Problem 11 · 2013 AMC 8
Medium
Ratios, Rates & Proportions
time-equals-distance-over-rate
Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
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Answer: D — 4 minutes.
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Hint 1 of 2
Same 2 miles every day, so time is decided entirely by speed: time = distance ÷ rate. Slower = longer, faster = shorter. The "always 4 mph" plan is your benchmark.
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Hint 2 of 2
You don't even need all three days. Friday was already 4 mph (no change). Monday at 5 mph was faster than 4, so it would actually take more time at 4 — the real saving lives only where he went slower than 4.
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Approach: compare each real day to the 4-mph benchmark
- Distance is fixed at 2 miles, so time = 2 ÷ rate. At 4 mph the benchmark is 2/4 hr = 30 min per day.
- Friday was 4 mph — already on benchmark, 0 difference. Monday at 5 mph took 2/5 hr = 24 min, which is 6 min less than 30 (so "always 4" would cost +6 there).
- Wednesday at 3 mph took 2/3 hr = 40 min, which is 10 min more than 30.
- Net time saved by going all-4-mph = +6 (Mon) − 10 (Wed) + 0 (Fri)… flip sign for "less time": 10 − 6 = 4 minutes less.
- Check the totals: actual 24 + 40 + 30 = 94 min; all-4 plan 3 × 30 = 90 min; 94 − 90 = 4. Same answer.
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