2019 AMC 8 Problems/Problem 10

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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 $21$ 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 $5/5 = 1$.

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
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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