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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 2==
 
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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>
 
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The median can be solved by finding the score present at the 50% mark, which is 90.<br>
 
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90-87 equals 3, which is (C).
 
  
 
==See Also==
 
==See Also==

Revision as of 11:34, 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, $10\%$ 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 1

Without loss of generality, let there be $100$ students who took the test. We have $10$ students score $70$ points, $35$ students score $80$ points, $30$ students score $90$ points and $25$ students score $100$ points.

The median is easy to find by simply eliminating members from each group. The median is $90$ points.

The mean is equal to \[\dfrac{700+2800+2700+2500}{100}=7+28+27+25=87\]

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

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