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

Revision as of 11:21, 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  which is .

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

@@ Line 9: / Line 9: @@
 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>.
-We substitute the median for <math> 10 </math> and the equation becomes <math> n+6=10 </math>. Subtract both sides by 6 and we get <math> n=4 </math>.
-<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>.
+We substitute the median for <math> 10 </math> and the equation becomes <math> n+6=10 </math>.
+Subtract both sides by 6 and we get <math> n=4 </math>.
+<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>.
 Substitute <math> n </math> for <math> 4 </math> and $ 4+7=\boxed{'''(B)'''11}.