Question

某人在下午 $6{:}00$ 刚过不久外出赴宴时，看到手表的时针与分针成 $110^\circ$ 角。他在下午 $7{:}00$ 前返回时，发现两根指针再次成 $110^\circ$ 角。他离开的分钟数为：

A man on his way to dinner shortly after $6{:}00$ p.m. observes that the hands of his watch form an angle of $110^\circ.$ Returning before $7{:}00$ p.m. he notices that again the hands of his watch form an angle of $110^\circ.$ The number of minutes that he has been away is:

$36\dfrac23$

$40$

$42$

$42.4$

$45$

答案：B

知识点：时钟相对速度

难度评级：1570

小提示：

在这段时间内，分针每分钟比时针多转 $5.5^\circ$

During the interval, the minute hand gains on the hour hand at $5.5^\circ$ per minute

大提示：

两次观察之间，带符号的角度差从 $110^\circ$ 变为 $-110^\circ$

Between the two observations, the signed separation changes from $110^\circ$ to $-110^\circ$

解答：

两次观察分别位于两针重合时刻的前后两侧。带符号的角度差变化了 $220^\circ$ 。由于分针每分钟比时针多转 $6-0.5=5.5^\circ$ ，所以经过的时间为 $\frac{220}{5.5}=40$ 分钟。

所以正确答案是 B。

The two observations lie on opposite sides of the instant when the hands coincide. Their signed angular separation changes by $220^\circ.$ Since the minute hand gains on the hour hand at $6-0.5=5.5^\circ$ per minute, the elapsed time is $\frac{220}{5.5}=40$ minutes.

Thus, the correct answer is B.

1962 AMC 12 第 35 题

答案：B

难度评级：1570

小提示：

大提示：

解答：

其他年份的第 35 题