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.
Contents
Problem
What is the sum of all real numbers for which the median of the numbers and is equal to the mean of those five numbers?
Solution 1
There are cases: is the median, is the median, and is the median. In all cases, the mean is .
For case 1, . This allows 6 to be the median because the set is .
For case 2, . This is impossible because the set is .
For case 3, . This is impossible because the set is .
Only case 1 yields a solution, , so the answer is .
Solution 2
The mean is .
There are 3 possibilities: either the median is 6, 8, or x.
Let's start with 6.
when and the sequence is -5, 4, 6, 8, 17 which has 6 as the median so we're good.
Now let the mean=8
when and the sequence is 4, 5, 6, 8, 17 which has median 6 so no go.
Finally we let the mean=x
and the sequence is 4, 6, 8, 8.75, 17 which has median 8 so no go.
So the only option for x is
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
SUB2PEWDS