2019 AMC 10A Problems/Problem 12

Revision as of 19:48, 9 February 2019 by Argonauts16 (talk | contribs) (Solution 1)
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 1

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 than $\mu$ because when calculating $\mu$ you include $29$, $30$, and $31$. Thus the answer is $d < \mu < M \implies \boxed{\textbf{(E)}}$

Solution 2

Notice that there are $365$ total entries, so the median has to be the $183\text{rd}$ one. Then, realize that $12 \cdot 15$ 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 $\boxed{\textbf{E}}$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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