# 2019 AMC 10B Problems/Problem 13

The following problem is from both the 2019 AMC 10B #13 and 2019 AMC 12B #7, so both problems redirect to this page.

## 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

The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.

There are three possibilities for the median: it is either $6$, $8$, or $x$.

Let's start with $6$. $\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.

Now let the median be $8$. $\frac{35+x}{5}=8$ gives $x=5$, so the sequence is $4, 5, 6, 8, 17$, which has median $6$, so this is not valid.

Finally we let the median be $x$. $\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75$, and the sequence is $4, 6, 8, 8.75, 17$, which has median $8$. This case is therefore again not valid.

Hence the only possible value of $x$ is $\boxed{\textbf{(A) }-5}.$

~IceMatrix

## See Also

 2019 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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 10 Problems and Solutions
 2019 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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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