Difference between revisions of "1988 AJHSME Problems/Problem 21"
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{{AJHSME box|year=1988|num-b=20|num-a=22}} | {{AJHSME box|year=1988|num-b=20|num-a=22}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 22:56, 4 July 2013
Problem
A fifth number, 
, is added to the set 
to make the mean of the set of five numbers equal to its median. The number of possible values of is
Solution
The possible medians after is added are 
, , or 
. Now we use casework.
Case 1: The median is
In this case, 
and ![\[\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2\]](http://latex.artofproblemsolving.com/d/3/8/d38966a455dc6d38005c45687be7978ac018ede9.png)
so this case contributes 
.
Case 2: The median is
We have 
and ![\[\frac{3+6+n+9+10}{5}=n \Rightarrow n=7\]](http://latex.artofproblemsolving.com/9/1/f/91ff2d93c085fc8854933d00ee1f1e3eada82b66.png)
so this case also contributes .
Case 3: The median is
We have 
and ![\[\frac{3+6+9+n+10}{5}=9 \Rightarrow 17\]](http://latex.artofproblemsolving.com/1/e/6/1e6be5876aa529ea728fec918581b25e38430ee4.png)
so this case adds .
In all there are 
possible values of .
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
