2025 AMC 8 第 19 题

先试着解答 2025 AMC 8 第 19 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2025 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

城镇 AA 和 BB 由一条长 1515 英里的直路连接。从 AA 到 BB 行驶时,每 55 英里限速变化一次:从 2525 到 4040 再到 2020 英里每小时。两辆车分别从 AA 和 BB 同时出发相向而行,并在每段路上都恰好按限速行驶。两车会在离城镇 AA 多少英里处相遇?
Two towns, AA and B,B, are connected by a straight road, 1515 miles long. Traveling from town AA to town B,B, the speed limit changes every 55 miles: from 2525 to 4040 to 2020 miles per hour (mph). Two cars, one at town AA and one at town B,B, start moving toward each other at the same time. They drive at exactly the speed limit in each portion of the road. How far from town A,A, in miles, will the two cars meet?
7.757.75
88
8.258.25
8.58.5
8.758.75
答案:D
知识点:相对速度路程、速度与时间
小提示:
求每辆车何时到达中间 55 英里路段
Find when each car reaches the middle 55-mile section
大提示:
两辆车都在中间路段后,它们速度相同
Once both cars are in the middle section, their speeds are equal
视频讲解:
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文字解答:
把道路看作左、中、右三段。每段长 55 英里。
从 AA 出发的车到达中间路段需要 525=15 \frac{5}{25} = \frac{1}{5} 小时。
从 BB 出发的车到达中间路段需要 520=14 \frac{5}{20} = \frac{1}{4} 小时。到那时,从 AA 出发的车已经在中间路段行驶了 14−15=120 \frac{1}{4} - \frac{1}{5} = \frac{1}{20} 小时。这辆从 AA 出发的车在中间路段已经行驶了 40×120=2 40 \times \frac{1}{20} = 2 英里。
于是中间路段中两车之间还剩 5−2=35 - 2 = 3 英里。
此时从 AA 出发的车距离 AA 为 5+2=75 + 2 = 7 英里。
由于两车在中间路段都以每小时 4040 英里的速度行驶,它们再各行 1.51.5 英里后相遇。因此从 AA 出发的车总共行驶 7+1.5=8.57 + 1.5 = 8.5 英里,选 D。
Think of the road as having three sections: left, middle, and right. Each section is 55 miles long.
The car from AA reaches the middle section in 525=15 \frac{5}{25} = \frac{1}{5} hours.
The car from BB reaches the middle section in 520=14 \frac{5}{20} = \frac{1}{4} hours. By that time, the car from AA has already driven in the middle section for 14−15=120 \frac{1}{4} - \frac{1}{5} = \frac{1}{20} hours. During this time, that car from AA has traveled 40×120=2 40 \times \frac{1}{20} = 2 miles in the middle section.
That leaves 5−2=35 - 2 = 3 miles between the two cars in the middle section.
At that moment, the car from AA is 5+2=75 + 2 = 7 miles from A.A.
Since the cars drive at the same speed of 4040 mph in the middle section, they meet after each driving 1.51.5 more miles. This takes the car from AA a total distance of 7+1.5=8.57 + 1.5 = 8.5 miles, which is choice D.
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