2000 AMC 12 第 14 题

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14.

当列表 10,2,5,2,4,2,x10, 2, 5, 2, 4, 2, x

的平均数、中位数和众数按递增顺序排列后,它们形成一个非常数等差数列。所有可能的实数 xx 的和是多少?

When the mean, median, and mode of the list 10,2,5,2,4,2,x10, 2, 5, 2, 4, 2, x

are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of x?x?

33

66

99

1717

2020

答案:E
知识点:平均数中位数(数据)众数分类讨论
难度评级:1840
解答:

六个固定数的和为 2525,所以平均数是 25+x7\dfrac{25 + x}{7},众数是 22。若 x2x \le 2,则 22 同时是中位数和众数,会迫使等差数列为常数,所以 x>2x \gt 2

情形 2<x<42 \lt x \lt 4 中位数为 xx。 要求 2,x,25+x72, x, \dfrac{25 + x}{7} 构成等差数列,得到此范围内唯一的值 x=3x = 3

情形 x4x \ge 4 中位数为 44, 等差数列 2,4,62, 4, 6 迫使平均数为 66, 所以 25+x7=6\dfrac{25 + x}{7} = 6, 得 x=17x = 17

所有可能值的和为 3+17=203 + 17 = 20

因此,正确答案是 E

The six fixed numbers sum to 25,25, so the mean is 25+x7,\dfrac{25 + x}{7}, and the mode is 2.2. If x2,x \le 2, then 22 is both median and mode, forcing a constant progression, so x>2.x \gt 2.

Case 2<x<42 \lt x \lt 4: the median is x.x. Requiring 2,x,25+x72, x, \dfrac{25 + x}{7} to form an arithmetic progression yields x=3x = 3 as the only value in this range.

Case x4x \ge 4: the median is 4,4, and the progression 2,4,62, 4, 6 forces the mean to be 6,6, so 25+x7=6,\dfrac{25 + x}{7} = 6, giving x=17.x = 17.

The sum of all possible values is 3+17=20.3 + 17 = 20.

Thus, the correct answer is E.

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