13. Pipes and Cistern problems

 13. Pipes and Cistern problems

Home page of Arithmetic Problems 

Below are 10 pipes and cisterns problems, each with a step-by-step solution procedure and detailed explanation. Pipes and cisterns problems are a subset of work rate problems, focusing on the rate at which pipes fill or empty tanks (cisterns). The key concept is work rate, where the rate is the fraction of the tank filled or emptied per unit of time (e.g., tanks per hour). The standard formulas are:
  • Work (tank fraction) = Rate × Time
  • Rate = Work / Time
  • Time = Work / Rate
  • Filling pipe rate: Positive (e.g., 1/10 tank/hour if it fills in 10 hours)
  • Emptying pipe/leak rate: Negative (e.g., -1/15 tank/hour if it empties in 15 hours)
  • Combined rate: Sum of individual rates (positive for filling, negative for emptying)
Work is typically normalized to 1 tank (full tank = 1). Problems may involve multiple filling pipes, emptying pipes, leaks, staggered starts, or partial filling. Units are usually hours or minutes for time, and the tank’s capacity is assumed as 1 unless specified. Each problem explores different scenarios for a comprehensive understanding.
Problem 1: Two Filling Pipes
Pipe A fills a tank in 12 hours, and Pipe B fills it in 15 hours. How long will they take to fill the tank together?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s time = 12 hours, rate = 1/12 tank/hour
    • Pipe B’s time = 15 hours, rate = 1/15 tank/hour
  2. Calculate combined rate: 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20 tank/hour
  3. Calculate time: Time = 1 / (3/20) = 20/3 ≈ 6.667 hours
  4. State result: They take approximately 6.67 hours.
Detailed Explanation: Each pipe’s rate is the reciprocal of its filling time. The combined rate is their sum, reflecting both pipes filling simultaneously. The time to fill one tank is the reciprocal of the combined rate, faster than either pipe alone due to increased efficiency. The LCM (60) simplifies fraction addition.
Answer: ≈ 6.67 hours
Problem 2: Filling Pipe and Leak
Pipe A fills a tank in 10 hours, and a leak empties it in 15 hours. How long to fill the tank if both are open?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/10 tank/hour
    • Leak’s rate = -1/15 tank/hour (negative for emptying)
  2. Calculate net rate: 1/10 - 1/15 = 3/30 - 2/30 = 1/30 tank/hour
  3. Calculate time: Time = 1 / (1/30) = 30 hours
  4. State result: The tank takes 30 hours to fill.
Detailed Explanation: The leak opposes the filling, so its rate is negative. The net rate is the difference, showing the effective filling rate. The time is longer than Pipe A alone (10 hours), as the leak slows the process. This introduces opposing rates, a common cistern problem feature.
Answer: 30 hours
Problem 3: Three Filling Pipes
Pipe A fills a tank in 8 hours, Pipe B in 12 hours, and Pipe C in 24 hours. How long to fill the tank if all are open?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/8 tank/hour
    • Pipe B’s rate = 1/12 tank/hour
    • Pipe C’s rate = 1/24 tank/hour
  2. Calculate combined rate: 1/8 + 1/12 + 1/24 = 3/24 + 2/24 + 1/24 = 6/24 = 1/4 tank/hour
  3. Calculate time: Time = 1 / (1/4) = 4 hours
  4. State result: They take 4 hours.
Detailed Explanation: The combined rate is the sum of all three pipes’ rates, simplified using the LCM (24). The time is the reciprocal, significantly less than any individual time due to the three pipes’ combined effort. This extends the two-pipe scenario to multiple contributors.
Answer: 4 hours
Problem 4: Staggered Start
Pipe A fills a tank in 6 hours, and Pipe B in 9 hours. Pipe A runs alone for 2 hours, then Pipe B joins. How long to fill the tank?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/6 tank/hour
    • Pipe B’s rate = 1/9 tank/hour
  2. Calculate A’s work in 2 hours: 1/6 × 2 = 2/6 = 1/3 tank
  3. Calculate remaining work: 1 - 1/3 = 2/3 tank
  4. Calculate combined rate: 1/6 + 1/9 = 3/18 + 2/18 = 5/18 tank/hour
  5. Calculate time for remaining work: Time = (2/3) / (5/18) = 2/3 × 18/5 = 12/5 = 2.4 hours
  6. Calculate total time: 2 + 2.4 = 4.4 hours
  7. State result: The tank takes 4.4 hours to fill.
Detailed Explanation: Pipe A’s initial work reduces the tank to be filled. The remaining work is done at the combined rate. Total time includes A’s solo period and the joint period, showing how staggered starts affect completion. Verification: A’s work (1/3) + combined work (2.4 × 5/18 = 2/3) = 1 tank.
Answer: 4.4 hours
Problem 5: Two Pipes and One Leak
Pipe A fills a tank in 15 hours, Pipe B in 20 hours, and a leak empties it in 30 hours. How long to fill the tank if all are open?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/15 tank/hour
    • Pipe B’s rate = 1/20 tank/hour
    • Leak’s rate = -1/30 tank/hour
  2. Calculate net rate: 1/15 + 1/20 - 1/30 = 4/60 + 3/60 - 2/60 = 5/60 = 1/12 tank/hour
  3. Calculate time: Time = 1 / (1/12) = 12 hours
  4. State result: The tank takes 12 hours to fill.
Detailed Explanation: The net rate combines the filling rates (positive) and the leak’s rate (negative). The LCM (60) simplifies the calculation. The time is longer than without the leak, showing its impact, but shorter than individual filling times due to two pipes.
Answer: 12 hours
Problem 6: Finding Leak’s Time
Pipe A fills a tank in 8 hours. With a leak, it takes 10 hours. How long would the leak take to empty a full tank?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/8 tank/hour
    • Combined time (A + leak) = 10 hours, net rate = 1/10 tank/hour
  2. Set up equation: A’s rate + Leak’s rate = Net rate
    • 1/8 + Leak’s rate = 1/10
    • Leak’s rate = 1/10 - 1/8 = 2/80 - 2.5/80 = -0.5/80 = -1/160 tank/hour
  3. Calculate leak’s time: Time = 1 / (1/160) = 160 hours
  4. State result: The leak takes 160 hours to empty the tank.
Detailed Explanation: The leak’s rate is negative, reducing the net filling rate. Subtracting A’s rate from the net rate gives the leak’s rate, and the reciprocal is the time to empty the tank. The large time reflects the leak’s slow effect, consistent with the small rate difference.
Answer: 160 hours
Problem 7: Partial Filling
Pipe A fills a tank in 20 hours, and Pipe B in 30 hours. They run together for 5 hours, then B stops. How long does A take to fill the rest?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/20 tank/hour
    • Pipe B’s rate = 1/30 tank/hour
  2. Calculate combined rate: 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12 tank/hour
  3. Calculate work in 5 hours: 1/12 × 5 = 5/12 tank
  4. Calculate remaining work: 1 - 5/12 = 7/12 tank
  5. Calculate A’s time for remaining work: Time = (7/12) / (1/20) = 7/12 × 20 = 35/3 ≈ 11.667 hours
  6. State result: Pipe A takes approximately 11.67 hours.
Detailed Explanation: The combined work reduces the tank to be filled. A then fills the remaining portion at its rate. The problem focuses on the time after B stops, but total time could be 5 + 11.67 = 16.67 hours if needed. This shows partial collaboration effects.
Answer: ≈ 11.67 hours
Problem 8: Alternating Operation
Pipe A fills a tank in 6 hours, and Pipe B in 8 hours. They run alternately for 1 hour each, starting with A. How long to fill the tank?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/6 tank/hour
    • Pipe B’s rate = 1/8 tank/hour
  2. Calculate work in one cycle (2 hours, A for 1 hour, B for 1 hour):
    • A’s work = 1/6 × 1 = 1/6 tank
    • B’s work = 1/8 × 1 = 1/8 tank
    • Cycle work = 1/6 + 1/8 = 4/24 + 3/24 = 7/24 tank
  3. Calculate cycles for most work: 3 cycles (6 hours) = 3 × 7/24 = 21/24 = 7/8 tank
  4. Calculate remaining work: 1 - 7/8 = 1/8 tank
  5. Next is A’s turn (hour 7): Time for 1/8 tank = (1/8) / (1/6) = 1/8 × 6 = 3/4 hour
  6. Calculate total time: 6 + 3/4 = 6.75 hours
  7. State result: The tank takes 6.75 hours to fill.
Detailed Explanation: Alternating pipes require calculating work per cycle (2 hours). Three cycles nearly fill the tank, and A finishes the small remainder. The fractional hour (3/4) reflects precise completion, showing how alternating schedules differ from simultaneous work.
Answer: 6.75 hours
Problem 9: Emptying a Full Tank
Pipe A fills a tank in 5 hours, and Pipe B empties it in 7 hours. If the tank is full and both pipes are open, how long to empty it?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/5 tank/hour (filling)
    • Pipe B’s rate = -1/7 tank/hour (emptying)
  2. Calculate net rate: 1/5 - 1/7 = 7/35 - 5/35 = 2/35 tank/hour (positive, so filling)
  3. Analyze: Since the net rate is positive, the tank won’t empty with both open; it will fill.
  4. Correct interpretation: Assume Pipe B alone empties the full tank.
    • Pipe B’s time = 7 hours (as given)
  5. State result: Pipe B takes 7 hours to empty the full tank.
Detailed Explanation: The problem’s phrasing suggests emptying, but both pipes open result in filling due to A’s higher rate. Assuming the intent is Pipe B alone (common in such problems), the answer is straightforward. If both were meant, the tank wouldn’t empty, highlighting the need to check net rates.
Answer: 7 hours (Pipe B alone)
Problem 10: Time to Fill a Fraction
Pipe A fills a tank in 4 hours, Pipe B in 6 hours, and a leak empties it in 12 hours. How long to fill 3/4 of the tank if all are open?
Step-by-Step Solution:
  1. Identify given values:
    • Pipe A’s rate = 1/4 tank/hour
    • Pipe B’s rate = 1/6 tank/hour
    • Leak’s rate = -1/12 tank/hour
  2. Calculate net rate: 1/4 + 1/6 - 1/12 = 3/12 + 2/12 - 1/12 = 4/12 = 1/3 tank/hour
  3. Calculate time for 3/4 tank: Time = (3/4) / (1/3) = 3/4 × 3 = 9/4 = 2.25 hours
  4. State result: It takes 2.25 hours to fill 3/4 of the tank.
Detailed Explanation: The net rate accounts for two filling pipes and the leak. The time to fill 3/4 of the tank is the desired work divided by the net rate, showing how partial tank problems scale linearly with work. Verification: In 2.25 hours, 1/3 × 2.25 = 3/4 tank.
Answer: 2.25 hours
Summary
These 10 pipes and cisterns problems cover two and three filling pipes, leaks, staggered starts, multiple pipes with leaks, finding leak times, partial filling, alternating operation, emptying, and filling fractions. Each uses work rate principles, with rates added for filling and subtracted for emptying. Step-by-step solutions ensure clarity, and detailed explanations highlight net rates, staggered effects, and pitfalls (e.g., misinterpreting emptying). These problems build a strong understanding of pipes and cisterns, applicable to other work rate scenarios. 

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