AMC 8 · 2022 · #19

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

mean-median-mode-rangegraph-readingsystematic-enumeration caseworkbound-inequality-then-enumerate ↑ Prerequisites: mean-median-mode-rangegraph-reading

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A class of 20 students took a test whose scores are shown on a dot plot. After regrading, the teacher added 5 extra points to some students, and the new median jumped to 85. We want the smallest possible number of students who received the 5-point bonus.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The dot plot is small enough to lay out every score as an ordered list (Tool #2), which lets us see exactly which positions are the 10th and 11th. Tool #9 (Easier Problem) is the key insight: instead of tracking many students, ask the simpler question "how many of the 20 scores need to be at least 85?". Because the median position is fixed (positions 10 and 11), we only need both of those slots to read 85. Tool #6 (Guess and Check) verifies that 4 works and 3 does not — a clean minimum-existence argument.

1STEP 1

List all 20 scores in order from the dot-plot counts, so the 10th and 11th slots — the ones that set the median — stand out.

65, 65₁-2, 70, 70₃-4, 75, 75, 75, 75₅-8, 80, 80, 80, 80, 80₉-13, 85, 85₁4-15, 90, 90, 90₁6-18, 95, 100

2STEP 2

With 20 scores the median is the mean of the 10th and 11th, both 80, so the original median is 80 — below our target of 85.

old median =

\frac{80 + 80}{2}

= 80

3STEP 3

Reframe it (Tool #9): for a median of 85, both middle slots must be 85, so 10 scores must reach 85; only 7 do, so we need 3 more.

scores already ≥ 85: 2 + 3 + 1 + 1 = 7. need: 10 - 7 = 3 more.

4STEP 4

Adding 5 to a 65, 70, or 75 gives 70, 75, 80 — still under 85; only an 80 becomes 85, so every useful bonus goes to an 80.

80 + 5 = 85 ✓, 75 + 5 = 80 ＜ 85, 70 + 5 = 75 ＜ 85, 65 + 5 = 70 ＜ 85

5STEP 5

Test 3 first: three 80s become 85, but that leaves an 80 at position 10 and 85 at 11, so the median is only 82.5 — 3 is not enough.

with 3 bonuses: positions 10 and 11 are 80, 85 → median =

\frac{80 + 85}{2}

= 82.5

6STEP 6

Try 4: four 80s become 85, so positions 10 and 11 are both 85 and the median is exactly 85 — since 3 failed, 4 is the minimum, choice (C).

with 4 bonuses: positions 10 and 11 are both 85 → median =

\frac{85 + 85}{2}

= 85 → (C)

Answer

The answer 4 sits in the middle of the choice list and is clearly the smallest value that works: 3 bonuses give a median of 82.5 (still short), 4 bonuses give exactly 85, and any larger number 5 or 6 would also work but is wasteful. Also note that the 5 students at 80 are the only viable targets, so the answer cannot exceed 5 — and 4 ≤ 5 is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 6 dot-plot and median skills you already know!

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