Difference between revisions of "2019 AMC 10B Problems/Problem 13"
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For case 2, <math>x=5</math>. This is an extraneous case because the set is <math>4,5,6,8,17</math>.<br> | For case 2, <math>x=5</math>. This is an extraneous case because the set is <math>4,5,6,8,17</math>.<br> | ||
For case 3, <math>x=\frac{35}{4}</math>. This is an extraneous case 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 an extraneous case 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>.<br> |
+ | <math>Q.E.D \blacksquare</math> | ||
+ | Solution by [[User:a1b2|a1b2]] | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}} | {{AMC10 box|year=2019|ab=B|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:17, 14 February 2019
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
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 an extraneous case because the set is
.
For case 3, . This is an extraneous case because the set is
.
Only case 1 yields a solution, , so the answer is
.
Solution by a1b2
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 |
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