AMC 8 · 2025 · #7

Grade 2 arithmetic

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

interval-arithmeticset-partitionmulti-digit-arithmetic complementary-countingidentify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

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Problem

Professor Xochi reports cumulative score counts on her exam: 5 students scored at least 95%, 13 scored at least 90%, 27 scored at least 85%, and 50 scored at least 80%. We want the number of students whose score is at least 80% but strictly less than 90%.

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 Change Focus / Count the Complement

Counting students directly in the 80-to-89% band is awkward because no row of the data names that band. But the band IS exactly the complement of " ≥ 90%" inside the larger group " ≥ 80%". Tool #16 (Change Focus / Complement) turns the question into a clean subtraction. Tool #7 (Identify Subproblems) helps us see the " ≥ 80%" group as one set that splits into two disjoint pieces — " ≥ 90%" and "80-to-89%" — so we know which two numbers to combine. Tool #3 (Eliminate Possibilities) is the meta-move that lets us throw out the 85% and 95% rows as irrelevant distractors before computing.

1STEP 1

Only the 80% and 90% rows touch the target band's edges — the 85% and 95% rows are decoys we set aside.

2STEP 2

See the 50 who scored at least 80% as one group splitting with no overlap into the at-least-90% students and the 80-to-89% band we want.

50_ ≥ 80% = 13_ ≥ 90% + N₈0 ≤ s ＜ 90

3STEP 3

Tool #16 (Complement): the band is everyone at least 80% minus those at least 90%, so just subtract to get 37.

N = 50 - 13 = 37

4STEP 4

Our N = 37 lands on answer choice (D).

37 → (D)

Answer

Sanity check the size: 37 is less than 50 (the whole ≥ 80% group) and bigger than 13 (the ≥ 90% subgroup) — both of which it must be, so the magnitude is in the right window. Also notice 37 + 13 = 50, which reproduces the cumulative count for ≥ 80%. Finally, the unused data is consistent — 27 scored ≥ 85%, and of those 13 scored ≥ 90%, leaving 14 in the 85-to-89% slice — so the 80-to-89% band (37 students) breaks into 80-to-84% (37 - 14 = 23) and 85-to-89% (14), all non-negative. Everything ticks.

💡Key takeaway

This AMC 8 problem only needs Grade 2 subtraction-within-100 you already know — 50 - 13 = 37!

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