Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 3 (direct calculation))
 
(20 intermediate revisions by 14 users not shown)
Line 7: Line 7:
 
<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>
  
−
==Solution==
+
==Solution 1==
−
===Solution 1===
 
  
−
First of all, <math>d</math> obviously has to smaller than <math>M</math> since when calculating <math>M</math> you must take into account the <math>29's</math>, <math>30's</math>, and <math>31s</math>. So we can eliminate <math>(B)</math> and <math>(C)</math>. The median, <math>\mu</math>, is <math>16</math>, but the mean (<math>M</math>) must be smaller than <math>16</math> since there are much less <math>29's</math>, <math>30's</math>, and <math>31s</math>. <math>d</math> is less that <math>\mu</math> because when calculating <math>\mu</math> you include <math>29</math>, <math>30</math>, and <math>31</math>.Thus the answer is <math>d < \mu < M \implies \boxed{\textbf{(E)}}</math>
+
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>.
  
−
===Solution 2===
+
==Solution 2==
  
−
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>.
+
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)==
 +
 
 +
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>.
 +
 
 +
==Video Solution 1==
 +
 
 +
https://youtu.be/3xVKbAVkHo8
 +
 
 +
Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/tzaUr4iVfgc
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_12&oldid=204795"