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

Latest revision as of 23:10, 26 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.

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?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

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. Therefore, the mean is $87$ and the median is $90$ .

Thus, the solution is $90-87=3\implies\boxed{\textbf{(C)} \ 3}$

Solution 2

The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $70$ or $80$ ).The mean is $10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87$ . So, our solution is $90-87=3\Rightarrow\boxed{3 \ \textbf{(C)} }$ ~sosiaops

Solution 3

The $\le 80$ -point scores make up $10\%+35\% = 45\% < 50\%$ of the scores, but the $\le 90$ -point scores make up $45\%+30\% = 75\% > 50\%$ of the scores, so the median is $90$ .

$10\%$ of scores were $70-90 = -20$ more than the median, $35\%$ were $-10$ more, $100\%-75\% = 25\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\left|-3\right| = \boxed{\textbf{(C)}\ 3}$ . ~emerald_block

Video Solution (CREATIVE THINKING)

https://youtu.be/OHF3WWRq0h4

~Education, the Study of Everything

Video Solution

https://youtu.be/Oe-QLPIuTeY

~savannahsolver

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.

@@ Line 1: / Line 1: @@
+{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #5]] and [[2014 AMC 10A Problems|2014 AMC 10A #5]]}}
 ==Problem==
 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)}}\ 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 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>
+==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==
+https://youtu.be/Oe-QLPIuTeY
+~savannahsolver
+==See Also==
+{{AMC10 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}}
+[[Category: Introductory Algebra Problems]]

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