AMC 8 1991 Problem 14
AMC8 1991 年第 14 题
逻辑推理（跨域）★★★☆☆
Several students are competing in a series of three races. A student earns 5 points for winning a race, 3 points for finishing second and 1 point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?
几名学生在三场系列赛中竞争。赢一场得 5 分，第二名得 3 分，第三名得 1 分。没有平局。一个学生在三场比赛中至少需要获得多少分，才能保证总分超过其他任何学生？
- A.9
- B.10
- C.11
- D.13
- E.15
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
It is possible for two students to get 11 points: one student could win the first two races and finish third in the third race (scoring 5+5+1=11 ), while another student could finish second in the first two races and win the third race (scoring 3+3+5=11 ).
To score 13 or more points, a student must win at least 2 races. So any other student could win at most 1 race and thus score at most 5 + 3 + 3 = 11 points. Therefore, 13 is the minimum score a student needs to guarantee outscoring all other students.
Note: it is impossible for a student to score exactly 12 points.
-stjwyl, edited by j314andrews
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