AMC 8 · 2011 · #10
Grade 6 algebraPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The 10 has to do three different jobs — tip, flat fare for the first 0.5 mile, and the per-0.1-mile charge after that — so Tool #7 (Identify Subproblems) cleanly peels them off one at a time: take out the tip, take out the flat fare, then ask "how far does the leftover money buy?" Tool #8 (Analyze the Units) turns the awkward0.20 per 0.1 mile rate into the friendlier $2 per mile, which makes the final division a one-line decimal computation rather than an algebra problem.
Set aside the tip first: you hand over the 8 for the meter.
Splitting the $10 into "tip" and "fare" is the Tool #7 move — handle one job at a time.
4.MD.A.2Identify SubproblemsPay the flat fare for the first mile: subtract 5.60 for the extra distance.
Subtracting decimals to the hundredths place is a Grade 5 arithmetic move.
5.NBT.B.7Identify SubproblemsConvert the rate: 2 per mile — a friendlier unit.
Scaling the top and bottom of the rate by 10 turns "per 0.1 mile" into "per mile" — a Grade 6 unit-rate move.
6.RP.A.3Analyze The UnitsAt 5.60 buys 2.8 extra miles (5.60 ÷ 2).
Dividing 5.60 by 2 is a direct Grade 5 decimal calculation.
5.NBT.B.7Identify SubproblemsAdd the first 0.5 mile back to the 2.8 extra: 0.5 + 2.8 = 3.3 miles — choice (C).
Re-combining the two subproblem pieces (first 0.5 mile + extra 2.8 mile) finishes the Tool #7 split.
5.NBT.B.7Identify SubproblemsBig AMC 8 word problems often shrink down to Grade 6 rate reasoning — once you split off the tip and the flat fare, only a tidy decimal division is left.