AMC 8 · 2025 · #14

Grade 6 arithmetic

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

mean-median-mode-rangelinear-equations-one-varmulti-digit-arithmetic guess-and-checkidentify-subproblems ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

📏 Medium solution 💡 2 insights

Problem

Start with the list 2, 6, 7, 7, 28. We insert one more number N so the new six-number list has a mean that is exactly twice the new median. Which of the five answer choices works?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

This is a multiple-choice problem with only five candidates, so Tool #3 (Eliminate) is the natural first move. Tool #7 (Identify Subproblems) splits the work into two clean pieces — "find the new median" and "find the new mean" — and once we see that the median is locked at 7 for every choice (since N ≥ 7), Tool #6 (Guess and Check) just plugs each candidate into the mean formula and keeps the one that lands on 2 × 7 = 14. No algebra is required, though algebra would also work.

1STEP 1

The list 2, 6, 7, 7, 28 is already sorted, and its sum is 50 — we reuse this total when finding the mean later.

2 + 6 + 7 + 7 + 28 = 50

2STEP 2

Every choice has N ≥ 7, so the middle pair is locked at 7 and 7 and the new median is 7.

median = (7 + 7)/2 = 7

3STEP 3

Since the new median is 7, the required new mean must be 2 × 7 = 14.

new mean = 2 × 7 = 14

4STEP 4

The six numbers sum to 50 + N with mean $\frac{50 + N}{6}$ , so setting it equal to 14 gives 50 + N = 84.

\frac{50 + N}{6}

= 14 ⟹ 50 + N = 84

5STEP 5

Subtracting 50 from both sides gives N = 34, so only choice (E) makes the mean exactly 14.

N = 84 - 50 = 34 → (E)

Answer

Sanity check: with N = 34, the new list is 2, 6, 7, 7, 28, 34. Middle two are still 7 and 7, so the median is 7. The sum is 50 + 34 = 84, the mean is 84 / 6 = 14, and 14 = 2 × 7. Everything matches. Also, the answer 34 is the largest choice, which makes sense — to pull the mean far above the median, we need a value that is significantly larger than 7.

💡Key takeaway

This AMC 8 problem only needs Grade 6 mean and median ideas — and the trick that the median stays at 7 — which you already know!

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