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2003 AMC 12B Problem 11

Problem 11 of 25IntermediateAlgebra

Cassandra sets her watch to the correct time at noon. At the actual time of 1:001{:}00 PM, she notices that her watch reads 12:5712{:}57 and 3636 seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads 10:0010{:}00 PM?

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Solution

In 6060 real minutes the watch advances only 5757 minutes 3636 seconds =57.6= 57.6 minutes. So when the watch shows tt minutes past noon, the real elapsed time is 6057.6t=2524t\dfrac{60}{57.6}t = \dfrac{25}{24}t minutes. The watch reads 10:0010{:}00 PM after 600600 recorded minutes, so the real elapsed time is 2524(600)=625 \frac{25}{24}(600) = 625 minutes =10= 10 hours 2525 minutes past noon. The actual time is 10:2510{:}25 PM. Thus, the correct answer is C.

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Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.