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Difference between revisions of "2001 AIME I Problems/Problem 2"

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== Solution ==
 
== Solution ==
Let x be the mean of S. Let a be the number of elements in S.
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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+1}{a+1}=x-13</cmath> and <cmath>\frac{ax+2001}{a+1}=x+27</cmath>

Revision as of 02:31, 15 March 2008

Problem

A finite set $\mathcal{S}$ of distinct real numbers has the following properties: the mean of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$. Find the mean of $\mathcal{S}$.

Solution

Let $x$ be the mean of $\mathcal{S}$. Let $a$ be the number of elements in $\mathcal{S}$. Then, \[\frac{ax+1}{a+1}=x-13\] and \[\frac{ax+2001}{a+1}=x+27\] \[\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}\] \[\frac{2000}{a+1}=40\] so \[2000=40(a+1)\] \[a=49\] We plug that into our very first formula, and get: \[\frac{49x+1}{50}=x-13\] \[49x+1=50x-650\] \[x=651\]

See Also

2001 AIME I (ProblemsAnswer KeyResources)
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
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