Difference between revisions of "2022 AMC 8 Problems/Problem 19"
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== Solution == | == Solution == | ||
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| − | + | We set up our cases as solution 1 showed, realizing that only the second case is possible. | |
| − | + | We notice that <math>13</math> students have scores under <math>85</math> currently and only <math>5</math> have scores over <math>85</math>. We find the median of these two numbers, getting: | |
| − | + | <cmath>13-5=8</cmath> | |
| + | <cmath>\frac{8}{2}=4</cmath> | ||
| + | <cmath>13-4=9</cmath> | ||
| − | < | + | Thus, we realize that <math>4</math> students must have their score increased by <math>5</math>. |
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| − | ==Video Solution== | + | So, the correct answer is <math>\boxed{\textbf{(C)}4}</math>. |
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/oUEa7AjMF2A?si=Mnek8vOzVX77YLIS&t=3389 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/eSueEOf15c8 | ||
| + | |||
| + | <i>~Education, the Study of Everything</i> | ||
| + | |||
| + | ==Video Solution 3== | ||
https://youtu.be/Ij9pAy6tQSg?t=1741 | https://youtu.be/Ij9pAy6tQSg?t=1741 | ||
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https://www.youtube.com/watch?v=VuiX0JcXR7Q | https://www.youtube.com/watch?v=VuiX0JcXR7Q | ||
| + | |||
| + | ~David | ||
| + | |||
| + | ==Video Solution 4== | ||
| + | https://youtu.be/hs6y4PWnoWg?t=294 | ||
| + | |||
| + | ~STEMbreezy | ||
| + | |||
| + | ==Video Solution 5== | ||
| + | https://youtu.be/jXx0fc-DgQM | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=18|num-a=20}} | {{AMC8 box|year=2022|num-b=18|num-a=20}} | ||
Latest revision as of 05:07, 3 November 2024
Contents
Problem
Mr. Ramos gave a test to his class of 
students. The dot plot below shows the distribution of test scores.
![[asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((3.5,1.5)); dot((3.5,2)); dot((3.5,2.5)); dot((3.5,3)); dot((4.5,1)); dot((4.5,1.5)); dot((5.5,1)); dot((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label("$65$", (0.5,-0.1)); label("$70$", (1.5,-0.1)); label("$75$", (2.5,-0.1)); label("$80$", (3.5,-0.1)); label("$85$", (4.5,-0.1)); label("$90$", (5.5,-0.1)); label("$95$", (6.5,-0.1)); label("$100$", (7.5,-0.1)); [/asy]](http://latex.artofproblemsolving.com/8/f/1/8f168ff6d1e7635e68e0139420bed9ecfc2e4993.png)
Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students 
extra points, which increased the median test score to 
. What is the minimum number of students who received extra points?
(Note that the median test score equals the average of the 
scores in the middle if the test scores are arranged in increasing order.)
Solution
We set up our cases as solution 1 showed, realizing that only the second case is possible.
We notice that 
students have scores under currently and only have scores over . We find the median of these two numbers, getting:
![\[13-5=8\]](http://latex.artofproblemsolving.com/a/e/a/aea3e8150f45440100b8d998970e1d8fd6d61859.png)
![\[\frac{8}{2}=4\]](http://latex.artofproblemsolving.com/d/b/0/db0178224a8588827d38ca0711bf889750cf0e66.png)
![\[13-4=9\]](http://latex.artofproblemsolving.com/8/d/1/8d127b922e0e675e4f34eebea490cbd3c3d38322.png)
Thus, we realize that 
students must have their score increased by .
So, the correct answer is 
.
Video Solution 1
https://youtu.be/oUEa7AjMF2A?si=Mnek8vOzVX77YLIS&t=3389
~Math-X
Video Solution 2
~Education, the Study of Everything
Video Solution 3
https://youtu.be/Ij9pAy6tQSg?t=1741
~Interstigation
https://www.youtube.com/watch?v=VuiX0JcXR7Q
~David
Video Solution 4
https://youtu.be/hs6y4PWnoWg?t=294
~STEMbreezy
Video Solution 5
~savannahsolver
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
