2019 AMC 10B Problems/Problem 13

Problem

What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?

$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$

Solution

There are $3$ cases: $6$ is the median, $8$ is the median, and $x$ is the median. In all cases, the mean is $7+\frac{x}{5}$ .
For case 1, $x=-5$ . This allows 6 to be the median because the set is $-5,4,6,8,17$ .
For case 2, $x=5$ . This is an extraneous case because the set is $4,5,6,8,17$ .
For case 3, $x=\frac{35}{4}$ . This is an extraneous case because the set is $4,6,8,\frac{35}{4},17$ .
Only case 1 yields a solution, $x=-5$ , so the answer is $\textbf{(A) } -5$ .

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 10 Problems and Solutions

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2019 AMC 10B Problems/Problem 13

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