Problem 19 · 2025 AMC 8
Hard
Ratios, Rates & Proportions
distance-speed-time
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Answer: D — 8.5 miles from A.
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Hint 1 of 2
Guess where they meet first: the middle section, where both happen to drive 40 mph. There, equal speeds mean they close the gap evenly — all the trickiness is in how unevenly they reach the middle.
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Hint 2 of 2
Time A to the middle: 5/25 = 1/5 hr. Time B to the middle: 5/20 = 1/4 hr. So A enters the middle 1/20 hr early — turn that head start into miles, then split what's left.
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Approach: find each car's position when both are in the equal-speed middle section
- Bet that they meet in the middle 40-mph section (then they go the same speed, so you only need positions). A covers its 5-mile left section in 5/25 = 1/5 hr; B covers its 5-mile right section in 5/20 = 1/4 hr. So A reaches the middle 1/4 − 1/5 = 1/20 hr before B.
- In that 1/20-hr head start, A drives 40 × 1/20 = 2 miles into the middle. So when B finally enters, A is at mile 7, B at mile 10 — a 3-mile gap, both now at 40 mph.
- Equal speeds split the gap evenly: each drives 1.5 more miles. A's meeting point is 7 + 1.5 = 8.5 miles from A.
- Why this transfers: when speeds match over the stretch where they meet, the closing is symmetric — so isolate the asymmetry (the unequal times to arrive), convert it to a distance, then share the remainder equally.
Another way — track total time to the meeting instant:
- They meet somewhere in the middle section at the same clock time t. Suppose A has gone a distance d into the middle and B has gone 5 − d into it (their middle distances fill the 5-mile section).
- A's time: 5/25 + d/40. B's time: 5/20 + (5 − d)/40. Set equal: 1/5 + d/40 = 1/4 + (5 − d)/40.
- Solving, 2d/40 = 1/4 − 1/5 + 5/40 = 1/20 + 1/8, giving d = 3.5. A's distance from A is 5 + 3.5 = 8.5 miles.
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