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

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== Problem ==
 
== Problem ==
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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>.
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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>.
  
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== Solution ==
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==Solution==
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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>.
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Then,
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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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<cmath>\frac{ax+1}{a+1}=x-13</cmath> and <cmath>\frac{ax+2001}{a+1}=x+27</cmath>
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<cmath>\begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*}</cmath>
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<cmath>\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1}</cmath>
 
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<cmath>\frac{2000}{a+1}=40</cmath> so <cmath>2000=40(a+1)</cmath>
 
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<cmath>a=49</cmath>
 
 
We plug that into our very first formula, and get:
 
We plug that into our very first formula, and get:
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<cmath>\frac{49x+1}{50}=x-13</cmath>
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<cmath>\begin{align*}\frac{49x+1}{50}&=x-13 \\
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<cmath>49x+1=50x-650</cmath>
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49x+1&=50x-650 \\
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<cmath>x=651</cmath>
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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
 +
<math>1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}</math>.
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/IziHKOubUI8?t=27
 +
 
 +
~ pi_is_3.14
  
 
== 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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 +
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:57, 4 November 2022

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 $1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}$.

Video Solution by OmegaLearn

https://youtu.be/IziHKOubUI8?t=27

~ pi_is_3.14

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

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