Difference between revisions of "2022 AMC 8 Problems/Problem 19"
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/Ij9pAy6tQSg?t=1741 | https://youtu.be/Ij9pAy6tQSg?t=1741 | ||
| + | ~Interstigation | ||
| − | + | https://www.youtube.com/watch?v=VuiX0JcXR7Q | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=18|num-a=20}} | {{AMC8 box|year=2022|num-b=18|num-a=20}} | ||
Revision as of 22:46, 28 October 2022
Contents
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
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 
.
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]](http://latex.artofproblemsolving.com/1/9/9/199e9138c545eeaa3b267478380ab149ac47c02b.png)
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1741 ~Interstigation
https://www.youtube.com/watch?v=VuiX0JcXR7Q
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 | ||
