Difference between revisions of "2019 AMC 10B Problems/Problem 13"
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For case 2, <math>x=5</math>. This is impossible because the set is <math>4,5,6,8,17</math>.<br> | For case 2, <math>x=5</math>. This is impossible because the set is <math>4,5,6,8,17</math>.<br> | ||
For case 3, <math>x=\frac{35}{4}</math>. This is impossible because the set is <math>4,6,8,\frac{35}{4},17</math>.<br> | For case 3, <math>x=\frac{35}{4}</math>. This is impossible because the set is <math>4,6,8,\frac{35}{4},17</math>.<br> | ||
− | Only case 1 yields a solution, <math>x=-5</math>, so the answer is <math>\textbf{(A) } -5</math>. | + | Only case 1 yields a solution, <math>x=-5</math>, so the answer is <math>\textbf{(A) } -5</math>. |
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==Solution 2== | ==Solution 2== |
Revision as of 11:18, 15 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.
Contents
[hide]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
--mguempel
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.