Episode 8 — Aptitude and Reasoning / 8.10 — Pipes and Cisterns
8.10.c Solved Examples
Problems are arranged in three tiers: Basic, Medium, and Advanced.
Basic Problems
Problem 1: Single Pipe Filling
A pipe can fill a tank in 15 hours. What fraction of the tank is filled in 5 hours?
Solution:
Rate of pipe = 1/15 per hour
In 5 hours: 5 x 1/15 = 5/15 = 1/3
Answer: 1/3 of the tank is filled.
Problem 2: Two Inlets Together
Pipe A can fill a tank in 10 hours and pipe B can fill it in 15 hours. How long do they take together?
Solution:
Rate of A = 1/10 per hour
Rate of B = 1/15 per hour
Combined rate = 1/10 + 1/15
LCM(10, 15) = 30:
= 3/30 + 2/30
= 5/30
= 1/6 per hour
Time to fill = 6 hours
Quick check: (10 x 15) / (10 + 15) = 150/25 = 6 hours
Answer: 6 hours
Problem 3: Outlet Pipe
A pipe can empty a full tank in 18 hours. How long to empty 2/3 of the tank?
Solution:
Rate of emptying = 1/18 per hour
Time to empty 2/3 = (2/3) / (1/18) = (2/3) x 18 = 12 hours
Answer: 12 hours
Problem 4: One Inlet + One Outlet
Pipe A fills a tank in 8 hours. Pipe B empties it in 12 hours. If both are opened on an empty tank, how long to fill it?
Solution:
Rate of A (inlet) = +1/8
Rate of B (outlet) = -1/12
Net rate = 1/8 - 1/12
LCM(8, 12) = 24:
= 3/24 - 2/24
= 1/24 per hour
Time to fill = 24 hours
Quick check: (8 x 12) / (12 - 8) = 96/4 = 24 hours
Answer: 24 hours
Problem 5: Three Inlets
Pipes A, B, and C can fill a tank in 6, 8, and 12 hours respectively. How long do all three take together?
Solution (LCM Method):
LCM(6, 8, 12) = 24 units (tank capacity)
Rate of A = 24/6 = 4 units/hr
Rate of B = 24/8 = 3 units/hr
Rate of C = 24/12 = 2 units/hr
Combined rate = 4 + 3 + 2 = 9 units/hr
Time = 24/9 = 8/3 hours = 2 hours 40 minutes
Answer: 2 hours 40 minutes
Problem 6: Finding Individual Time from Combined Rate
Two pipes together fill a tank in 12 hours. If one pipe alone fills it in 20 hours, how long does the other pipe take alone?
Solution:
Combined rate = 1/12
Rate of pipe 1 = 1/20
Rate of pipe 2 = 1/12 - 1/20
LCM(12, 20) = 60:
= 5/60 - 3/60
= 2/60
= 1/30
Pipe 2 alone takes 30 hours.
Answer: 30 hours
Problem 7: Capacity in Litres
A tank is filled by a pipe that delivers 5 litres per minute. If the tank takes 2 hours to fill, what is its capacity?
Solution:
Time = 2 hours = 120 minutes
Rate = 5 litres/min
Capacity = 5 x 120 = 600 litres
Answer: 600 litres
Medium Problems
Problem 8: Two Inlets + One Outlet
Pipe A fills a tank in 10 hours, pipe B fills it in 12 hours, and pipe C empties it in 15 hours. If all three are opened simultaneously, how long to fill the tank?
Solution (LCM Method):
LCM(10, 12, 15) = 60 units
Rate of A = 60/10 = 6 units/hr (+)
Rate of B = 60/12 = 5 units/hr (+)
Rate of C = 60/15 = 4 units/hr (-)
Net rate = 6 + 5 - 4 = 7 units/hr
Time = 60/7 hours = 8 hours 34 minutes (approx)
Answer: 60/7 hours (8 hours 34 minutes approximately)
Problem 9: Finding Leak Rate
A pipe can fill a tank in 5 hours. Due to a leak, the tank is filled in 7 hours. In how many hours can the leak empty a full tank?
Solution:
Rate of pipe = 1/5
Net rate (pipe + leak) = 1/7
Rate of leak = 1/5 - 1/7
= 7/35 - 5/35
= 2/35
Leak empties full tank in = 35/2 = 17.5 hours
Shortcut: (5 x 7) / (7 - 5) = 35/2 = 17.5 hours
Answer: 17.5 hours (17 hours 30 minutes)
Problem 10: Delayed Opening
Pipe A fills a tank in 16 hours. Pipe B fills it in 24 hours. A is opened first. After 4 hours, B is also opened. How long from the beginning does it take to fill the tank?
Solution (LCM Method):
LCM(16, 24) = 48 units
Rate of A = 48/16 = 3 units/hr
Rate of B = 48/24 = 2 units/hr
First 4 hours (A alone):
Work = 3 x 4 = 12 units
Remaining = 48 - 12 = 36 units
A + B together: rate = 3 + 2 = 5 units/hr
Time for remaining = 36/5 = 7.2 hours
Total time = 4 + 7.2 = 11.2 hours = 11 hours 12 minutes
Answer: 11 hours 12 minutes
Problem 11: Alternate Operation (Two Inlets)
Pipe A fills a tank in 10 hours. Pipe B fills it in 15 hours. They are opened alternately for 1 hour each, starting with A. How long to fill the tank?
Solution (LCM Method):
LCM(10, 15) = 30 units
Rate of A = 30/10 = 3 units/hr
Rate of B = 30/15 = 2 units/hr
Work per cycle (2 hours) = 3 + 2 = 5 units
Complete cycles needed: 30/5 = 6 cycles exactly
Total time = 6 x 2 = 12 hours
Answer: 12 hours
Problem 12: Alternate Operation (Does Not Divide Evenly)
Pipe A fills a tank in 12 hours. Pipe B fills it in 18 hours. Opened alternately for 1 hour each, starting with A. Find the time.
Solution (LCM Method):
LCM(12, 18) = 36 units
Rate of A = 36/12 = 3 units/hr
Rate of B = 36/18 = 2 units/hr
Work per cycle (2 hours) = 3 + 2 = 5 units
Complete cycles: 36/5 = 7 remainder 1
7 complete cycles = 14 hours
Work done = 7 x 5 = 35 units
Remaining = 36 - 35 = 1 unit
Next turn: pipe A (rate = 3 units/hr)
Time = 1/3 hour = 20 minutes
Total = 14 hours 20 minutes
Answer: 14 hours 20 minutes
Problem 13: Partially Full Tank
A tank is 3/5 full. An inlet pipe fills the entire tank in 10 hours. An outlet pipe empties it in 15 hours. Both are opened. How long to fill the tank completely?
Solution:
Remaining to fill = 1 - 3/5 = 2/5
Rate of inlet = 1/10
Rate of outlet = -1/15
Net rate = 1/10 - 1/15 = 3/30 - 2/30 = 1/30 per hour
Time = (2/5) / (1/30) = (2/5) x 30 = 12 hours
Answer: 12 hours
Problem 14: Three Equations, Three Pipes
Pipes A and B together fill a tank in 8 hours. B and C together fill it in 12 hours. A and C together fill it in 24 hours. How long does each pipe take alone?
Solution:
1/A + 1/B = 1/8 ... (i)
1/B + 1/C = 1/12 ... (ii)
1/A + 1/C = 1/24 ... (iii)
Add all three:
2(1/A + 1/B + 1/C) = 1/8 + 1/12 + 1/24
LCM(8, 12, 24) = 24:
= 3/24 + 2/24 + 1/24
= 6/24
= 1/4
So: 1/A + 1/B + 1/C = 1/8
From (i): 1/C = 1/8 - 1/8 = 0?
Wait -- let me recheck. 1/A + 1/B + 1/C = 1/8.
1/C = 1/8 - (1/A + 1/B) = 1/8 - 1/8 = 0
This means pipe C contributes nothing. Let me recheck the addition.
2(1/A + 1/B + 1/C) = 1/8 + 1/12 + 1/24 = 3/24 + 2/24 + 1/24 = 6/24 = 1/4
1/A + 1/B + 1/C = 1/8
From (i): 1/A + 1/B = 1/8
So 1/C = 1/8 - 1/8 = 0. C = infinity (C does nothing).
From (ii): 1/B + 1/C = 1/12, and 1/C = 0, so 1/B = 1/12, B = 12 hours.
From (i): 1/A = 1/8 - 1/12 = 3/24 - 2/24 = 1/24, A = 24 hours.
Verify (iii): 1/A + 1/C = 1/24 + 0 = 1/24. Correct!
Answer: A = 24 hours, B = 12 hours, C = infinite (C has zero rate)
Let me redo this with more typical numbers:
Revised Problem 14: Pipes A and B together fill a tank in 12 hours. B and C together fill it in 15 hours. A and C together fill it in 20 hours. Find the time each pipe takes alone.
Solution:
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(12, 15, 20) = 60:
= 5/60 + 4/60 + 3/60
= 12/60
= 1/5
1/A + 1/B + 1/C = 1/10
From (ii): 1/A = 1/10 - 1/15 = 3/30 - 2/30 = 1/30 --> A = 30 hours
From (iii): 1/B = 1/10 - 1/20 = 2/20 - 1/20 = 1/20 --> B = 20 hours
From (i): 1/C = 1/10 - 1/12 = 6/60 - 5/60 = 1/60 --> C = 60 hours
Verify: 1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12 (matches)
1/20 + 1/60 = 3/60 + 1/60 = 4/60 = 1/15 (matches)
1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20 (matches)
Answer: A = 30 hours, B = 20 hours, C = 60 hours
Problem 15: Outlet Opened After Some Time
An inlet pipe fills a tank in 10 hours. After 5 hours, an outlet pipe is opened that can empty the full tank in 15 hours. How long from the start does it take to fill the tank?
Solution (LCM Method):
LCM(10, 15) = 30 units
Rate of inlet = 30/10 = 3 units/hr
Rate of outlet = 30/15 = 2 units/hr
First 5 hours (inlet alone):
Work done = 3 x 5 = 15 units
Remaining = 30 - 15 = 15 units
Both pipes open:
Net rate = 3 - 2 = 1 unit/hr
Time for remaining = 15/1 = 15 hours
Total = 5 + 15 = 20 hours
Answer: 20 hours
Problem 16: Finding When a Pipe Was Closed
Pipes A and B can fill a tank in 12 and 16 hours respectively. Both are opened together. After how many hours should B be closed so that the tank is full in exactly 9 hours?
Solution (LCM Method):
LCM(12, 16) = 48 units
Rate of A = 48/12 = 4 units/hr
Rate of B = 48/16 = 3 units/hr
Let B be closed after t hours.
A works for all 9 hours. B works for t hours.
Total work = 4 x 9 + 3 x t = 48
36 + 3t = 48
3t = 12
t = 4 hours
B should be closed after 4 hours.
Answer: After 4 hours
Advanced Problems
Problem 17: Alternate Operation with Inlet and Outlet
Pipe A fills a tank in 8 hours (inlet). Pipe B empties it in 12 hours (outlet). They are opened alternately for 1 hour each, starting with A. The tank is initially empty. How long to fill it?
Solution (LCM Method):
LCM(8, 12) = 24 units
Rate of A = 24/8 = 3 units/hr (fills)
Rate of B = 24/12 = 2 units/hr (empties)
Per cycle (2 hours):
Hour 1 (A): +3 units
Hour 2 (B): -2 units
Net per cycle = +1 unit
Cycles needed: 24 units / 1 unit per cycle = 24 cycles
But wait -- we need to check if the tank fills mid-cycle.
After each cycle of 2 hours, the tank has gained 1 unit.
But at the end of each A-turn (odd hours), the tank is higher.
After cycle n (2n hours):
Total filled = n units (net)
At the end of A's turn in cycle n (at hour 2n-1):
Filled = (n-1) + 3 = n + 2 units
We need: when does the tank reach 24 units at the end of an A-turn?
After A's turn in cycle n: tank = (n-1) + 3 = n + 2 units
Set n + 2 = 24 --> n = 22
At the end of cycle 21 (42 hours): tank = 21 units
Hour 43 (A works): tank = 21 + 3 = 24 units -- FULL!
The tank fills at hour 43, during A's turn.
Let me verify more carefully:
After hour 1 (A): 0 + 3 = 3
After hour 2 (B): 3 - 2 = 1
After hour 3 (A): 1 + 3 = 4
After hour 4 (B): 4 - 2 = 2
...
Pattern after each complete cycle (2 hrs): +1 unit net
After 2n hours: n units
After 44 hours (22 complete cycles): 22 units
Hour 45 (A): 22 + 3 = 25 > 24
So A fills the remaining during hour 45.
At start of hour 45: 22 units
Need: 24 - 22 = 2 more units
A's rate = 3 units/hr
Time = 2/3 hour = 40 minutes
Total = 44 hours + 40 minutes = 44 hours 40 minutes
Answer: 44 hours 40 minutes
Problem 18: Multiple Pipes with Complex Scheduling
A tank has three inlet pipes A, B, C that fill it in 6, 8, and 12 hours respectively. They are opened in a rotating pattern: A for 1 hour, then B for 1 hour, then C for 1 hour, and repeat. How long to fill the tank?
Solution (LCM Method):
LCM(6, 8, 12) = 24 units
Rate of A = 24/6 = 4 units/hr
Rate of B = 24/8 = 3 units/hr
Rate of C = 24/12 = 2 units/hr
Work per cycle (3 hours) = 4 + 3 + 2 = 9 units
Complete cycles: 24/9 = 2 remainder 6
2 complete cycles = 6 hours, work = 18 units
Remaining = 24 - 18 = 6 units
Hour 7 (A's turn): A fills 4 units. Total = 22 units. Remaining = 2 units.
Hour 8 (B's turn): B fills at 3 units/hr. Need 2 units.
Time = 2/3 hour = 40 minutes
Total = 7 hours + 40 minutes = 7 hours 40 minutes
Answer: 7 hours 40 minutes
Problem 19: Two Leaks
A pipe fills a tank in 6 hours. There are two leaks. With only leak 1, the pipe takes 8 hours. With only leak 2, the pipe takes 9 hours. How long does the pipe take with both leaks present?
Solution:
Rate of pipe = 1/6
With leak 1: net rate = 1/8
Rate of leak 1 = 1/6 - 1/8 = 4/24 - 3/24 = 1/24
With leak 2: net rate = 1/9
Rate of leak 2 = 1/6 - 1/9 = 3/18 - 2/18 = 1/18
With both leaks:
Net rate = 1/6 - 1/24 - 1/18
LCM(6, 24, 18) = 72:
= 12/72 - 3/72 - 4/72
= 5/72
Time = 72/5 = 14.4 hours = 14 hours 24 minutes
Answer: 14 hours 24 minutes
Problem 20: Finding Capacity from Rate Information
Pipe A delivers water at 12 litres per minute and can fill a tank in 2 hours. Pipe B can fill the same tank in 3 hours. If both pipes are opened together, in how many minutes will 1,200 litres be filled?
Solution:
Capacity = 12 litres/min x 120 min = 1440 litres
Rate of A = 1440 / 120 = 12 litres/min
Rate of B = 1440 / 180 = 8 litres/min
Combined rate = 12 + 8 = 20 litres/min
Time to fill 1200 litres = 1200 / 20 = 60 minutes
Answer: 60 minutes
Problem 21: Pipe Efficiency Ratio
Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both pipes are opened simultaneously, after how many hours should B be closed so that the tank is full in exactly 30 hours?
Solution (LCM Method):
LCM(36, 45) = 180 units
Rate of A = 180/36 = 5 units/hr
Rate of B = 180/45 = 4 units/hr
A works for all 30 hours. B works for t hours.
5 x 30 + 4 x t = 180
150 + 4t = 180
4t = 30
t = 7.5 hours
B should be closed after 7.5 hours (7 hours 30 minutes).
Answer: After 7 hours 30 minutes
Problem 22: Inlet, Outlet, and Leak
A tank has an inlet pipe that fills it in 10 hours and an outlet pipe that empties it in 20 hours. There is also a leak that empties the full tank in 40 hours. If all three are open, how long to fill the tank from empty?
Solution (LCM Method):
LCM(10, 20, 40) = 40 units
Rate of inlet = 40/10 = 4 units/hr (+)
Rate of outlet = 40/20 = 2 units/hr (-)
Rate of leak = 40/40 = 1 unit/hr (-)
Net rate = 4 - 2 - 1 = 1 unit/hr
Time = 40/1 = 40 hours
Answer: 40 hours
Problem 23: Filling and Draining Alternately
A tank has an inlet that fills it in 6 hours. First, the inlet runs for 1 hour, then it is closed and an outlet (empties in 8 hours) runs for 1 hour. This cycle repeats. How long to fill the tank?
Solution (LCM Method):
LCM(6, 8) = 24 units
Rate of inlet = 24/6 = 4 units/hr
Rate of outlet = 24/8 = 3 units/hr
Per cycle (2 hours):
Hour 1 (inlet): +4 units
Hour 2 (outlet): -3 units
Net per cycle = +1 unit
After n complete cycles (2n hours): n units filled
After 5 complete cycles (10 hours): 5 x 1 = 5 units
Hmm, this will take very long. Let me check when the inlet fills it during its hour.
After n complete cycles: n units
In the next hour (inlet): tank goes from n to n + 4
Tank is full (24 units) when: n + 4 >= 24 --> n >= 20
After 20 complete cycles (40 hours): 20 units
Hour 41 (inlet): 20 + 4 = 24 units -- FULL!
Total = 41 hours
But wait -- does the tank reach 24 during an inlet hour earlier?
After 20 cycles (40 hrs): 20 units. Inlet hour: 20 + 4 = 24. Yes, full at hour 41.
Actually, we should check: does the tank overflow during an inlet hour before cycle 20 completes?
After n cycles: n units. During next inlet hour: fills 4 more. If n + 4 >= 24, full.
n >= 20 is the first time. So hour 41 is correct.
Actually, let me reconsider: the net gain per cycle is 1 unit, starting from 0. After the inlet hour of each cycle:
End of inlet in cycle k: (k-1) + 4 = k + 3
End of outlet in cycle k: k + 3 - 3 = k
So after cycle k: k units.
Set k + 3 >= 24 (tank fills during inlet hour of cycle k):
k >= 21
End of cycle 20: 20 units.
Inlet hour of cycle 21 (hour 41): starts at 20, fills 4 units.
Time to fill remaining 4: 4/4 = 1 hour.
Tank reaches 24 at end of hour 41.
Total = 41 hours
Answer: 41 hours
Problem 24: Working Backwards from Combined Information
When pipe A and B work together, they fill a tank in 6 hours. When pipe A and C work together, they fill it in 10 hours. When A works alone, it takes 15 hours. Find the time for B and C working together.
Solution:
1/A + 1/B = 1/6 ... (i)
1/A + 1/C = 1/10 ... (ii)
1/A = 1/15
From (i): 1/B = 1/6 - 1/15 = 5/30 - 2/30 = 3/30 = 1/10
B = 10 hours
From (ii): 1/C = 1/10 - 1/15 = 3/30 - 2/30 = 1/30
C = 30 hours
B + C together: 1/10 + 1/30 = 3/30 + 1/30 = 4/30 = 2/15
Time = 15/2 = 7.5 hours = 7 hours 30 minutes
Answer: 7 hours 30 minutes
Problem 25: The Classic "Cistern with Multiple Operations" Problem
A cistern can be filled by pipe A in 4 hours and by pipe B in 6 hours. When full, it can be emptied by pipe C in 8 hours. If all three pipes are opened at 6 AM, at what time will the cistern be full?
Solution (LCM Method):
LCM(4, 6, 8) = 24 units
Rate of A = 24/4 = 6 units/hr (+)
Rate of B = 24/6 = 4 units/hr (+)
Rate of C = 24/8 = 3 units/hr (-)
Net rate = 6 + 4 - 3 = 7 units/hr
Time = 24/7 hours = 3 hours + 3/7 hours
= 3 hours + (3/7 x 60) minutes
= 3 hours + 25.7 minutes
= 3 hours 25 minutes 43 seconds (approx)
Starting at 6:00 AM:
Full at approximately 9:26 AM
For exact answer: 24/7 hours after 6 AM
= 6 AM + 3 hours 25 minutes 43 seconds
= 9:25:43 AM (approximately 9:26 AM)
Answer: 24/7 hours after 6 AM (approximately 9:26 AM)
Problem 26: Efficiency and Fraction of Work
Pipes A and B fill a tank in 8 hours and 12 hours respectively. They are opened together. After 3 hours, pipe A is turned off. How long does B take to fill the rest?
Solution (LCM Method):
LCM(8, 12) = 24 units
Rate of A = 24/8 = 3 units/hr
Rate of B = 24/12 = 2 units/hr
First 3 hours (A + B):
Work = (3 + 2) x 3 = 15 units
Remaining = 24 - 15 = 9 units
B alone: Time = 9/2 = 4.5 hours = 4 hours 30 minutes
Total time = 3 + 4.5 = 7.5 hours = 7 hours 30 minutes
Answer: B takes 4 hours 30 minutes more. Total time is 7 hours 30 minutes.
Problem 27: Proportional Filling
Two pipes A and B can fill a tank in 20 hours and 30 hours respectively. Both are opened together. What fraction of the tank is filled by pipe A?
Solution:
Rate of A = 1/20
Rate of B = 1/30
Since both run for the same duration, the fraction each fills is proportional to its rate.
Ratio of rates = 1/20 : 1/30 = 30 : 20 = 3 : 2
Fraction filled by A = 3/(3+2) = 3/5
Fraction filled by B = 2/(3+2) = 2/5
Answer: Pipe A fills 3/5 of the tank.
Problem 28: Increasing Number of Pipes
A pipe fills 1/4 of a tank in 1 hour. After that, a second identical pipe is also opened. After another hour, a third identical pipe is opened, and so on. How long to fill the entire tank?
Solution:
Each pipe fills 1/4 of the tank per hour.
Hour 1: 1 pipe --> fills 1/4
Hour 2: 2 pipes --> fills 2/4 = 1/2
Hour 3: 3 pipes --> fills 3/4
After hour 1: filled = 1/4
After hour 2: filled = 1/4 + 2/4 = 3/4
After hour 3: filled = 3/4 + 3/4 = 6/4 > 1
So the tank overflows during hour 3.
At start of hour 3: 3/4 filled. Remaining = 1/4.
3 pipes work at 3/4 per hour.
Time = (1/4) / (3/4) = 1/3 hour = 20 minutes.
Total = 2 hours + 20 minutes = 2 hours 20 minutes.
Answer: 2 hours 20 minutes
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