Problem 8 · AMC 8 Stretch
Stretch
Logic & Word Problems
Counting & ProbabilityArithmetic & Operations
logical-reasoningorganizing-datapattern-recognition
A senator must meet five groups, one at a time; while a group is being seen, everyone in later groups waits. Group 1 has 4 members and meets 20 min; group 2 has 8 members, 10 min; group 3 has 5 members, 30 min; group 4 has 10 members, 15 min; group 5 has 6 members, 25 min. In what order should he call the groups to make the total waiting time of all people as small as possible?
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Answer: G2, G4, G5, G1, G3 (increasing time-per-member)
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Hint 1 of 4
This is like the dentist problem, but each 'patient' is a whole group of people who all wait together.
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Hint 2 of 4
A group that takes a long time but has few people isn't so bad. What matters is the time PER PERSON in the group.
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Hint 3 of 4
Compute time divided by members for each group.
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Approach: Schedule by smallest time-per-member first
- A group of \(g\) people meeting for \(t\) minutes acts like \(g\) people who each weigh \(t/g\) minutes. So compare time per member, then go smallest-first.
- Time per member: \(G_1 = 20/4 = 5\); \(G_2 = 10/8 = 1.25\); \(G_3 = 30/5 = 6\); \(G_4 = 15/10 = 1.5\); \(G_5 = 25/6 \approx 4.17\).
- Smallest to largest: \(G_2 (1.25), G_4 (1.5), G_5 (4.17), G_1 (5), G_3 (6)\).
- So the best order is G2, G4, G5, G1, G3. (Swapping any neighbor pair shows the earlier group should have the smaller time-per-person, which forces this whole order.)
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