Difference between revisions of "2019 AMC 8 Problems/Problem 10"
(→Problem 10) |
|||
(2 intermediate revisions by 2 users not shown) | |||
Line 36: | Line 36: | ||
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) 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) 21</math>. The median and mean both change, so the answer is <math>\boxed{\textbf{(B)}}</math> | 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) 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) 21</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 == | ||
+ | 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== | ||
+ | https://youtu.be/8d0VqXncuXY | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 11:21, 7 February 2022
Contents
Problem 10
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?
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 1
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 . 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
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
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.