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 $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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