Problem 17 · 2015 AMC 8
Medium
Algebra & Patterns
distance-speed-time
Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive him 18 miles per hour faster and gets him to school in 12 minutes. How far in miles is it to school?
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Answer: D — 9 miles.
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Hint 1 of 2
The two trips cover the same distance — that's the anchor. Write that one distance two ways (each as speed × time) and set them equal; the matching distances kill the unknown distance and leave you solving for the speed.
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Hint 2 of 2
Turn the minutes into hours so speed × time gives miles: 20 min = 1/3 h, 12 min = 1/5 h. Then solve s · (1/3) = (s + 18) · (1/5).
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Approach: the distance is the same both days — equate the two expressions
- Let the rush-hour speed be s mph. Same distance both days: s · (1/3) = (s + 18) · (1/5).
- Clear fractions (×15): 5s = 3(s + 18) ⇒ 2s = 54 ⇒ s = 27.
- Distance = 27 · (1/3) = 9 miles.
- Why this transfers: 'two ways to cover the same distance' is a classic setup — equate speed×time for both and the shared quantity drops out.
Another way — fixed distance → speed is inversely proportional to time:
- For a fixed distance, faster means proportionally less time: the time ratio 20 : 12 = 5 : 3 flips to a speed ratio of 3 : 5.
- So speeds are 3 and 5 'parts'; the 18-mph boost is the 5 − 3 = 2-part gap, making 1 part = 9 mph. Rush-hour speed = 3 parts = 27 mph.
- Distance = 27 mph × (1/3 h) = 9 miles — same answer with almost no algebra.
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