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

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

@@ Line 46: / Line 46: @@
 == Solution ==
-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.
+We set up our cases as solution 1 showed, realizing that only the second case is possible.
-In progress please do not edit
+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:
-== Solution 2 ==
+<cmath>13-5=8</cmath>
-Before Mr. Ramos added scores, the median was <math>\frac{80+80}{2}=80</math>. There are two cases now:
+<cmath>\frac{8}{2}=4</cmath>
+<cmath>13-4=9</cmath>
-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.
+Thus, we realize that <math>4</math> students must have their score increased by <math>5</math>.
-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>.
+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}}

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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