Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 12B Problems/Problem 9"

m (tagged empty)
(→‎Solution 2)
 
(25 intermediate revisions by 7 users not shown)
Line 1: Line 1:
−
{{empty}}
+
{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #9]] and [[2005 AMC 10B Problems|2005 AMC 10B #19]]}}
 
== Problem ==
 
== Problem ==
 +
On a certain math exam, <math>10\%</math> of the students got <math>70</math> points, <math>25\%</math> got <math>80</math> points, <math>20\%</math> got <math>85</math> points, <math>15\%</math> got <math>90</math> points, and the rest got <math>95</math> points.  What is the difference between the [[mean]] and the [[median]] score on this exam?
 +
 +
<math>\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5 </math>
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
To begin, we see that the remaining <math>30\%</math> of the students got <math>95</math> points. Assume that there are <math>20</math> students; we see that <math>2</math> students got <math>70</math> points, <math>5</math> students got <math>80</math> points, <math>4</math> students got <math>85</math> points, <math>3</math> students got <math>90</math> points, and <math>6</math> students got <math>95</math> points. The median is <math>85</math>, since the <math>10^{\text{th}}</math> and <math>11^{\text{th}}</math> terms are both <math>85</math>. The mean is <math>\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86</math>. The difference between the mean and median, therefore, is <math>\boxed{\textbf{(B)}\ 1}</math>.
 +
 
 +
== Solution 2==
 +
The remaining <math>30\%</math> of the students got <math>95</math> points.
 +
The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86</math>.
 +
The score greater than or equal to <math>50\%</math> of other scores is the median. Since <math>35\%</math> scored <math>80</math> or lower and the next <math>20\%</math> scored <math>85</math>, the median is <math>85</math>. The difference between the mean and the median is <math>\boxed{\textbf{(B)}\ 1}</math>.
 +
 
 +
~mobius247
 +
 
 
== See also ==
 
== See also ==
−
* [[2005 AMC 12B Problems/Problem 8 | Previous problem]]
+
{{AMC10 box|year=2005|ab=B|num-b=18|num-a=20}}
−
* [[2005 AMC 12B Problems/Problem 10 | Next problem]]
+
{{AMC12 box|year=2005|ab=B|num-b=8|num-a=10}}
−
* [[2005 AMC 12B Problems]]
+
{{MAA Notice}}

Latest revision as of 15:35, 16 December 2021

The following problem is from both the 2005 AMC 12B #9 and 2005 AMC 10B #19, so both problems redirect to this page.

Problem

On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?

$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$

Solution

To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$, since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$. The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$. The difference between the mean and median, therefore, is $\boxed{\textbf{(B)}\ 1}$.

Solution 2

The remaining $30\%$ of the students got $95$ points. The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$. The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$, the median is $85$. The difference between the mean and the median is $\boxed{\textbf{(B)}\ 1}$.

~mobius247

See also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_9&oldid=167802"