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

Revision as of 09:31, 8 January 2021

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

Video Solution 1

https://youtu.be/3xVKbAVkHo8

Education, the Study of Everything

Video Solution 2

https://youtu.be/tzaUr4iVfgc

~savannahsolver

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 17: / Line 17: @@
 ==Solution 3 (direct calculation)==
-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 14.7</math>. Thus the answer is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
+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>.
 ==Video Solution 1==

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