AMC 8 · 2002 · #24
Grade 6 rate-ratioPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The count of fruit is never specified, which is Tool #11's tell: the percent must be invariant under that count, so any convenient count gives the right answer. Tool #9 (Solve an Easier Problem) lets us pick the friendliest count — 6 of each fruit, the LCM of 3 and 2 — so both juice totals are whole numbers and the percent is a one-line fraction. No algebra needed.
Per fruit: one pear gives oz of juice, one orange gives 4 oz (8 oz from 2).
Dividing total juice by number of fruit is the Grade 6 unit-rate move.
6.RP.A.2Solve An Easier Related ProblemPick a friendly equal count — 6 of each, the LCM of 3 and 2 — so both juice totals stay whole.
The Grade 6 LCM picks the smallest common count that clears both denominators.
6.NS.B.4Work BackwardsAt 6 each: 6 pears give 16 oz pear juice (2×8), 6 oranges give 24 oz orange juice (3×8).
Scaling a known 3-pear batch by 2 and a 2-orange batch by 3 — Grade 6 ratio reasoning.
6.RP.A.3Solve An Easier Related ProblemPear share = 16 over the 40 oz total, times 100 — that gives 40%.
Part divided by whole, times 100, is the Grade 6 "find a percent" recipe.
6.RP.A.3Solve An Easier Related ProblemEach orange out-juices each pear, so pears should be the smaller share. Pick 6 of each fruit to get whole-number ounces — 16 oz pear vs 24 oz orange — and the pear share is = 40%, answer (B).