AMC 10 · 2007 · #3
Grade 6 arithmeticA college student drove his compact car 120 miles home for the weekend and averaged 30 miles per gallon. On the return trip the student drove his parents' SUV and averaged only 20 miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
Pick an answer.
AMC 10 2007 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: A student drives $120$ miles home getting $30$ miles per gallon, then drives $120$ miles back getting $20$ miles per gallon. Find the average gas mileage, in miles per gallon, for the whole round trip.
Givens: One-way distance is $120$ miles, so the round trip covers the same $120$ miles twice; The trip home averaged $30$ miles per gallon; The return trip averaged $20$ miles per gallon; Answer choices: (A) $22$, (B) $24$, (C) $25$, (D) $26$, (E) $28$
Unknowns: The average miles per gallon for the entire round trip
Understand
Restated: A student drives $120$ miles home getting $30$ miles per gallon, then drives $120$ miles back getting $20$ miles per gallon. Find the average gas mileage, in miles per gallon, for the whole round trip.
Givens: One-way distance is $120$ miles, so the round trip covers the same $120$ miles twice; The trip home averaged $30$ miles per gallon; The return trip averaged $20$ miles per gallon; Answer choices: (A) $22$, (B) $24$, (C) $25$, (D) $26$, (E) $28$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Miles per gallon is a rate — miles divided by gallons — so Tool #8 (Analyze the Units) tells you that overall mileage must be total miles over total gallons, not the average of $30$ and $20$. Tool #7 (Identify Subproblems) splits the work into finding the gas each leg burns before combining. Tool #3 (Eliminate Possibilities) flags the trap answer: naively averaging the two mileages gives $(30+20)/2=25$, which is choice (C) and is wrong.
Execute — Answer: B
6.RP.A.3 Step 1 Gas used driving home
- On the way home the car covers $120$ miles at $30$ miles per gallon.
- Gallons used is miles divided by miles-per-gallon, so $120\div30=4$ gallons.
💡 Dividing the miles by the mileage cancels the miles and leaves the gallons.
6.RP.A.3 Step 2 Gas used driving back
- On the return the SUV covers the same $120$ miles but at only $20$ miles per gallon.
- Gallons used is $120\div20=6$ gallons — more gas for the same distance because the mileage is worse.
💡 Worse mileage means the same distance eats more gallons.
4.OA.A.3 Step 3 Total miles and total gallons
- Add the two legs.
- Total distance is $120+120=240$ miles.
- Total gas is $4+6=10$ gallons.
- These two totals are what the round-trip mileage compares.
💡 The whole trip is just the two legs stacked end to end, in both miles and gallons.
6.RP.A.3 Step 4 Average mileage for the round trip
- Average mileage is total miles divided by total gallons: $240\div10=24$ miles per gallon.
- That is choice (B).
- Note it is not $(30+20)/2=25$ — the return leg burns more gallons, so it pulls the overall mileage below the halfway point, ruling out the trap answer (C).
💡 Overall mileage is all the miles shared out over all the gallons, and the thirstier leg counts for more gallons.
6.RP.A.3 On the way home the car covers $120$ miles at $30$ miles per gallon. Gallons use 6.RP.A.3 On the return the SUV covers the same $120$ miles but at only $20$ miles per gal 4.OA.A.3 Add the two legs. Total distance is $120+120=240$ miles. Total gas is $4+6=10$ g 6.RP.A.3 Average mileage is total miles divided by total gallons: $240\div10=24$ miles pe Review
Reasonableness: The answer $24$ lands between the two mileages $20$ and $30$, which it must, and it sits below the plain midpoint $25$. That is exactly right: the low-mileage return leg uses more gallons ($6$ versus $4$), so it carries more weight and drags the average toward $20$. A quick check confirms the totals: $4\text{ gal}+6\text{ gal}=10\text{ gal}$ carries the car $240$ miles, and $240/10=24$.
Alternative: Because the two distances are equal, the round-trip mileage is the harmonic mean of $30$ and $20$: $\dfrac{2}{\frac{1}{30}+\frac{1}{20}} = \dfrac{2}{\frac{2}{60}+\frac{3}{60}} = \dfrac{2}{\frac{5}{60}} = \dfrac{120}{5} = 24$, giving (B) directly without choosing a specific distance.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Treating miles per gallon as a rate: dividing miles by mileage to get gallons on each leg, then dividing total miles by total gallons to get the round-trip mileage.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Adding the two legs to get $240$ total miles and $10$ total gallons before the final division.)
⭐ Average mileage is all your miles divided by all your gallons, not the average of the two speeds — so the gas-guzzling leg pulls it down.
⭐ Average mileage is all your miles divided by all your gallons, not the average of the two speeds — so the gas-guzzling leg pulls it down.
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