AMC 8 · 2017 · #13

Grade 2 logicarithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

logical-deductionmental-arithmetic logical-deductionidentify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Peter, Emma, and Kyler play chess games against each other (no draws). Peter has $4$ wins and $2$ losses; Emma has $3$ wins and $3$ losses; Kyler has $3$ losses. How many games did Kyler win?

Givens: Peter: $4$ wins, $2$ losses; Emma: $3$ wins, $3$ losses; Kyler: $3$ losses (wins unknown); Every chess game has exactly one winner and one loser (no draws); Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$

Unknowns: The number of games Kyler won, $W_K$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #6 Guess and Check

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.

Execute — Answer: B

#1 Draw a Diagram 1.OA.A.1 Step 1

Picture every game as one arrow.
Each chess game between two of the three friends produces one winner (the head of the arrow) and one loser (the tail).
So if we drew every game played among them, the number of arrowheads (total wins) would equal the number of arrowtails (total losses).

$$\text{Total Wins} \;=\; \text{Total Losses}$$

💡 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 Draw a Diagram 1.OA.A.2 Step 2

Add up all the losses the problem tells us.
Peter lost $2$, Emma lost $3$, Kyler lost $3$, so the group's total loss count is $2 + 3 + 3 = 8$.

$$2 + 3 + 3 = 8 \text{ total losses}$$

💡 Putting three small loss counts together is exactly a Grade 1 "three whole numbers, sum within $20$" word problem.

#1 Draw a Diagram 1.OA.A.1 Step 3

Add up the wins we already know.
Peter won $4$ and Emma won $3$, giving $4 + 3 = 7$ wins so far.
Kyler's wins $W_K$ are still missing from this total.

$$4 + 3 + W_K = 7 + W_K \text{ total wins}$$

💡 Counting the wins we already know and labeling the missing one as $W_K$ is the Grade 1 "add-to with unknown" setup.

#1 Draw a Diagram 2.OA.A.1 Step 4

Use the invariant from Step 1: total wins must equal total losses.
So $7 + W_K$ must equal $8$, meaning Kyler's wins are the missing addend that turns $7$ into $8$.

$$7 + W_K = 8 \;\Rightarrow\; W_K = 8 - 7 = 1$$

💡 "What plus $7$ makes $8$?" is the Grade 2 unknown-addend word problem — a one-step take-from.

#6 Guess and Check 2.OA.A.1 Step 5

Match the number to the answer choices.
Kyler won $1$ game, which is choice (B).
Quick guess-and-check on the other choices confirms only $W_K = 1$ makes wins ($7 + W_K$) equal losses ($8$).

$$W_K = 1 \;\Rightarrow\; \textbf{(B)}$$

💡 Plugging each answer choice back into $7 + W_K = 8$ is a Grade 2 check of a one-step equation.

[1] #1 1.OA.A.1 Picture every game as one arrow. Each chess game between two of the three friend

[2] #1 1.OA.A.2 Add up all the losses the problem tells us. Peter lost $2$, Emma lost $3$, Kyler

[3] #1 1.OA.A.1 Add up the wins we already know. Peter won $4$ and Emma won $3$, giving $4 + 3 =

[4] #1 2.OA.A.1 Use the invariant from Step 1: total wins must equal total losses. So $7 + W_K$

[5] #6 2.OA.A.1 Match the number to the answer choices. Kyler won $1$ game, which is choice (B).

Review

Reasonableness: The answer $W_K = 1$ is between $0$ and $4$, well inside the answer-choice range. Sanity check the totals: wins $= 4 + 3 + 1 = 8$ and losses $= 2 + 3 + 3 = 8$ — they match, as they must when every game produces exactly one of each. Also, Kyler has more losses ($3$) than wins ($1$), which fits the story that the other two friends collectively won more games against him.

Alternative: Tool #2 (Make a Systematic List) works too: list candidate game tallies between each pair (Peter-Emma, Peter-Kyler, Emma-Kyler) so that Peter ends with $4$ wins / $2$ losses and Emma with $3$ / $3$. One valid distribution is Peter $3$-$0$ vs Kyler, $1$-$2$ vs Emma; Emma $2$-$1$ vs Kyler. That forces Kyler $0+1 = 1$ win, matching answer (B).

CCSS standards used (min grade 2)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Setting up the win/loss picture as a one-step add-to situation and recognizing that each game adds $1$ to a win and $1$ to a loss.)
1.OA.A.2 Solve word problems involving three whole numbers whose sum is within 20 (Adding the three loss counts $2 + 3 + 3 = 8$ to get the group's total losses.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Solving the one-step equation $7 + W_K = 8$ for Kyler's wins and verifying the answer against the choices.)

⭐ This AMC 8 problem only needs Grade 2 addition and subtraction word-problem skills you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 2)

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