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

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==Solution 2==
 
==Solution 2==
  
As in Solution 1, we find that the median is <math>16</math>. Then, looking at the modes <math>(1-28)</math>, we realize that even if we were to have <math>12</math> of each, their median would remain the same, being <math>14.5</math>. As for the mean, we note that the mean of the first <math>28</math> is simply the same as the median of them, which is <math>14.5</math>. Hence, since we in fact have <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s, the mean has to be higher than <math>14.5</math>. 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). By comparing these values, the answer is <math>\boxed{\textbf{(E)}}</math>.
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As in Solution 1, we find that the median is <math>16</math>. Then, looking at the modes <math>(1-28)</math>, we realize that even if we were to have <math>12</math> of each, their median would remain the same, being <math>14.5</math>. As for the mean, we note that the mean of the first <math>28</math> is simply the same as the median of them, which is <math>14.5</math>. Hence, since we in fact have <math>29</math>'s, <math>30</math>'s, and <math>31</math>'s, the mean has to be higher than <math>14.5</math>. 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). By comparing these values, the answer is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
  
 
==Solution 3 (direct calculation)==
 
==Solution 3 (direct calculation)==

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

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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