AMC 8 · 1999 · #22
Grade 6 rate-ratioProblem
In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?
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: Two trades are given: $3$ fish trade for $2$ loaves of bread, and $1$ loaf of bread trades for $4$ bags of rice. Using only these trade rates, how many bags of rice is one fish worth?
Givens: $3$ fish $=$ $2$ loaves of bread (same trade value); $1$ loaf of bread $=$ $4$ bags of rice (same trade value); Answer choices: (A) $\tfrac{3}{8}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{4}$, (D) $2\tfrac{2}{3}$, (E) $3\tfrac{1}{3}$
Unknowns: The value of $1$ fish in bags of rice
Understand
Restated: Two trades are given: $3$ fish trade for $2$ loaves of bread, and $1$ loaf of bread trades for $4$ bags of rice. Using only these trade rates, how many bags of rice is one fish worth?
Givens: $3$ fish $=$ $2$ loaves of bread (same trade value); $1$ loaf of bread $=$ $4$ bags of rice (same trade value); Answer choices: (A) $\tfrac{3}{8}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{4}$, (D) $2\tfrac{2}{3}$, (E) $3\tfrac{1}{3}$
Plan
Primary tool: #3 Set Up an Equation
Secondary: #4 Introduce a Variable
Each trade statement is a value equation, so Tool #3 (Set Up an Equation) turns the two sentences into $3F = 2B$ and $B = 4R$ directly. Tool #4 (Introduce a Variable) names the values of fish, bread, and rice with single letters so substitution is mechanical. Bread shows up in both equations, which makes it the bridge — replace $B$ in the first equation with $4R$ from the second, and the bread variable disappears, leaving fish written in terms of rice. Then dividing by $3$ gives the value of one fish.
Execute — Answer: D
6.EE.A.2 Step 1 - Name the value of one of each item with a single letter.
- This lets each trade statement become an equation.
💡 Giving the three unknowns short names turns the word problem into algebra in one move.
6.RP.A.3 Step 2 - Translate each trade into an equation.
- "$3$ fish $=$ $2$ loaves" and "$1$ loaf $=$ $4$ bags of rice" become equations of equal value.
💡 A trade is just an equation between two equal values — write what is on each side of the trade and set them equal.
6.EE.A.2 Step 3 - Eliminate bread by substitution.
- The second equation tells us a loaf $B$ is the same value as $4R$, so we can replace $B$ in the first equation with $4R$.
💡 Bread appears in both equations, so it can be swapped out. After the swap, only fish and rice remain.
6.EE.B.7 Step 4 Solve for one fish by dividing both sides by $3$, then write the result as a mixed number.
💡 If $3$ fish are worth $8$ bags of rice, then one fish is worth one-third of $8$ bags, which is $2\tfrac{2}{3}$ bags.
6.EE.A.2 Name the value of one of each item with a single letter. This lets each trade st 6.RP.A.3 Translate each trade into an equation. "$3$ fish $=$ $2$ loaves" and "$1$ loaf $ 6.EE.A.2 Eliminate bread by substitution. The second equation tells us a loaf $B$ is the 6.EE.B.7 Solve for one fish by dividing both sides by $3$, then write the result as a mix Review
Reasonableness: Check the size. One fish should be worth quite a bit of rice, because fish trade for bread which already trades for $4$ bags of rice. From $3F = 2B$, one fish is worth $\tfrac{2}{3}$ of a loaf of bread, and $\tfrac{2}{3}$ of a loaf is $\tfrac{2}{3} \times 4 = \tfrac{8}{3} = 2\tfrac{2}{3}$ bags of rice — answer (D). The small-fraction choices (A) $\tfrac{3}{8}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{3}{4}$ each describe one bag of rice in fish, not one fish in bags of rice — they are the reciprocal trap. Choice (E) $3\tfrac{1}{3}$ comes from mixing up the $2$ and $3$ in the first trade ($2F = 3B$ instead of $3F = 2B$).
Alternative: Tool #9 (Solve an Easier Problem): scale up to whole numbers first. Take $3$ fish: they trade for $2$ loaves, which trade for $2 \times 4 = 8$ bags of rice. So $3$ fish are worth $8$ bags, and one fish is worth $\tfrac{8}{3} = 2\tfrac{2}{3}$ bags — answer (D). This skips variables entirely by working with whole quantities through the bread bridge.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the values of one fish, one loaf, and one bag of rice with the variables $F$, $B$, $R$ so the trade statements become equations.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Reading each trade as an equal-value rate ($3$ fish per $2$ loaves; $1$ loaf per $4$ bags of rice) and chaining the rates through the common item, bread.)6.EE.B.7Solve real-world and mathematical problems by writing and solving one-variable equations of the form $px = q$ (Solving $3F = 8R$ for $F$ by dividing both sides by $3$ to get $F = \tfrac{8}{3}R = 2\tfrac{2}{3}R$.)
⭐ Bread is the bridge between fish and rice. Three fish trade for two loaves, and two loaves trade for $8$ bags of rice — so one fish is worth $\tfrac{8}{3} = 2\tfrac{2}{3}$ bags. Answer (D).
⭐ Bread is the bridge between fish and rice. Three fish trade for two loaves, and two loaves trade for $8$ bags of rice — so one fish is worth $\tfrac{8}{3} = 2\tfrac{2}{3}$ bags. Answer (D).
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