1991 AMC 8 第 23 题

先试着解答 1991 AMC 8 第 23 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。

23.

毕达哥拉斯高中的乐队有 100100 名女生和 8080 名男生。毕达哥拉斯高中的管弦乐队有 8080 名女生和 100100 名男生。有 6060 名女生同时属于乐队和管弦乐队。总共有 230230 名学生属于乐队或管弦乐队或两者都属于。属于乐队但不属于管弦乐队的男生人数是
The Pythagoras High School band has 100100 female and 8080 male members. The Pythagoras High School orchestra has 8080 female and 100100 male members. There are 6060 females who are members in both band and orchestra. Altogether, there are 230230 students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is
1010
2020
3030
5050
7070
答案:A
知识点:容斥原理韦恩图
小提示:
先用 100+80−60100 + 80 - 60 求属于至少一个团体的女生人数
First find how many females are in at least one group, using 100+80−60100 + 80 - 60
大提示:
用总人数 230230 减去女生人数,得到属于至少一个团体的男生人数;再用容斥原理求同时属于两个团体的男生人数
Subtract the females from the 230230 total to get the males in at least one group, then use inclusion–exclusion to find how many males are in both groups
解答:
属于乐队或管弦乐队的女生人数为 100+80−60=120100 + 80 - 60 = 120。所以属于至少一个团体的男生人数为 230−120=110230 - 120 = 110。
乐队有 8080 名男生,管弦乐队有 100100 名男生,因此两个团体都有的男生人数是 80+100−110=7080 + 100 - 110 = 70。属于乐队但不属于管弦乐队的男生有 80−70=1080 - 70 = 10 人。
所以正确答案是 A。
Females in band or orchestra: 100+80−60=120.100 + 80 - 60 = 120. So males in at least one group: 230−120=110.230 - 120 = 110.
With 8080 males in band and 100100 in orchestra, the males in both are 80+100−110=70.80 + 100 - 110 = 70. Hence males in band but not orchestra: 80−70=10.80 - 70 = 10.
Thus, the correct answer is A .
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