2003 AMC 12B Problems/Problem 11

Problem

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?

$\text {(A) 10:22 PM and 24 seconds} \qquad \text {(B) 10:24 PM} \qquad \text {(C) 10:25 PM} \qquad \text {(D) 10:27 PM} \qquad \text {(E) 10:30 PM}$

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 $\Rightarrow \boxed{\textbf{(C)}\ \text{10:25 PM}}$ .

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

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2003 AMC 12B Problems/Problem 11

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