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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

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?
10:2210{:}22 PM and 2424 seconds
10:2410{:}24 PM
10:2510{:}25 PM
10:2710{:}27 PM
10:3010{:}30 PM
Answer: C
Concepts:rateratio and proportion
Small hint:
In 6060 real minutes the watch records only 57.657.6 minutes
Big hint:
If the watch shows tt minutes past noon, the real time is 6057.6 t=2524 t\dfrac{60}{57.6}\,t = \dfrac{25}{24}\,t
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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