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 | ||
− | < | + | <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> |
We plug that into our very first formula, and get: | We plug that into our very first formula, and get: | ||
− | < | + | <cmath>\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*}</cmath> |
== See Also == | == See Also == |
Revision as of 16:32, 13 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:
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.