2013 AMC 8 第 16 题

先试着解答 2013 AMC 8 第 16 题,然后核对你的答案与视频与精心整理的解答,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 2013 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。

16.

Fibonacci Middle School 的一些学生参加社区服务项目。88 年级学生与 66 年级学生的比为 5:35:388 年级学生与 77 年级学生的比为 8:58:5。参加该项目的学生最少可能有多少人?

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 88th-graders to 66th-graders is 5:3,5:3, and the ratio of 88th-graders to 77th-graders is 8:5.8:5. What is the smallest number of students that could be participating in the project?

1616

4040

5555

7979

8989

答案:E
知识点:比与比例最小公倍数
难度评级:1370
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文字解答:

88 年级学生人数必须同时是 5588 的倍数,所以最少为 4040 这时 66 年级人数为 4035=2440 \cdot \dfrac{3}{5} = 2477 年级人数为 4058=2540 \cdot \dfrac{5}{8} = 25

因此学生总数为 40+24+25=8940 + 24 + 25 = 89

所以正确答案是 E

The number of 88th-graders must be a multiple of 55 and 8,8, which means that it is at least 40.40. Using this number, we get that the number of 66th-graders is 4035=24.40 \cdot \dfrac{3}{5} = 24. Similarly, the number of 77th-graders is 4058=25.40 \cdot \dfrac{5}{8} = 25.

The total number of students is therefore 40+24+25=89.40 + 24 + 25 = 89.

Thus, E is the correct answer.

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