2019 AMC 10A Problems/Problem 12

The following problem is from both the 2019 AMC 10A #12 and 2019 AMC 12A #7, so both problems redirect to this page.

Problem

Melanie computes the mean $\mu$ , the median $M$ , and the modes of the $365$ values that are the dates in the months of $2019$ . Thus her data consist of $12$ $1\text{s}$ , $12$ $2\text{s}$ , . . . , $12$ $28\text{s}$ , $11$ $29\text{s}$ , $11$ $30\text{s}$ , and $7$ $31\text{s}$ . Let $d$ be the median of the modes. Which of the following statements is true?

$\textbf{(A) } \mu < d < M \qquad\textbf{(B) } M < d < \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d < M < \mu \qquad\textbf{(E) } d < \mu < M$

Solution

Solution 1

First of all, $d$ obviously has to smaller than $M$ since when calculating $M$ you must take into account the $29's$ , $30's$ , and $31s$ . So we can eliminate $(B)$ and $(C)$ . The median, $\mu$ , is $16$ , but the mean ( $M$ ) must be smaller than $16$ since there are much less $29's$ , $30's$ , and $31s$ . $d$ is less than $\mu$ because when calculating $\mu$ you include $29$ , $30$ , and $31$ . Thus the answer is $d < \mu < M \implies \boxed{\textbf{(E)}}$

Solution 2

Notice that there are $365$ total entries, so the median has to be the $183\text{rd}$ one. Then, realize that $12 \cdot 15$ is $180$ , so $16$ has to be the median (because $16$ is from $181$ to $192$ ). Then, look at the modes $(1-28)$ and realize that even if you have $12$ of each, the median of those remains the same and you have $14.5$ . When trying to find the mean, you realize that the mean of the first $28$ is simply the same as the median of them, which is $14.5$ . Then, when you see $29$ 's, $30$ 's, and $31$ 's, you realize that the mean has to be higher. On the other hand, since there are fewer $29$ 's, $30$ 's, and $31$ 's than the rest of the numbers, the mean has to be lower than $16$ (the median). Then, you compare those values and you get the answer, which is $\boxed{\textbf{(E)}}$ .

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 10 Problems and Solutions

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2019 AMC 10A Problems/Problem 12

Contents

Problem

Solution

Solution 1

Solution 2

See Also