# 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

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