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

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A finite [[set]] <math>\mathcal{S}</math> of distinct real numbers has the following properties: the [[arithmetic mean|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>.
 
A finite [[set]] <math>\mathcal{S}</math> of distinct real numbers has the following properties: the [[arithmetic mean|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>.
  
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== Solution ==
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==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, the given tells us that <math>\frac{ax+1}{a+1}=x-13</math> and <math>\frac{ax+2001}{a+1}=x+27</math>. Subtracting, we have  
 
Then, the given tells us that <math>\frac{ax+1}{a+1}=x-13</math> and <math>\frac{ax+2001}{a+1}=x+27</math>. Subtracting, we have  
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49x+1&=50x-650 \\
 
49x+1&=50x-650 \\
 
x&=\boxed{651}.\end{align*}</cmath>
 
x&=\boxed{651}.\end{align*}</cmath>
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==Solution 2==
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Since this is a weighted average problem, the mean of <math>S</math> is <math>\frac{13}{27}</math> as far from <math>1</math> as it is from <math>2001</math> Thus, the mean of <math>S</math> is
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<cmath>1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}</cmath>.
  
 
== See Also ==
 
== See Also ==

Revision as of 14:02, 29 March 2018

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, the given tells us that $\frac{ax+1}{a+1}=x-13$ and $\frac{ax+2001}{a+1}=x+27$. Subtracting, we have \begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*} We plug that into our very first formula, and get: \begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651}.\end{align*}

Solution 2

Since this is a weighted average problem, the mean of $S$ is $\frac{13}{27}$ as far from $1$ as it is from $2001$ Thus, the mean of $S$ is \[1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}\].

See Also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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