2007 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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Concepts:median (data)mean casework

Difficulty rating: 1500

15.

The set $\{3,6,9,10\}$ is augmented by a fifth element $n,$ not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of $n?$

$7$

$9$

$19$

$24$

$26$

Solution:

The mean is $\dfrac{28+n}{5}.$

If $n\lt 6,$ the median is $6,$ so $28+n=30$ and $n=2.$

If $6\lt n\lt 9,$ the median is $n,$ so $28+n=5n$ and $n=7.$

If $n\gt 9,$ the median is $9,$ so $28+n=45$ and $n=17.$

The sum of all possible values is $2+7+17=26.$

Thus, the correct answer is E.

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