AMC10 2012 A

AMC10 2012 A · Q13

AMC10 2012 A · Q13. It mainly tests Fractions, Averages (mean).

An iterative average of the numbers 1, 2, 3, 4, and 5 is computed in 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?

对数字 1、2、3、4 和 5 进行迭代平均的计算方式如下。将五个数字按某种顺序排列。先求前两个数的平均数，然后将该平均数与第三个数的平均数，再将结果与第四个数的平均数，最后与第五个数的平均数。使用此过程可能得到的最大值与最小值之差是多少？

(A) \frac{31}{16} \frac{31}{16}

(B) 2 2

(D) 3 3

(E) \frac{65}{16} \frac{65}{16}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): If the numbers are arranged in the order $a, b, c, d, e$, then the iterative average is $$\frac{\frac{\frac{\frac{a+b}{2}+c}{2}+d}{2}+e}{2}=\frac{a+b+2c+4d+8e}{16}.$$ The largest value is obtained by letting $(a,b,c,d,e)=(1,2,3,4,5)$ or $(2,1,3,4,5)$, and the smallest value is obtained by letting $(a,b,c,d,e)=(5,4,3,2,1)$ or $(4,5,3,2,1)$. In the former case the iterative average is $65/16$, and in the latter case the iterative average is $31/16$, so the desired difference is $$\frac{65}{16}-\frac{31}{16}=\frac{34}{16}=\frac{17}{8}.$$

答案（C）：如果这些数按顺序 $a,b,c,d,e$ 排列，则迭代平均数为 $$\frac{\frac{\frac{\frac{a+b}{2}+c}{2}+d}{2}+e}{2}=\frac{a+b+2c+4d+8e}{16}.$$ 当 $(a,b,c,d,e)=(1,2,3,4,5)$ 或 $(2,1,3,4,5)$ 时取最大值；当 $(a,b,c,d,e)=(5,4,3,2,1)$ 或 $(4,5,3,2,1)$ 时取最小值。前者的迭代平均数为 $65/16$，后者为 $31/16$，因此所求差为 $$\frac{65}{16}-\frac{31}{16}=\frac{34}{16}=\frac{17}{8}.$$

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