2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 12 Easy

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}, \cdots, 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? (2019 AMC 10A Problem, Question#12)

  • A.

    \mu<d<M

  • B.

    M<d<\mu

  • C.

    d=M=\mu

  • D.

    d<M<\mu

  • E.

    d<\mu<M

Answer:E

First of all, d obviously has to be smaller than M, since when calculating M, we must take into account the 29\text{s}, 30\text{s}, and 31\text{s}. So we can eliminate choices B and C. Since there are 365 total entries, the median, M, must be the 183\text{rd} one, at which point we note that 12\cdot15 is 180, so 16 has to be the median (because 183 is between 12\cdot15+1=181 and 12\cdot16=192). Now, the mean, \mu, must be smaller than 16, since there are many fewer 29\text{s}, 30\text{s}, and 31\text{s}. d is less than \mu ,because when calculating \mu, we would include 29, 30, and 31. Thus the answer is d<\mu<M.

As in Solution 1, we find that the median is 16. Then, looking at the modes (1-28) , we realize that even if we were to have 12 of each, their median would remain the same, being 14.5. As for the mean, we note that the mean of the first 28 is simply the same as the median of them, which is 14.5. Hence, since we in fact have 29's, 30's, and 31's, the mean has to be higher than 14.5. 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). By comparing these values, the answer is \text{E}.

We can solve this problem simply by carefully calculating each of the values, which turn out to be M=16, d=14.5, and \mu\approx15.7. Thus the answer is \text{E}

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