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 the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School 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.

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

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Mrs. Germain's class: $11$ students taking the AMC 8; Mr. Newton's class: $8$ students taking the AMC 8; Mrs. Young's class: $9$ students taking the AMC 8; Answer choices: (A) $26$, (B) $27$, (C) $28$, (D) $29$, (E) $30$

Unknowns: The total number of math students at Euclid Middle School taking the AMC 8

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #14 Sanity Check

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.

Execute — Answer: C

#7 Identify Subproblems 2.OA.A.1 Step 1

List the count from each class as given.
Each teacher's class contributes one whole-number subtotal to the grand total.

$$\text{Germain} = 11,\quad \text{Newton} = 8,\quad \text{Young} = 9$$

💡 Pulling the three numbers out of the word problem is exactly the Grade 2 "represent a word problem with numbers" skill.

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

Add the first two classes.
Group $11 + 8$ first to keep the numbers small.

$$11 + 8 = 19$$

💡 Adding two two-digit numbers within $100$ is the Grade 2 fluency standard.

#7 Identify Subproblems 2.NBT.B.5 Step 3

Add the third class to the running total.
Use the "make a ten" idea: $19 + 1 = 20$, then $20 + 8 = 28$.

$$19 + 9 = 28$$

💡 Decomposing $9$ as $1 + 8$ to make a friendly $20$ is a standard Grade 2 mental-math move.

#14 Sanity Check 2.OA.A.1 Step 4

Match the total to the answer choices.
$28$ is choice (C).

$$11 + 8 + 9 = 28 \;\Rightarrow\; \textbf{(C)}$$

💡 Confirming the computed total is one of the listed options is the final "does this answer the question?" check.

[1] #7 2.OA.A.1 List the count from each class as given. Each teacher's class contributes one wh

[2] #7 2.NBT.B.5 Add the first two classes. Group $11 + 8$ first to keep the numbers small.

[3] #7 2.NBT.B.5 Add the third class to the running total. Use the "make a ten" idea: $19 + 1 = 2

[4] #14 2.OA.A.1 Match the total to the answer choices. $28$ is choice (C).

Review

Reasonableness: Each class has between $8$ and $11$ students, so the total of three classes should be between $3 \times 8 = 24$ and $3 \times 11 = 33$. 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.

Alternative: Tool #6 (Guess and Check) on the choices: subtract any candidate from the running total $11 + 8 = 19$ and check whether what is left equals Mrs. Young's $9$ students. Only (C) $28$ gives $28 - 19 = 9$, matching the third class exactly. The other choices ($26, 27, 29, 30$) leave $7, 8, 10, 11$, none of which equal $9$.

CCSS standards used (min grade 2)

2.OA.A.1 Use addition and subtraction within $100$ to solve one- and two-step word problems (Translating the three-class story into the addition expression $11 + 8 + 9$ and confirming the final total answers the question.)
2.NBT.B.5 Fluently add and subtract within $100$ using place value and properties of operations (Carrying out the additions $11 + 8 = 19$ and $19 + 9 = 28$ to find the total number of students.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 2)

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