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

Latest revision as of 16:12, 18 October 2025

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 $4+7=\boxed{\textbf{(E) }11}$ .

~Thesmartgreekmathdude

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

@@ 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{\textbf{(E) }11} </math>.
+==Video Solution by Daily Dose of Math==
+https://youtu.be/GQNnGl3nFok?si=wnPOHufsNE5QbaXY
+~Thesmartgreekmathdude
+== See Also ==
+{{AMC10 box|year=2001|before=First<br />Question|num-a=2}}
+{{MAA Notice}}
+[[Category: Introductory Algebra Problems]]