AMC 8 · 2002 · #10
Grade 6 arithmetic
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Multiply each country's price per stamp by its '70s count to get its cost.
Cents-per-stamp times stamps cancels "stamps" and gives cents — exactly the cost we need country by country.
6.RP.A.3Analyze The UnitsAdd the four country costs to get the total: 233 cents.
All cents add up to one grand total — the numerator of the average.
6.SP.B.5Identify SubproblemsAdd the four counts from the table to get the total: 43 stamps.
All stamps count once — the denominator of the average.
6.SP.B.5Identify SubproblemsDivide 233 by 43: about 5.42 cents per stamp, so the closest choice is (E).
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.SP.B.5Analyze The UnitsA 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).