Pipes & Cisterns
Similar to Time & Work, but with a twist: Negative Work. Master Inlet (Filling) vs Outlet (Emptying) logic and 'Leakage' problems.
1. Inlet vs Outlet (Sign Convention)
Direction matters.
- Inlet (Fill): Positive Work (+).
- Outlet (Empty): Negative Work (-).
- Net Rate: Rate(In) - Rate(Out).
Example:
Q: A fills in 10h, B empties in 15h. Together?
Solution: Total=30. A=+3, B=-2. Net=+1. Time = 30/1 = 30h.
2. Leak at Bottom
If a leak increases filling time.
- Method: Find Leak Efficiency by subtraction.
- . Usually Net is slower.
Example:
Q: Pipe A fills in 4h. Due to leak, takes 5h.
Solution: TW=20. A=+5. Net(A-L)=+4. Why? Slower.
. Leak Time = 20/1 = 20h.
. Leak Time = 20/1 = 20h.
3. Capacity Calculation
To find Total Liters.
- Logic: Capacity = Rate of Pipe (L/min) Time taken by THAT Pipe (min).
- First find Time of Pipe X using Ratios, then multiply by its tangible rate.
Example:
Q: Leak empties in 10h. Inlet adds 4L/min. Full tank empties in 15h.
Solution: Leak=-3, Net=-2. Inlet=+1. Inlet Time=30h. .
4. Alternating Pipes
Positive and Negative cycles.
- Trap: Tank fills only at the END of a positive cycle. Do not divide directly.
- Hack: Subtract last positive work from Total. Divide rest by Cycle. Then add last step manually.
Example:
Q: A(+3) 1h, B(-2) 1h. TW=10.
Solution: Cycle(2h) = +1. Target .
7 cycles (14h) 7 filled.
Next hour A adds 3 Full. Total 15h.
7 cycles (14h) 7 filled.
Next hour A adds 3 Full. Total 15h.
5. Multiple Pipes Logic
Complex combinations.
- Rule: Sum all efficiencies algebraically.
- If Net Eff is negative, tank will EMPTY (Total Work = Initial Fills).
Example:
Q: A(+5), B(+4), C(-10). Full tank. Open all.
Solution: Net = 5+4-10 = -1.
Time = Capacity / 1. Tank empties.
Time = Capacity / 1. Tank empties.
6. Time Lag (Opening Differently)
Pipes open at different times.
- Method: Calculate partial work done before all open.
- Subtract from Total. Solve rest with Combined Efficiency.
7. Diameter & Flow Rule
Flow rate depends on pipe Cross-Section Area.
- Rule: Rate (square of diameter).
- If diameter doubles (), Flow becomes . Time becomes .
Example:
Q: Pipe A () takes 60m. Pipe B (). Time?
Solution: Dia Ratio 1:3. Area Ratio 1:9. Flow Ratio 1:9.
Time Ratio 9:1.
Time B = min.
Time Ratio 9:1.
Time B = min.
ЁЯОУ
Course4All Editorial Board
Verified ExpertSubject Matter Experts
Comprising experienced educators and curriculum specialists dedicated to providing accurate, exam-aligned preparation material.
Pattern: 2026 Ready
Updated: Weekly
Found an issue or have a suggestion?
Help us improve! Report bugs or suggest new features on our Telegram group.