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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? (2003 AMC 12B Problems, Question \#11) 
Overseas Competition->Knowledge Point->Calculation Modules
For every $60$ minutes that pass by in actual time, $57+\frac{36}{60}=57.6$ minutes pass by on Cassandra\textquotesingle 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 $$ \Rightarrow \text { (C) 10:25} $$ PM. 