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

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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #9]] and [[2005 AMC 10B Problems|2005 AMC 10B #19]]}}
 
{{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?
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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>
  
−
<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5 </math>
 
 
== 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{\mathrm{(B)}\ 1}</math>.
+
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==
 
== Solution 2==
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The remaining <math>30\%</math> of the students got <math>95</math> points
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The remaining <math>30\%</math> of the students got <math>95</math> points.
−
The mean is  
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The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86</math>.
−
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{\mathrm{(B)}\ 1}</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>.
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 +
~mobius247
  
 
== See also ==
 
== See also ==

Latest revision as of 14:35, 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

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