Question

将 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ 按某种顺序排列。先求前两个数的平均数，再把这个平均数与第三个数求平均数，继续这样依次与第四个、第五个数求平均数。最终结果的最大可能值与最小可能值之差是多少？

An iterative average of the numbers $1,$ $2,$ $3,$ $4,$ and $5$ is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

$\dfrac{31}{16}$

$2$

$\dfrac{17}{8}$

$3$

$\dfrac{65}{16}$

答案：C

知识点：平均数代数变形最优化

难度评级：1600

解答：

设排列顺序为 $a, b, c, d, e.$

则迭代平均数为 $\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$ $= \dfrac{a + b + 2c + 4d + 8e}{16}.$

要使结果最小，应按排列，得到 $5, 4, 3, 2, 1,$ $\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$

要使结果最大，则反过来排列，得到 $\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$

两者之差为 $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$

所以正确答案是 C。

Let the order of the numbers be $a, b, c, d, e.$

Then the iterative average is $\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$ $= \dfrac{a + b + 2c + 4d + 8e}{16}.$

To minimize this, we make the order $5, 4, 3, 2, 1,$ which gives us a sum of $\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$

To maximize it, we have to reverse this order to get an average of $\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$

The difference between these is $\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$

Thus, C is the correct answer.

2012 AMC 10A 第 13 题

答案：C

难度评级：1600

解答：

其他年份的第 13 题