2019 AMC 10B Problems/Problem 13

Revision as of 11:18, 15 February 2019 by Xmidnightfirex (talk | contribs) (Solution 2)
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.

Let's start with 6.

$\frac{35+x}{5}=6$ 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 $\boxed{-5}.$

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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