AMC 8 · 1999 · #12
Grade 6 rate-ratioPick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Because the actual game count is not given but the ratio 11:4 is, the loss percent must be the same for every team with that ratio. Tool #9 (Solve an Easier Related Problem) says: pick the simplest team that fits — exactly 11 wins and 4 losses, for 15 games total. Then the loss fraction is , no variable needed. Tool #1 (Draw a Diagram) keeps it concrete: a bar of 15 equal blocks with 4 shaded as losses lets you see the fraction before converting to a percent.
Pick the smallest schedule matching 11:4 — 11 wins and 4 losses, so 11 + 4 = 15 total games; scaling up keeps the same loss fraction.
A ratio 11:4 is a Grade 6 "for every 11 wins, 4 losses" statement. Taking exactly one copy of that block is the simplest case that still obeys the rule.
6.RP.A.1Solve An Easier Related ProblemDraw 15 equal blocks: 11 marked W (win), 4 marked L (loss). The loss fraction is the shaded part, of the bar.
Splitting the bar into 15 equal pieces and counting 4 of them is the Grade 3 definition of as 4 copies of .
3.NF.A.1Draw A DiagramTurn the fraction into a percent: × 100% = %, and 400 ÷ 15 = 26.6%.
Percent means "per 100". Multiplying the fraction by 100 rescales the same ratio to a per-100 form, which is the Grade 6 percent move.
6.RP.A.3Solve An Easier Related ProblemRound 26.6% up — the first decimal is 6 (≥ 5), so it becomes 27% → (B).
Grade 5 rounding rule: look at the next digit; 6 ≥ 5, so the integer part goes up by one.
5.NBT.A.4Solve An Easier Related ProblemWhen a ratio fixes the answer, swap the unknown game count for the smallest matching schedule (11 wins, 4 losses, 15 total). Then "4 out of 15" turns into about 27% with a quick divide-and-round.