2014 AMC 8 第 24 题

先试着解答 2014 AMC 8 第 24 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 2014 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。

24.

某天 Beverage Barn 卖出了 252252 罐汽水,购买者共有 100100 位,并且每位顾客至少买了一罐。当天每位顾客购买罐数的中位数最大可能是多少?
One day the Beverage Barn sold 252252 cans of soda to 100100 customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?
2.5 2.5
3.0 3.0
3.5 3.5
4.0 4.0
4.5 4.5
答案:C
知识点:中位数(数据)最优化
小提示:
中位数是按顺序排列后第 5050 项和第 5151 项的平均数
The median is the average of the 5050th and 5151st ordered purchases.
大提示:
让前 4949 位顾客购买数尽可能小
Make the first 4949 purchases as small as possible.
解答:
将 100100 位顾客的购买数从小到大排列。中位数是第 5050 项和第 5151 项的平均数。为了最大化它,让前 4949 项尽可能小,都等于 11。
如果第 5050 项至少为 44,那么总数至少为 49+51⋅4=25349+51\cdot4=253,超过二百五十二。所以第 5050 项至多为 33。
这个最大值可以达到:取 4949 项等于 11,第 5050 项等于 33,最后 5050 项都等于 44。总数为 49+3+50⋅4=25249+3+50\cdot4=252,中位数为 3+42=3.5\frac{3+4}{2}=3.5。
所以正确答案是 C。
Order the 100100 purchases from least to greatest. The median is the average of the 5050th and 5151st entries. To maximize it, make the first 4949 entries as small as possible, so set them all equal to 11.
If the 5050th entry were at least 44, then the total would be at least 49+51⋅4=25349+51\cdot4=253, too many cans. So the 5050th entry is at most 33.
This maximum is attainable: use 4949 entries equal to 11, the 5050th entry equal to 33, and the final 5050 entries equal to 44. The total is 49+3+50⋅4=25249+3+50\cdot4=252, and the median is 3+42=3.5\frac{3+4}{2}=3.5.
Thus, C is the correct answer.
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