AMC 8 · 2003 · #9
Grade 6 rate-ratio
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The final price hides behind a chain of small questions, so Tool #7 (Break into Subproblems) is the cleanest fit. Split the work into four sub-jobs: (a) area of one Art cookie, (b) total dough area for Art's batch, (c) area of one Roger cookie, and (d) how many cookies Roger gets from the same dough. Tool #1 (Draw a Diagram) is already done for us by the problem's figures — we just read off the bases, heights, and side lengths to feed each area formula. Once Roger's cookie count is known, matching Art's revenue is one division.
One Art cookie is a trapezoid with bases 5 in and 3 in and height 3 in, so its area is 12 in².
Grade 6 area-of-trapezoid formula: average the two parallel bases, then multiply by the height.
6.G.A.1Draw A DiagramArt bakes 12 such cookies, so his dough covers 144 in² of area.
Equal thickness means dough volume is proportional to total surface area, so 144 in² is the shared dough budget for every friend.
6.G.A.1Identify SubproblemsOne Roger cookie is a 4 in × 2 in rectangle, so its area is 8 in².
Grade 6 rectangle area: length times width.
6.G.A.1Draw A DiagramThe same 144 in² split into 8 in² pieces gives Roger 18 cookies.
Same dough, smaller cookies, so Roger gets more pieces — 18 ＞ 12 is the expected direction.
6.RP.A.3Identify SubproblemsArt earns 12 × 60 = 720 cents, so Roger's 18 cookies must each cost 720 ÷ 18 = 40 cents.
Roger sells more cookies than Art, so each must cost less than 60 cents — 40 fits.
6.RP.A.3Identify SubproblemsWhen everyone uses the same dough, swap cookies for square inches. Convert Art's batch to area, see how many of Roger's smaller cookies fit, then split Art's money across them — each Roger cookie is 40 cents.