Problem 3 · 2015 AMC 8
Easy
Ratios, Rates & Proportions
distance-speed-time
Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of 10 miles per hour. Jack walks to the pool at a constant speed of 4 miles per hour. How many minutes before Jack does Jill arrive?
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Answer: D — 9 minutes.
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Hint 1 of 2
The distance is the same for both (1 mile), so don't compute any distance — the only thing that differs is how long each takes. Find both times, then subtract.
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Hint 2 of 2
Reading 'miles per hour' as 'miles per 60 minutes' turns time into a one-step division: minutes for one mile = 60 ÷ speed.
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Approach: same distance — just compare the two travel times
- Both travel 1 mile, so all that matters is each person's time. At v mph one mile takes 60/v minutes (since 1 hour = 60 min).
- Jill: 60/10 = 6 min. Jack: 60/4 = 15 min.
- Jill arrives 15 − 6 = 9 minutes earlier.
- Sanity check: Jack is slower, so he takes longer — the slower person's time should be the bigger number, and 15 > 6. ✓
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