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2022 AMC 8 Problems/Problem 19

Revision as of 07:33, 1 January 2023 by Savannahsolver (talk | contribs)

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

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 $\boxed{\textbf{(C) }4}$.

A diagram to visualize better what I explained here:

[asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram —> sry I only have 999,999,999 dollars no 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((4.5,1)); dot((4.5,1.5)); dot((4.5,2)); dot((4.5,2.5)); dot((4.5,3)); dot((4.5,3.5)); 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]

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1741

~Interstigation

https://www.youtube.com/watch?v=VuiX0JcXR7Q

~David

Video Solution

https://youtu.be/hs6y4PWnoWg?t=294

~STEMbreezy

Video Solution

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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