AMC 8 · 2002 · #24

Grade 6 rate-ratio
rateratio-proportionpercentagefraction-arithmetic identify-subproblemsdimensional-analysis ↑ Prerequisites: fraction-arithmeticratio-proportion
📏 Short solution 💡 2 insights

Problem

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

Pick an answer.

(A)
30
(B)
40
(C)
50
(D)
60
(E)
70
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Toolkit + CCSS Solution

Understand

Restated: A juicer gives $8$ ounces of pear juice from $3$ pears and $8$ ounces of orange juice from $2$ oranges. Miki blends an equal number of pears and oranges. What percent of the blend is pear juice?

Givens: $3$ pears yield $8$ oz of pear juice, so the rate is $\tfrac{8}{3}$ oz per pear; $2$ oranges yield $8$ oz of orange juice, so the rate is $\tfrac{8}{2} = 4$ oz per orange; The blend uses the same number of pears as oranges; Answer choices: (A) $30$, (B) $40$, (C) $50$, (D) $60$, (E) $70$

Unknowns: The percent of the blend that is pear juice

Understand

Restated: A juicer gives $8$ ounces of pear juice from $3$ pears and $8$ ounces of orange juice from $2$ oranges. Miki blends an equal number of pears and oranges. What percent of the blend is pear juice?

Givens: $3$ pears yield $8$ oz of pear juice, so the rate is $\tfrac{8}{3}$ oz per pear; $2$ oranges yield $8$ oz of orange juice, so the rate is $\tfrac{8}{2} = 4$ oz per orange; The blend uses the same number of pears as oranges; Answer choices: (A) $30$, (B) $40$, (C) $50$, (D) $60$, (E) $70$

Plan

Primary tool: #9 Solve an Easier Problem

Secondary: #11 Find an Invariant

The count of fruit is never specified, which is Tool #11's tell: the percent must be invariant under that count, so any convenient count gives the right answer. Tool #9 (Solve an Easier Problem) lets us pick the friendliest count — $6$ of each fruit, the LCM of $3$ and $2$ — so both juice totals are whole numbers and the percent is a one-line fraction. No algebra needed.

Execute — Answer: B

#9 Solve an Easier Problem 6.RP.A.2 Step 1
  • Find the per-fruit juice rates.
  • The juicer makes $8$ oz from $3$ pears, so each pear gives $\tfrac{8}{3}$ oz.
  • It makes $8$ oz from $2$ oranges, so each orange gives $\tfrac{8}{2} = 4$ oz.
$$\text{pear rate} = \dfrac{8}{3} \text{ oz/pear}, \quad \text{orange rate} = 4 \text{ oz/orange}$$

💡 Dividing total juice by number of fruit is the Grade 6 unit-rate move.

#11 Find an Invariant 6.NS.B.4 Step 2
  • Pick a friendly equal count.
  • Choose $n = 6$, the LCM of $3$ and $2$, so neither rate produces a fraction.
  • The percent will not depend on $n$, so we just want clean arithmetic.
$$n = \operatorname{lcm}(3, 2) = 6 \text{ pears and } 6 \text{ oranges}$$

💡 The Grade 6 LCM picks the smallest common count that clears both denominators.

#9 Solve an Easier Problem 6.RP.A.3 Step 3
  • Compute each juice total at $n = 6$.
  • From $6$ pears: $6 \div 3 = 2$ batches of $8$ oz, so $16$ oz of pear juice.
  • From $6$ oranges: $6 \div 2 = 3$ batches of $8$ oz, so $24$ oz of orange juice.
$$\text{pear juice} = \tfrac{6}{3} \cdot 8 = 16 \text{ oz}, \quad \text{orange juice} = \tfrac{6}{2} \cdot 8 = 24 \text{ oz}$$

💡 Scaling a known $3$-pear batch by $2$ and a $2$-orange batch by $3$ — Grade 6 ratio reasoning.

#9 Solve an Easier Problem 6.RP.A.3 Step 4

Form the pear-juice percent of the blend.

$$\dfrac{16}{16 + 24} \times 100 = \dfrac{16}{40} \times 100 = 40\% \;\Rightarrow\; \textbf{(B)}$$

💡 Part divided by whole, times $100$, is the Grade 6 "find a percent" recipe.

[1] #9 6.RP.A.2 Find the per-fruit juice rates. The juicer makes $8$ oz from $3$ pears, so each
[2] #11 6.NS.B.4 Pick a friendly equal count. Choose $n = 6$, the LCM of $3$ and $2$, so neither
[3] #9 6.RP.A.3 Compute each juice total at $n = 6$. From $6$ pears: $6 \div 3 = 2$ batches of $
[4] #9 6.RP.A.3 Form the pear-juice percent of the blend.

Review

Reasonableness: Check invariance with a different count. Take $n = 1$: pear juice $= \tfrac{8}{3}$ oz, orange juice $= 4$ oz, so the pear share is $\dfrac{8/3}{8/3 + 4} = \dfrac{8/3}{20/3} = \dfrac{8}{20} = 40\%$ — same answer, confirming the count did not matter. The sign also fits intuition: each orange gives more juice ($4$ oz) than each pear ($\tfrac{8}{3} \approx 2.67$ oz), so pears should make up less than half the blend, and $40\%$ is comfortably under $50\%$.

Alternative: Tool #4 (Introduce a Variable): let $n$ be the equal number of each fruit. Pear juice $= \tfrac{8n}{3}$, orange juice $= 4n$. Then $\dfrac{8n/3}{8n/3 + 4n} = \dfrac{8/3}{8/3 + 4} = \dfrac{8}{8 + 12} = \dfrac{8}{20} = 40\%$. The $n$ cancels — that algebraic cancellation is the reason a single convenient count was enough in the main path.

CCSS standards used (min grade 6)

  • 6.RP.A.2 Understand the concept of a unit rate $a/b$ associated with a ratio $a:b$ with $b \neq 0$, and use rate language in the context of a ratio relationship (Converting $8$ oz from $3$ pears into the unit rate $\tfrac{8}{3}$ oz/pear, and $8$ oz from $2$ oranges into $4$ oz/orange.)
  • 6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to $100$ and the least common multiple of two whole numbers less than or equal to $12$ (Choosing $n = \operatorname{lcm}(3, 2) = 6$ as the friendliest equal count so both juice totals are whole numbers.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including finding a percent of a quantity as a rate per $100$ (Scaling the per-batch yields to $6$ pears and $6$ oranges and computing $\tfrac{16}{40} \times 100 = 40\%$.)

⭐ Each orange out-juices each pear, so pears should be the smaller share. Pick $6$ of each fruit to get whole-number ounces — $16$ oz pear vs $24$ oz orange — and the pear share is $\tfrac{16}{40} = 40\%$, answer (B).

⭐ Each orange out-juices each pear, so pears should be the smaller share. Pick $6$ of each fruit to get whole-number ounces — $16$ oz pear vs $24$ oz orange — and the pear share is $\tfrac{16}{40} = 40\%$, answer (B).