60 min.
Tap C fills 10 L water in 20 min.
work done by all the taps together in 2 h = \(\frac{20}{24}\) x120 + \(\frac{40}{60}\) x120 + \(\frac{10}{20}\) x120 = 240 L
Capacity of the tank is 240 L
Practice Questions
Qies 1: Pipe A basically used as inlet pipe and pipe B is used as outlet pipe. Pipes A and B both are opened simultaneously, all the time. When pipe A fills the tank and B empty the tank, it will take double the time than when both the pipe fill the tank. When pipe B is used for filling the tank, its efficiency remains constant. What is the ratio of efficiency of pipe A and pipe B respectively?
- 3 : 1
- 5 : 2
- 1 : 3
- 3 : 2
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Ans: (A)
Efficiency when both pipes used to fill = A + B
Efficiency when pipe A is used to fill and pipe B is used to empty the tank = A − B
(By componendo and dividendo)
\(\frac{A+B}{A-B}= \frac{2}{1}\)
\(\frac{A}{B}= \frac{3}{1}\)
Thus, the ratio of efficiency of pipe A and B = 3 : 1.
Ques 2: Pipe A can fill the tank in 4 hours, while pipe B can fill it in 6 hours working separately. Pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. What is the time difference between both the cases, to fill the tank fully?
- 48 min
- 54 min
- 30 min
- none of these
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Ans: (D)
Time taken to fill the half tank by A and B = \(\frac{50}{41.66}= \frac{6}{5}\) hours
Time taken by A, B and C to fill rest half of the tank = \(\frac{50}{16.66} = 3\ hours\)
Total time = \(\frac{6}{5}\) + 3= 4hours 12 minutes
2nd case:-
Time taken to fill (3/4)th of the tank = \(\frac{75}{41.66}= \frac{9}{5} \ hours\)
Time taken to fill the rest (1/4)th tank by A, B and C = \(\frac{25}{16.66}= \frac{3}{2} \ hours\)
Ques 3: Two pipes A and B can fill a tank in 24 hours and 120/7 hours respectively. Harihar opened the pipes A and B to fill an empty tank and some times later he closed the taps A and B, when the tank was supposed to be full. After that it was found that the tank was emptied in 2.5 hours because an outlet pipe C connected to the tank was open from the beginning. If Harihar closed the pipe C instead of closing pipes A and B the remaining tank would have been filled in :
- 2 hours
- 8 hours
- 6 hours
- 4 hours
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Ans: (B)
Efficiency of inlet pipe A = 4.16% (\(\frac{100}{24}\))
Efficiency of inlet pipe B = 5.83% (\(\frac{100}{120/7}\))
Efficiency of A and B together = 10%∴(time = 10 hours) Now, if the efficiency of outlet pipe be x % then in 10 hours the capacity of tank which will be filled = 10 × (10 − x ) Now, since this amount of water is being emptied by C at x % per hour, then
\(\frac{10(10-x)}{x}\) = 2.5 hours
X= 8 %
Therefore in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in = \(\frac{80}{10}\) = 8 hours.
Ques 4: Pipe A can fill a tank in 12 hours and pipe B can fill it in 15 hours, separately. A third pipe C can empty it in 20 hours. Initially pipe A was opened, after one hour pipe B was opened and then after 1 hour when pipe B was opened pipe C was also opened. In how many hours the tank will be full?
- 9\(\frac{2}{3}\) hours
- 6 \(\frac{2}{3}\)hours
- 10 hours
- None of these
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Ans: (A)
Efficiency of pipe A = 8.33%
Efficiency of pipe B = 6.67%
Efficiency of pipe C = 5%
When tap C was opened pipe A filled 16.66% capacity
When tap C was opened pipe B filled 6.67% capacity
Therefore rest capacity of the tank to be filled = 100 − 23.34 = 76.66%
Now, the net efficiency of A, B and C = 10%
Hence, pipes A, B and C will take = 76.66/10 =7.66 =7(2/3) hours
∴ Total time = 2hours + 7 (2/3)hours = 9 (2/3) hours
Ques 5: (Both 5 & 6) A tank has an inlet and outlet pipe. The inlet pipe fills the tank completely in 2 hours when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged.
If both pipes are opened simultaneously at a time when the tank was one-third filled, when will the tank fill thereafter?
- 3/2 hours
- 2/3 hour
- 2 hours
- 1\(\frac{2}{3}\) hours
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Ans: (C)
Efficiency of inlet pipe = 50%
Efficiency of outlet pipe = 16.66%
Net efficiency of pipes A and B = 33.33%
Capacity of tank to be filled up = 66.66%
Hence, required time = 66.66/33.33 = 2 hours
Ques 6: If there is a leakage also which is capable of draining out the liquid from the tank at half of the rate of outlet pipe, then what is the time taken to fill the empty tank when both the pipes are opened?
- 3 hours
- 3\(\frac{2}{3}\) hours
- 4 hours
- None of these
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Ans: (C)
Rate of leakage = 8.33% per hour
Net efficiency = 50 − (16.66 + 8.33) = 25%
Time required = 100/25 = 4 hours
How to approach Pipes and Cistern question in CAT exam
- Be thorough with the concepts of Ratio and proportion and Time and Work.
- Conversion of Fraction to percentage and vice-versa is very crucial to this chapter.
- Practicing more and more question will give u a hang of different types of questions asked in CAT.
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