AMC 8 · 2002 · #10
Grade 6 arithmeticProblem
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and
France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)
The average price of his '70s stamps is closest to
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Juan paid Brazil and France $6$ cents per stamp, Peru $4$ cents per stamp, and Spain $5$ cents per stamp. In the '70s column of his table he has $12$ Brazil, $12$ France, $6$ Peru, and $13$ Spain stamps. Find the average price per stamp for his '70s stamps, rounded to the closest choice.
Givens: Prices per stamp: Brazil $6$¢, France $6$¢, Peru $4$¢, Spain $5$¢; '70s counts from the table: Brazil $12$, France $12$, Peru $6$, Spain $13$; Answer choices: (A) $3.5$ cents, (B) $4$ cents, (C) $4.5$ cents, (D) $5$ cents, (E) $5.5$ cents
Unknowns: The average price per stamp across all of Juan's '70s stamps
Understand
Restated: Juan paid Brazil and France $6$ cents per stamp, Peru $4$ cents per stamp, and Spain $5$ cents per stamp. In the '70s column of his table he has $12$ Brazil, $12$ France, $6$ Peru, and $13$ Spain stamps. Find the average price per stamp for his '70s stamps, rounded to the closest choice.
Givens: Prices per stamp: Brazil $6$¢, France $6$¢, Peru $4$¢, Spain $5$¢; '70s counts from the table: Brazil $12$, France $12$, Peru $6$, Spain $13$; Answer choices: (A) $3.5$ cents, (B) $4$ cents, (C) $4.5$ cents, (D) $5$ cents, (E) $5.5$ cents
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The unit of the answer is "cents per stamp," which Tool #8 (Analyze the Units) turns into a recipe: multiply (cents/stamp) by (stamps) to get cents for each country, add those to get total cents, then divide by total stamps to cancel "stamps" and leave "cents per stamp." Tool #7 (Identify Subproblems) breaks the bookkeeping into two clean subproblems — total cost and total count — that are computed independently and then divided. Together they keep a four-country weighted average from turning into a tangle.
Execute — Answer: E
6.RP.A.3 Step 1 - Subproblem 1a: cost of each country's '70s stamps.
- Multiply price per stamp by the count in the '70s column.
💡 Cents-per-stamp times stamps cancels "stamps" and gives cents — exactly the cost we need country by country.
6.SP.B.5 Step 2 - Subproblem 1b: total cost.
- Add the four country costs.
💡 All cents add up to one grand total — the numerator of the average.
6.SP.B.5 Step 3 - Subproblem 2: total number of '70s stamps.
- Add the four counts from the table.
💡 All stamps count once — the denominator of the average.
6.SP.B.5 Step 4 - Combine the two subproblems.
- Divide total cost by total stamps to get the average price per stamp, then pick the closest choice.
💡 Cents divided by stamps cancels to cents per stamp. The value $5.42$ sits between $5$ and $5.5$, but $|5.42 - 5.5| = 0.08 < |5.42 - 5| = 0.42$, so $5.5$ wins.
6.RP.A.3 Subproblem 1a: cost of each country's '70s stamps. Multiply price per stamp by t 6.SP.B.5 Subproblem 1b: total cost. Add the four country costs. 6.SP.B.5 Subproblem 2: total number of '70s stamps. Add the four counts from the table. 6.SP.B.5 Combine the two subproblems. Divide total cost by total stamps to get the averag Review
Reasonableness: Sanity check via bounds. Every '70s stamp costs between $4$¢ (Peru) and $6$¢ (Brazil/France), so the average must lie strictly between $4$ and $6$. That immediately rules out (A) $3.5$ and any value outside this band. Most '70s stamps cost $5$¢ or $6$¢ ($12 + 12 + 13 = 37$ of $43$), with only $6$ Peru stamps at $4$¢, so the average should sit above $5$ and below $6$ — closer to the $6$ end than the $4$ end. The computed $5.42$ matches that picture, and $5.5$ is the closest choice.
Alternative: Tool #6 (Guess and Check) on the choices: at average $= 5.5$ cents over $43$ stamps, the total would be $5.5 \times 43 = 236.5$¢; at average $= 5$ cents the total would be $5 \times 43 = 215$¢. The actual total $233$¢ is much closer to $236.5$ than to $215$ (gap of $3.5$ vs. $18$), so $5.5$ wins, confirming (E).
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Multiplying the rate cents-per-stamp by the count of stamps for each country to get cents: $12 \times 6 = 72$, $6 \times 4 = 24$, $13 \times 5 = 65$.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Using the definition of mean (total $\div$ count) on the combined data: $233 \text{ cents} \div 43 \text{ stamps} \approx 5.42$ cents per stamp.)
⭐ A weighted average is just total cost divided by total count. Add up what Juan paid for his '70s stamps ($233$¢) and divide by how many he has ($43$) to get about $5.42$¢ — the closest choice is $5.5$¢, answer (E).
⭐ A weighted average is just total cost divided by total count. Add up what Juan paid for his '70s stamps ($233$¢) and divide by how many he has ($43$) to get about $5.42$¢ — the closest choice is $5.5$¢, answer (E).