2019 AMC 10B Problem 13

Question

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?

-5

0

5

\dfrac{15}{4}

\dfrac{35}{4}

Case on whether

x<6

,

6\le x\le8

, or

x>8

In each case the median has a simple expression

The mean is

7+\frac{x}{5}

. If

x<6

, the median is

6

, so

7+\frac{x}{5}=6

gives the valid value

x=-5

.

If

6\le x\le8

, the median is

x

, but

7+\frac{x}{5}=x

gives

x=\frac{35}{4}

, outside this interval. If

x>8

, the median is

8

, but

7+\frac{x}{5}=8

gives

x=5

, again outside the required range.

Therefore, the only possible

x

is

-5,

making the sum

-5.

Thus, the answer is A .

Problem 13 in other years