Difference between revisions of "2001 AMC 10 Problems/Problem 1"
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<math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math> | <math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math> | ||
| − | == | + | == Solution == |
The median of the list is <math> 10 </math>, and there are <math> 9 </math> numbers in the list, so the median must be the 5th number from the left, which is <math> n+6 </math>. | The median of the list is <math> 10 </math>, and there are <math> 9 </math> numbers in the list, so the median must be the 5th number from the left, which is <math> n+6 </math>. | ||
Revision as of 22:18, 9 December 2020
Problem
The median of the list
is 
. What is the mean?
Solution
The median of the list is , and there are 
numbers in the list, so the median must be the 5th number from the left, which is 
.
We substitute the median for and the equation becomes 
.
Subtract both sides by 6 and we get 
.
.
The mean of those numbers is 
which is 
.
Substitute 
for 
and 
.
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 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. 
