2002 AMC 10A 第 12 题

先试着解答 2002 AMC 10A 第 12 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 10A 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。

12.

Earl E. Bird 先生每天早上正好在 8:008{:}00 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?
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、速度与时间一次方程
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:
两种情况下路程相同:40(t+0.05)=60(t−0.05)40(t+0.05)=60(t-0.05)。
The distance is the same both ways: 40(t+0.05)=60(t−0.05)40(t+0.05)=60(t-0.05)
解答:
设准时所需时间为 tt 小时。题中的 33 分钟等于 0.050.05 小时,所以 40(t+0.05)=60(t−0.05)40(t+0.05)=60(t-0.05)。化简得 40t+2=60t−340t+2=60t-3,从而 t=0.25t=0.25。
路程为 40(0.30)=1240(0.30)=12 英里,因此准时速度为 120.25=48\dfrac{12}{0.25}=48 英里每小时。
所以正确答案是 B。
Let tt hours be the on-time travel time. Since 33 minutes is 0.050.05 hours, 40(t+0.05)=60(t−0.05).40(t+0.05)=60(t-0.05). Then 40t+2=60t−3,40t+2=60t-3, so t=0.25.t=0.25.
The distance is 40(0.30)=1240(0.30)=12 miles, so the required speed is 120.25=48\dfrac{12}{0.25}=48 mph.
Thus, the correct answer is B.
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