Difference between revisions of "2001 AIME I Problems/Problem 2"
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== Problem == | == Problem == | ||
| − | A finite | + | A finite set <math>\mathcal{S}</math> of distinct real numbers has the following properties: the mean of <math>\mathcal{S}\cup\{1\}</math> is <math>13</math> less than the mean of <math>\mathcal{S}</math>, and the mean of <math>\mathcal{S}\cup\{2001\}</math> is <math>27</math> more than the mean of <math>\mathcal{S}</math>. Find the mean of <math>\mathcal{S}</math>. |
== Solution == | == Solution == | ||
Let <math>x</math> be the mean of <math>\mathcal{S}</math>. Let <math>a</math> be the number of elements in <math>\mathcal{S}</math>. | Let <math>x</math> be the mean of <math>\mathcal{S}</math>. Let <math>a</math> be the number of elements in <math>\mathcal{S}</math>. | ||
| − | Then, | + | Then, |
| − | < | + | <cmath>\frac{ax+1}{a+1}=x-13</cmath> and <cmath>\frac{ax+2001}{a+1}=x+27</cmath> |
| + | <cmath>\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}</cmath> | ||
| + | <cmath>\frac{2000}{a+1}=40</cmath> so <cmath>2000=40(a+1)</cmath> | ||
| + | <cmath>a=49</cmath> | ||
We plug that into our very first formula, and get: | We plug that into our very first formula, and get: | ||
| − | < | + | <cmath>\frac{49x+1}{50}=x-13</cmath> |
| − | 49x+1 | + | <cmath>49x+1=50x-650</cmath> |
| − | x | + | <cmath>x=\boxed{651}</cmath> |
== See Also == | == See Also == | ||
{{AIME box|year=2001|n=I|num-b=1|num-a=3}} | {{AIME box|year=2001|n=I|num-b=1|num-a=3}} | ||
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Revision as of 20:50, 13 March 2009
Problem
A finite set 
of distinct real numbers has the following properties: the mean of 
is 
less than the mean of , and the mean of 
is 
more than the mean of . Find the mean of .
Solution
Let 
be the mean of . Let 
be the number of elements in .
Then,
![\[\frac{ax+1}{a+1}=x-13\]](http://latex.artofproblemsolving.com/f/0/9/f09865e863288b03a2aa67f6a5131186c8904197.png)
and ![\[\frac{ax+2001}{a+1}=x+27\]](http://latex.artofproblemsolving.com/1/8/2/182c37ee9d12b2e116d4713f0b2911e442bea89e.png)
![\[\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}\]](http://latex.artofproblemsolving.com/5/b/2/5b263a908527a16e89e2a26f5d18253351500f3b.png)
![\[\frac{2000}{a+1}=40\]](http://latex.artofproblemsolving.com/e/7/d/e7d394251897b554997add7fc6d236587e64a106.png)
so ![\[2000=40(a+1)\]](http://latex.artofproblemsolving.com/f/9/0/f9066c95126ed103c8637d351c6468b4bff18fad.png)
![\[a=49\]](http://latex.artofproblemsolving.com/1/c/4/1c45a9d9b7f91ee97c61079399d3e2d3b4ad5f2d.png)
We plug that into our very first formula, and get:
![\[\frac{49x+1}{50}=x-13\]](http://latex.artofproblemsolving.com/6/a/3/6a3481fd66eb015f40c012d6bccd59dd84a2b631.png)
![\[49x+1=50x-650\]](http://latex.artofproblemsolving.com/7/0/6/7068477c4e6c6404afc9ffc97028ffa957b16f69.png)
![\[x=\boxed{651}\]](http://latex.artofproblemsolving.com/5/8/0/5805de6fea38d238529dbc515c4404c8bd129b9a.png)
See Also
| 2001 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
