Difference between revisions of "2019 AMC 8 Problems/Problem 10"
(→Solution 1) |
|||
| (5 intermediate revisions by 5 users not shown) | |||
| Line 1: | Line 1: | ||
| − | ==Problem | + | ==Problem== |
The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually <math>21</math> participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made? | The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually <math>21</math> participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made? | ||
| Line 33: | Line 33: | ||
<math>\textbf{(E) }</math>The mean increases by <math>5</math> and the median increases by <math>5</math>. | <math>\textbf{(E) }</math>The mean increases by <math>5</math> and the median increases by <math>5</math>. | ||
| − | ==Solution | + | ==Solution== |
| − | On Monday, <math>20</math> people come. On Tuesday, <math>26</math> people come. On Wednesday, <math>16</math> people come. On Thursday, <math>22</math> people come. Finally, on Friday, <math>16</math> people come. <math>20+26+16+22+16=100</math>, so the mean is <math>20</math>. The median is <math>(16, 16, 20, 22, 26)</math> <math>20</math>. The coach figures out that actually <math>21</math> people come on Wednesday. The new mean is <math>21</math>, while the new median is <math>(16, 20, 21, 22, 26)</math> <math>21</math>. Also, the median increases by | + | On Monday, <math>20</math> people come. On Tuesday, <math>26</math> people come. On Wednesday, <math>16</math> people come. On Thursday, <math>22</math> people come. Finally, on Friday, <math>16</math> people come. <math>20+26+16+22+16=100</math>, so the mean is <math>20</math>. The median is <math>(16, 16, 20, 22, 26)</math> <math>20</math>. The coach figures out that actually <math>21</math> people come on Wednesday. The new mean is <math>21</math>, while the new median is <math>(16, 20, 21, 22, 26)</math> <math>21</math>. Also, the median increases by <math>1</math> because now the median is <math>21</math> instead of <math>20</math>. The median and mean both change, so the answer is <math>\boxed{\textbf{(B)}}</math>. |
Another way to compute the change in mean is to notice that the sum increased by <math>5</math> with the correction. So, the average increased by <math>5/5 = 1</math>. | Another way to compute the change in mean is to notice that the sum increased by <math>5</math> with the correction. So, the average increased by <math>5/5 = 1</math>. | ||
| − | |||
| − | |||
==Video Solution by Math-X (First fully understand the problem!!!)== | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
| Line 65: | Line 63: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
| − | ==Video Solution by The Power of Logic( | + | ==Video Solution by The Power of Logic(1 to 25 Full Solution)== |
https://youtu.be/Xm4ZGND9WoY | https://youtu.be/Xm4ZGND9WoY | ||
Latest revision as of 09:31, 9 November 2024
Contents
Problem
The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually 
participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?
![[asy] unitsize(2mm); defaultpen(fontsize(8bp)); real d = 5; real t = 0.7; real r; int[] num = {20,26,16,22,16}; string[] days = {"Monday","Tuesday","Wednesday","Thursday","Friday"}; for (int i=0; i<30; i=i+2) { draw((i,0)--(i,-5*d),gray); }for (int i=0; i<5; ++i) { r = -1*(i+0.5)*d; fill((0,r-t)--(0,r+t)--(num[i],r+t)--(num[i],r-t)--cycle,gray); label(days[i],(-1,r),W); }for(int i=0; i<32; i=i+4) { label(string(i),(i,1)); }label("Number of students at soccer practice",(14,3.5)); [/asy]](http://latex.artofproblemsolving.com/e/1/7/e1796ef723acbb9a4cfeacf57e69202cddf3e0d3.png)
The mean increases by 
and the median does not change.
The mean increases by and the median increases by .
The mean increases by and the median increases by 
.
The mean increases by and the median increases by .
The mean increases by and the median increases by .
Solution
On Monday, 
people come. On Tuesday, 
people come. On Wednesday, 
people come. On Thursday, 
people come. Finally, on Friday, people come. 
, so the mean is . The median is 
. The coach figures out that actually people come on Wednesday. The new mean is , while the new median is 
. Also, the median increases by because now the median is instead of . The median and mean both change, so the answer is 
.
Another way to compute the change in mean is to notice that the sum increased by with the correction. So, the average increased by 
.
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=Al-ATwscPJRaKEtF&t=3094
~Math-X
The Learning Royal : https://youtu.be/8njQzoztDGc
Video Solution 2
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=vkm1ZXuuQcc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=11
Video Solution 3
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=2171
~ pi_is_3.14
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic(1 to 25 Full Solution)
~Hayabusa1
See Also
| 2019 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
