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> | ||
− | + | ==Solution 1== | |
− | |||
− | First of all, <math>d</math> obviously has to smaller than <math>M</math> since when calculating <math>M</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== | |
− | + | 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 07:56, 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.
Contents
Problem
Melanie computes the mean , the median
, and the modes of the
values that are the dates in the months of
. Thus her data consist of
,
, . . . ,
,
,
, and
. Let
be the median of the modes. Which of the following statements is true?
Solution 1
First of all, obviously has to be smaller than
, since when calculating
, we must take into account the
s,
s, and
s. So we can eliminate choices
and
. Since there are
total entries, the median,
, must be the
one, at which point we note that
is
, so
has to be the median (because
is between
and
). Now, the mean,
, must be smaller than
, since there are many fewer
s,
s, and
s.
is less than
, because when calculating
, we would include
,
, and
. Thus the answer is
.
Solution 2
As in Solution 1, we find that the median is . Then, looking at the modes
, we realize that even if we were to have
of each, their median would remain the same, being
. As for the mean, we note that the mean of the first
is simply the same as the median of them, which is
. Hence, since we in fact have
's,
's, and
's, the mean has to be higher than
. On the other hand, since there are fewer
's,
's, and
's than the rest of the numbers, the mean has to be lower than
(the median). By comparing these values, the answer is
.
Solution 3 (direct calculation)
We can solve this problem simply by carefully calculating each of the values, which turn out to be ,
, and
. Thus the answer is
.
Video Solution 1
Education, the Study of Everything
Video Solution 2
~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.