The Rooftop Tank
A rooftop tank is fed by two intake pipes and drained by a third. Pipe A alone fills the empty tank in 6 hours. Pipe B alone fills it in 9 hours. Drain C alone empties a full tank in 18 hours. All three run at steady rates and none is ever partly closed. The caretaker, Mr. Lim, needs the tank full when the building's three-hour afternoon supply cutoff ends, and he wants to know how the tank behaves under different combinations before he commits to a schedule.
1. If the tank starts empty and all three openings are left fully open, how long until the tank is full? Does the answer support Mr. Lim's fear that the drain makes filling impossible?
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Work in tank-per-hour rates.
- A: 1/6 tank per hour
- B: 1/9 tank per hour
- C (draining): −1/18 tank per hour
Combine over a common denominator of 18:
3/18 + 2/18 − 1/18 = 4/18 = 2/9 tank per hour
The net rate is positive, so the tank does fill. Time = 1 ÷ (2/9) = 4.5 hours.
Answer: 4.5 hours. Mr. Lim's fear is unfounded - the two intakes together add 5/18 per hour, far more than the 1/18 the drain removes. The drain slows filling to two-thirds of the A+B rate (4.5 h instead of 3.6 h) but cannot reverse it.
2. Mr. Lim instead runs A and B together for 2 hours, then shuts off A and opens C, leaving B and C running. How long in total until the tank is full?
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Phase 1 (0–2 h), A + B open: Rate = 1/6 + 1/9 = 3/18 + 2/18 = 5/18 per hour. In 2 hours: 2 × 5/18 = 10/18 = 5/9 of the tank is filled. Remaining: 1 − 5/9 = 4/9.
Phase 2, B + C open: Rate = 1/9 − 1/18 = 2/18 − 1/18 = 1/18 per hour. Time = (4/9) ÷ (1/18) = (4/9) × 18 = 8 hours.
Total = 2 + 8 = 10 hours.
This is the judgement call in the problem: the net rate after the switch is only 1/18 per hour, so the slow final stretch dominates the total. A student who adds the two rates without recomputing the remaining fraction will get a wrong answer.
3. Mr. Lim wants the tank to fill in exactly 3 hours, with B and C open the whole time. What fill-time must pipe A have for this to work? Is such a pipe realistic given the other two?
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Set the target rate. Filling in 3 hours requires 1/3 tank per hour.
Rate already supplied by B and C: 1/9 − 1/18 = 1/18 per hour.
Rate A must supply: 1/3 − 1/18 = 6/18 − 1/18 = 5/18 per hour.
Fill time for A alone = 1 ÷ (5/18) = 18/5 = 3.6 hours.
Answer: A would need to fill the tank alone in 3.6 hours.
Sanity check: 3.6 h is faster than either existing intake (6 h and 9 h) but not absurdly so - roughly 1.7 times B's speed. The key structural point is that the drain C removes only 1/18 per hour, so nearly all of the required 1/3 must come from A and B together; asking one pipe to carry 5/18 on its own is a demanding specification, not a routine one.