Difference between revisions of "2019 AMC 10A Problems/Problem 12"
(→Problem) |
(→Solution) |
||
Line 9: | Line 9: | ||
==Solution== | ==Solution== | ||
First of all, <math>d</math> obviously has to smaller than <math>M</math> since when calculating <math>M</math> you most 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 you realize that 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>. 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 most 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 you realize that 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>. Thus the answer is <math>d < \mu < M \implies \boxed{(E)}</math> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
==See Also== | ==See Also== |
Revision as of 19:19, 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
First of all, obviously has to smaller than since when calculating you most take into account the , , and . So we can eliminate and . The median, , is , but you realize that the mean () must be smaller than since there are much less , , and . Thus the answer 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.
Solution 1:
Notice that there are 365 total entries, so the median has to be the 183rd one. Then, realize that 12 * 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 E. Edit: Hello can i move this to solution