Difference between revisions of "2019 AMC 10B Problems/Problem 13"
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<math>\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}</math> | <math>\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}</math> | ||
| + | ==Solution== | ||
| + | There are <math>3</math> cases: <math>6</math> is the median, <math>8</math> is the median, and <math>x</math> is the median. In all cases, the mean is <math>7+\frac{x}{5}</math>.<br> | ||
| + | For case 1, <math>x=-5</math>. This allows 6 to be the median because the set is <math>-5,4,6,8,17</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> | ||
| + | Only case 1 yields a solution, <math>x=-5</math>, so the answer is <math>\textbf{(A) } -5</math>. | ||
==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 13:16, 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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
