2002 AMC 12A 第 11 题

先试着解答 2002 AMC 12A 第 11 题,然后核对你的答案与精心整理的解答,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 2002 AMC 12A 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。

11.

Earl E. Bird 先生每天上午 8:008{:}00 整离家上班。当他的平均速度为每小时 4040 英里时,会晚到三分钟;当平均速度为每小时 6060 英里时,会早到三分钟。Bird 先生应以多少英里每小时的平均速度行驶,才能准时到达工作地点?
Mr. Earl E. Bird leaves his house for work at exactly 8:008{:}00 A.M. every morning. When he averages 4040 miles per hour, he arrives at his workplace three minutes late. When he averages 6060 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?
4545
4848
5050
5555
5858
答案:B
知识点:路程、速度与时间方程组
小提示:
设准时所需行驶时间为 tt 小时;三分钟是 0.050.05 小时。
Let tt be the on-time travel time in hours; three minutes is 0.050.05 hours
大提示:
列方程 40(t+0.05)=60(t−0.05)40(t+0.05) = 60(t-0.05),求出 tt,再求路程。
Set 40(t+0.05)=60(t−0.05),40(t+0.05) = 60(t-0.05), solve for t,t, then find the distance
解答:
设准时所需时间为 tt 小时。三分钟是 0.050.05 小时,所以 40(t+0.05)=60(t−0.05)40(t+0.05) = 60(t-0.05)。
得 40t+2=60t−340t + 2 = 60t - 3,所以 t=0.25t = 0.25。路程为 40(0.25+0.05)=1240(0.25+0.05) = 12 英里,所需速度为 120.25=48\dfrac{12}{0.25} = 48 英里每小时。
因此,正确答案是 B。
Let tt be the time in hours to arrive on time. Since three minutes is 0.050.05 hours, 40(t+0.05)=60(t−0.05).40(t+0.05) = 60(t-0.05).
This gives 40t+2=60t−3,40t + 2 = 60t - 3, so t=0.25.t = 0.25. The distance is 40(0.25+0.05)=1240(0.25+0.05) = 12 miles, and the required speed is 120.25=48\dfrac{12}{0.25} = 48 miles per hour.
Thus, the correct answer is B.
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