AMC 10 · 2006 · #1
Grade 3 arithmeticSandwiches at Joe's Fast Food cost 3$ each and sodas cost2each.Howmanydollarswillitcosttopurchase5sandwichesand8$ sodas?
Pick an answer.
AMC 10 2006 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: Each sandwich costs $\$3$ and each soda costs $\$2$. Find the total cost, in dollars, of buying $5$ sandwiches and $8$ sodas.
Givens: One sandwich costs $\$3$; One soda costs $\$2$; The order is $5$ sandwiches and $8$ sodas; Answer choices: (A) $31$, (B) $32$, (C) $33$, (D) $34$, (E) $35$
Unknowns: The total number of dollars the whole order costs
Understand
Restated: Each sandwich costs $\$3$ and each soda costs $\$2$. Find the total cost, in dollars, of buying $5$ sandwiches and $8$ sodas.
Givens: One sandwich costs $\$3$; One soda costs $\$2$; The order is $5$ sandwiches and $8$ sodas; Answer choices: (A) $31$, (B) $32$, (C) $33$, (D) $34$, (E) $35$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
This is a money problem where each item has a price in dollars-per-item, so Tool #8 (Analyze the Units) says: multiply how many by the price each to get dollars. Tool #7 (Identify Subproblems) splits the order into two easy pieces — the cost of the sandwiches and the cost of the sodas — which are then added. Tool #3 (Eliminate Possibilities) gives a fast reality check: the choices are all near $30$, and the two subtotals are $15$ and $16$, so the answer must sit at the low end.
Execute — Answer: A
3.OA.A.3 Step 1 Cost of the sandwiches
- Five sandwiches at $\$3$ each is an equal-groups multiplication: $5$ groups of $\$3$.
- Multiply the count by the price to get $5 \times 3 = 15$ dollars for all the sandwiches.
💡 Same price repeated is just multiplication: how many times how much each.
3.OA.A.3 Step 2 Cost of the sodas
- Eight sodas at $\$2$ each is another equal-groups multiplication: $8$ groups of $\$2$.
- Multiply to get $8 \times 2 = 16$ dollars for all the sodas.
💡 Count the sodas, multiply by the price each, and you have their share of the bill.
2.NBT.B.5 Step 3 Add the two subtotals
- The whole bill is the sandwiches plus the sodas, so add the two subtotals: $15 + 16 = 31$ dollars.
- That matches choice (A).
- Because the parts were only $\$15$ and $\$16$, no total near $34$ or $35$ is possible, which rules out (D) and (E) and confirms the low end.
💡 The total bill is simply every part of the order added together.
3.OA.A.3 Five sandwiches at $\$3$ each is an equal-groups multiplication: $5$ groups of $ 3.OA.A.3 Eight sodas at $\$2$ each is another equal-groups multiplication: $8$ groups of 2.NBT.B.5 The whole bill is the sandwiches plus the sodas, so add the two subtotals: $15 + Review
Reasonableness: A rough estimate confirms it: $5$ sandwiches and $8$ sodas is $13$ items, and prices sit between $\$2$ and $\$3$, so the bill should land between $13 \times 2 = 26$ and $13 \times 3 = 39$ dollars — and $31$ sits right in that window. The two subtotals $\$15$ and $\$16$ are each just under half of the total, and $15 + 16 = 31$ checks out, so (A) is solid while (D) $34$ and (E) $35$ are too high.
Alternative: Add the items first, then price them separately. Money can be regrouped: $\$15 + \$16 = \$15 + \$15 + \$1 = \$30 + \$1 = \$31$. Splitting $16$ into $15 + 1$ makes a friendly $\$30$, then the leftover dollar gives $\$31$, the same answer (A).
CCSS standards used (min grade 3)
3.OA.A.3Solve multiplication and division word problems within 100 (Turning "5 sandwiches at $3 each" and "8 sodas at $2 each" into $5 \times 3 = 15$ and $8 \times 2 = 16$.)2.NBT.B.5Fluently add and subtract within 100 (Adding the two subtotals $15 + 16 = 31$ to get the full bill.)
⭐ For a shopping bill, multiply each item's count by its price, then add the pieces together.
⭐ For a shopping bill, multiply each item's count by its price, then add the pieces together.
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