AMC 10 · 2019 · #7

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

Write down the day-of-month number for every day of 2019: twelve copies each of 1 through 28, eleven copies each of 29 and 30, and seven copies of 31. Order the mean, the median, and the median of the modes of these 365 numbers from smallest to largest.

Pick an answer.

(A)

$\mu < d < M$

(B)

$M < d < \mu$

(C)

$d = M =\mu$

(D)

$d < M < \mu$

(E)

$d < \mu < M$

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

Check the total

They do add up to 365.

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

2STEP 2

Median of the modes

With twenty-eight modes, take their median.

modes = {1, 2, …, 28}, d = (14 + 15)/2 = 14.5

3STEP 3

Find the median

Counting to the 183rd value gives 16.

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

4STEP 4

The mean sits below the median

Large values are under-represented, pulling the mean down.

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

5STEP 5

The mean sits above the mode median

The days beyond 28 push the mean upward.

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

6STEP 6

Order the three

The order is mode median, then mean, then median.

d = 14.5 ＜ μ ＜ 16 = M → (E)

Answer

d ＜ μ ＜ M

Direct computation confirms the chain. Sum of all 365 values is 12 · (28 · 29)/2 + 11 · 29 + 11 · 30 + 7 · 31 = 12 · 406 + 319 + 330 + 217 = 4872 + 866 = 5738, so μ = 5738/365 ≈ 15.72. Then 14.5 ＜ 15.72 ＜ 16 — exactly d ＜ μ ＜ M.

💡Key takeaway

This AMC 12 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 = (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).