1991 AMC 8 第 14 题

先试着解答 1991 AMC 8 第 14 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1991 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

14.

几名学生参加三场赛跑。每场比赛第一名得 55 分,第二名得 33 分,第三名得 11 分。没有并列名次。一个学生在三场比赛中至少要得到多少分,才能保证他的总分高于任何其他学生?
Several students are competing in a series of three races. A student earns 55 points for winning a race, 33 points for finishing second, and 11 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?
99
1010
1111
1313
1515
答案:D
知识点:极端原理
小提示:
要保证领先,要考虑最坏情况:另一个学生也尽可能得高分
To be guaranteed the lead, plan for the worst case where another student does as well as possible
大提示:
依次检查可能的三场总分;对每个总分,判断剩余名次是否允许另一名学生追平或超过它
Check the attainable three-race totals in increasing order; for each one, see whether the remaining places allow a rival to tie or exceed it
解答:
总分 1111 并不能保证第一,例如 5+5+15+5+1 可能被另一个学生也得到 1111 分。
但如果一个学生得到 5+5+3=135+5+3 = 13 分,那么剩下的名次使任何其他学生最多得到 3+3+5=113+3+5 = 11 分。因此 1313 分能保证领先。
所以正确答案是 D。
A total of 1111 (for example 5+5+15+5+1) does not guarantee first place, since another student could also reach 11.11.
But if one student scores 5+5+3=13,5+5+3 = 13, the remaining places give every other student at most 3+3+5=11.3+3+5 = 11. So 1313 points guarantees the lead.
Thus, the correct answer is D .
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.