Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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==Solution== | ==Solution== | ||
| + | [[WLOG]] 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. The median is easy to find by simply eliminating members from each group. The median is <math>90</math> points. The mean is just <math>\dfrac{700+2800+2700+2500}{100}=7+28+27+25=87</math>. The difference is <math>90-87=\boxed{\textbf{(C)}\ 3}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 23:34, 6 February 2014
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
WLOG 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 just 
. The difference is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
