AMC 8 · 2009 · #14
Grade 6 rate-ratioProblem
Austin and Temple are 50 miles apart along Interstate 35. Bonnie drove from Austin to her daughter's house in Temple, averaging 60 miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged 40 miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?
Pick an answer.
AMC 8 2009 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: Austin and Temple are $50$ miles apart. Bonnie drives the $50$ miles from Austin to Temple at an average speed of $60$ mph, then rides a bus back the same $50$ miles to Austin at an average speed of $40$ mph. What is her average speed over the entire $100$-mile round trip, in mph?
Givens: One-way distance Austin $\leftrightarrow$ Temple $= 50$ miles; Outbound (Austin $\to$ Temple) average speed $= 60$ mph; Return (Temple $\to$ Austin) average speed $= 40$ mph; Same route both directions, so round-trip distance $= 2 \times 50 = 100$ miles; Answer choices: (A) $46$, (B) $48$, (C) $50$, (D) $52$, (E) $54$ (mph)
Unknowns: Average speed for the round trip in miles per hour
Understand
Restated: Austin and Temple are $50$ miles apart. Bonnie drives the $50$ miles from Austin to Temple at an average speed of $60$ mph, then rides a bus back the same $50$ miles to Austin at an average speed of $40$ mph. What is her average speed over the entire $100$-mile round trip, in mph?
Givens: One-way distance Austin $\leftrightarrow$ Temple $= 50$ miles; Outbound (Austin $\to$ Temple) average speed $= 60$ mph; Return (Temple $\to$ Austin) average speed $= 40$ mph; Same route both directions, so round-trip distance $= 2 \times 50 = 100$ miles; Answer choices: (A) $46$, (B) $48$, (C) $50$, (D) $52$, (E) $54$ (mph)
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The trap here is averaging $60$ and $40$ to get $50$ — that ignores units. Tool #8 (Analyze the Units) keeps us honest: "mph" is miles per hour, so the only correct average speed is $\dfrac{\text{total miles}}{\text{total hours}}$. Tool #7 (Identify Subproblems) splits the trip into the two legs so we can compute each leg's time from $\text{time} = \text{distance} / \text{speed}$, then add to get the total time before dividing into the total distance.
Execute — Answer: B
4.MD.A.2 Step 1 - Find the time for the outbound leg.
- Bonnie covers $50$ miles at $60$ mph, so the time is distance divided by speed.
💡 Splitting the round trip into two legs and using distance $\div$ speed for each is a Grade 4 distance/time word-problem move.
4.MD.A.2 Step 2 Find the time for the return leg the same way: $50$ miles at $40$ mph.
💡 Notice the return leg takes longer ($\tfrac{5}{4} > \tfrac{5}{6}$) because the speed is lower — the slower leg spends more hours on the road.
5.NF.A.1 Step 3 - Add the two leg times to get the total time.
- Use a common denominator of $12$ to add the fractions.
💡 Adding fractions with unlike denominators by finding a common denominator is the Grade 5 fraction-addition standard.
4.MD.A.2 Step 4 - Compute the total round-trip distance.
- Same $50$-mile route in both directions, so the total is $2 \times 50 = 100$ miles.
💡 The units stay "miles" — we are just summing the distance traveled.
6.RP.A.3 Step 5 - Apply the definition of average speed: total distance divided by total time.
- Dividing by $\tfrac{25}{12}$ is the same as multiplying by $\tfrac{12}{25}$.
💡 Computing miles per hour as a unit rate from total miles and total hours is Grade 6 rate reasoning.
4.MD.A.2 Find the time for the outbound leg. Bonnie covers $50$ miles at $60$ mph, so the 4.MD.A.2 Find the time for the return leg the same way: $50$ miles at $40$ mph. 5.NF.A.1 Add the two leg times to get the total time. Use a common denominator of $12$ to 4.MD.A.2 Compute the total round-trip distance. Same $50$-mile route in both directions, 6.RP.A.3 Apply the definition of average speed: total distance divided by total time. Div Review
Reasonableness: The naive average $\tfrac{60 + 40}{2} = 50$ mph would only be correct if Bonnie spent equal time at each speed. But she spent equal distance at each speed, so the slower $40$-mph leg ate up more hours and pulled the average below $50$. The answer $48$ mph is just under $50$, which matches that intuition. It also sits between $40$ and $60$, as any sensible average must.
Alternative: Tool #9 (Find a Formula): for a round trip with equal distances at speeds $a$ and $b$, the average speed is the harmonic mean $\dfrac{2ab}{a+b}$. Plugging in $a = 60$, $b = 40$ gives $\dfrac{2 \cdot 60 \cdot 40}{60 + 40} = \dfrac{4800}{100} = 48$ mph, matching choice (B). This formula is exactly what the total-distance-over-total-time calculation produces when distances are equal, so it is a fast shortcut once you know it.
CCSS standards used (min grade 6)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Computing each leg's travel time as $\text{distance} \div \text{speed}$ and the total round-trip distance $2 \times 50 = 100$ miles.)5.NF.A.1Add and subtract fractions with unlike denominators (Adding the two leg times $\tfrac{5}{6} + \tfrac{5}{4} = \tfrac{25}{12}$ hours by finding a common denominator of $12$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Computing the average speed as the unit rate $\overline{v} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{100}{25/12} = 48$ mph.)
⭐ Average speed isn't just the average of two speeds — it's total miles divided by total hours, a Grade 6 rate idea you already use!
⭐ Average speed isn't just the average of two speeds — it's total miles divided by total hours, a Grade 6 rate idea you already use!
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