Difference between revisions of "2019 AMC 10A Problems/Problem 12"

Revision as of 15:31, 31 May 2020

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 $\boxed{\textbf{(E) } d < \mu < M}$ .

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

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

@@ Line 18: / Line 18: @@
 We can solve this problem simply by carefully calculating each of the values, which turn out to be <math>M=16</math>, <math>d=14.5</math>, and <math>\mu \approx 15.7</math>. Thus the answer is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
-==Solution 4 (instant 1 second solve)==
-All we have to do is to notice that 1-27 occur every month, so all of these are the modes. Hence, the median of the modes would be about around 14. We note that this is clearly smaller than all of the other options, and the only option where <math>\mu</math> is the smallest is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
--srisainandan6
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