AMC 8 · 2002 · #18
Grade 6 rate-ratioalgebraPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The unknown day-9 time is the only mystery quantity, so Tool #4 (Introduce a Variable) names it x and turns the average into a single equation (75 · 5 + 90 · 3 + x)/9 = 85. Tool #7 (Identify Subproblems) splits the bookkeeping into clean pieces — required total, known total, missing piece — so the arithmetic stays simple. Together they reduce a word problem about averages to one subtraction.
Convert every time to minutes: 1 hr 15 min = 75 min, 1 hr 30 min = 90 min — one unit avoids a common slip.
Hours-to-minutes is the Grade 4 unit-conversion move: 1 hr = 60 min, then add the extra minutes.
4.MD.A.1Identify SubproblemsSubproblem 1 — the required total: a 9-day average of 85 needs 765 minutes in all.
Average × count = total — the Grade 6 mean formula run in reverse to find the target sum.
6.SP.B.5Identify SubproblemsSubproblem 2 — minutes already skated: five 75s plus three 90s make 645 minutes over days 1–8.
Two multiplications plus an addition give the running total after 8 days.
5.NBT.B.5Identify SubproblemsName day 9's minutes x: since 645 + x = 765, subtracting gives x = 120 minutes.
Naming the unknown and balancing the equation is the Grade 6 "solve p + x = q" move.
6.EE.B.7Use Matrix LogicTurn 120 minutes back into hours: 120 ÷ 60 = 2 hr, which is choice (E).
Dividing 120 by 60 gives 2 hours exactly — the reverse of the conversion in step 1.
4.MD.A.1Identify SubproblemsA target average sets the required total: 85 × 9 = 765 minutes. Subtract the 645 minutes already skated and only 120 minutes — exactly 2 hours — are left for day 9, answer (E).