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AMC 8 · 2005 · #22

Grade 7 rate-ratio
Archetype Multiplicative Ratio with Integrality Constraints
percentageratio-proportionratefraction-arithmetic identify-subproblemsratio-proportion ↑ Prerequisites: percentageratio-proportion
📏 Medium solution 💡 3 insights

Problem

A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs 50% more than the small size and contains 20% less detergent than the large size. The large size contains twice as much detergent as the small size and costs 30% more than the medium size. Rank the three sizes from best to worst buy.

Pick an answer.

(A)
SML
(B)
LMS
(C)
MSL
(D)
LSM
(E)
MLS
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Toolkit + CCSS Solution

Understand

Restated: A detergent comes in small (S), medium (M), and large (L) boxes. The medium costs $50\%$ more than the small and holds $20\%$ less detergent than the large. The large holds twice as much as the small and costs $30\%$ more than the medium. Rank the three sizes from best buy to worst buy.

Givens: Cost: M is $50\%$ more than S, so $\text{cost}(M) = 1.5 \cdot \text{cost}(S)$; Cost: L is $30\%$ more than M, so $\text{cost}(L) = 1.3 \cdot \text{cost}(M)$; Size: M holds $20\%$ less than L, so $\text{size}(M) = 0.8 \cdot \text{size}(L)$; Size: L holds twice as much as S, so $\text{size}(L) = 2 \cdot \text{size}(S)$; Answer choices: (A) SML, (B) LMS, (C) MSL, (D) LSM, (E) MLS

Unknowns: The ordering of S, M, L from lowest to highest price per ounce

Understand

Restated: A detergent comes in small (S), medium (M), and large (L) boxes. The medium costs $50\%$ more than the small and holds $20\%$ less detergent than the large. The large holds twice as much as the small and costs $30\%$ more than the medium. Rank the three sizes from best buy to worst buy.

Givens: Cost: M is $50\%$ more than S, so $\text{cost}(M) = 1.5 \cdot \text{cost}(S)$; Cost: L is $30\%$ more than M, so $\text{cost}(L) = 1.3 \cdot \text{cost}(M)$; Size: M holds $20\%$ less than L, so $\text{size}(M) = 0.8 \cdot \text{size}(L)$; Size: L holds twice as much as S, so $\text{size}(L) = 2 \cdot \text{size}(S)$; Answer choices: (A) SML, (B) LMS, (C) MSL, (D) LSM, (E) MLS

Plan

Primary tool: #9 Try a Simpler Case (WLOG)

Secondary: #4 Introduce a Variable, #7 Break into Subproblems

The question asks only for a ranking, so the actual prices and sizes do not matter — only their ratios do. Tool #9 (Try a Simpler Case) lets us pick convenient round numbers for the small box without loss of generality, then derive the other two boxes from the percentage rules. Tool #4 (Introduce a Variable) backs this up: if you set the small box at cost $c$ and size $s$, every other quantity is a fixed multiple of $c$ or $s$, so the ratios cost/size only depend on those multiples. Picking $c = \$1$ and $s = 5$ oz keeps the arithmetic clean.

Execute — Answer: E

#9 Try a Simpler Case (WLOG) 6.RP.A.3 Step 1
  • Pick convenient numbers for the small box.
  • Let the small box cost $\$1$ and hold $5$ oz.
  • Any other choice would give the same final ranking because the problem fixes every box as a percentage of the small one.
$\text{cost}(S) = \$1, \quad \text{size}(S) = 5 \text{ oz}$

💡 Grade 6 ratio reasoning: rankings depend on ratios, so any convenient baseline works.

#4 Introduce a Variable 7.RP.A.3 Step 2
  • Build the large box from the small box.
  • The large holds twice as much as the small, so size $= 2 \times 5 = 10$ oz.
  • The large's cost comes later from the medium, so set its size first.
$$\text{size}(L) = 2 \cdot 5 = 10 \text{ oz}$$

💡 "Twice as much" is the cleanest ratio relation — apply it directly.

#4 Introduce a Variable 7.RP.A.3 Step 3
  • Build the medium box.
  • Cost is $50\%$ more than the small, so multiply by $1.5$.
  • Size is $20\%$ less than the large, so multiply by $0.8$.
$\text{cost}(M) = 1.5 \cdot \$1 = \$1.50, \quad \text{size}(M) = 0.8 \cdot 10 = 8 \text{ oz}$

💡 Grade 7 percent reasoning: "$50\%$ more" means multiply by $1.5$; "$20\%$ less" means multiply by $0.8$.

#4 Introduce a Variable 7.RP.A.3 Step 4
  • Fill in the large box's cost.
  • The large costs $30\%$ more than the medium, so multiply $\$1.50$ by $1.3$.
$\text{cost}(L) = 1.3 \cdot \$1.50 = \$1.95$

💡 Same percent move: "$30\%$ more" scales by $1.3$.

#7 Break into Subproblems 6.RP.A.2 Step 5
  • Compute the unit price (cost per ounce) for each box.
  • Lower is better.
$\dfrac{\$1}{5\text{ oz}} = \$0.200/\text{oz}, \quad \dfrac{\$1.50}{8\text{ oz}} = \$0.1875/\text{oz}, \quad \dfrac{\$1.95}{10\text{ oz}} = \$0.195/\text{oz}$

💡 Grade 6 unit rate: divide cost by size to compare apples to apples.

#7 Break into Subproblems 6.RP.A.3 Step 6
  • Rank the unit prices from lowest to highest.
  • Medium ($\$0.1875$) is cheapest per ounce, then large ($\$0.195$), then small ($\$0.200$).
  • Best to worst buy is M, L, S.
$$0.1875 < 0.195 < 0.200 \;\Rightarrow\; M < L < S \;\Rightarrow\; \textbf{(E)}\ \text{MLS}$$

💡 The cheapest cost-per-ounce is the "best buy" — order from smallest to largest.

[1] #9 6.RP.A.3 Pick convenient numbers for the small box. Let the small box cost $\$1$ and hold
[2] #4 7.RP.A.3 Build the large box from the small box. The large holds twice as much as the sma
[3] #4 7.RP.A.3 Build the medium box. Cost is $50\%$ more than the small, so multiply by $1.5$.
[4] #4 7.RP.A.3 Fill in the large box's cost. The large costs $30\%$ more than the medium, so mu
[5] #7 6.RP.A.2 Compute the unit price (cost per ounce) for each box. Lower is better.
[6] #7 6.RP.A.3 Rank the unit prices from lowest to highest. Medium ($\$0.1875$) is cheapest per

Review

Reasonableness: Try a different baseline to confirm the ranking is invariant. Let the small cost $\$2$ and hold $10$ oz. Then large $= 20$ oz and costs $1.3 \cdot (1.5 \cdot \$2) = \$3.90$; medium costs $\$3.00$ and holds $0.8 \cdot 20 = 16$ oz. Unit prices: S $= \$0.200$, M $= \$3.00/16 = \$0.1875$, L $= \$3.90/20 = \$0.195$. Same order: M $<$ L $<$ S. Also, M being the cheapest per ounce passes the sanity check that prices grew by $50\%$ but size grew by $60\%$ ($5 \to 8$ oz) from S to M.

Alternative: Tool #4 (Introduce a Variable) without picking numbers. Let $\text{cost}(S) = c$ and $\text{size}(S) = s$. Then $\text{cost}(M) = 1.5c$, $\text{cost}(L) = 1.3 \cdot 1.5c = 1.95c$, $\text{size}(L) = 2s$, $\text{size}(M) = 0.8 \cdot 2s = 1.6s$. Unit prices: S $= c/s$, M $= 1.5c/1.6s = 0.9375 \cdot (c/s)$, L $= 1.95c/2s = 0.975 \cdot (c/s)$. So in units of $c/s$: M $= 0.9375$, L $= 0.975$, S $= 1$. Same ranking MLS, confirming (E).

CCSS standards used (min grade 7)

  • 6.RP.A.2 Understand the concept of a unit rate associated with a ratio (Dividing cost by ounces for each box to get a comparable cost-per-ounce.)
  • 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Picking a convenient baseline for the small box and ranking the three unit prices.)
  • 7.RP.A.3 Use proportional relationships to solve multistep ratio and percent problems (Translating "$50\%$ more", "$30\%$ more", "$20\%$ less", and "twice as much" into multiplicative scale factors $1.5$, $1.3$, $0.8$, $2$.)

⭐ When the question is "which is the best buy?", only cost per ounce matters — pick easy numbers for the small box, follow the percentages to the medium and large, then divide.

⭐ When the question is "which is the best buy?", only cost per ounce matters — pick easy numbers for the small box, follow the percentages to the medium and large, then divide.

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