Difference between revisions of "2014 AMC 10A Problems/Problem 5"
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On an algebra quiz, <math>10\%</math> of the students scored <math>70</math> points, <math>35\%</math> scored <math>80</math> points, <math>30\%</math> scored <math>90</math> points, and the rest scored <math>100</math> points. What is the difference between the mean and median score of the students' scores on this quiz? | On an algebra quiz, <math>10\%</math> of the students scored <math>70</math> points, <math>35\%</math> scored <math>80</math> points, <math>30\%</math> scored <math>90</math> points, and the rest scored <math>100</math> points. What is the difference between the mean and median score of the students' scores on this quiz? | ||
− | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> |
+ | ==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 | ||
+ | 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> | |
− | |||
− | The median | + | ==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> ~sosiaops | ||
− | + | ==Video Solution== | |
+ | https://youtu.be/Oe-QLPIuTeY | ||
− | + | ~savannahsolver | |
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==See Also== | ==See Also== | ||
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{{AMC12 box|year=2014|ab=A|num-b=4|num-a=6}} | {{AMC12 box|year=2014|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] |
Latest 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
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 ~sosiaops
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.