AMC 10 · 2018 · #2
Grade 6 rate-ratioPick an answer.
AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a rate problem (distance = speed × time), so Tool #8 (Analyze the Units) leads: convert each 30-minute leg to 1/2 hour so that mph × hours gives miles cleanly. Tool #7 (Identify Subproblems) supports it — break the 96-mile total into three leg-distances, find the two known legs, and the missing third leg is forced by subtraction. Then one more unit step turns that distance back into a speed.
Match the units
Speeds are in mph but each leg lasts 30 minutes, which is exactly 1/2 hour.
A speed in mph only multiplies cleanly when the time is in hours, so turn the 30 minutes into half an hour first.
4.MD.A.1Analyze The UnitsDistance of the first two legs
Distance = speed × time: 60 × 1/2 = 30 and 65 × 1/2 = 32.5, so the first two legs cover 62.5 miles.
At a steady speed, distance is just speed scaled by how long you drive — half an hour means half the miles-per-hour number.
6.RP.A.3Analyze The UnitsDistance left for the last leg
All three legs must total 96 miles, so the last leg is 96 - 62.5 = 33.5 miles.
The total distance is fixed, so whatever the first two legs do not cover is exactly what the last leg must cover.
5.NBT.B.7Identify SubproblemsTurn distance back into speed
Dividing 33.5 miles by the half hour is the same as doubling: 33.5 × 2 = 67 mph, choice (D).
If you cover 33.5 miles in half an hour, in a whole hour you would cover twice as much — so the unit rate is 67 mph.
6.RP.A.2Analyze The UnitsTurn every 30-minute leg into half an hour, add up the miles you know, and whatever distance is missing from 96 tells you the last leg's speed.
- Match the units
- Distance of the first two legs
- Distance left for the last leg
- Turn distance back into speed