Difference between revisions of "2001 AIME I Problems/Problem 2"
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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>. | 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 | ||
| − | <center | + | <center>\begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49</center> |
We plug that into our very first formula, and get: | We plug that into our very first formula, and get: | ||
| − | <center | + | <center>\begin{align*}\frac{49x+1}{50}&=x-13 \\ |
49x+1&=50x-650 \\ | 49x+1&=50x-650 \\ | ||
| − | x&=\boxed{651}.\end{align*} | + | x&=\boxed{651}.\end{align*}</center> |
== See Also == | == See Also == | ||
Revision as of 22:23, 4 March 2015
Problem
A finite set 
of distinct real numbers has the following properties: the mean of 
is 
less than the mean of , and the mean of 
is 
more than the mean of . Find the mean of .
Solution
Let 
be the mean of . Let 
be the number of elements in .
Then, the given tells us that 
and 
. Subtracting, we have
We plug that into our very first formula, and get:
49x+1&=50x-650 \\
x&=\boxed{651}.\end{align*}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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
