2021 AMC 10B Spring 第 19 题

先试着解答 2021 AMC 10B Spring 第 19 题,然后核对你的答案与视频与精心整理的解答,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 2021 AMC 10B Spring 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.

SS 是一个有限的正整数集合。

如果从 SS 中移除 SS 中最大的整数,剩余整数的平均值为 3232。如果再从 SS 中移除最小的整数,剩余整数的平均值为 3535。如果随后把最大的整数放回 SS,平均值升高到 4040。原集合 SS 中最大的整数比最小的整数大 7272

集合 SS 中所有整数的平均值是多少?

Suppose that SS is a finite set of positive integers.

If the greatest integer in SS is removed from S,S, then the average value (arithmetic mean) of the integers remaining is 32.32. If the least integer in SS is also removed, then the average value of the integers remaining is 35.35. If the greatest integer is then returned to the set, the average value of the integers rises to 40.40. The greatest integer in the original set SS is 7272 greater than the least integer in S.S.

What is the average value of all the integers in the set S?S?

36.236.2

36.436.4

36.636.6

36.836.8

3737

答案:D
知识点:平均数方程组
难度评级:1540
小提示:

用总和、最小值、最大值和集合大小写出三个平均数方程。

Write equations for the three averages using the total sum, least value, greatest value, and set size

大提示:

将两个分母为 n1n-1 的平均数方程相减,利用差为 7272

Subtract the two averages with denominator n1n-1 to use the difference 7272

视频讲解:
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文字解答:

设所有整数之和为 ss,最大数为 gg,最小数为 ll,集合 SS 的大小为 nn

两个含 n1n-1 个元素的集合的平均数给出 sl=40(n1),sg=32(n1) \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1) \end{aligned}\text{。} 两式相减并利用 gl=72g-l=72,得 72=8(n1)72=8(n-1),所以 n=10n=10

同时去掉最大数和最小数后,还剩 88 个数,其和为 835=2808\cdot35=280。另一方面,只去掉最小数时剩下的和为 940=3609\cdot40=360,所以 g=360280=80g=360-280=80。于是 l=8072=8l=80-72=8,原来的总和为 s=280+80+8=368s=280+80+8=368

平均值为 sn=36810=36.8\frac{s}{n}=\frac{368}{10}=36.8

所以答案是 D

Let the sum of all the integers be s,s, the greatest number be g,g, the least number be l,l, and the size of SS be n.n.

The two averages of sets with n1n-1 elements give sl=40(n1),sg=32(n1). \begin{aligned} s-l&=40(n-1),\\ s-g&=32(n-1). \end{aligned} Subtracting and using gl=72g-l=72 gives 72=8(n1),72=8(n-1), so n=10.n=10.

Removing both extremes leaves 88 numbers with sum 835=280.8\cdot35=280. On the other hand, removing only the least number leaves a sum of 940=360,9\cdot40=360, so g=360280=80.g=360-280=80. Hence l=8072=8,l=80-72=8, and the original sum is s=280+80+8=368.s=280+80+8=368.

The required average is sn=36810=36.8.\frac{s}{n}=\frac{368}{10}=36.8.

Thus, the answer is D .

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