AMC 8 · 2008 · #11

Grade 4 counting

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

set-partitioncomplementary-countingmulti-digit-arithmetic identify-subproblemscomplementary-counting ↑ Prerequisites: multi-digit-arithmeticset-partition

📏 Short solution 💡 2 insights

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Problem

Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2008 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.

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Toolkit + CCSS Solution

Understand

Restated: All $39$ eighth graders own a dog, a cat, or both. $20$ own a dog and $26$ own a cat. How many own both?

Givens: Total students: $39$; Students with a dog: $20$; Students with a cat: $26$; Every student owns at least one of the two pets; Answer choices: (A) $7$, (B) $13$, (C) $19$, (D) $39$, (E) $46$

Unknowns: The number of students who own both a dog and a cat

Understand

Restated: All $39$ eighth graders own a dog, a cat, or both. $20$ own a dog and $26$ own a cat. How many own both?

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

Tool #1 (Draw a Diagram) is the cleanest move: sketch two overlapping circles (a Venn diagram) for the dog-owners and the cat-owners. The picture shows immediately that the overlap is double-counted when you add $20 + 26$. Tool #7 (Identify Subproblems) then turns the picture into two simple sentences: "dog total $+$ cat total counts every student, plus the overlap one extra time" and "$20 + 26 - \text{both} = 39$." That single subtraction gives the answer.

Execute — Answer: A

#1 Draw a Diagram 2.MD.D.10 Step 1

Draw two overlapping circles: one for dog-owners (size $20$), one for cat-owners (size $26$).
Their union is the whole class of $39$ students.
Label the overlap region "both."

$$\text{dog circle} = 20,\;\text{cat circle} = 26,\;\text{union} = 39$$

💡 Drawing two overlapping circles and labeling each region is the Grade 2 "picture graph" idea — a visual way to sort the same kids into two groups.

#7 Identify Subproblems 2.NBT.B.5 Step 2

Add the dog count and the cat count.
This sum counts every student who owns at least one pet, but it counts the both-pet students twice (once in each circle).

$$20 + 26 = 46$$

💡 Adding two two-digit numbers is Grade 2 arithmetic. The key insight is what the sum means, not the sum itself.

#7 Identify Subproblems 4.OA.A.3 Step 3

The true number of students is $39$, but the sum gave $46$.
The extra $46 - 39 = 7$ comes from counting the both-pet students one extra time.
So the overlap has $7$ students.

$$46 - 39 = 7 \;\Rightarrow\; \text{both} = 7$$

💡 This is the Grade 4 "solve word problems using the four operations" move: the difference between the over-count and the real total is exactly the overlap.

#7 Identify Subproblems 4.OA.A.3 Step 4

Check the answer choices.
$7$ matches choice (A).

$$7 = \textbf{(A)}$$

💡 Always read the answer letter off the choice list — small habit, saves bubble mistakes.

[1] #1 2.MD.D.10 Draw two overlapping circles: one for dog-owners (size $20$), one for cat-owners

[2] #7 2.NBT.B.5 Add the dog count and the cat count. This sum counts every student who owns at l

[3] #7 4.OA.A.3 The true number of students is $39$, but the sum gave $46$. The extra $46 - 39 =

[4] #7 4.OA.A.3 Check the answer choices. $7$ matches choice (A).

Review

Reasonableness: Plug $7$ back in. If $7$ students own both pets, then $20 - 7 = 13$ own only a dog and $26 - 7 = 19$ own only a cat. Adding the three disjoint groups: $13 + 19 + 7 = 39$, exactly the class size. The answer is consistent. It is also clearly in the right ballpark: the overlap must be at most $20$ (can't exceed the smaller group), and the sum $20 + 26 = 46$ exceeds $39$ by $7$, so $7$ is forced.

Alternative: Tool #4 (Introduce a Variable) gives the algebra version: let $x$ be the number of students with both pets. Then dog-only $= 20 - x$, cat-only $= 26 - x$, and adding the three disjoint groups gives $(20 - x) + (26 - x) + x = 39$, so $46 - x = 39$ and $x = 7$. This is the Principle of Inclusion-Exclusion written as an equation, but the Venn picture makes it visible without algebra.

CCSS standards used (min grade 4)

2.MD.D.10 Draw a picture graph and a bar graph to represent a data set (Drawing two overlapping circles (a Venn diagram) to sort the $39$ students into dog-only, cat-only, and both regions.)
2.NBT.B.5 Fluently add and subtract within 100 (Adding $20 + 26 = 46$ and subtracting $46 - 39 = 7$ — both are two-digit operations well within Grade 2 fluency.)
4.OA.A.3 Solve multistep word problems using the four operations (Recognizing that the over-count $46 - 39$ equals the size of the overlap, then translating that multi-step reasoning into the final subtraction.)

⭐ When you add the two groups, the kids in both groups get counted twice — so the gap between $20 + 26$ and the real total $39$ IS the overlap. That's all this AMC 8 problem asks.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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