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Difference between revisions of "2005 AMC 12B Problems/Problem 9"
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<math>\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(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. The smallest case is that there are <math>20</math> students, because <math>\gcd\ (10,25,20,15,30)=5</math>, and <math>5\%^{-1}=20</math>. 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 in an ordered sequence 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 in an ordered sequence 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>.
 +
 
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2005|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2005|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2005|ab=B|num-b=8|num-a=10}}
 
{{AMC12 box|year=2005|ab=B|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:34, 26 November 2013

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?

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

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 in an ordered sequence 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 $\boxed{\mathrm{(B)}\ 1}$.

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

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