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Difference between revisions of "2014 AMC 10A Problems/Problem 5"

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Thus, the difference between the median and the mean is equal to <math>90-87=\boxed{\textbf{(C)}\ 3}</math>
 
Thus, the difference between the median and the mean is equal to <math>90-87=\boxed{\textbf{(C)}\ 3}</math>
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(Solution/revised by armalite46)
 
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(<math>\LaTeX</math> fixed by TrueshotBarrage)
 
  
 
==See Also==
 
==See Also==

Revision as of 13:25, 26 January 2015

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, $10\%$ of the students scored $70$ points, $35\%$ scored $80$ points, $30\%$ scored $90$ points, and the rest scored $100$ 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 $20$ students(the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points.

The median can be obtained by eliminating members from each group. The median is $90$ points.

The mean is equal to the total number of points divided by the number of people, which gives $87$

Thus, the difference between the median and the mean is equal to $90-87=\boxed{\textbf{(C)}\ 3}$

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
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. AMC logo.png

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