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

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<math>\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</math>
 
<math>\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</math>
  
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==Solution 1==
  
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First of all, <math>d</math> obviously has to be smaller than <math>M</math>, since when calculating <math>M</math>, we must take into account the <math>29</math>s, <math>30</math>s, and <math>31</math>s. So we can eliminate choices <math>B</math> and <math>C</math>. Since there are <math>365</math> total entries, the median, <math>M</math>, must be the <math>183\text{rd}</math> one, at which point we note that <math>12 \cdot 15</math> is <math>180</math>, so <math>16</math> has to be the median (because <math>183</math> is between <math>12 \cdot 15 + 1 = 181</math> and <math>12 \cdot 16 = 192</math>). Now, the mean, <math>\mu</math>, must be smaller than <math>16</math>, since there are many fewer <math>29</math>s, <math>30</math>s, and <math>31</math>s. <math>d</math> is less than <math>\mu</math>, because when calculating <math>\mu</math>, we would include <math>29</math>, <math>30</math>, and <math>31</math>. Thus the answer is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
  
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Someone please add a solution??
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==Solution 2==
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==Solution==
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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>.
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==Solution 3 (direct calculation)==
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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 = \frac{\sum_{n=1}^{30} 12n - 29 - 30 +31*7}{365} \approx 15.7</math>. Thus the answer is <math>\boxed{\textbf{(E) } d < \mu < M}</math>.
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==Video Solution 1==
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https://youtu.be/3xVKbAVkHo8
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Education, the Study of Everything
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==Video Solution 2==
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https://youtu.be/tzaUr4iVfgc
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2019|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2019|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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Solution 1:
 
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Notice that there are 365 total entries, so the median has to be the 183rd one. Then, realize that 12 * 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 E.
 

Latest revision as of 06:56, 20 November 2023

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 $\mu = \frac{\sum_{n=1}^{30} 12n - 29 - 30 +31*7}{365} \approx 15.7$. 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

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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