1988 AMC 8 第 21 题

先试着解答 1988 AMC 8 第 21 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 1988 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。

21.

在数集 {3,6,9,10}\{3, 6, 9, 10\} 中加入第五个数 nn,使这五个数的平均数等于它们的中位数。nn 的可能取值个数是
A fifth number, n,n, is added to the set of numbers {3,6,9,10}\{3, 6, 9, 10\} to make the mean of the set of five numbers equal to its median. The number of possible values for nn is
11
22
33
44
多于 44
more than 44
答案:C
知识点:平均数中位数(数据)分类讨论
小提示:
五个数的平均数是 28+n5\dfrac{28 + n}{5};中位数取决于 nn 在排序后的位置
The mean of the five numbers is 28+n5;\dfrac{28 + n}{5}; the median depends on where nn lands in the sorted order
大提示:
分情况讨论:中位数是 66、nn 本身或 99,并在每种情况下利用方程“平均数 == 中位数”求解
Split into cases: the median is 6,6, is nn itself, or is 9,9, and solve the equation mean == median in each
解答:
五个数的和是 28+n28 + n,平均数是 28+n5\dfrac{28 + n}{5}。中位数是排序后的中间值,依 nn 的大小可能是 66、nn 或 99。
如果中位数是 66:28+n5=6\dfrac{28 + n}{5} = 6,得 n=2n = 2,确实小于 66。如果中位数是 nn:28+n5=n\dfrac{28 + n}{5} = n,得 n=7n = 7,确实在 66 和 99 之间。如果中位数是 99:28+n5=9\dfrac{28 + n}{5} = 9,得 n=17n = 17,确实大于 99。
每种情况都给出一个有效值,所以 nn 可以是 22、77 或 1717,共有三个可能值。
所以正确答案是 C。
The five numbers have sum 28+n28 + n and mean 28+n5.\dfrac{28 + n}{5}. The median is the middle value when sorted, which is 6,6, n,n, or 99 depending on the size of n.n.
If the median is 66: 28+n5=6\dfrac{28 + n}{5} = 6 gives n=2,n = 2, which is indeed less than 6.6. If the median is nn: 28+n5=n\dfrac{28 + n}{5} = n gives n=7,n = 7, which is between 66 and 9.9. If the median is 99: 28+n5=9\dfrac{28 + n}{5} = 9 gives n=17,n = 17, which is indeed greater than 9.9.
Each case yields a valid value, so nn can be 2,2, 7,7, or 1717 — three possible values.
Thus, the correct answer is C .
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