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Difference between revisions of "2022 AMC 8 Problems/Problem 19"

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<math>\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad</math>
 
<math>\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad</math>
  
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== Solution 1==
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== Solution ==
  
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The median is the average of the <math>10^\text{th}</math> and <math>11^\text{th}</math> scores. In order to minimize the number of changed scores, the <math>10^\text{th}</math> score must be <math>85</math>. It is easiest to move the people whose scores are <math>80</math> since it only takes <math>1</math> move. There are currently <cmath></cmath>
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We set up our cases as solution 1 showed, realizing that only the second case is possible.  
  
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In progress please do not edit
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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:
  
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== Solution 2 ==
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<cmath>13-5=8</cmath>
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Before Mr. Ramos added scores, the median was <math>\frac{80+80}{2}=80</math>. There are two cases now:
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<cmath>\frac{8}{2}=4</cmath>
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<cmath>13-4=9</cmath>
  
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Case #<math>1</math>: The middle two scores are <math>80</math> and <math>90</math>. To do this, we firstly suppose that the two students who got <math>85</math> are awarded the extra <math>5</math> points. We then realize that this case will have a lot of students who receive the extra points, therefore we reject this case.
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Thus, we realize that <math>4</math> students must have their score increased by <math>5</math>.
  
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Case #<math>2</math>: The middle two scores are both <math>85</math>. To do this, we simply need to suppose that some of the students who got <math>80</math> are awarded the extra <math>5</math> points. Note that there are <math>8</math> students who got <math>75</math> or less. Therefore there must be only <math>1</math> student who got <math>80</math> so that the middle two scores are both <math>85</math>. Therefore our answer is <math>\boxed{\textbf{(C) }4}</math>.
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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
 +
 
 +
~Interstigation
 +
 
 +
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==
 +
{{AMC8 box|year=2022|num-b=18|num-a=20}}

Latest revision as of 05:07, 3 November 2024

Problem

Mr. Ramos gave a test to his class of $20$ 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]

Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$. What is the minimum number of students who received extra points?

(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)

$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$

Solution

We set up our cases as solution 1 showed, realizing that only the second case is possible.

We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$. We find the median of these two numbers, getting:

\[13-5=8\] \[\frac{8}{2}=4\] \[13-4=9\]

Thus, we realize that $4$ students must have their score increased by $5$.

So, the correct answer is $\boxed{\textbf{(C)}4}$.

Video Solution 1

https://youtu.be/oUEa7AjMF2A?si=Mnek8vOzVX77YLIS&t=3389

~Math-X

Video Solution 2

https://youtu.be/eSueEOf15c8

~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

https://youtu.be/jXx0fc-DgQM

~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
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