Question

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 $\mathcal{S}.$

Answer

Answer: 651

Concepts:mean system of equations

Difficulty rating: 1840

Solution:

Let $\mathcal{S}$ have $n$ elements with mean $x,$ so the elements sum to $nx.$ The two conditions say $\frac{nx + 1}{n + 1} = x - 13$ and $\frac{nx + 2001}{n + 1} = x + 27.$

Clearing denominators, $nx + 1 = (n+1)x - 13(n+1)$ gives $x - 13(n+1) = 1,$ and $nx + 2001$ $= (n+1)x + 27(n+1)$ gives $x + 27(n+1) = 2001.$ Subtracting the first equation from the second yields $40(n+1) = 2000,$ so $n + 1 = 50.$

Then $x = 1 + 13 \cdot 50 = 651.$

2001 AIME I Problem 2

Answer: 651

Difficulty rating: 1840

Solution:

Problem 2 in Other Years