Difference between revisions of "2019 AMC 10B Problems/Problem 13"

Revision as of 22:37, 16 February 2019

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

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?

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

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

2019 AMC 10B (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

SUB2PEWDS

@@ Line 25: / Line 25: @@
 <math>\frac{35+x}{5}=x \implies 35+x=5x \implies x=\frac{35}{4}=8.75.</math> and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go.
-So the only option for x is <math>\boxed{-5}.</math>
+So the only option for x is <math>\boxed{\textbf{(A) }-5}.</math>
 ==See Also==