AMC 8 · 2014 · #5
Easy mode Grade 6Problem
Margie's car can drive 32 miles using 1 gallon of gas.
Gas costs $4 for each gallon.
Margie has \textdollar20 to spend on gas.
How many miles can she drive with that gas?
Pick an answer.
AMC 8 2014 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: Margie's car gets $32$ miles for every gallon of gas. Gas costs $\$4$ per gallon. If she spends $\$20$ on gas, how many miles can she drive?
Givens: Fuel efficiency = $32$ miles per gallon; Price of gas = $\$4$ per gallon; Money available = $\$20$; Answer choices: (A) $64$, (B) $128$, (C) $160$, (D) $320$, (E) $640$ (miles)
Unknowns: The total number of miles Margie can drive on $\$20$ worth of gas
Understand
Restated: Margie's car gets $32$ miles for every gallon of gas. Gas costs $\$4$ per gallon. If she spends $\$20$ on gas, how many miles can she drive?
Givens: Fuel efficiency = $32$ miles per gallon; Price of gas = $\$4$ per gallon; Money available = $\$20$; Answer choices: (A) $64$, (B) $128$, (C) $160$, (D) $320$, (E) $640$ (miles)
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units
The question "how many miles" links three quantities — dollars, gallons, and miles — so we use Tool #7 (Identify Subproblems) to split the work into two clean steps: first turn $\$20$ into gallons, then turn gallons into miles. Tool #8 (Analyze the Units) keeps us honest at each step: $\$ \div (\$/\text{gal}) = \text{gal}$, and $\text{gal} \times (\text{mi}/\text{gal}) = \text{mi}$. The units cancel correctly, so we know the operations are right.
Execute — Answer: C
4.NBT.B.6 Step 1 Subproblem 1: figure out how many gallons $\$20$ buys. Each gallon costs $\$4$, so divide total money by price per gallon.
💡 Dividing dollars by dollars-per-gallon cancels the dollars and leaves gallons. The arithmetic ($20 \div 4 = 5$) is a Grade 4 division fact.
4.NBT.B.5 Step 2 - Subproblem 2: turn those $5$ gallons into miles.
- The car covers $32$ miles for each gallon, so multiply gallons by the miles-per-gallon rate.
💡 The second subproblem reuses the answer to the first. Multiplying gallons by miles-per-gallon cancels gallons and leaves miles — exactly the Grade 4 multi-digit multiplication skill.
6.RP.A.3 Step 3 - Combine the two subproblems to read off the final answer.
- $160$ miles matches choice (C).
💡 The whole chain $\$ \to \text{gal} \to \text{mi}$ is a unit-rate problem: at $\$4$/gal and $32$ mi/gal, every dollar buys $8$ miles, and $\$20 \times 8 = 160$.
4.NBT.B.6 Subproblem 1: figure out how many gallons $\$20$ buys. Each gallon costs $\$4$, 4.NBT.B.5 Subproblem 2: turn those $5$ gallons into miles. The car covers $32$ miles for e 6.RP.A.3 Combine the two subproblems to read off the final answer. $160$ miles matches ch Review
Reasonableness: Quick sanity check: $1$ dollar buys $\tfrac{1}{4}$ gallon, and $\tfrac{1}{4}$ gallon goes $\tfrac{32}{4} = 8$ miles. So $\$1 = 8$ mi, and $\$20 = 20 \times 8 = 160$ mi. Same answer, no setup needed. The magnitude is right too — a small car driving a couple hundred miles on $\$20$ is realistic at $\$4$/gal.
Alternative: Tool #6 (Guess and Check) on the answer choices: the miles per dollar must be $32/4 = 8$, so the total miles must equal $20 \times 8 = 160$. Of the five choices only (C) $160$ fits; (A) $64$ would mean $3.2$ mi/$\$$, (B) $128$ would mean $6.4$ mi/$\$$, (D) $320$ would mean $16$ mi/$\$$, and (E) $640$ would mean $32$ mi/$\$$ — none match the $8$ mi/$\$$ rate the problem forces.
CCSS standards used (min grade 6)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors (Dividing $\$20$ by $\$4$ per gallon to get $5$ gallons.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Multiplying $5$ gallons by $32$ miles per gallon to get $160$ miles.)4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Setting up the dollars-to-gallons-to-miles word-problem chain in a single real-world scenario.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Combining the two rates ($\$4$/gal and $32$ mi/gal) into a single unit rate ($8$ mi/$\$$) and applying it to $\$20$.)
⭐ This AMC 8 problem only needs Grade 4 division and multiplication, with a Grade 6 "rate" idea on top — split it into two small steps and the answer falls out.
⭐ This AMC 8 problem only needs Grade 4 division and multiplication, with a Grade 6 "rate" idea on top — split it into two small steps and the answer falls out.
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