Problem 6 · AMC 8 Stretch
Core
Ratios, Rates & Proportions
Algebra & Patterns
symmetryconsidering-extreme-caseslogical-reasoning
Two hikers need to reach a village 12 km away as fast as possible. They walk at 4 km/h and ride a bike at 12 km/h, but they have only one bike, carrying one rider. Using the smartest plan, what is the shortest time (in hours) for BOTH of them to arrive?
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Answer: 2 hours (average speed 6 km/h)
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Hint 1 of 4
The two hikers are identical, so the fair, symmetric plan is to share the bike equally and have both arrive at the same moment.
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Hint 2 of 4
Hiker 1 rides, then leaves the bike on the road and walks the rest. Hiker 2 walks until he reaches the parked bike, then rides. Where should the bike be dropped so they tie?
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Hint 3 of 4
Drop the bike exactly at the halfway point, 6 km. Then each rides 6 km and walks 6 km. Compute the time for one hiker: \(\tfrac{\text{riding}}{12} + \tfrac{\text{walking}}{4}\).
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Approach: Symmetric bike-sharing: each rides half, walks half
- Since both hikers have the same speeds, the smartest plan is symmetric: split the biking equally so they arrive together.
- Hiker 1 rides the first 6 km, leaves the bike, and walks the last 6 km. Hiker 2 walks the first 6 km, finds the parked bike, and rides the last 6 km. Each rides 6 km and walks 6 km, so by symmetry they arrive together.
- For either hiker: \(T = \tfrac{6}{12} + \tfrac{6}{4} = 0.5 + 1.5 = 2\) hours.
- So both arrive in 2 hours. (Average speed \(= \tfrac{12}{2} = 6\) km/h, the harmonic mean of 4 and 12.)
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