AMC 10 · 2004 · #1
Grade 4 arithmeticEach row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?
Pick an answer.
AMC 10 2004 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: Every row in an amphitheater holds $33$ seats. The rows numbered $12$ through $22$ are all reserved for one club. Find the total number of reserved seats.
Givens: Each row has $33$ seats.; Rows $12$ through $22$ are reserved, and this range includes both row $12$ and row $22$.; Answer choices: (A) $297$, (B) $330$, (C) $363$, (D) $396$, (E) $726$
Unknowns: The total number of seats in the reserved rows.
Understand
Restated: Every row in an amphitheater holds $33$ seats. The rows numbered $12$ through $22$ are all reserved for one club. Find the total number of reserved seats.
Givens: Each row has $33$ seats.; Rows $12$ through $22$ are reserved, and this range includes both row $12$ and row $22$.; Answer choices: (A) $297$, (B) $330$, (C) $363$, (D) $396$, (E) $726$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The problem splits into two clean subproblems (Tool #7): first count how many rows run from $12$ to $22$, then multiply that count by the $33$ seats in each row. The one real trap is the inclusive count — writing $22-12=10$ forgets that row $12$ itself is reserved. Tool #3 (Eliminate Possibilities) shows every wrong choice is a specific miscount: $330=10\times33$ (off by one), $297=9\times33$, $396=12\times33$, and $726=22\times33$.
Execute — Answer: C
4.OA.A.3 Step 1 Count the reserved rows
- To count the whole numbers from $12$ to $22$ when both ends are included, subtract the endpoints and then add $1$: $22-12=10$, and $10+1=11$.
- The extra $1$ is there because both row $12$ and row $22$ are reserved, not just the gap between them.
- So there are $11$ reserved rows.
💡 Counting a run of numbers that includes both ends means subtract the ends, then add one for the starting row.
4.NBT.B.5 Step 2 Multiply rows by seats per row
- Each of the $11$ rows holds $33$ seats, and the rows are all the same size, so the total is $11$ groups of $33$.
- Compute it as $11\times33 = 10\times33 + 1\times33 = 330 + 33 = 363$.
💡 Equal groups multiply: number of rows times seats in each row gives the total seats.
4.OA.A.3 Step 3 Check against the trap choices
- The total $363$ is choice (C).
- Every nearby choice matches a specific miscount: $330$ (B) uses only $10$ rows, dropping row $12$; $297$ (A) uses $9$ rows; $396$ (D) uses $12$ rows; and $726$ (E) is $22\times33$, as if every row from $1$ to $22$ were reserved.
- Only the correct inclusive count of $11$ rows gives $363$, so the answer is (C).
💡 Each wrong choice is a specific off-by-one or over-count, so nailing the row count pins the answer.
4.OA.A.3 To count the whole numbers from $12$ to $22$ when both ends are included, subtra 4.NBT.B.5 Each of the $11$ rows holds $33$ seats, and the rows are all the same size, so t 4.OA.A.3 The total $363$ is choice (C). Every nearby choice matches a specific miscount: Review
Reasonableness: A quick estimate: about a dozen rows of about $33$ seats is roughly $12\times33\approx396$, so the total should sit a little under $400$. The value $363$ fits, while $726$ (nearly double) and $297$ (too few rows) do not. The exact inclusive count is $11$ rows, and $11\times33=363$ confirms choice (C).
Alternative: Factor the seats: $33 = 3\times11$, so $11\times33 = 11\times11\times3 = 121\times3 = 363$. Reaching $363$ by this different multiplication route confirms (C) without re-adding $33$ eleven times.
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Interpreting "rows 12 through 22" as an inclusive count of 11 rows and checking the total against the trap answers.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $11\times33=363$, the total seats across the 11 reserved rows.)
⭐ To count rows from a first number through a last number, subtract them and add one — then multiply by how many seats each row holds.
⭐ To count rows from a first number through a last number, subtract them and add one — then multiply by how many seats each row holds.
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