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)
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2019 AMC 8 Problems/Problem 10

Problem 10

Solution 1

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