2014 AMC 12B Problem 11

Attempt Problem 11 of the 2014 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 2014 AMC 12B solutions, or check the answer key.

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11.

A list of 1111 positive integers has a mean of 10,10, a median of 9,9, and a unique mode of 8.8. What is the largest possible value of an integer in the list?

2424

3030

3131

3333

3535

Answer: E
Concepts:meanmedian (data)modeoptimization
Difficulty rating: 1690
Solution:

The list sums to 1110=110.11 \cdot 10 = 110. To maximize one entry, minimize the sum of the other ten.

Sorted, the sixth number must be 99 (the median), and 88 must appear more often than any other value. Trying 88 three times, the smallest possible ten numbers are 1,1,8,8,8,9,9,10,10,11, 1,1,8,8,8,9,9,10,10,11, which sum to 7575 and keep 88 the unique mode.

The largest entry is then 11075=35.110 - 75 = 35.

Thus, the correct answer is E.

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