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

📘 View easy version →

Problem

On the most recent exam on Prof. Xochi's class,

$5$ students earned a score of at least $95\%$ ,

$13$ students earned a score of at least $90\%$ ,

$27$ students earned a score of at least $85\%$ ,

$50$ students earned a score of at least $80\%$ ,

How many students earned a score of at least $80\%$ and 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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 $90\%$.

Givens: $5$ students scored $\ge 95\%$; $13$ students scored $\ge 90\%$; $27$ students scored $\ge 85\%$; $50$ students scored $\ge 80\%$; Answer choices: (A) $8$, (B) $14$, (C) $22$, (D) $37$, (E) $45$

Unknowns: The number of students whose score $s$ satisfies $80\% \le s < 90\%$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

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 "$\ge 90\%$" inside the larger group "$\ge 80\%$". Tool #16 (Change Focus / Complement) turns the question into a clean subtraction. Tool #7 (Identify Subproblems) helps us see the "$\ge 80\%$" group as one set that splits into two disjoint pieces — "$\ge 90\%$" 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.

Execute — Answer: D

#3 Eliminate Possibilities 2.OA.A.1 Step 1

Identify which two pieces of data are actually needed.
The target band $80 \le s < 90$ has a lower edge at $80\%$ and an upper edge at $90\%$.
The $85\%$ and $95\%$ rows sit strictly inside the bands above and below the cutoff we care about, so they don't pin down anyone in the target band.
We can set them aside.

💡 Picking out only the data that touches the boundary of the band is the same kind of word-problem reading kids do in Grade 2.

#7 Identify Subproblems 1.OA.B.4 Step 2

Picture the $50$ students who scored at least $80\%$ as one big group.
That group splits cleanly into two non-overlapping subgroups: the ones who ALSO scored at least $90\%$, and the ones who did NOT — i.e., the $80$-to-$89\%$ band we want.

$$\underbrace{50}_{\ge 80\%} \;=\; \underbrace{13}_{\ge 90\%} \;+\; \underbrace{N}_{80\le s<90}$$

💡 Splitting a set into two parts with no overlap is exactly the part-part-whole picture from Grade 1 unknown-addend problems.

#16 Change Focus / Count the Complement 2.NBT.B.5 Step 3

Use Tool #16 (Complement): the band we want is everyone in the $\ge 80\%$ group EXCEPT those in the $\ge 90\%$ group.
So we subtract the smaller cumulative count from the larger.

$$N = 50 - 13 = 37$$

💡 Subtracting two two-digit numbers within $100$ is the Grade 2 fluency standard — no algebra needed.

#3 Eliminate Possibilities 2.NBT.A.4 Step 4

Match $N = 37$ to the answer choices and confirm the correct letter.

$$37 \;\Rightarrow\; \textbf{(D)}$$

💡 Comparing our number to the listed choices is the same number-comparison move from Grade 2.

[1] #3 2.OA.A.1 Identify which two pieces of data are actually needed. The target band $80 \le s

[2] #7 1.OA.B.4 Picture the $50$ students who scored at least $80\%$ as one big group. That grou

[3] #16 2.NBT.B.5 Use Tool #16 (Complement): the band we want is everyone in the $\ge 80\%$ group

[4] #3 2.NBT.A.4 Match $N = 37$ to the answer choices and confirm the correct letter.

Review

Reasonableness: Sanity check the size: $37$ is less than $50$ (the whole $\ge 80\%$ group) and bigger than $13$ (the $\ge 90\%$ 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 $\ge 80\%$. Finally, the unused data is consistent — $27$ scored $\ge 85\%$, and of those $13$ scored $\ge 90\%$, 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.

Alternative: Tool #12 (Venn / nested sets) gives the same picture visually: draw three nested ovals for $\ge 80$, $\ge 85$, $\ge 90$ with the sizes $50$, $27$, $13$, then label each annular ring by subtraction — the $80$-to-$89\%$ ring is $50 - 13 = 37$, no calculation needed for any of the other rings.

CCSS standards used (min grade 2)

1.OA.B.4 Understand subtraction as an unknown-addend problem (Seeing the relation $50 = 13 + N$ as a part-part-whole equation where the unknown $N$ is the missing part of the $\ge 80\%$ group.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Reading the cumulative "at least" language and identifying which two numbers ($50$ and $13$) the problem actually depends on.)
2.NBT.B.5 Fluently add and subtract within 100 (Performing the single subtraction $50 - 13 = 37$ that produces the answer.)
2.NBT.A.4 Compare two three-digit numbers using symbols (Matching the computed value $37$ to the answer choices to select letter (D).)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 2)

More like this