2001 AIME I Problems/Problem 2

Revision as of 22:03, 21 December 2007 by Serialk11r (talk | contribs) (Solution)


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 $\mathcal{S}$.


Let x be the mean of S. Let a be the number of elements in 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
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