# 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

There are $365$ values which means that the median is the $183$rd biggest value. Because there are $12$ of each of the first $28$ numbers, the median is $183/12=15.25$.The mean of these numbers is a little under $16$ because there are only $7$ $31$'s. The median of the modes is the median of $1$ to $28$, yielding 14.5. Because $14.5<15.25<16$, the answer is $B$

-Lcz

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 