2020 AMC 8 第 20 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

一位科学家穿过森林时,把排成一行的 55 棵树的高度记录为整数。她观察到每棵树要么是右边那棵树的两倍高,要么是右边那棵树的一半高。不幸的是,雨水让笔记中的一些数据丢失了。她的笔记如下,空白表示缺失数字。根据观察,科学家能够恢复丢失数据。树的平均高度是多少米?

T100mT211mT300mT400mT500mT00.2m\begin{array}{|c|c|}\hline T_1&\underline{\phantom{00}}\,\mathrm m\\T_2&11\,\mathrm m\\T_3&\underline{\phantom{00}}\,\mathrm m\\T_4&\underline{\phantom{00}}\,\mathrm m\\T_5&\underline{\phantom{00}}\,\mathrm m\\\hline\overline T&\underline{\phantom{00}}.2\,\mathrm m\\\hline\end{array}

A scientist walking through a forest recorded as integers the heights of 55 trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?

T100mT211mT300mT400mT500mT00.2m\begin{array}{|c|c|}\hline T_1&\underline{\phantom{00}}\,\mathrm m\\T_2&11\,\mathrm m\\T_3&\underline{\phantom{00}}\,\mathrm m\\T_4&\underline{\phantom{00}}\,\mathrm m\\T_5&\underline{\phantom{00}}\,\mathrm m\\\hline\overline T&\underline{\phantom{00}}.2\,\mathrm m\\\hline\end{array}

22.222.2

24.224.2

33.233.2

35.235.2

37.237.2

答案:B
知识点:分类讨论平均数
难度评级:1370
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文字解答:

22 棵树高 1111 米。由于相邻两棵树高度相差因子 22,且高度都是整数,第 11 棵和第 33 棵都必须高 2222 米。

前三棵树总高为 22+11+22=5522+11+22=55。第 44 棵树可能是 (11,22)(11,22) 米或 (44,22)(44,22) 米。若第 (44,88)(44,88) 棵是 米,则第 55 棵可能是 米或 米,后一种不是整数,前一种的平均数以 结尾,与表格不符。

若第 17.617.6 米,则第 棵可能是 米或 米,对应平均数分别为 24.224.237.437.4。符合表中 .2.2 结尾的选项是 24.224.2

正确答案是 B

Tree 22 is 1111 meters tall. Since all heights are integers and neighboring trees differ by a factor of 2,2, trees 11 and 33 must both be 2222 meters tall.

The first three trees total 22+11+22=5522+11+22=55 meters. The possible integer pairs for trees 44 and 55 are (11,22),(11,22), (44,22),(44,22), and (44,88).(44,88).

These give averages 17.6,17.6, 24.2,24.2, and 37.4,37.4, respectively. The notebook shows that the average ends in .2,.2, so it must be 24.2.24.2.

Thus, the correct answer is B.

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