AMC 10 · 2019 · #12

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

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

📏 Medium solution 💡 3 insights

Problem

Melanie lists the day-of-month number for every day of 2019 — that is 12 copies of each of 1, 2, …, 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 μ be the mean, M the median, and d the median of the modes of these 365 numbers. Order μ, M, d from smallest to largest.

Pick an answer.

(A)

$\mu < d < M$

(B)

$M < d < \mu$

(C)

$d = M = \mu$

(D)

$d < M < \mu$

(E)

$d < \mu < M$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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 183rd 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.

1STEP 1

Frequency table: 1–28 appear 12 times each, 29 and 30 eleven times, 31 seven times — total 365, matching 2019's days.

12 · 28 + 11 · 2 + 7 = 336 + 22 + 7 = 365

2STEP 2

The modes are the values with the top count of 12 — exactly 1 through 28. For that even list of 28, d = $\frac{14+15}{2}$ = 14.5.

modes = {1, 2, …, 28}, d =

\frac{14 + 15}{2}

= 14.5

3STEP 3

The median sits at position $\frac{365+1}{2}$ = 183. Cumulative count reaches 180 after value 15, and 16 fills positions 181–192, so M = 16.

12 · 15 = 180 ＜ 183 ≤ 192 = 12 · 16 → M = 16

4STEP 4

If dates 1–31 were equally frequent the mean would equal 16 = M; the data is short on 29, 30, 31, dragging the mean down, so μ ＜ 16.

x(1, …, 31) = 16 and high values are under-represented → μ ＜ 16 = M

5STEP 5

The mean of 1–28 alone is 14.5 = d. The real data piles 29s, 30s, 31s on top — all above 14.5 — which lifts the mean, so μ ＞ 14.5.

x(1, …, 28) = 14.5, extras 29, 30, 31 ＞ 14.5 → μ ＞ d = 14.5

6STEP 6

Chain them: d = 14.5 ＜ μ ＜ 16 = M, i.e. d ＜ μ ＜ M, choice (E); the other four contradict the data.

d = 14.5 ＜ μ ＜ 16 = M → (E)

Answer

d ＜ μ ＜ M

Direct computation confirms the chain. Sum of all 365 values is 12 ·

\frac{28 · 29}{2}

+ 11 · 29 + 11 · 30 + 7 · 31 = 12 · 406 + 319 + 330 + 217 = 4872 + 866 = 5738, so μ =

\frac{5738}{365}

≈ 15.72. Then 14.5 ＜ 15.72 ＜ 16 — exactly d ＜ μ ＜ M.

💡Key takeaway

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 183rd 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 μ drops just below 16 — to about 15.72. That gives d ＜ μ ＜ M, answer (E).

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