Problem 17 · 2014 AMC 8
Medium
Ratios, Rates & Proportions
distance-speed-time
George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first 12 mile at a speed of only 2 miles per hour. At how many miles per hour must George run the last 12 mile in order to arrive just as school begins today?
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Answer: B — 6 mph.
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Hint 1 of 2
Time is the fixed budget here, not speed. He must arrive at the usual moment, so his total travel time is locked at the normal value — figure that out first.
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Hint 2 of 2
Subtract the time the slow first half ate up; whatever's left is all he has to cover the last half-mile. Speed = distance ÷ that leftover time.
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Approach: time is the fixed budget — subtract what's spent
- Normal trip: 1 mile at 3 mph takes 1/3 hr. That's the total time budget he must hit today.
- Slow first half: (1/2 mile) ÷ (2 mph) = 1/4 hr used up.
- Time left for the second half-mile: 1/3 − 1/4 = 4/12 − 3/12 = 1/12 hr.
- Required speed = (1/2 mile) ÷ (1/12 hr) = 6 mph.
- Why this transfers: in "arrive on time" problems, hold time constant and treat it as a budget. Don't average the speeds — dawdling on the first half costs disproportionately more time, which is why he must nearly triple his pace, not just speed up a little.
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