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 1)
Line 10: Line 10:
 
===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 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 than <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>
  
 
===Solution 2===
 
===Solution 2===

Revision as of 18:48, 9 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 smaller than $M$ since when calculating $M$ you must take into account the $29's$, $30's$, and $31s$. So we can eliminate $(B)$ and $(C)$. The median, $\mu$, is $16$, but the mean ($M$) must be smaller than $16$ since there are much less $29's$, $30's$, and $31s$. $d$ 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 $\boxed{\textbf{E}}$.

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=101624"