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

m (Solution 2)
m (Solution 2)
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==Solution 2==
 
==Solution 2==
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
+
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>.
 
<math>1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}</math>.
  

Revision as of 08:40, 27 August 2020

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 (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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