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Difference between revisions of "2019 AMC 8 Problems/Problem 10"

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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>.
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 +
== Video Solution ==
 +
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=vkm1ZXuuQcc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=11
  
 
==See Also==
 
==See Also==

Revision as of 12:50, 23 April 2021

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 $21$ 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]

$\textbf{(A) }$The mean increases by $1$ and the median does not change.

$\textbf{(B) }$The mean increases by $1$ and the median increases by $1$.

$\textbf{(C) }$The mean increases by $1$ and the median increases by $5$.

$\textbf{(D) }$The mean increases by $5$ and the median increases by $1$.

$\textbf{(E) }$The mean increases by $5$ and the median increases by $5$.

Solution 1

On Monday, $20$ people come. On Tuesday, $26$ people come. On Wednesday, $16$ people come. On Thursday, $22$ people come. Finally, on Friday, $16$ people come. $20+26+16+22+16=100$, so the mean is $20$. The median is $(16, 16, 20, 22, 26) 20$. The coach figures out that actually $21$ people come on Wednesday. The new mean is $21$, while the new median is $(16, 20, 21, 22, 26) 21$. The median and mean both change, so the answer is $\boxed{\textbf{(B)}}$ Another way to compute the change in mean is to notice that the sum increased by $5$ with the correction. So the average increased by $5/5 = 1$.

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=vkm1ZXuuQcc&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=11

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

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