Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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Without loss of generality, let there be <math>100</math> students who took the test. We have <math>10</math> students score <math>70</math> points, <math>35</math> students score <math>80</math> points, <math>30</math> students score <math>90</math> points and <math>25</math> students score <math>100</math> points. | Without loss of generality, let there be <math>100</math> students who took the test. We have <math>10</math> students score <math>70</math> points, <math>35</math> students score <math>80</math> points, <math>30</math> students score <math>90</math> points and <math>25</math> students score <math>100</math> points. | ||
Revision as of 10:34, 9 February 2014
- The following problem is from both the 2014 AMC 12A #5 and 2014 AMC 10A #5, so both problems redirect to this page.
Problem
On an algebra quiz, 
of the students scored 
points, 
scored 
points, 
scored 
points, and the rest scored 
points. What is the difference between the mean and median score of the students' scores on this quiz?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}}\ 4\qquad\textbf{(E)}\ 5$ (Error compiling LaTeX. Unknown error_msg)
Solution
Without loss of generality, let there be students who took the test. We have 
students score points, 
students score points, 
students score points and 
students score points.
The median is easy to find by simply eliminating members from each group. The median is points.
The mean is equal to ![\[\dfrac{700+2800+2700+2500}{100}=7+28+27+25=87\]](http://latex.artofproblemsolving.com/e/e/0/ee0787b1d34e3248ffa35d1e9beeca9831bb4122.png)
Thus, the difference between the median and the mean is equal to 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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. 
