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

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

Solution 2:

First of all, $d$ obviously has to smaller than $M$ since when calculating $M$ you must take into account the $29's$, $30's$, and $31s$. So we can eliminate $(B)$ and $(C)$. The median, $\mu$, is $16$, but the mean ($M$) must be smaller than $16$ since there are much less $29's$, $30's$, and $31s$. $d$ is less that $\mu$ because when calculating $\mu$ you include $29$, $30$, and $31$.Thus the answer is $d < \mu < M \implies \boxed{(E)}$

Solution 2:

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.