Difference between revisions of "2001 AMC 12 Problems/Problem 4"
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Revision as of 20:02, 3 July 2013
- The following problem is from both the 2001 AMC 12 #4 and 2001 AMC 10 #16, so both problems redirect to this page.
Problem
The mean of three numbers is 
more than the least of the numbers and 
less than the greatest. The median of the three numbers is 
. What is their
sum?
Solution
Let 
be the mean of the three numbers. Then the least of the numbers is 
and the greatest is 
. The middle of the three numbers is the median, 5. So
![$\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$](http://latex.artofproblemsolving.com/c/a/c/cac4efd76d95dce2367e45a25811bc7142c1a3b8.png)
, which implies that 
.
Hence, the sum of the three numbers is 
, and the answer is 
.
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 | |
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
