Competition · AMC preparation · step 4 of 4
AMC 8 · 2009 · #14
Grade 6 rate-ratioPick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 (total miles)/(total hours). Tool #7 (Identify Subproblems) splits the trip into the two legs so we can compute each leg's time from time = distance / speed, then add to get the total time before dividing into the total distance.
Find the outbound time
Outbound leg: 50 miles at 60 mph gives time = distance ÷ speed = hr.
Splitting the round trip into two legs and using distance ÷ speed for each is a Grade 4 distance/time word-problem move.
4.MD.A.2Identify SubproblemsFind the return time
Return leg: 50 miles at 40 mph gives time = hr, longer because the speed is lower.
Notice the return leg takes longer (5/4 ＞ 5/6) because the speed is lower — the slower leg spends more hours on the road.
4.MD.A.2Identify SubproblemsAdd the two times
Add the leg times using a common denominator of 12: + = hr total.
Adding fractions with unlike denominators by finding a common denominator is the Grade 5 fraction-addition standard.
5.NF.A.1Identify SubproblemsFind the round-trip distance
Total distance: the same 50-mile route twice, so 2 × 50 = 100 miles.
The units stay "miles" — we are just summing the distance traveled.
4.MD.A.2Analyze The UnitsDivide distance by time
Average speed = total distance ÷ total time = 100 ÷ = 100 × = 48 mph → (B).
Computing miles per hour as a unit rate from total miles and total hours is Grade 6 rate reasoning.
The round trip's average speed is the total distance divided by the total time it took, which is not the same as the plain average of the two leg speeds.
▸ Why?
Average speed names the single steady speed that would carry Bonnie over the whole distance in the whole time she actually spent, and a steady speed is just distance divided by time.
▸ Why?
A steady speed held for some time covers a distance equal to that speed times the time, so recovering the speed means dividing the distance back out by the time.
▸ Why?
The whole time is the outbound time plus the return time, and the whole distance is the two legs' distances added together, because the trip is exactly those two legs with nothing left out and nothing counted twice.
▸ Why?
Each leg's time is its own distance divided by its own speed, so the slower leg quietly runs up more hours even though both legs cover the same miles, which is why weighting the two speeds equally would be wrong.
▸ Why?
Since distance equals speed times time for a steady speed, time is distance divided by speed, and dividing the same distance by a smaller speed gives a larger time.
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!
- Find the outbound time
- Find the return time
- Add the two times
- Find the round-trip distance
- Divide distance by time
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