Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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The mean can solved by the following. 10% of 70 is 7, 35% of 80 is 28, 30% of 90 is 27, and (100% - 10% - 35% - 30%) = 25%. 25% of 100 is 25. 7 + 28 + 27 + 25 = 87. <br> | The mean can solved by the following. 10% of 70 is 7, 35% of 80 is 28, 30% of 90 is 27, and (100% - 10% - 35% - 30%) = 25%. 25% of 100 is 25. 7 + 28 + 27 + 25 = 87. <br> | ||
The median can be solved by finding the score present at the 50% mark, which is 90.<br> | The median can be solved by finding the score present at the 50% mark, which is 90.<br> | ||
− | 90-87 equals 3, which is (C) 3 | + | 90-87 equals 3, which is \boxed{\textbf{(C)}\ 3}$ |
==See Also== | ==See Also== |
Revision as of 20:50, 7 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.
Contents
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 1
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
Solution 2
The mean can solved by the following. 10% of 70 is 7, 35% of 80 is 28, 30% of 90 is 27, and (100% - 10% - 35% - 30%) = 25%. 25% of 100 is 25. 7 + 28 + 27 + 25 = 87.
The median can be solved by finding the score present at the 50% mark, which is 90.
90-87 equals 3, which is \boxed{\textbf{(C)}\ 3}$
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.