AMC 10 · 2019 · #12

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangeweighted-averagepattern-recognition identify-subproblemspattern-recognition ↑ Prerequisites: mean-median-mode-range

📏 Medium solution 💡 3 insights

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?

$\textbf{(A) } \mu < d < M \qquad\textbf{(B) } M < d < \mu \qquad\textbf{(C) } d = M =\mu \qquad\textbf{(D) } d < M < \mu \qquad\textbf{(E) } d < \mu < M$

Pick an answer.

(A)

$\mu < d < M$

(B)

$M < d < \mu$

(C)

$d = M = \mu$

(D)

$d < M < \mu$

(E)

$d < \mu < M$

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Toolkit + CCSS Solution

Understand

Restated: Melanie lists the day-of-month number for every day of $2019$ — that is $12$ copies of each of $1, 2, \dots, 28$ (every month has those days), $11$ copies each of $29$ and $30$ (every month except February), and $7$ copies of $31$ (the seven $31$-day months). Let $\mu$ be the mean, $M$ the median, and $d$ the median of the modes of these $365$ numbers. Order $\mu, M, d$ from smallest to largest.

Givens: $365$ total entries: $12$ each of $1, 2, \dots, 28$; $11$ each of $29, 30$; $7$ copies of $31$; $\mu = $ mean of the $365$ numbers; $M = $ median of the $365$ numbers; $d = $ median of the modes; Answer choices order $\mu, M, d$

Unknowns: Which of $\mu < d < M$, $M < d < \mu$, $d = M = \mu$, $d < M < \mu$, $d < \mu < M$ holds

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities

Tool #2 (Systematic List): write out the frequency table (each value with its count) so the median position and the mode set are read off directly. Tool #15 (Reorganize): keep the totals as a cumulative-count column so the $183$rd entry (median) is found by scanning. The mean compares to the median by noting that the only entries pulling the mean down are the under-represented $29, 30, 31$. Tool #3 eliminates the four false orderings.

Execute — Answer: E

#2 Make a Systematic List 6.SP.B.5 Step 1

Build the frequency table.
Values $1$ through $28$ each appear $12$ times; values $29$ and $30$ each appear $11$ times; value $31$ appears $7$ times.
Check the total: $12 \cdot 28 + 11 \cdot 2 + 7 = 336 + 22 + 7 = 365$.
Good — this matches $365$ days in $2019$.

$$12 \cdot 28 + 11 \cdot 2 + 7 = 336 + 22 + 7 = 365$$

💡 Grade 6 data summary: lay the counts out so every later question is just a table lookup.

#2 Make a Systematic List 6.SP.A.3 Step 2

Find the modes.
The modes are the values that appear most often.
Every value in $\{1, 2, \dots, 28\}$ appears $12$ times, and that is the largest count, so the modes are exactly $1, 2, \dots, 28$.
Find $d$, the median of these $28$ modes — for an even-length list, average the $14$th and $15$th values: $\frac{14 + 15}{2} = 14.5$.

$$\text{modes} = \{1, 2, \dots, 28\},\quad d = \frac{14 + 15}{2} = 14.5$$

💡 Grade 6 measure of center: the median of an even-length list is the average of the two middle entries.

#15 Organize Information in More Ways 6.SP.A.3 Step 3

Find the median $M$ of all $365$ numbers.
The median position is $\frac{365+1}{2} = 183$ — the $183$rd entry of the sorted list.
Use the cumulative count: after value $15$, the cumulative total is $12 \cdot 15 = 180$, and value $16$ occupies positions $181$ to $192$.
The $183$rd position is therefore $16$, so $M = 16$.

$$12 \cdot 15 = 180 < 183 \le 192 = 12 \cdot 16 \;\Rightarrow\; M = 16$$

💡 Grade 6 median: a cumulative-count strip tells you which value the middle slot lands on.

#15 Organize Information in More Ways 6.SP.A.3 Step 4

Compare the mean $\mu$ to $M = 16$.
If every value $1, 2, \dots, 31$ had the same count, the mean of $1$ to $31$ would be $16$ — exactly the median.
The actual data under-represents the high end ($29, 30$ short by $1$ each; $31$ short by $5$).
Removing high values pulls the mean down, so $\mu < 16 = M$.

$$\bar{x}(1, \dots, 31) = 16 \;\text{and high values are under-represented} \;\Rightarrow\; \mu < 16 = M$$

💡 Grade 6: dropping copies of the largest values drags the mean below the median.

#15 Organize Information in More Ways 6.SP.A.3 Step 5

Compare $\mu$ to $d = 14.5$.
The mean of $1$ to $28$ alone is $14.5$.
The full data adds $11$ copies of $29$, $11$ of $30$, and $7$ of $31$ on top of those $1$-to-$28$ values — every extra entry is larger than $14.5$, which lifts the mean.
Hence $\mu > 14.5 = d$.

$$\bar{x}(1, \dots, 28) = 14.5,\;\text{extras } 29, 30, 31 > 14.5 \;\Rightarrow\; \mu > d = 14.5$$

💡 Grade 6: adding entries above $14.5$ pulls the mean above $14.5$.

#3 Eliminate Possibilities 6.NS.C.7 Step 6

Combine: $d = 14.5 < \mu < 16 = M$.
This is the ordering $d < \mu < M$, choice $(E)$.
Eliminate the others — they each force the mean above $M$ or push $d$ above $\mu$, neither of which the data supports.

$$d = 14.5 < \mu < 16 = M \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 6 ordering decimals/integers: chain the two comparisons into a single inequality.

[1] #2 6.SP.B.5 Build the frequency table. Values $1$ through $28$ each appear $12$ times; value

[2] #2 6.SP.A.3 Find the modes. The modes are the values that appear most often. Every value in

[3] #15 6.SP.A.3 Find the median $M$ of all $365$ numbers. The median position is $\frac{365+1}{2

[4] #15 6.SP.A.3 Compare the mean $\mu$ to $M = 16$. If every value $1, 2, \dots, 31$ had the sam

[5] #15 6.SP.A.3 Compare $\mu$ to $d = 14.5$. The mean of $1$ to $28$ alone is $14.5$. The full d

[6] #3 6.NS.C.7 Combine: $d = 14.5 < \mu < 16 = M$. This is the ordering $d < \mu < M$, choice $

Review

Reasonableness: Direct computation confirms the chain. Sum of all $365$ values is $12 \cdot \tfrac{28 \cdot 29}{2} + 11 \cdot 29 + 11 \cdot 30 + 7 \cdot 31 = 12 \cdot 406 + 319 + 330 + 217 = 4872 + 866 = 5738$, so $\mu = \frac{5738}{365} \approx 15.72$. Then $14.5 < 15.72 < 16$ — exactly $d < \mu < M$.

Alternative: Tool #6 (Guess and Check) on the answer choices. Note immediately that the modes set $\{1, \dots, 28\}$ skews low, so $d$ must be the smallest of the three. That kills choices $(A), (B), (C)$ at once, leaving $(D)$ vs $(E)$. Then ask whether the mean exceeds $M = 16$ or sits below it — since the higher values $29, 30, 31$ are under-represented, the mean must dip below $16$, ruling out $(D)$ and leaving $(E)$.

CCSS standards used (min grade 6)

6.NS.C.7 Understand ordering and absolute value of rational numbers (Chaining $14.5 < \mu < 16$ into the single ordering $d < \mu < M$.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Computing the mean, median, and median-of-modes, and comparing them.)
6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Setting up the frequency table for the $365$ daily values.)

⭐ This AMC 10 problem only needs Grade 6 measures of center you already know! Every value from $1$ to $28$ shows up $12$ times — all are modes — so $d = \frac{14 + 15}{2} = 14.5$. The $183$rd entry of the sorted $365$ values lives at $16$, so $M = 16$. The mean of $1$ to $31$ would be $16$, but the dataset is short of $29, 30, 31$, so $\mu$ drops just below $16$ — to about $15.72$. That gives $d < \mu < M$, answer $(E)$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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