2016 AMC 12A 第 11 题

先试着解答 2016 AMC 12A 第 11 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 2016 AMC 12A 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。

11.

某夏令营的 100100 名学生每人都会唱歌、跳舞或表演。有些学生有不止一种才艺,但没有学生三种才艺都会。有 4242 名学生不会唱歌,6565 名学生不会跳舞,2929 名学生不会表演。有多少名学生正好有两种才艺?
Each of the 100100 students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are 4242 students who cannot sing, 6565 students who cannot dance, and 2929 students who cannot act. How many students have two of these talents?
1616
2525
3636
4949
6464
答案:E
知识点:补集计数容斥原理双重计数
小提示:
分别数会唱歌、会跳舞、会表演的人数。
Count how many students can do each activity
大提示:
把三个人数相加时,两种才艺的学生被数了两次;减去 100100。
Adding the three counts tallies two-talent students twice; subtract 100100
解答:
会唱歌、跳舞、表演的人数分别为 100−42=58100-42=58、100−65=35100-65=35、100−29=71100-29=71,总和为 58+35+71=16458+35+71=164。
因为没有人三种才艺都会,每个学生有一种或两种才艺,所以只有一种才艺的学生被数一次,有两种才艺的学生被数两次。因此,被重复计算的人数为 164−100=64164-100=64。
所以正确答案是 E。
The numbers who can sing, dance, and act are 100−42=58,100-42=58, 100−65=35,100-65=35, and 100−29=71,100-29=71, respectively, for a total of 58+35+71=164.58+35+71=164.
Since no student has all three talents, each student has one or two talents, so single-talent students are counted once and two-talent students are counted twice. The number counted twice is 164−100=64.164-100=64.
Thus, the correct answer is E.
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