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

(Solution)
 
Line 19: Line 19:
  
 
Substitute <math> n </math> for <math> 4 </math> and <math> 4+7=\boxed{\textbf{(E) }11} </math>.
 
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 ==
 
== See Also ==

Latest revision as of 15:08, 15 July 2024

Problem

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?

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

Solution

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

Video Solution by Daily Dose of Math

https://youtu.be/GQNnGl3nFok?si=wnPOHufsNE5QbaXY

~Thesmartgreekmathdude

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
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. AMC logo.png