AMC 8 · 2008 · #11
Easy mode Grade 4Problem
A class at Lincoln Middle School has 39 eighth-grade students. Every student has at least one pet. The pet is either a dog, a cat, or both.
20 of the students have a dog. 26 of the students have a cat.
Some students show up in both groups — they have a dog and a cat.
How many students have both a dog and a cat?
Pick an answer.
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.
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?
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) $46$
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
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."
💡 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.
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).
💡 Adding two two-digit numbers is Grade 2 arithmetic. The key insight is what the sum means, not the sum itself.
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.
💡 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.
4.OA.A.3 Step 4 - Check the answer choices.
- $7$ matches choice (A).
💡 Always read the answer letter off the choice list — small habit, saves bubble mistakes.
2.MD.D.10 Draw two overlapping circles: one for dog-owners (size $20$), one for cat-owners 2.NBT.B.5 Add the dog count and the cat count. This sum counts every student who owns at l 4.OA.A.3 The true number of students is $39$, but the sum gave $46$. The extra $46 - 39 = 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.10Draw 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.5Fluently 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.3Solve 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.
⭐ 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.
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