AMC 10 · 2013 · #8
Grade 6 rate-ratioRay's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline?
Pick an answer.
AMC 10 2013 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: Two cars drive the same distance. One car goes $40$ miles on each gallon, the other goes $10$ miles on each gallon. Find the miles-per-gallon rate for the two cars taken together, meaning the total miles they drive divided by the total gallons they burn.
Givens: Ray's car: $40$ miles per gallon; Tom's car: $10$ miles per gallon; Ray and Tom each drive the same number of miles; Answer choices: (A) $10$, (B) $16$, (C) $25$, (D) $30$, (E) $40$
Unknowns: The combined miles-per-gallon rate for both cars together
Understand
Restated: Two cars drive the same distance. One car goes $40$ miles on each gallon, the other goes $10$ miles on each gallon. Find the miles-per-gallon rate for the two cars taken together, meaning the total miles they drive divided by the total gallons they burn.
Givens: Ray's car: $40$ miles per gallon; Tom's car: $10$ miles per gallon; Ray and Tom each drive the same number of miles; Answer choices: (A) $10$, (B) $16$, (C) $25$, (D) $30$, (E) $40$
Plan
Primary tool: #8 Analyze the Units
Secondary: #9 Solve an Easier Related Problem
The whole problem lives in the units "miles per gallon." Tool #8 (Analyze the Units) fixes the meaning: a combined rate is total miles divided by total gallons, so the job is to add up the miles and add up the gallons. To keep the gallons whole and avoid fractions, Tool #9 (Solve an Easier Related Problem) picks one convenient distance that both $40$ and $10$ divide evenly, works the numbers there, and trusts that equal distances give the same rate no matter which distance is chosen.
Execute — Answer: B
6.RP.A.2 Step 1 Set up the combined rate
- Read the units carefully.
- Miles per gallon means miles divided by gallons, so the combined rate for both cars is the total miles they drive divided by the total gallons they use.
- The tempting move is to average $40$ and $10$ to get $25$, but that ignores that the slow car burns far more gas over the same distance, so its low rate should count more.
💡 A rate is one thing per another thing, so combining rates means totalling each thing separately, not averaging the rates.
6.RP.A.3 Step 2 Pick one convenient distance
- Choose a distance that both $40$ and $10$ divide evenly, so the gallons come out whole.
- Let each car drive $40$ miles.
- Ray's car burns $40 \div 40 = 1$ gallon, and Tom's car burns $40 \div 10 = 4$ gallons.
- Tom's car uses four times as much gas to cover the very same distance.
💡 Turning gas mileage into gallons for a picked distance replaces an abstract rate with countable amounts you can add.
6.RP.A.3 Step 3 Add the totals and divide
- Now total each thing.
- The two cars together drive $40 + 40 = 80$ miles and burn $1 + 4 = 5$ gallons.
- Divide to get the combined rate: $80 \div 5 = 16$ miles per gallon.
- This matches choice (B).
- Any other equal distance, such as $80$ or $200$ miles each, scales both totals by the same factor and gives $16$ again.
💡 Dividing the summed miles by the summed gallons gives the honest overall rate because it weighs each car by the gas it actually used.
6.RP.A.2 Read the units carefully. Miles per gallon means miles divided by gallons, so th 6.RP.A.3 Choose a distance that both $40$ and $10$ divide evenly, so the gallons come out 6.RP.A.3 Now total each thing. The two cars together drive $40 + 40 = 80$ miles and burn Review
Reasonableness: The combined rate must land between the two given rates, $10$ and $40$, so choices (A) $10$ and (E) $40$ are out. Because the thirsty $10$-mpg car burns most of the gas, the combined rate should sit much closer to $10$ than to $40$; $16$ fits that, while the plain average $25$ (choice C) sits right in the middle and wrongly ignores the uneven gas use.
Alternative: Skip the numbers and use the definition directly. For a distance $d$ each, total miles $= 2d$ and total gallons $= \dfrac{d}{40} + \dfrac{d}{10} = \dfrac{d}{40} + \dfrac{4d}{40} = \dfrac{5d}{40} = \dfrac{d}{8}$. Then the rate is $2d \div \dfrac{d}{8} = 16$, the harmonic mean of $40$ and $10$, and the $d$ cancels — confirming the distance never mattered.
CCSS standards used (min grade 6)
6.RP.A.2Understand the concept of a unit rate and use rate language (Reading "miles per gallon" as miles divided by gallons and defining the combined rate as total miles over total gallons.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Converting each car's mileage into gallons for a chosen distance, then dividing summed miles by summed gallons to get $80 \div 5 = 16$.)
⭐ To combine two rates, add up the total miles and the total gallons and then divide — never just average the two rates, because the gas-guzzler counts for more.
⭐ To combine two rates, add up the total miles and the total gallons and then divide — never just average the two rates, because the gas-guzzler counts for more.
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