Problem 8 · AMC 8 Stretch
Core
Arithmetic & Operations
Algebra & Patterns
make-a-running-total-tabletrack-the-minimum-and-maximum
Fuel flows steadily into a tank at \(2{,}000\) liters per hour. The day is split into six \(4\)-hour periods. During those periods the tank uses \(6{,}000\), \(13{,}500\), \(7{,}300\), \(10{,}000\), \(8{,}000\), and \(3{,}200\) liters, in that order. Each day repeats the same pattern. What is the capacity (in liters) of the smallest tank that can always keep at least \(200\) liters of fuel inside?
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Answer: 7,000 liters
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Hint 1 of 4
First, how much fuel flows IN during one 4-hour period? It's \(4\times2000=8000\) liters every period.
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Hint 2 of 4
For each period, the net change is (inflow \(8000\)) minus (that period's usage). Make a table and keep a running total, starting from some unknown amount \(x\) at the beginning of the day.
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Hint 3 of 4
After all six periods, find the lowest running total and the highest running total. The lowest must stay at or above 200; that tells you the smallest starting amount \(x\).
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Approach: Running-total table, then bound by the lowest and highest levels
- Each 4-hour period brings in \(4\times2000=8000\) liters. Let \(x\) be the amount at the start of the day; the net change in a period is \(8000\) minus the usage.
- Track the running total:
Period Usage Net (8000−usage) Tank after 1 6000 +2000 x+2000 2 13500 −5500 x−3500 3 7300 +700 x−2800 4 10000 −2000 x−4800 5 8000 0 x−4800 6 3200 +4800 x - The lowest the tank ever gets is \(x-4800\). To keep at least 200 liters: \(x-4800\ge200\Rightarrow x\ge5000\).
- Using the smallest allowed start \(x=5000\), the highest the tank ever gets is \(x+2000=7000\). The tank must hold that peak, so the smallest workable capacity is \(7000\) liters.
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