AMC12 2003 B

AMC12 2003 B · Q11

AMC12 2003 B · Q11. It mainly tests Interest / growth (simple), Arithmetic misc.

Cassandra sets her watch to the correct time at noon. At the actual time of 1:00 PM, she notices that her watch reads 12:57 and 36 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:00 PM?

Cassandra 在中午将她的手表调到正确时间。在实际时间下午 1:00 时，她注意到她的手表显示 12:57 和 36 秒。假设她的手表以恒定速率走慢，当她的手表第一次显示晚上 10:00 时，实际时间是几点？

(A) 10:22 PM and 24 seconds 晚上10:22分24秒

(B) 10:24 PM 晚上10:24

(D) 10:27 PM 晚上10:27

(E) 10:30 PM 晚上10:30

Answer

Correct choice: (C)

正确答案：（C）

Solution

For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra's watch. When her watch first reads, 10:00 pm, $10(60)=600$ minutes have passed by on her watch. Setting up a proportion, \[\frac{57.6}{60}=\frac{600}{x}\] where $x$ is the number of minutes that have passed by in actual time. Solve for $x$ to get $625$ minutes, or $10$ hours and $25$ minutes $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$.

实际时间每经过 $60$ 分钟，Cassandra 的手表只走过 $57+\frac{36}{60}=57.6$ 分钟。当她的手表第一次显示晚上 10:00 时，她的手表从中午起共走过 $10(60)=600$ 分钟。设实际时间经过了 $x$ 分钟，则有比例 \[\frac{57.6}{60}=\frac{600}{x}\] 解得 $x=625$ 分钟，即 $10$ 小时 $25$ 分钟 $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$。

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