AMC 8 · 2019 · #9
Grade 8 geometry-3drate-ratioPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question "ratio of two volumes" naturally splits into three subproblems (Tool #7): (i) compute V_A, (ii) compute V_F, (iii) form and simplify the ratio. Tool #8 (Analyze the Units) keeps us honest about diameter-vs-radius (the formula wants r, not d) and reminds us that both volumes are in cm³, so π and cm³ cancel in the ratio. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net: once we see Felicia's can is wider where it counts (radius is squared) but only half as tall, we expect V_F > V_A, which already eliminates (C), (D), (E).
The formula V = π r² h needs the radius, so halve each diameter: r_A = 3 cm, r_F = 6 cm.
Knowing that the radius is half the diameter is a Grade 4 measurement fact.
4.MD.A.1Analyze The UnitsSquare Alex's radius and multiply by his height: V_A = π·9·12 = 108π cm³.
Applying the cylinder volume formula V = π r² h is exactly the Grade 8 "volumes of cylinders" standard.
8.G.C.9Identify SubproblemsDo the same for Felicia (r_F = 6, h_F = 6): V_F = π·36·6 = 216π cm³.
Same cylinder-volume formula, second application — Grade 8 again.
8.G.C.9Identify SubproblemsDivide the volumes; π and cm³ cancel, leaving = .
Setting up a part-to-part comparison as a fraction is the Grade 6 ratio concept.
6.RP.A.1Analyze The UnitsSince 216 = 2 × 108, the fraction reduces to , so the ratio is 1 : 2 — choice (B).
Recognizing as equivalent to is Grade 4 equivalent-fractions reasoning, which then pinpoints choice (B).
4.NF.A.1Eliminate PossibilitiesThis AMC 8 problem really only needs the Grade 8 cylinder volume formula V = π r² h — once you plug in, π cancels and the ratio simplifies in one step!