2003 AMC 12B Problem 11

Attempt Problem 11 of the 2003 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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11.

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 3636 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?

10:22 PM and 2424 seconds

10:24 PM

10:25 PM

10:27 PM

10:30 PM

Answer: C
Concepts:rateratio and proportion
Difficulty rating: 1390
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 10:00 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 10:25 PM.

Thus, the correct answer is C.

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