# 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

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 $\boxed{(C)4}$.

~Math-X

## Video Solution 2

~Education, the Study of Everything

~Interstigation

~David

~STEMbreezy

~savannahsolver

## See Also

 2022 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 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