2014 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2014 AMC 12B, 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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Concepts:mean median (data)mode optimization

Difficulty rating: 1690

11.

A list of $11$ positive integers has a mean of $10,$ a median of $9,$ and a unique mode of $8.$ What is the largest possible value of an integer in the list?

$24$

$30$

$31$

$33$

$35$

Solution:

The list sums to $11 \cdot 10 = 110.$ To maximize one entry, minimize the sum of the other ten.

Sorted, the sixth number must be $9$ (the median), and $8$ must appear more often than any other value. Trying $8$ three times, the smallest possible ten numbers are $1,1,8,8,8,9,9,10,10,11,$ which sum to $75$ and keep $8$ the unique mode.

The largest entry is then $110 - 75 = 35.$

Thus, the correct answer is E.

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