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

(Created page with "{{duplicate|2019 AMC 10A #12 and 2019 AMC 12A #7}} ==Problem== Melanie computes the mean <math>\mu</math>, the median <ma...")
 
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==Solution==
 
==Solution==
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There are <math>365</math> values which means that the median is the <math>183</math>rd biggest value. Because there are <math>12</math> of each of the first <math>28</math> numbers, the median is <math>183/12=15.25</math>.The mean of these numbers is a little under <math>16</math> because there are only <math>7</math> <math>31</math>'s. The median of the modes is the median of <math>1</math> to <math>28</math>, yielding 14.5. Because <math>14.5<15.25<16</math>, the answer is <cmath>B</cmath>
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-Lcz
  
 
==See Also==
 
==See Also==

Revision as of 18:15, 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

There are $365$ values which means that the median is the $183$rd biggest value. Because there are $12$ of each of the first $28$ numbers, the median is $183/12=15.25$.The mean of these numbers is a little under $16$ because there are only $7$ $31$'s. The median of the modes is the median of $1$ to $28$, yielding 14.5. Because $14.5<15.25<16$, the answer is \[B\]

-Lcz

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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