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

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.

Page

Toolbox

Search

2019 AMC 8 Problems/Problem 10

Problem 10

Solution 1

See Also