Difference between revisions of "2007 AMC 12A Problems/Problem 15"
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*Median is <math>6</math>: Then <math>n \le 6</math> and <math>\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2</math>. | *Median is <math>6</math>: Then <math>n \le 6</math> and <math>\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2</math>. | ||
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*Median is <math>9</math>: Then <math>n \ge 9</math> and <math>\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17</math>. | *Median is <math>9</math>: Then <math>n \ge 9</math> and <math>\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17</math>. | ||
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*Median is <math>n</math>: Then <math>6 < n < 9</math> and <math>\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7</math>. | *Median is <math>n</math>: Then <math>6 < n < 9</math> and <math>\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7</math>. | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 13:52, 7 August 2017
Problems
The set 
is augmented by a fifth element 
, not equal to any of the other four. The median of the resulting set is equal to its mean. What is the sum of all possible values of ?
Solution
The median must either be 
or . Casework:
- Median is
: Then 
and 
.
: Then 
and 
.- Median is
: Then 
and 
.
: Then 
and 
.- Median is
: Then 
and 
.
and 
.All three cases are valid, so our solution is 
.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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. 
