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> | ||
| − | == Solution == | + | == Solution 1== |
| − | 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. | + | 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> |
In progress please do not edit | In progress please do not edit | ||
Revision as of 17:55, 28 January 2022
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 1
The median is the average of the 
and 
scores. In order to minimize the number of changed scores, the score must be . It is easiest to move the people whose scores are 
since it only takes 
move. There are currently ![\[\]](http://latex.artofproblemsolving.com/5/7/f/57f1406b65f9f2e6a365b7356da731780b42cceb.png)
In progress please do not edit
Solution 2
Before Mr. Ramos added scores, the median was 
. There are two cases now:
Case #: The middle two scores are and 
. To do this, we firstly suppose that the two students who got are awarded the extra points. We then realize that this case will have a lot of students who receive the extra points, therefore we reject this case.
Case #: The middle two scores are both . To do this, we simply need to suppose that some of the students who got are awarded the extra points. Note that there are 
students who got 
or less. Therefore there must be only student who got so that the middle two scores are both . Therefore our answer is 
.
