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 $\boxed{\textbf{(A) }-5}.$

Video Solution

https://youtu.be/mXvetCMMzpU?t=448

~IceMatrix

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 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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