1999 AMC 8 第 2 题

先试着解答 1999 AMC 8 第 2 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 1999 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。

2.

1010 点整时,钟表两根指针形成的较小角的度数是多少?
What is the degree measure of the smaller angle formed by the hands of a clock at 1010 o’clock?
3030
4545
6060
7575
9090
答案:C
知识点:时钟
小提示:
每相邻两个小时刻度相差 30∘30^\circ
Each hour mark is 30∘30^\circ apart
大提示:
1010 点整时,两根指针相隔两个小时刻度
At 1010 o’clock, the hands are two hour marks apart
解答:
1010 点整时,时针指向 1010,分针指向 1212。
两根指针相隔整个圆的 16\frac{1}{6}。所以角度为 360∘⋅16=60∘。 360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}\text{。}
所以正确答案是 C。
At 1010 o’clock, we have that the hour hand is at 1010 and the minute hand is at 12.12.
This means that the hands are a 16\frac{1}{6} of the entire circle apart. This is equal to 360∘⋅16=60∘. 360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}.
Thus, C is the correct answer.
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