Episode 8 — Aptitude and Reasoning / 8.10 — Pipes and Cisterns
8.10.a Concepts and Formulas
1. The Pipes and Cisterns Analogy
In Work and Time problems, workers do positive work to complete a task. Pipes and Cisterns is exactly the same framework with one twist: some agents (outlet pipes, leaks) do negative work -- they undo progress by emptying the tank.
Work and Time Pipes and Cisterns
----------- ------------------
Worker ---> Pipe
Job / Task ---> Cistern (Tank)
Worker completes job ---> Inlet pipe fills the tank
Outlet pipe / Leak empties the tank
Rate of work = 1/n ---> Rate of filling = 1/n
Combined rate (all +) ---> Net rate = (inlet rates) - (outlet rates)
Key insight: An inlet pipe is a "positive worker" and an outlet pipe is a "negative worker." The net rate of filling is the algebraic sum of all individual rates.
2. Basic Definitions
2.1 Cistern (Tank)
A cistern is a container or tank that holds water (or any liquid). Its total volume is the "work" to be done. We usually take the total capacity as 1 unit (the whole tank) unless using the LCM method.
2.2 Inlet Pipe
An inlet pipe is a pipe that fills the cistern. It does positive work.
If an inlet pipe can fill a tank in X hours:
Rate of filling = 1/X of the tank per hour
2.3 Outlet Pipe
An outlet pipe (or drain) is a pipe that empties the cistern. It does negative work.
If an outlet pipe can empty a tank in Y hours:
Rate of emptying = 1/Y of the tank per hour
(This is treated as -1/Y when calculating net rate)
2.4 Leak
A leak is an unintentional outlet -- a crack or hole in the tank that lets water escape. It works exactly like an outlet pipe mathematically.
3. Fundamental Formulas
3.1 Single Inlet Pipe
If pipe A can fill a tank in X hours:
Rate of A = 1/X per hour
Time to fill = X hours
Worked Example:
Pipe A can fill a tank in 12 hours. What fraction of the tank does it fill in 1 hour?
Rate of A = 1/12 per hour
In 1 hour, A fills 1/12 of the tank.
In 3 hours, A fills 3/12 = 1/4 of the tank.
3.2 Single Outlet Pipe
If pipe B can empty a full tank in Y hours:
Rate of B = 1/Y per hour (emptying)
Time to empty = Y hours
Worked Example:
An outlet pipe can empty a full tank in 8 hours. What fraction remains after 3 hours?
Rate of emptying = 1/8 per hour
In 3 hours, emptied = 3/8 of the tank
Remaining = 1 - 3/8 = 5/8 of the tank
4. Two Pipes Working Together
4.1 Two Inlet Pipes
If pipe A fills in X hours and pipe B fills in Y hours, working together:
Combined rate = 1/X + 1/Y = (X + Y) / XY
Time to fill together = XY / (X + Y)
Worked Example:
Pipe A fills a tank in 6 hours and pipe B fills it in 12 hours. How long do they take together?
Rate of A = 1/6 per hour
Rate of B = 1/12 per hour
Combined rate = 1/6 + 1/12
= 2/12 + 1/12
= 3/12
= 1/4 per hour
Time to fill = 1 / (1/4) = 4 hours
Alternatively: (6 x 12) / (6 + 12) = 72 / 18 = 4 hours
4.2 One Inlet + One Outlet
If inlet A fills in X hours and outlet B empties in Y hours:
Net rate = 1/X - 1/Y = (Y - X) / XY
Case 1: Y > X (inlet is slower than outlet at emptying, i.e., filling wins)
Time to fill = XY / (Y - X)
Case 2: X > Y (outlet empties faster than inlet fills)
The tank will never fill. It will empty.
Time to empty a full tank = XY / (X - Y)
Case 3: X = Y (rates are equal)
Net rate = 0. Tank level stays constant.
Worked Example:
Pipe A fills a tank in 10 hours. Pipe B empties it in 15 hours. If both are opened together, how long to fill the tank?
Rate of A (inlet) = +1/10 per hour
Rate of B (outlet) = -1/15 per hour
Net rate = 1/10 - 1/15
= 3/30 - 2/30
= 1/30 per hour (positive, so tank fills)
Time to fill = 30 hours
Worked Example (outlet faster):
Pipe A fills a tank in 20 hours. Pipe B empties it in 12 hours. Both are opened on a full tank. When does it empty?
Net rate = 1/20 - 1/12
= 3/60 - 5/60
= -2/60
= -1/30 per hour (negative, so tank empties)
Time to empty = 30 hours
5. Three or More Pipes
5.1 General Formula
For n pipes, where some are inlets (+) and some are outlets (-):
Net rate = (sum of inlet rates) - (sum of outlet rates)
Net rate = (1/A + 1/B + ...) - (1/P + 1/Q + ...)
where A, B, ... are inlet times
and P, Q, ... are outlet times
Time to fill = 1 / Net rate (if net rate is positive)
Time to empty = 1 / |Net rate| (if net rate is negative)
5.2 Three Pipes: Two Inlets + One Outlet
Net rate = 1/A + 1/B - 1/C
Time to fill = 1 / (1/A + 1/B - 1/C)
= ABC / (BC + AC - AB)
Worked Example:
Pipe A fills in 10 hrs, pipe B fills in 15 hrs, pipe C empties in 20 hrs. All three are opened. How long to fill the tank?
Net rate = 1/10 + 1/15 - 1/20
LCM of 10, 15, 20 = 60
= 6/60 + 4/60 - 3/60
= 7/60 per hour
Time to fill = 60/7 = 8 hours 34 minutes (approx 8.57 hours)
5.3 Three Pipes: One Inlet + Two Outlets
Net rate = 1/A - 1/B - 1/C
Worked Example:
An inlet fills a tank in 8 hours. Two outlets empty it in 12 hours and 24 hours respectively. If all three are open, what happens?
Net rate = 1/8 - 1/12 - 1/24
LCM of 8, 12, 24 = 24
= 3/24 - 2/24 - 1/24
= 0/24
= 0
The tank level remains constant -- it will neither fill nor empty.
6. Pipes with Leaks
A leak is mathematically identical to an outlet pipe. The typical problem pattern:
Without leak: pipe fills tank in X hours --> rate = 1/X
With leak: pipe fills tank in Y hours --> net rate = 1/Y
Rate of leak = (rate without leak) - (net rate with leak)
= 1/X - 1/Y
Time for leak alone to empty full tank = 1 / (1/X - 1/Y) = XY / (Y - X)
Worked Example:
A pipe can fill a tank in 6 hours. Due to a leak at the bottom, the tank is filled in 8 hours. How long does the leak alone take to empty a full tank?
Rate of pipe = 1/6 per hour
Net rate (pipe + leak) = 1/8 per hour
Rate of leak = 1/6 - 1/8
= 4/24 - 3/24
= 1/24 per hour
The leak alone empties the full tank in 24 hours.
Worked Example:
A cistern is filled by a pipe in 10 hours. A leak empties the full cistern in 15 hours. If both the pipe and leak are active, how long to fill the cistern?
Rate of pipe = 1/10
Rate of leak = -1/15
Net rate = 1/10 - 1/15 = 3/30 - 2/30 = 1/30
Time to fill = 30 hours
7. Alternate Pipe Operation
Sometimes pipes are operated alternately -- pipe A runs for some time, then pipe B runs, and so on in a cycle.
7.1 Pipes Opened Alternately (Each for 1 Hour)
Step 1: Find work done in one complete cycle.
Step 2: Find how many complete cycles are needed.
Step 3: Handle the remaining work in the last (partial) cycle.
Worked Example:
Pipe A fills a tank in 12 hours. Pipe B fills it in 18 hours. They are opened alternately for 1 hour each, starting with A. How long does it take to fill the tank?
Rate of A = 1/12 per hour
Rate of B = 1/18 per hour
In one cycle (2 hours): A works for 1 hr + B works for 1 hr
Work in one cycle = 1/12 + 1/18
= 3/36 + 2/36
= 5/36
Number of complete cycles for 1 full tank:
36/5 = 7.2 cycles
In 7 complete cycles (14 hours):
Work done = 7 x 5/36 = 35/36
Remaining = 1 - 35/36 = 1/36
Next turn is pipe A (hour 15):
A fills 1/12 per hour
Time for A to fill 1/36 = (1/36) / (1/12) = 12/36 = 1/3 hour = 20 minutes
Total time = 14 hours + 20 minutes = 14 hours 20 minutes
7.2 Pipes Opened for Different Durations in Each Cycle
If pipe A runs for 'a' minutes and pipe B runs for 'b' minutes alternately:
Work per cycle = a/X + b/Y (where X, Y are full times for A and B)
Cycle length = a + b minutes
8. Pipes Opened at Different Times
Sometimes pipes are not opened simultaneously. One pipe starts first, and another joins later.
Worked Example:
Pipe A can fill a tank in 20 hours and pipe B can fill it in 30 hours. A is opened first and B is opened after 4 hours. How long does it take to fill the tank from the start?
Rate of A = 1/20 per hour
Rate of B = 1/30 per hour
In the first 4 hours (only A works):
Work done = 4 x 1/20 = 4/20 = 1/5
Remaining work = 1 - 1/5 = 4/5
After 4 hours, both A and B work together:
Combined rate = 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12 per hour
Time for remaining 4/5 = (4/5) / (1/12) = (4/5) x 12 = 48/5 = 9.6 hours
Total time = 4 + 9.6 = 13.6 hours = 13 hours 36 minutes
9. Capacity-Based Problems
Some problems ask you to find the actual capacity of the cistern (in litres), not just the time.
If a pipe fills at R litres per minute and takes T minutes to fill the tank:
Capacity = R x T
Worked Example:
Two pipes together fill a tank in 6 hours. If one pipe fills 10 litres per minute and the other fills 5 litres per minute, find the capacity of the tank.
Combined flow rate = 10 + 5 = 15 litres per minute
Time = 6 hours = 6 x 60 = 360 minutes
Capacity = 15 x 360 = 5,400 litres
Worked Example:
A cistern has a capacity of 720 litres. Pipe A fills it at 20 litres/min and pipe B empties it at 12 litres/min. Starting from empty, how long to fill?
Net rate = 20 - 12 = 8 litres per minute
Time = 720 / 8 = 90 minutes = 1 hour 30 minutes
10. Finding Efficiency / Rate of an Unknown Pipe
Worked Example:
Pipes A and B together fill a tank in 12 hours. Pipes B and C together fill it in 15 hours. Pipes A and C together fill it in 20 hours. How long does each pipe take alone?
Given:
1/A + 1/B = 1/12 ... (i)
1/B + 1/C = 1/15 ... (ii)
1/A + 1/C = 1/20 ... (iii)
Add all three:
2(1/A + 1/B + 1/C) = 1/12 + 1/15 + 1/20
LCM of 12, 15, 20 = 60:
= 5/60 + 4/60 + 3/60 = 12/60 = 1/5
So: 1/A + 1/B + 1/C = 1/10
Now find each:
1/C = 1/10 - 1/12 = 6/60 - 5/60 = 1/60 --> C = 60 hours
1/A = 1/10 - 1/15 = 3/30 - 2/30 = 1/30 --> A = 30 hours
1/B = 1/10 - 1/20 = 2/20 - 1/20 = 1/20 --> B = 20 hours
Verification:
1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12 (matches equation i)
11. Partially Filled / Partially Empty Tank
Sometimes the tank is not empty at the start.
If the tank is already P fraction full:
Remaining to fill = (1 - P)
Time = (1 - P) / (Net rate)
Worked Example:
A tank is 2/5 full. Pipe A can fill the entire tank in 10 hours. How long to fill the remaining portion?
Remaining = 1 - 2/5 = 3/5
Rate of A = 1/10 per hour
Time = (3/5) / (1/10) = (3/5) x 10 = 6 hours
Worked Example:
A tank is 3/4 full. An outlet empties the full tank in 12 hours. How long to empty the tank?
Amount to empty = 3/4
Rate of outlet = 1/12 per hour
Time = (3/4) / (1/12) = (3/4) x 12 = 9 hours
12. Variable Rate Problems
Sometimes a pipe's rate changes over time, or different pipes are opened at different stages.
Worked Example:
A pipe can fill a tank in 20 hours. After half the tank is filled, three more identical pipes are opened. How long does it take to fill the tank completely?
Rate of one pipe = 1/20 per hour
Time to fill first half = (1/2) / (1/20) = 10 hours
After half is filled, 4 pipes work together:
Combined rate = 4 x 1/20 = 4/20 = 1/5 per hour
Time for second half = (1/2) / (1/5) = 5/2 = 2.5 hours
Total time = 10 + 2.5 = 12.5 hours = 12 hours 30 minutes
13. Summary of All Formulas
1. Inlet rate = 1/X (fills in X hours)
2. Outlet rate = 1/Y (empties in Y hours)
3. Two inlets (X, Y) = XY / (X + Y) hours to fill
4. Inlet + Outlet (X < Y) = XY / (Y - X) hours to fill
5. Inlet + Outlet (X > Y) = XY / (X - Y) hours to empty
6. Net rate (multiple pipes) = Sum(inlet rates) - Sum(outlet rates)
7. Time to fill/empty = 1 / |Net rate|
8. Leak rate = 1/X - 1/Y
(pipe fills in X hrs alone, takes Y hrs with leak, Y > X)
9. Leak empties full tank in = XY / (Y - X) hours
10. Capacity = Flow rate x Time
11. Partially full (fraction P)
Time to fill rest = (1 - P) / Net rate
12. Alternate operation: find work per cycle, then count cycles