Revision as of 12:34, 18 April 2020

Problem 10

sfgd

$\textbf{(A) }$ 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 $5/5 = 1$ .

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.

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 =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 <math>21</math> participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?
+sfgd
-<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>
 <math>\textbf{(A) }</math>The mean increases by <math>1</math> and the median does not change.

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

Revision as of 12:34, 18 April 2020

Problem 10

Solution 1

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