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Algebra Difficulty 3.2 AMC 10/12 Find the answer

A finite set S\mathcal{S} of distinct real numbers has the following properties: the mean of S∪{1}\mathcal{S}\cup\{1\} is 1313 less than the mean of S\mathcal{S}, and the mean of S∪{2001}\mathcal{S}\cup\{2001\} is 2727 more than the mean of S\mathcal{S}. Find the mean of S\mathcal{S}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let xx be the mean of S\mathcal{S}. Let aa be the number of elements in S\mathcal{S}.
Then, the given tells us that ax+1a+1=x−13\frac{ax+1}{a+1}=x-13 and ax+2001a+1=x+27\frac{ax+2001}{a+1}=x+27. Subtracting, we have
ax+2001a+1−40=ax+1a+1⟹2000a+1=40⟹a=49\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:
49x+150=x−1349x+1=50x−650x=651.\begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651}.\end{align*}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.