Difference between revisions of "2001 AIME I Problems/Problem 2"
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== Problem == | == Problem == | ||
| + | A finite [[set]] <math>\mathcal{S}</math> of distinct real numbers has the following properties: the [[arithmetic mean|mean]] of <math>\mathcal{S}\cup\{1\}</math> is <math>13</math> less than the mean of <math>\mathcal{S}</math>, and the mean of <math>\mathcal{S}\cup\{2001\}</math> is <math>27</math> more than the mean of <math>\mathcal{S}</math>. Find the mean of <math>\mathcal{S}</math>. | ||
| − | == Solution == | + | ==Solution== |
| + | 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 | ||
| + | <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: | ||
| + | <cmath>\begin{align*}\frac{49x+1}{50}&=x-13 \\ | ||
| + | 49x+1&=50x-650 \\ | ||
| + | x&=\boxed{651}.\end{align*}</cmath> | ||
| − | {{AIME box|year=2001|num-b=1|num-a=3}} | + | ==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 | ||
| + | <math>1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}</math>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/IziHKOubUI8?t=27 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | == See Also == | ||
| + | {{AIME box|year=2001|n=I|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 06:57, 4 November 2022
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:
Solution 2
Since this is a weighted average problem, the mean of 
is 
as far from 
as it is from 
. Thus, the mean of is
.
Video Solution by OmegaLearn
https://youtu.be/IziHKOubUI8?t=27
~ pi_is_3.14
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. 
