Difference between revisions of "2022 AMC 8 Problems/Problem 19"
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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: | 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: | ||
| − | <math>13-5=8< | + | <math>13-5=</math>8<math> |
| − | <math>\frac{8}{2}=4< | + | </math>\frac{8}{2}=4<math> |
| − | <math>13-4=9< | + | </math>13-4=9<math> |
| − | Thus, we realize that <math>4< | + | Thus, we realize that </math>4<math> students must have their score increased by </math>5$. |
So, the correct answer is \boxed{(C)4}. | So, the correct answer is \boxed{(C)4}. | ||
Revision as of 17:41, 9 December 2023
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 2
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:
8$$ (Error compiling LaTeX. Unknown error_msg)\frac{8}{2}=4$$ (Error compiling LaTeX. Unknown error_msg)13-4=9
4
5$.
So, the correct answer is \boxed{(C)4}.
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=Mnek8vOzVX77YLIS&t=3389
~Math-X
Video Solution (🚀50 seconds🚀)
~Education, the Study of Everything
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1741
~Interstigation
https://www.youtube.com/watch?v=VuiX0JcXR7Q
~David
Video Solution
https://youtu.be/hs6y4PWnoWg?t=294
~STEMbreezy
Video Solution
~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 | ||
