AMC 8 · 2010 · #1
Easy mode Grade 2Problem
Three math teachers at Euclid Middle School are signing up students for the AMC .
Mrs. Germain has students taking it. Mr. Newton has students taking it. Mrs. Young has students taking it.
How many students at the school are taking the contest in total?
Pick an answer.
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
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) $30$
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
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.
💡 Pulling the three numbers out of the word problem is exactly the Grade 2 "represent a word problem with numbers" skill.
2.NBT.B.5 Step 2 - Add the first two classes.
- Group $11 + 8$ first to keep the numbers small.
💡 Adding two two-digit numbers within $100$ is the Grade 2 fluency standard.
2.NBT.B.5 Step 3 - Add the third class to the running total.
- Use the "make a ten" idea: $19 + 1 = 20$, then $20 + 8 = 28$.
💡 Decomposing $9$ as $1 + 8$ to make a friendly $20$ is a standard Grade 2 mental-math move.
2.OA.A.1 Step 4 - Match the total to the answer choices.
- $28$ is choice (C).
💡 Confirming the computed total is one of the listed options is the final "does this answer the question?" check.
2.OA.A.1 List the count from each class as given. Each teacher's class contributes one wh 2.NBT.B.5 Add the first two classes. Group $11 + 8$ first to keep the numbers small. 2.NBT.B.5 Add the third class to the running total. Use the "make a ten" idea: $19 + 1 = 2 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.1Use 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.5Fluently 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!
⭐ This AMC 8 problem only needs Grade 2 addition within $100$ — just add the three class sizes!
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