AMC 8 · 2010 · #1

Grade 2 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

multi-digit-arithmeticmental-arithmetic identify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

At Euclid Middle School, three math teachers — Mrs. Germain, Mr. Newton, and Mrs. Young — each have students signed up for the AMC 8. The three classes have 11, 8, and 9 students respectively. Find the total number of math students at the school who are taking the contest.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The total is naturally split into three disjoint pieces — one per teacher — so Tool #7 (Identify Subproblems) just means counting each class separately and then adding. Because there is no overlap between classes, the total is simply the sum 11 + 8 + 9. Tool #14 (Sanity Check) is used at the end to confirm the sum lands in the choice range and to double-check by re-grouping the addends.

1STEP 1

Read each class size straight from the problem: 11, 8, and 9.

Germain = 11, Newton = 8, Young = 9

2STEP 2

Add the first two classes: 11 + 8 gives 19.

11 + 8 = 19

3STEP 3

Add the last class: 19 + 9 makes 28 (make a ten: 19 + 1 = 20, then + 8).

19 + 9 = 28

4STEP 4

The total 28 matches answer choice (C).

11 + 8 + 9 = 28 → (C)

Answer

Each class has between 8 and 11 students, so the total of three classes should be between 3 × 8 = 24 and 3 × 11 = 33. Our answer 28 sits in that range. Re-adding in a different order — 11 + 9 = 20, then 20 + 8 = 28 — gives the same total, which confirms the arithmetic.

💡Key takeaway

This AMC 8 problem only needs Grade 2 addition within 100 — just add the three class sizes!

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