Pipes and Cisterns — Study Notes

Overview

Pipes and cisterns is a staple time-and-work variant in Railway Group D mathematics. Instead of workers completing jobs, you deal with pipes filling or emptying tanks. The core principle is identical: work done = rate × time. A pipe's rate is expressed as "tank per hour" (or minute), and questions test your ability to combine multiple pipes working together or alternately, handle leaks, and compute net filling or emptying time.

Expect 1–2 questions on this topic in the exam. Mastery requires comfort with unitary method, LCM technique for assumed tank capacity, and sign conventions (inlet positive, outlet negative). Most problems are direct application once you set up the rate equation correctly. The tricky part is tracking which pipes are filling and which are draining, especially when both operate simultaneously or in cycles.

Focus on three skill areas: single-pipe time calculations, combined pipe scenarios (all inlets/outlets together), and alternating pipe schedules. With consistent practice, you can solve these in under 90 seconds per question.

Key Concepts

• **Rate of work analogy**: If a pipe fills a tank in *n* hours, its rate is 1/n tank per hour. If it empties in *m* hours, its rate is –1/m tank per hour.
• **Net rate for combined pipes**: Add individual rates algebraically. Inlet rates are positive, outlet (leak/drain) rates are negative. Net rate = (sum of inlet rates) – (sum of outlet rates).
• **LCM method for assumed capacity**: When pipes fill in *a*, *b*, *c* hours, assume tank capacity = LCM(a, b, c) litres. Then each pipe's efficiency is capacity/time in litres per hour. This avoids fractions in intermediate steps.
• **Time to fill/empty together**: If net rate is *R* tank/hour, time = 1/R hours. If net rate is negative, the tank empties; if positive, it fills.
• **Partial work in cycles**: When pipes operate alternately (e.g. A for 1 hour, B for 1 hour, repeat), calculate work done in one complete cycle, then see how many cycles are needed.
• **Pipe opened/closed mid-way**: Compute work done in each phase separately. Track remaining capacity after each phase and apply the next pipe's rate to that remainder.
• **Leak or waste pipe**: Treat as an outlet. If a leak empties the full tank in *L* hours, its rate is –1/L.
• **Multiple inlets and outlets**: Sum all inlet rates, sum all outlet rates, then subtract. The sign of the result tells you if the tank is filling or emptying.

Formulas / Key Facts

1. **Single pipe time**: If pipe fills in *T* hours, rate = 1/T tank/hour. If it empties in *T* hours, rate = –1/T tank/hour. 2. **Combined rate formula**: Net rate = (1/A + 1/B + …) – (1/X + 1/Y + …), where A, B are inlet times and X, Y are outlet times. 3. **Time for combined work**: Time = 1 / (Net rate). Units must match (hours, minutes, etc.). 4. **LCM capacity method**: Capacity = LCM(T₁, T₂, …). Efficiency of pipe filling in Tᵢ hours = Capacity / Tᵢ litres per hour. 5. **Work done = Rate × Time**. For partial filling/emptying, fraction filled = (rate × time). 6. **Cycle work**: If A and B alternate every hour, work in 2-hour cycle = (1/A + 1/B). Number of cycles = Tank capacity / cycle work. Add any remaining time for incomplete cycle. 7. **Capacity with leak**: If a full tank is emptied by a leak in *L* hours while an inlet fills in *F* hours, effective filling time = 1 / (1/F – 1/L). 8. **Partial opening**: If a pipe is opened for *t* hours only, work done = (1/T) × t, where T is the time to fill/empty the full tank alone.

Worked Examples

**Example 1**: Pipe A fills a tank in 6 hours, pipe B fills in 8 hours. Both opened together; find time to fill the tank.

• **Solution**:
• Rate of A = 1/6 tank/hour.
• Rate of B = 1/8 tank/hour.
• Combined rate = 1/6 + 1/8 = (4 + 3)/24 = 7/24 tank/hour.
• Time = 1 / (7/24) = 24/7 hours = 3 hours 25.7 minutes ≈ **3 hours 26 minutes**.

**Example 2**: Pipe C empties a full tank in 12 hours. Pipes A (fills in 6 hours) and C work together; find time to fill the empty tank.

• **Solution**:
• Rate of A = 1/6, rate of C = –1/12 (outlet).
• Net rate = 1/6 – 1/12 = (2 – 1)/12 = 1/12 tank/hour.
• Time = 1 / (1/12) = **12 hours**.

**Example 3** (LCM method): Pipe X fills in 4 hours, pipe Y in 5 hours, pipe Z (outlet) empties in 20 hours. All opened together; find time to fill.

• **Solution**:
• LCM(4, 5, 20) = 20 litres (assumed capacity).
• Efficiency of X = 20/4 = 5 L/hour.
• Efficiency of Y = 20/5 = 4 L/hour.
• Efficiency of Z = –20/20 = –1 L/hour (drains).
• Net efficiency = 5 + 4 – 1 = 8 L/hour.
• Time = 20/8 = **2.5 hours = 2 hours 30 minutes**.

**Example 4** (Alternating pipes): Pipe P fills in 10 hours, pipe Q fills in 15 hours. They are opened alternately for 1 hour each, starting with P. When will the tank be full?

• **Solution**:
• In 2-hour cycle: P works 1 hour → 1/10 filled; Q works 1 hour → 1/15 filled.
• Cycle work = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6 tank per 2 hours.
• After 6 cycles (12 hours): 6 × 1/6 = 1 tank filled.
• **Answer: 12 hours** (tank exactly full at end of 6th cycle).

Common Mistakes

1. **Sign error on outlets**: Forgetting to subtract the outlet rate or treating leak as positive. *Fix*: Always assign negative rate to any pipe that empties/leaks, then add algebraically.

2. **Unit mismatch**: Mixing hours and minutes without conversion (e.g. one pipe in 90 minutes, another in 2 hours). *Fix*: Convert all times to the same unit before calculating rates.

3. **Inverting the final answer**: Computing net rate correctly but forgetting time = 1/(net rate), writing the rate itself as the answer. *Fix*: Remember the rate is fraction of tank per hour; flip it to get hours to complete 1 tank.

4. **Ignoring partial cycles**: In alternating pipe problems, stopping at full cycles and missing that the last partial hour may complete the tank. *Fix*: After full cycles, check remaining work and see if the next pipe (in order) finishes it in less than 1 hour.

5. **Assuming simultaneous means continuous**: In some questions, pipes are opened together for part of the time, then one is closed. *Fix*: Break the problem into phases — calculate work done in each phase separately, track the remaining fraction, then apply the next phase's rate to that remainder.

Quick Reference

• **Single pipe rate**: 1/T tank per hour if fills in T hours; –1/T if empties in T hours.
• **Combined rate** = sum of all inlet rates minus sum of all outlet rates.
• **Time together** = 1 / (combined rate). Flip the rate fraction.
• **LCM trick**: Assume capacity = LCM of all pipe times to avoid fraction arithmetic.
• **Alternating pipes**: Calculate one full cycle work, divide 1 by cycle work to find number of cycles, then handle the remainder.
• **Always check units**: Convert minutes ↔ hours before computing. 1 hour = 60 minutes.