Difference between revisions of "2001 AIME I Problems/Problem 2"
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==Solution 2== | ==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 | + | 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>. | <math>1 + \frac{13}{13 + 27}(2001 - 1) = \boxed{651}</math>. | ||
Revision as of 07:40, 27 August 2020
Contents
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
.
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. 
