Competition · AMC preparation · step 4 of 4
AMC 8 · 2015 · #3
Grade 6 rate-ratioPick an answer.
AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a rate problem with time = distance / speed. The catch is that distance is in miles and speed is in mph, so dividing gives hours — but the answer must be in minutes. Tool #8 (Analyze the Units) keeps the conversion honest: mile/(mile/hour) = hour, then hour × 60 = minutes. Tool #7 (Identify Subproblems) splits the work into three clean pieces — Jill's time, Jack's time, and the difference — so each piece is a single short calculation.
Find Jill's travel time
Jill's time: divide 1 mile by 10 mph to get hour, then multiply by 60 to reach 6 minutes.
Dividing miles by miles-per-hour cancels "miles" and leaves "hours" — Grade 6 unit-rate reasoning.
Jill's one-mile trip at a steady 10 miles per hour takes 6 minutes.
▸ Why?
The trip first works out to 1/10 of an hour, before we rename that in minutes.
▸ Why?
At a steady 10 miles each hour, the miles covered after h hours are h equal groups of 10 miles, one group per hour, so the distance is 10 × h.
▸ Why?
To pull h back out of 10 × h = 1 mile, we undo the multiply-by-10 by dividing, giving h = 1/10 hour.
▸ Why?
One tenth of an hour is 6 minutes, since an hour is a fixed 60 minutes and a tenth of 60 is 6.
Find Jack's travel time
Jack's time, same method: divide 1 mile by 4 mph to get hour, then multiply by 60 to reach 15 minutes.
Same unit-rate move as Jill's step — once the method works for one rider, it works for both.
6.RP.A.3Analyze The UnitsSubtract the two times
Both start together, so the head start equals the time gap: 15 - 6 = 9 minutes → (D).
"Difference of two times" is the Grade 4 distance/time word-problem move — solving the third subproblem to finish the job.
4.MD.A.2Identify SubproblemsThis AMC 8 problem only needs the Grade 6 rate rule "time = distance ÷ speed" plus a minute-to-hour conversion you learned in Grade 5.
- Find Jill's travel time
- Find Jack's travel time
- Subtract the two times
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