Difference between revisions of "2001 AMC 10 Problems/Problem 1"

Revision as of 12:21, 16 March 2011

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$ .

We substitute the median for $10$ and the equation becomes $n+6=10$ .

Subtract both sides by 6 and we get $n=4$ .

$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$ .

The mean of those numbers is $\frac{9n-63}{9}$ which is $n+7$ .

Substitute $n$ for $4$ and $4+7=\boxed{'''(B)'''11}$ .

@@ Line 16: / Line 16: @@
 <math> n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63 </math>.
- The mean of those numbers is <math> \frac{9n-63}{9} </math> which is <math> n+7 </math>.
+The mean of those numbers is <math> \frac{9n-63}{9} </math> which is <math> n+7 </math>.
-Substitute <math> n </math> for <math> 4 </math> and $ 4+7=\boxed{'''(B)'''11}.
+Substitute <math> n </math> for <math> 4 </math> and <math> 4+7=\boxed{'''(B)'''11} </math>.