Difference between revisions of "2019 AMC 10A Problems/Problem 12"
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===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{(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 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> |
===Solution 2=== | ===Solution 2=== |
Revision as of 19:48, 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 , 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
Solution 1
First of all, obviously has to smaller than since when calculating you must take into account the , , and . So we can eliminate and . The median, , is , but the mean () must be smaller than since there are much less , , and . is less that because when calculating you include , , and .Thus the answer is
Solution 2
Notice that there are total entries, so the median has to be the one. Then, realize that is , so has to be the median (because is from to ). Then, look at the modes and realize that even if you have of each, the median of those remains the same and you have . When trying to find the mean, you realize that the mean of the first is simply the same as the median of them, which is . Then, when you see 's, 's, and 's, you realize that the mean has to be higher. 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). Then, you compare those values and you get the answer, which is .
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.