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

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Notice that there are <math>365</math> total entries, so the median has to be the <math>183\text{rd}</math> one. Then, realize that <math>12 \cdot 15</math> is <math>180</math>, so <math>16</math> has to be the median (because <math>16</math> is from <math>181</math> to <math>192</math>). Then, look at the modes <math>(1-28)</math> and realize that even if you have <math>12</math> of each, the median of those remains the same and you have <math>14.5</math>. When trying to find the mean, you realize that the mean of the first <math>28</math> is simply the same as the median of them, which is <math>14.5</math>. Then, when you see <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s, you realize that the mean has to be higher. On the other hand, since there are fewer <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s than the rest of the numbers, the mean has to be lower than <math>16</math> (the median). Then, you compare those values and you get the answer, which is <math>\boxed{\textbf{(E)}}</math>.
 
Notice that there are <math>365</math> total entries, so the median has to be the <math>183\text{rd}</math> one. Then, realize that <math>12 \cdot 15</math> is <math>180</math>, so <math>16</math> has to be the median (because <math>16</math> is from <math>181</math> to <math>192</math>). Then, look at the modes <math>(1-28)</math> and realize that even if you have <math>12</math> of each, the median of those remains the same and you have <math>14.5</math>. When trying to find the mean, you realize that the mean of the first <math>28</math> is simply the same as the median of them, which is <math>14.5</math>. Then, when you see <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s, you realize that the mean has to be higher. On the other hand, since there are fewer <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s than the rest of the numbers, the mean has to be lower than <math>16</math> (the median). Then, you compare those values and you get the answer, which is <math>\boxed{\textbf{(E)}}</math>.
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===Solution 3===
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Bash out each of the values. I'm not going to show work here, but you should get <math>M = 16</math> , <math>d = 14.5</math> , <math>\mu = 15.7205479</math> , so the answer is obviously <math>\boxed{\textbf{(E)}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:47, 11 February 2019

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 be smaller than $M$, since when calculating $M$, you must take into account the $29$s, $30$s, and $31$s. So we can eliminate $(B)$ and $(C)$. The median, $M$, is $16$, but 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 $\mu$, 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 $\boxed{\textbf{(E)}}$.

Solution 3

Bash out each of the values. I'm not going to show work here, but you should get $M = 16$ , $d = 14.5$ , $\mu = 15.7205479$ , so the answer is obviously $\boxed{\textbf{(E)}}$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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