2002 AMC 8 Problem 18

Attempt Problem 18 of the 2002 AMC 8 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 2002 AMC 8 solutions, or check the answer key.

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

18.

Gage skated 11 hr 1515 min each day for 55 days and 11 hr 3030 min each day for 33 days. How long would he have to skate the ninth day in order to average 8585 minutes of skating each day for the entire time?
11 hr
11 hr 1010 min
11 hr 2020 min
11 hr 4040 min
22 hr
Answer: E
Concepts:meanunit conversion
Small hint:
Convert all skating times to minutes.
Big hint:
Find the total minutes needed for a 99-day average of 85.85.
Solution:
Gage has skated a total of 5⋅75+3⋅90=645 5 \cdot 75 + 3 \cdot 90 = 645 minutes.
For an average of 8585 minutes over 99 days, Gage must have skated a total of 85⋅9=765 85 \cdot 9 = 765 minutes.
This means that Gage must skate 765−645=120 765 - 645 = 120 minutes on the last day. Note that 120120 minutes is the same as 22 hours.
Thus, E is the correct answer.
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