A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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Toolkit + CCSS Solution
Understand
Restated: Six players play a round-robin singles tournament — each pair plays exactly one game, no ties. The win counts so far are Helen $4$, Ines $3$, Janet $2$, Kendra $2$, Lara $2$. Find how many games Monica won.
Givens: There are $6$ players, each pair plays exactly one game; Every game has a winner and a loser (no ties); Wins: Helen $= 4$, Ines $= 3$, Janet $= 2$, Kendra $= 2$, Lara $= 2$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The number of games Monica won
Understand
Restated: Six players play a round-robin singles tournament — each pair plays exactly one game, no ties. The win counts so far are Helen $4$, Ines $3$, Janet $2$, Kendra $2$, Lara $2$. Find how many games Monica won.
Givens: There are $6$ players, each pair plays exactly one game; Every game has a winner and a loser (no ties); Wins: Helen $= 4$, Ines $= 3$, Janet $= 2$, Kendra $= 2$, Lara $= 2$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #11 Find an Invariant
Secondary: #7 Identify Subproblems
The key fact that pins down Monica's wins without knowing the matchup details is an invariant: in a no-tie round-robin, every game produces exactly one win, so the total wins across all players equals the total games played. Tool #11 (Find an Invariant) captures that. Tool #7 (Identify Subproblems) breaks the work into two small pieces — (1) count the total games, (2) subtract the five known win counts — so the arithmetic stays clean.
Execute — Answer: C
#7 Identify Subproblems 5.NBT.B.5Step 1
Count the total number of games.
Each of the $6$ players plays the other $5$, which counts every game twice (once from each player's side).
💡 The invariant turns the question into a single subtraction: total minus the part we know gives the part we want.
[1]
#7 5.NBT.B.5Count the total number of games. Each of the $6$ players plays the other $5$, wh
[2]
#11 6.EE.A.2Apply the invariant. Every game adds exactly $1$ win to somebody's tally, so the
[3]
#7 6.EE.B.7Add the five known win counts and subtract from $15$ to isolate Monica's wins.
Review
Reasonableness: Check the count of games a different way: pair every player with every other player. The pairs are (H,I), (H,J), (H,K), (H,L), (H,M), (I,J), (I,K), (I,L), (I,M), (J,K), (J,L), (J,M), (K,L), (K,M), (L,M) — exactly $15$ pairs, matching $\tfrac{6 \times 5}{2} = 15$. Then the wins $4 + 3 + 2 + 2 + 2 + 2 = 15$ line up with the $15$ games. Also, each player played $5$ games, so each individual win count must be between $0$ and $5$ — Monica's $2$ fits comfortably and is consistent with the other players' tallies.
Alternative: Tool #13 (Convert to Algebra): let $M$ be Monica's wins. Total wins equals total games, so $4 + 3 + 2 + 2 + 2 + M = 15$, giving $13 + M = 15$ and $M = 2$. Same answer (C) — algebra just names the unknown and solves a one-step equation, which is what the invariant approach already does informally.
CCSS standards used (min grade 6)
5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing the total number of games as $\tfrac{6 \times 5}{2} = 15$ from the round-robin pairing count.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing the invariant "sum of all six players' wins equals total games" as an expression that includes Monica's unknown win count.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Solving $13 + M = 15$ to get Monica's wins $M = 2$.)
⭐ In a no-tie round-robin, every game adds exactly one win to the scoreboard — so the wins always sum to the number of games played. Once you count $15$ games, Monica's wins are just $15$ minus the others.
⭐ In a no-tie round-robin, every game adds exactly one win to the scoreboard — so the wins always sum to the number of games played. Once you count $15$ games, Monica's wins are just $15$ minus the others.
