2007 AMC 12A Problems/Problem 15

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Problems

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 $n$ ?

$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 19\qquad \mathrm{(D)}\ 24\qquad \mathrm{(E)}\ 26$

Solution

The median must either be $6, 9,$ or $n$ . Casework:

Median is $6$ : Then $n \le 6$ and $\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2$ .
Median is $9$ : Then $n \ge 9$ and $\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17$ .
Median is $n$ : Then $6 < n < 9$ and $\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7$ .

All three cases are valid, so our solution is $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$ .

2007 AMC 12A (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
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2007 AMC 12A Problems/Problem 15

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Problems

Solution

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