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

(Created page with '== Problem == The median of the list <math> n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15 </math> is <math> 10 </math>. What is the mean? <math> \textbf{(A) }4\q…')
 
 
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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>.
−
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>.
−
Substitute <math> n </math> for <math> 4 </math> and $ 4+7=\boxed{'''(B)'''11}.
+
 
 +
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 <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}}

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 (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

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