AMC 8 · 1999 · #22

Grade 6 rate-ratio
ratio-proportionunit-conversionfraction-arithmeticsystems-of-equations ratio-proportiondimensional-analysisidentify-subproblems ↑ Prerequisites: ratio-proportionfraction-arithmetic
📏 Short solution 💡 2 insights

Problem

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.

(A)
$\frac{3}{8}$
(B)
$\frac{1}{2}$
(C)
$\frac{3}{4}$
(D)
$2 rac{2}{3}$
(E)
$3 rac{1}{3}$
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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

#4 Introduce a Variable 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.
$$F = \text{value of } 1 \text{ fish}, \quad B = \text{value of } 1 \text{ loaf of bread}, \quad R = \text{value of } 1 \text{ bag of rice}$$

💡 Giving the three unknowns short names turns the word problem into algebra in one move.

#3 Set Up an Equation 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.
$$3F = 2B \quad \text{and} \quad B = 4R$$

💡 A trade is just an equation between two equal values — write what is on each side of the trade and set them equal.

#3 Set Up an Equation 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 $4R$.
$$3F = 2B = 2(4R) = 8R$$

💡 Bread appears in both equations, so it can be swapped out. After the swap, only fish and rice remain.

#3 Set Up an Equation 6.EE.B.7 Step 4

Solve for one fish by dividing both sides by $3$, then write the result as a mixed number.

$$F = \dfrac{8R}{3} = \dfrac{8}{3}R = 2\tfrac{2}{3}R \;\Rightarrow\; \textbf{(D)}$$

💡 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.

[1] #4 6.EE.A.2 Name the value of one of each item with a single letter. This lets each trade st
[2] #3 6.RP.A.3 Translate each trade into an equation. "$3$ fish $=$ $2$ loaves" and "$1$ loaf $
[3] #3 6.EE.A.2 Eliminate bread by substitution. The second equation tells us a loaf $B$ is the
[4] #3 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.2 Write, 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.3 Use 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.7 Solve 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).