Difference between revisions of "2010 AMC 12B Problems/Problem 10"
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Revision as of 01:15, 22 December 2020
- The following problem is from both the 2010 AMC 12B #10 and 2010 AMC 10B #14, so both problems redirect to this page.
Contents
Problem 10
The average of the numbers 
and 
is 
. What is ?
Solution
We first sum the first 
numbers: 
. Then, we know that the sum of the series is 
. There are 
terms, so we can divide this sum by and set it equal to :
![\[\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x\]](http://latex.artofproblemsolving.com/d/e/e/dee34095c95d4eb8d0faffde82a81b4071561d9e.png)
Using difference of squares:
![\[x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}\]](http://latex.artofproblemsolving.com/5/d/3/5d32530090ee83aa2d0b78c9093d51789be3b4f0.png)
Thus, the answer is 
.
Video Solution
https://youtu.be/vYXz4wStBUU?t=413
~IceMatrix
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
