AMC 10 · 2007 · #3
Grade 6 rate-ratioPick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Gas used driving home
Home leg: 120 miles at 30 miles per gallon, so gallons = 120÷30 = 4 gallons.
Dividing the miles by the mileage cancels the miles and leaves the gallons.
6.RP.A.3Analyze The UnitsGas used driving back
Return leg: the same 120 miles at only 20 miles per gallon, so 120÷20 = 6 gallons — worse mileage, more gas.
Worse mileage means the same distance eats more gallons.
6.RP.A.3Analyze The UnitsTotal miles and total gallons
Stack the legs: total distance 120+120 = 240 miles, total gas 4+6 = 10 gallons.
The whole trip is just the two legs stacked end to end, in both miles and gallons.
4.OA.A.3Identify SubproblemsAverage mileage for the round trip
Total miles ÷ total gallons = 240÷10 = 24 miles per gallon, choice (B) — not the naive midpoint 25.
Overall mileage is all the miles shared out over all the gallons, and the thirstier leg counts for more gallons.
Overall mileage is all the miles shared out over all the gallons, so the thirstier leg counts for more.
▸ Why?
An average is a total shared over a count, so both totals must be built before dividing.
▸ Why?
At a steady mileage the fuel used is the distance divided by that rate, so each leg brings its own gallons.
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.
- Gas used driving home
- Gas used driving back
- Total miles and total gallons
- Average mileage for the round trip