AMC 8 · 2019 · #1
Grade 5 arithmeticPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question has two natural subproblems (Tool #7): (a) find the largest whole number of sandwiches that fits in the 1 drinks the leftover money can buy. Subproblem (a) is a perfect fit for Tool #6 (Guess and Check) on the small candidates 6 vs 7 sandwiches — much friendlier than dividing 4.50, which would force decimal division. Tool #6 also lets us verify directly that no other answer choice is reachable.
Test near the budget: 6 × 27.00 fits, but 7 × 31.50 goes over — so the most is 6 sandwiches.
Multiplying a whole number by a price like $4.50 is Grade 5 decimal multiplication to hundredths — no division needed.
5.NBT.B.7Guess And CheckSubtract the sandwich cost from the budget: 27.00 = $3.00 left over.
Subtracting money to find what is left over is a Grade 4 money word-problem move.
4.MD.A.2Identify SubproblemsEach drink is 3.00 buys 3 ÷ 1 = 3 drinks.
Splitting a total amount into equal $1 groups is Grade 3 division within 100.
3.OA.A.3Identify SubproblemsAdd the two counts: 6 sandwiches + 3 drinks = 9 items in all → (D).
Adding two small whole-number counts to get a total is Grade 2 addition within 100.
2.OA.A.1Identify SubproblemsThis AMC 8 problem only needs Grade 5 decimal multiplication — like figuring out the cost of 6 sandwiches at $4.50 — that you already know!