Competition · AMC preparation · step 4 of 4

AMC 8 · 1999 · #22

Grade 6 rate-ratio

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Chained Ratio Linking

ratio-proportionunit-conversionfraction-arithmeticsystems-of-equations ratio-proportiondimensional-analysisidentify-subproblems ↑ Prerequisites: ratio-proportionfraction-arithmetic

📏 Short solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

$\frac{3}{8}$

(B)

$\frac{1}{2}$

(C)

$\frac{3}{4}$

(D)

$2\frac{2}{3}$

(E)

$3\frac{1}{3}$

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Set Up an Equation

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.

1STEP 1

Name each item with a letter

Give each item's value a single letter so every trade becomes an equation.

F = value of 1 fish, B = value of 1 loaf of bread, R = value of 1 bag of rice

2STEP 2

Turn each trade into an equation

Write each trade as an equation of equal value: 3 fish for 2 loaves, and 1 loaf for 4 bags of rice.

3F = 2B and B = 4R

3STEP 3

Substitute to remove bread

Swap bread out: a loaf equals 4 rice, so 3 fish equal 2×4 = 8 bags of rice.

3F = 2B = 2(4R) = 8R

4STEP 4

Divide by 3 and write a mixed number

Divide by 3: one fish is worth $\frac{8}{3}$ = 2 $\frac{2}{3}$ bags of rice — choice (D).

F = 8R/3 = 8/3R = 2 2/3R → (D)

Answer

2 2/3

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

\frac{2}{3}

of a loaf of bread, and

\frac{2}{3}

of a loaf is

\frac{2}{3}

× 4 =

\frac{8}{3}

= 2

\frac{2}{3}

bags of rice — answer (D). The small-fraction choices (A)

\frac{3}{8}

, (B)

\frac{1}{2}

, (C)

\frac{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

\frac{1}{3}

comes from mixing up the 2 and 3 in the first trade (2F = 3B instead of 3F = 2B).

💡Key takeaway

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 $\frac{8}{3}$ = 2 $\frac{2}{3}$ bags. Answer (D).

Name each item with a letter
Turn each trade into an equation
Substitute to remove bread
Divide by 3 and write a mixed number

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Name each item with a letter

Turn each trade into an equation

Substitute to remove bread

Divide by 3 and write a mixed number

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