Problem 6 · AMC 8 Stretch
Core
Logic & Word Problems
Counting & ProbabilityArithmetic & Operations
organizing-datalogical-reasoning
Six patients A, B, C, D, E, F wait at a dentist. Their treatment times are A = 15 min, B = 30 min, C = 10 min, D = 10 min, E = 20 min, F = 5 min. The dentist wants the total waiting time of all patients added together to be as small as possible. What is that smallest possible total waiting time (in minutes)?
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Answer: 145 minutes (treat shortest first: F, C, D, A, E, B)
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Hint 1 of 4
The first patient seen waits 0 minutes. The second waits through the first one's treatment. Everyone waits through everyone seen before them.
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Hint 2 of 4
Whoever goes FIRST makes all 5 others wait through their treatment. So a long treatment early on is very costly.
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Hint 3 of 4
To keep the total small, put the shortest treatments first and the longest last.
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Approach: Shortest-job-first scheduling
- The first patient's time is waited through by all 5 others, the second's by 4, then 3, 2, 1, 0. To make the total small, the biggest counts should multiply the smallest times — so treat the shortest patient first.
- The times in order are 5, 10, 10, 15, 20, 30, which is F, C, D, A, E, B.
- Total waiting time = \(5(5) + 4(10) + 3(10) + 2(15) + 1(20) + 0(30) = 25 + 40 + 30 + 30 + 20 = 145\) minutes.
- So the minimum total is 145 minutes (2 h 25 min). The two 10-minute patients can swap, so the best order is not unique.
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