AMC 8 · 1999 · #12
Grade 6 rate-ratioProblem
The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is 11/4. To the nearest whole percent, what percent of its games did the team lose?
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A team has wins-to-losses ratio $11:4$ with no ties. Rounded to the nearest whole percent, what percent of all its games did the team lose?
Givens: Ratio of games won to games lost is $\dfrac{11}{4}$; No ties, so every game is either a win or a loss; Answer is the loss percentage rounded to the nearest whole percent; Answer choices: (A) $24\%$, (B) $27\%$, (C) $36\%$, (D) $45\%$, (E) $73\%$
Unknowns: The percent of all games that the team lost, rounded to the nearest whole percent
Understand
Restated: A team has wins-to-losses ratio $11:4$ with no ties. Rounded to the nearest whole percent, what percent of all its games did the team lose?
Givens: Ratio of games won to games lost is $\dfrac{11}{4}$; No ties, so every game is either a win or a loss; Answer is the loss percentage rounded to the nearest whole percent; Answer choices: (A) $24\%$, (B) $27\%$, (C) $36\%$, (D) $45\%$, (E) $73\%$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #1 Draw a Diagram
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 $\tfrac{4}{15}$, 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.
Execute — Answer: B
6.RP.A.1 Step 1 - Pick the easiest schedule that matches the ratio $11:4$.
- The smallest whole-number choice is $11$ wins and $4$ losses, giving $11 + 4 = 15$ total games.
- Any other matching schedule (like $22$ wins and $8$ losses) just scales both numbers equally, so the loss fraction does not change.
💡 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.
3.NF.A.1 Step 2 - Draw the $15$ games as a bar of $15$ equal blocks: $11$ marked W (win) and $4$ marked L (loss).
- The loss fraction is the shaded portion, $\tfrac{4}{15}$ of the whole bar.
💡 Splitting the bar into $15$ equal pieces and counting $4$ of them is the Grade 3 definition of $\tfrac{4}{15}$ as $4$ copies of $\tfrac{1}{15}$.
6.RP.A.3 Step 3 - Convert the fraction to a percent by multiplying by $100\%$.
- Divide $400$ by $15$ to get the decimal percent.
💡 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.
5.NBT.A.4 Step 4 - Round $26.\overline{6}\%$ to the nearest whole percent.
- The first decimal digit is $6$, which is $\ge 5$, so round up to $27\%$.
💡 Grade 5 rounding rule: look at the next digit; $6 \ge 5$, so the integer part goes up by one.
6.RP.A.1 Pick the easiest schedule that matches the ratio $11:4$. The smallest whole-numb 3.NF.A.1 Draw the $15$ games as a bar of $15$ equal blocks: $11$ marked W (win) and $4$ m 6.RP.A.3 Convert the fraction to a percent by multiplying by $100\%$. Divide $400$ by $15 5.NBT.A.4 Round $26.\overline{6}\%$ to the nearest whole percent. The first decimal digit Review
Reasonableness: Size check: more games were won than lost, so the loss percent must be under $50\%$. That kills (D) $45\%$ as borderline and (E) $73\%$ outright. Also $\tfrac{4}{15}$ is close to $\tfrac{4}{16} = \tfrac{1}{4} = 25\%$, so the answer should sit just above $25\%$ — eliminating (A) $24\%$ (below $25\%$) and (C) $36\%$ (too far above). That leaves (B) $27\%$, matching the calculation. Cross-check with the general team of $11k$ wins and $4k$ losses: loss fraction $= \tfrac{4k}{15k} = \tfrac{4}{15}$, same answer for every $k$, confirming the easier-case shortcut was valid.
Alternative: Tool #13 (Convert to Algebra): let the team have $11k$ wins and $4k$ losses for some positive integer $k$. Total games $= 15k$, loss fraction $= \tfrac{4k}{15k} = \tfrac{4}{15}$, which is $\tfrac{4}{15} \times 100\% \approx 26.67\%$, rounding to $27\%$. Same answer; the algebraic version makes it explicit that $k$ cancels out, but Tool #9 reaches the answer with one less symbol.
CCSS standards used (min grade 6)
6.RP.A.1Understand the concept of a ratio and use ratio language to describe a ratio relationship (Reading $11:4$ as "for every $11$ wins there are $4$ losses" and picking the smallest matching schedule of $11$ wins and $4$ losses.)3.NF.A.1Understand a fraction $1/b$ as the quantity formed by $1$ part when a whole is partitioned into $b$ equal parts (Treating the $15$ games as $15$ equal pieces and counting the $4$ losses as the fraction $\tfrac{4}{15}$ of the whole.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems, including finding a percent of a quantity (Converting the loss fraction $\tfrac{4}{15}$ to a percent by multiplying by $100$ to get $\tfrac{400}{15}\% = 26.\overline{6}\%$.)5.NBT.A.4Use place value understanding to round decimals to any place (Rounding $26.\overline{6}\%$ to the nearest whole percent: the tenths digit is $6$, so the integer rounds up to $27$.)
⭐ When 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.
⭐ When 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.
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