AMC 8 1997 Problem 14
AMC8 1997 年第 14 题
统计与数据图表★★★☆☆
There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?
有五个正整数组成的集合，其平均数为 5，中位数为 5，唯一的众数为 8。该集合中最大整数与最小整数之差是多少？
- A.3
- B.5
- C.6
- D.7
- E.8
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 do, and so our answer is 8-1=7.
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