1992 AMC 8 第 19 题

先试着解答 1992 AMC 8 第 19 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 1992 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.

一条州际公路上第 55 个出口和第 2626 个出口之间的距离是 118118 英里。如果任意两个出口之间至少相距 55 英里,那么在第 55 个和第 2626 个出口之间,相邻两个出口之间最多可以相距多少英里?
The distance between the 55th and 2626th exits on an interstate highway is 118118 miles. If any two exits are at least 55 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 55th and 2626th exits?
88
1313
1818
4747
9898
答案:C
知识点:极端原理
小提示:
从第 55 个出口到第 2626 个出口之间有 2121 个间隔。
Between the 55th and 2626th exits there are 2121 gaps
大提示:
要让一个间隔尽可能大,就让其余 2020 个间隔都取允许的最小值 55 英里。
To make one gap as large as possible, make the other 2020 gaps as small as allowed, 55 miles each
解答:
从第 55 个出口到第 2626 个出口,共有 26−5=2126 - 5 = 21 个相邻出口间隔,每个至少 55 英里。
要最大化其中一个间隔,就让其他 2020 个间隔都等于 55 英里,共占 20×5=10020 \times 5 = 100 英里。剩下的间隔是 118−100=18118 - 100 = 18 英里。
所以正确答案是 C。
From the 55th exit to the 2626th exit there are 26−5=2126 - 5 = 21 gaps between consecutive exits, each at least 55 miles.
To maximize one gap, make the other 2020 gaps exactly 55 miles, using 20×5=10020 \times 5 = 100 miles. The remaining gap is 118−100=18118 - 100 = 18 miles.
Thus, the correct answer is C .
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