AMC 8 · 2019 · #1

Easy mode Grade 5

Problem

Imagine you have $30.00$ to spend at a sandwich shop. One sandwich costs$ 4.50 $. One soft drink costs$ $1.00$ .

Ike and Mike want to buy as many sandwiches as they can first. Then they will use whatever money is left to buy soft drinks.

How many things do they end up buying in total, counting both sandwiches and soft drinks?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Ike and Mike have $\$30.00$ total. Each sandwich costs $\$4.50$ and each soft drink costs $\$1.00$. They buy as many sandwiches as they can afford, then spend whatever is left on soft drinks. How many items (sandwiches + drinks) do they take home in all?

Givens: Budget: $\$30.00$ total; Sandwich price: $\$4.50$ each; Soft drink price: $\$1.00$ each; Rule: buy max sandwiches first, then use leftover money for drinks; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$

Unknowns: The total number of items (sandwiches plus soft drinks) they will buy

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #6 Guess and Check

The question has two natural subproblems (Tool #7): (a) find the largest whole number of sandwiches that fits in the $\$30$ budget, and (b) figure out how many $\$1$ drinks the leftover money can buy. Subproblem (a) is a perfect fit for Tool #6 (Guess and Check) on the small candidates $6$ vs $7$ sandwiches — much friendlier than dividing $\$30 \div \$4.50$, which would force decimal division. Tool #6 also lets us verify directly that no other answer choice is reachable.

Execute — Answer: D

#6 Guess and Check 5.NBT.B.7 Step 1

Subproblem A — Find the maximum number of sandwiches.
Test the two natural candidates near the budget.
Try $6$ sandwiches: $6 \times \$4.50 = \$27.00$, which is within the $\$30$ budget. Try $7$ sandwiches: $7 \times \$4.50 = \$31.50$, which exceeds the budget. So $6$ is the largest whole number that fits.

$6 \times \$4.50 = \$27.00 \le \$30 \;\checkmark \quad 7 \times \$4.50 = \$31.50 > \$30 \;\times$

💡 Multiplying a whole number by a price like $\$4.50$ is Grade 5 decimal multiplication to hundredths — no division needed.

#7 Identify Subproblems 4.MD.A.2 Step 2

Find the leftover money after buying the sandwiches.
Subtract the sandwich cost from the starting budget: $\$30.00 - \$27.00 = \$3.00$.

$\$30.00 - \$27.00 = \$3.00$

💡 Subtracting money to find what is left over is a Grade 4 money word-problem move.

#7 Identify Subproblems 3.OA.A.3 Step 3

Subproblem B — Find how many soft drinks the leftover $\$3.00$ buys at $\$1.00$ each.
Since each drink costs exactly one dollar, the count equals the dollar amount: $3 \div 1 = 3$ drinks.

$\$3.00 \div \$1.00 = 3 \text{ drinks}$

💡 Splitting a total amount into equal $\$1$ groups is Grade 3 division within $100$.

#7 Identify Subproblems 2.OA.A.1 Step 4

Combine the two subproblems to get the total number of items.
Add the sandwich count and the drink count.

$$6 \text{ sandwiches} + 3 \text{ drinks} = 9 \text{ items} \;\Rightarrow\; \textbf{(D)}$$

💡 Adding two small whole-number counts to get a total is Grade 2 addition within $100$.

[1] #6 5.NBT.B.7 Subproblem A — Find the maximum number of sandwiches. Test the two natural candi

[2] #7 4.MD.A.2 Find the leftover money after buying the sandwiches. Subtract the sandwich cost

[3] #7 3.OA.A.3 Subproblem B — Find how many soft drinks the leftover $\$3.00$ buys at $\$1.00$

[4] #7 2.OA.A.1 Combine the two subproblems to get the total number of items. Add the sandwich c

Review

Reasonableness: Check the money: $6$ sandwiches at $\$4.50$ cost $\$27.00$, plus $3$ drinks at $\$1.00$ cost $\$3.00$, for a total of $\$30.00$ exactly — every dollar is spent. The total of $9$ items is choice (D), which sits in the middle of the answer choices and matches the everyday feel of "a few sandwiches and a few drinks". A small sanity check: if they had bought one more sandwich, they would have needed $\$31.50 > \$30$, so $6$ is indeed the cap.

Alternative: Tool #3 (Eliminate Possibilities) using the choices: with at most $\$30$, the most items happens when most are cheap drinks. Even $10$ drinks would cost only $\$10$, but the rule says they must buy as many sandwiches as possible first. Trying $6$ sandwiches + $k$ drinks: cost $= 27 + k$, fits when $k \le 3$, so $k = 3$ and total items $= 9$. Choices (A)-(C) require fewer than $6$ sandwiches (violates the "max sandwiches" rule) and (E) needs $\$31.50$ for $7$ sandwiches alone. Only (D) survives.

CCSS standards used (min grade 5)

2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Adding the sandwich count and drink count to get the final total of $6 + 3 = 9$ items.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Dividing the $\$3$ leftover by the $\$1$ drink price to count $3$ drinks.)
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Subtracting the sandwich spending from the starting budget ($\$30.00 - \$27.00 = \$3.00$) to find the leftover money.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Multiplying whole numbers of sandwiches by the price $\$4.50$ (e.g. $6 \times \$4.50 = \$27.00$, $7 \times \$4.50 = \$31.50$) to find the largest count that fits the budget.)

⭐ This AMC 8 problem only needs Grade 5 decimal multiplication — like figuring out the cost of $6$ sandwiches at $\$4.50$ — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

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