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2019 AMC 10A Problems/Problem 12

Revision as of 14:31, 31 May 2020 by Mathwinner1 (talk | contribs)
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 1

First of all, $d$ obviously has to be smaller than $M$, since when calculating $M$, we must take into account the $29$s, $30$s, and $31$s. So we can eliminate choices $B$ and $C$. Since there are $365$ total entries, the median, $M$, must be the $183\text{rd}$ one, at which point we note that $12 \cdot 15$ is $180$, so $16$ has to be the median (because $183$ is between $12 \cdot 15 + 1 = 181$ and $12 \cdot 16 = 192$). Now, the mean, $\mu$, must be smaller than $16$, since there are many fewer $29$s, $30$s, and $31$s. $d$ is less than $\mu$, because when calculating $\mu$, we would include $29$, $30$, and $31$. Thus the answer is $\boxed{\textbf{(E) } d < \mu < M}$.

Solution 2

As in Solution 1, we find that the median is $16$. Then, looking at the modes $(1-28)$, we realize that even if we were to have $12$ of each, their median would remain the same, being $14.5$. As for the mean, we note that the mean of the first $28$ is simply the same as the median of them, which is $14.5$. Hence, since we in fact have $29$'s, $30$'s, and $31$'s, the mean has to be higher than $14.5$. 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). By comparing these values, the answer is $\boxed{\textbf{(E) } d < \mu < M}$.

Solution 3 (direct calculation)

We can solve this problem simply by carefully calculating each of the values, which turn out to be $M=16$, $d=14.5$, and $\mu \approx 15.7$. Thus the answer is $\boxed{\textbf{(E) } d < \mu < M}$.

See Also

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

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