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> ~sosiaops | ||
+ | |||
+ | ==Solution 3== | ||
+ | The <math>\le 80</math>-point scores make up <math>10\%+35\% = 45\% < 50\%</math> of the scores, but the <math>\le 90</math>-point scores make up <math>45\%+30\% = 75\% > 50\%</math> of the scores, so the median is <math>90</math>. | ||
+ | |||
+ | <math>10\%</math> of scores were <math>70-90 = -20</math> more than the median, <math>35\%</math> were <math>-10</math> more, <math>100\%-75\% = 25\%</math> were <math>10</math> more, and the rest were equal. This means that the mean score is <math>10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3</math> more than the median, so their difference is <math>\left|-3\right| = \boxed{\textbf{(C)}\ 3}</math>. | ||
+ | ~[[User:emerald_block|emerald_block]] | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/OHF3WWRq0h4 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
==Video Solution== | ==Video Solution== |
Latest revision as of 00:10, 27 June 2023
- 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?
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
Solution 3
The -point scores make up
of the scores, but the
-point scores make up
of the scores, so the median is
.
of scores were
more than the median,
were
more,
were
more, and the rest were equal. This means that the mean score is
more than the median, so their difference is
.
~emerald_block
Video Solution (CREATIVE THINKING)
https://youtu.be/OHF3WWRq0h4
~Education, the Study of Everything
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.