# Difference between revisions of "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 1

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 impossible because the set is $4,5,6,8,17$.
For case 3, $x=\frac{35}{4}$. This is impossible because the set is $4,6,8,\frac{35}{4},17$.
Only case 1 yields a solution, $x=-5$, so the answer is $\textbf{(A) } -5$.

## Solution 2

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

There are 3 possibilities: either the median is 6, 8, or x.

$\frac{35+x}{5}=6$ when $x=-5$ and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.

Now let the mean=8

$\frac{35+x}{5}=8$ when $x=5$ and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.

Finally we let the mean=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 so no go.

So the only option for x is $\boxed{-5}.$

--mguempel