2022 AMC 8 Problems/Problem 19

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

The median is the average of the $10^\text{th}$ and $11^\text{th}$ scores. In order to minimize the number of changed scores, the $10^\text{th}$ score must be $85$ . It is easiest to move the people whose scores are $80$ since it only takes $1$ move.

In progress please do not edit

Solution 2

Before Mr. Ramos added scores, the median was $\frac{80+80}{2}=80$ . There are two cases now:

Case # $1$ : The middle two scores are $80$ and $90$ . To do this, we firstly suppose that the two students who got $85$ are awarded the extra $5$ points. We then realize that this case will have a lot of students who receive the extra points, therefore we reject this case.

Case # $2$ : The middle two scores are both $85$ . To do this, we simply need to suppose that some of the students who got $80$ are awarded the extra $5$ points. Note that there are 8 students who got $75$ or less. Therefore there must be only $1$ student who got $80$ so that the middle two scores are both $85$ . Therefore our answer is $\boxed{\textbf{(C) }4}$ .

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2022 AMC 8 Problems/Problem 19

Problem

Solution

Solution 2