2007 AMC 12A Problem 15

Attempt Problem 15 of the 2007 AMC 12A 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 2007 AMC 12A solutions, or check the answer key.

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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

The mean of the five numbers is

\dfrac{28+n}{5}

Split into cases: the median is

6

when

n\lt 6,

it is

n

when

6\lt n\lt 9,

and

9

when

n\gt 9

The mean is

\dfrac{28+n}{5}.

n\lt 6,

the median is

6,

28+n=30

and

n=2.

6\lt n\lt 9,

the median is

n,

28+n=5n

and

n=7.

n\gt 9,

the median is

9,

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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