AMC 8 · 2017 · #13
Grade 2 logicarithmeticPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
If we draw each game as an arrow from the loser to the winner, every arrow contributes exactly one win and exactly one loss. The diagram makes the key invariant visible at a glance: the pile of wins and the pile of losses are the same size. Tool #1 (Draw a Diagram) turns an abstract counting argument into something you can literally see. Tool #6 (Guess and Check) lets us verify the answer against the multiple-choice list by plugging it back into the win-equals-loss condition.
Draw each game as one arrow: it makes one win and one loss, so total wins always equal total losses.
Drawing each game as one arrow makes it obvious that one win and one loss are created together — a Grade 1 add-to / take-from picture.
1.OA.A.1Draw A DiagramAdd every loss the problem gives: 2 + 3 + 3 = 8 total losses.
Putting three small loss counts together is exactly a Grade 1 "three whole numbers, sum within 20" word problem.
1.OA.A.2Draw A DiagramAdd the wins we know: Peter's 4 + Emma's 3 = 7, with Kyler's W_K still missing.
Counting the wins we already know and labeling the missing one as W_K is the Grade 1 "add-to with unknown" setup.
1.OA.A.1Draw A DiagramUse total wins = total losses: 7 + W_K = 8, so W_K = 8 - 7 = 1.
"What plus 7 makes 8?" is the Grade 2 unknown-addend word problem — a one-step take-from.
2.OA.A.1Draw A DiagramKyler won 1 game — choice (B); testing the other choices in 7 + W_K = 8 confirms only this one works.
Plugging each answer choice back into 7 + W_K = 8 is a Grade 2 check of a one-step equation.
2.OA.A.1Guess And CheckThis AMC 8 problem only needs Grade 2 addition and subtraction word-problem skills you already know!
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