Pipes and Cisterns

Module 04 — Lesson 0004 · In Depth · 35 min

A tank is a work-rate problem with signs — inlets do positive work (+), outlets/leaks do negative work (−). The LCM method turns fractions 1/x into clean integer rates so you never solve with fractions under time pressure.

1. The Basic Concept: Positive vs Negative Work

Fundamental rule: If A fills in x hours and B empties in y hours:

Rate A = +1/x, Rate B = −1/y, Net = 1/x − 1/y

2. Methods to Solve

Method 1: The LCM (Capacity) Method — Recommended

Instead of fractions like 1/20, assume a concrete capacity:

  1. Assume capacity: LCM of all time values (in units/liters).
  2. Find flow rate: Rate = Capacity / Time — assign negative to outlets.
  3. Net rate: sum the signed rates.
  4. Time: Capacity / Net rate
Worked: Inlet A fills in 10 min, Outlet B empties in 15 min.
Capacity LCM(10,15)=30 units → A=+3 u/min (30/10), B=−2 u/min (30/15) → net +1 u/min → time to fill =30/1=30 min.
Fractional method would be 1/10−1/15=1/30 → 30 min — same answer, but LCM keeps everything integer.

Method 2: Equation Method (For Variable Times)

If A fills in x and B in x+10, together in T:

1/x + 1/(x+10) = 1/T → solve quadratic for x. Use LCM method when times are fixed numbers; use equation when one time is expressed in terms of another.

3. Handling Complex Scenarios

A. Leakage Problems

Leak = outlet. Normally fills in 10h but takes 12h due to leak →

1/10 − Rate(Leak) = 1/12Rate(Leak)=1/10−1/12=1/60 → leak empties full tank in 60h.

Pattern: "takes longer" → subtract leak rate; "empties in y hours alone" → that's the leak's time.

B. Alternating Pipes — Cycle Trap

Pipes open one after another (e.g., A 1 min, then B 1 min):

Safe approach: Subtract the positive work of the last minute from total, compute cycles for the remainder, then handle last fill manually.
Worked: A +5u/min, B −2u/min, capacity 30. Cycle (2 min)=+3u. Naive 30/3=10 cycles=20 min is wrong if +5 fills at 27→32 and overshoots. Correct: reserve last A (+5): need 25 in cycles → 25/3=8 remainder 1 → 8×2=16 min + last 1 min of A (+5 but only need 5 to reach 30) → total 17 min, not 20. Always reserve the last positive chunk.

C. Pipes Opened at Different Times

A opens at 8:00, B at 9:00:

  1. Work done by A alone in 1h.
  2. Remaining capacity = Total − that work.
  3. Divide remaining by combined rate (A+B) → time after 9:00.

D. Efficiency Changes (Variable Rate)

If rate changes (e.g., "drops by 50%" after some time):

  1. Compute work in normal-efficiency period.
  2. Compute new rate for remaining period.
  3. Sum times for each phase.

4. Advanced Formulas

Capacity from Flow Rate

If outlet pumps 10 L/min and LCM method says it empties full tank in T minutes → total capacity = T × 10 L. Convert rate×time to actual volume.

5. Summary of Variables

Symbol Meaning
C Capacity (LCM of times, in units)
R Flow rate (units/min)
+ve / −ve Filling / emptying sign
T Total time

A fills +1/10, B empties −1/15 → net?

LCM method: A 10m, B 15m → capacity?

In that LCM, rates are?

Normally fills in 10h, with leak takes 12h → leak empties in?

Alternating A(+5) then B(−2), capacity 30 — blind cycle division fails because?

A (8am) + B (9am) → first hour work by?

Outlet pumps 10 L/min, empties full tank in 30 min → capacity?

Notes

Pipes :- Basics :-

Inlet ⇒ + ; Outlet/leak ⇒ − ; net = 1/x −1/y ; +→fills, −→empties

LCM Method :-

Capacity ⇒ LCM(times) ; Rate ⇒ Capacity/Time (sign matters) ; Time ⇒ Capacity/net

(eg A10,B15→30 units→+3,−2→net+1→30 min)

Complex :-

a) Leak ⇒ 1/10−x=1/12 → x=1/60 → empties 60h

b) Alternating ⇒ cycle=+ sum but reserve last +fill; don’t blind divide

c) Staggered ⇒ solo work → remaining/COMBINED

d) Variable rate ⇒ phase-wise work then sum times

Capacity from flow :-

Capacity ⇒ T(empty)×rate (eg 30×10=300L)

Primary source: none pinned yet — drop your preferred video/book resource into RESOURCES.md and it will be linked here. Ask me anything that's unclear.

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