1959 AMC 12 Problem 13

Attempt Problem 13 of the 1959 AMC 12 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 1959 AMC 12 solutions, or check the answer key.

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

13.

The arithmetic mean (average) of a set of 5050 numbers is 38.38. If two numbers, namely, 4545 and 55,55, are discarded, the mean of the remaining set of numbers is:
36.536.5
3737
37.237.2
37.537.5
37.5237.52
Answer: D
Concepts:mean
Small hint:
Recover the original sum from the mean and number of entries
Big hint:
Subtract 45+5545+55 and divide by the 4848 remaining entries
Solution:
The original sum is 50⋅38=1900.50\cdot38=1900. After removing 4545 and 55,55, the remaining sum is 1800,1800, so the new mean is 180048=37.5. \frac{1800}{48}=37.5.
Therefore, the correct answer is D.
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