Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Without loss of generality, let there be <math>20</math> students (the least whole number possible) who took the test. We have <math>2</math> students score <math>70</math> points, <math>7</math> students score <math>80</math> points, <math>6</math> students score <math>90</math> points and <math>5</math> students score <math>100</math> points. Therefore, the mean | Without loss of generality, let there be <math>20</math> students (the least whole number possible) who took the test. We have <math>2</math> students score <math>70</math> points, <math>7</math> students score <math>80</math> points, <math>6</math> students score <math>90</math> points and <math>5</math> students score <math>100</math> points. Therefore, the mean | ||
is <math>87</math> and the median is <math>90</math>. | is <math>87</math> and the median is <math>90</math>. | ||
Thus, the solution is <cmath>90-87=3\implies\boxed{\textbf{(C)} \ 3}</cmath> | Thus, the solution is <cmath>90-87=3\implies\boxed{\textbf{(C)} \ 3}</cmath> | ||
| + | |||
| + | ==Solution 2== | ||
| + | The percentage who scored <math>100</math> points is <math>100\%-(10\%+35\%+30\%)=100\%-75\%=25\%</math>. Now, we need to find the median, which is the score that splits the upper and lower <math>50\%</math>.The lower <math>10\%+35\%=45\%</math> scored <math>70</math> or <math>80</math> points, so the median is <math>90</math> (since the upper <math>25\%</math> is <math>100</math> points and the lower <math>45\%</math> is <math>70</math> or <math>80</math>).The mean is <math>10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87</math>. | ||
| + | So, our solution is <math>90-87=3\Rightarrow\boxed{3 \ \textbf{(C)} }</math> | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 22:24, 18 January 2021
- 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?
Solution 1
Without loss of generality, let there be 
students (the least whole number possible) who took the test. We have 
students score points, 
students score points, 
students score points and 
students score points. Therefore, the mean
is 
and the median is .
Thus, the solution is ![\[90-87=3\implies\boxed{\textbf{(C)} \ 3}\]](http://latex.artofproblemsolving.com/9/c/c/9cc0a8bdca61ca93db41147a74ef754c1822910c.png)
Solution 2
The percentage who scored points is 
. Now, we need to find the median, which is the score that splits the upper and lower 
.The lower 
scored or points, so the median is (since the upper 
is points and the lower 
is or ).The mean is 
.
So, our solution is 
Video Solution
~savannahsolver
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. 
