AMC 8 · 2002 · #9
Grade 4 arithmetic
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The table has 16 cells but the question only cares about a small rectangle of them: two rows (Brazil, Peru) by two columns (50s, 60s). Tool #2 (Make an Organized List) pulls out exactly those four cells with the right per-stamp price attached. Tool #7 (Break into Subproblems) then handles them one country at a time: total Brazil stamps from those decades times 6 cents, total Peru stamps times 4 cents, then add. Splitting by country is the right cut because price changes with country, not with decade.
"South American" keeps Brazil and Peru; "before the 70s" keeps the 50s and 60s — so from the table, Brazil 4 and 7, Peru 6 and 4.
Grade 3 scaled data tables: read off only the cells the question asks for, ignore the rest.
3.MD.B.3Make A Systematic ListAdd Brazil's two decade counts and multiply by its 6-cent price to get 66 cents.
Add first, then multiply: one country, one price, so a single multiplication finishes the Brazil subtotal.
4.OA.A.3Identify SubproblemsDo the same for Peru at its 4-cent price to get 40 cents.
Peru is a separate subproblem because its per-stamp price differs from Brazil's.
4.OA.A.3Identify SubproblemsAdd the two subtotals — 106 cents — and convert to dollars and cents by dividing by 100.
Grade 4 money: 100 cents make $1, so 106 cents is one dollar and six cents.
4.MD.A.2Identify SubproblemsBig tables hide small questions. Use the wording to throw out everything you don't need (France, Spain, the 70s and 80s), then handle each country at its own price. Two short multiplications and an add — 11 × 6 + 10 × 4 = 106 cents = $1.06 — finish the problem.