Difference between revisions of "2005 AMC 12B Problems/Problem 9"
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== Solution 2== | == Solution 2== | ||
| − | The remaining <math>30\%</math> of the students got <math>95</math> points | + | 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</math> | + | The mean is equal to <math>10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = </math>86<math></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>. | 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>. | ||
Revision as of 08:37, 1 June 2021
- The following problem is from both the 2005 AMC 12B #9 and 2005 AMC 10B #19, so both problems redirect to this page.
Contents
Problem
On a certain math exam, 
of the students got 
points, 
got 
points, 
got 
points, 
got 
points, and the rest got 
points. What is the difference between the mean and the median score on this exam?
Solution
To begin, we see that the remaining 
of the students got points. Assume that there are 
students; we see that 
students got points, 
students got points, 
students got points, 
students got points, and 
students got points. The median is , since the 
and 
terms are both . The mean is 
. The difference between the mean and median, therefore, is 
.
Solution 2
The remaining of the students got points.
The mean is equal to 
86$$ (Error compiling LaTeX. Unknown error_msg)
To begin, we see that the remaining of the students got points. Assume that there are students; we see that students got points, students got points, students got points, students got points, and students got points. The median is , since the and terms are both . The mean is . The difference between the mean and median, therefore, is .
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. 
