AMC 8 · 2014 · #16

Grade 7 counting

Archetype Arithmetic Expression Decomposition

combinations-basicmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: combinations-basicmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?

$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

Pick an answer.

(A)

(B)

(C)

(D)

144

(E)

160

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Toolkit + CCSS Solution

Understand

Restated: The "Middle School Eight" conference has $8$ teams. Inside the conference, every team plays every other team twice ($1$ home game and $1$ away game). On top of that, each team also plays $4$ games against opponents outside the conference. Count the total number of games in one season that involve at least one Middle School Eight team.

Givens: $8$ teams in the conference; Each pair of conference teams plays $2$ games (home and away); Each conference team plays $4$ non-conference games; Answer choices: (A) $60$, (B) $88$, (C) $96$, (D) $144$, (E) $160$

Unknowns: The total number of games in the season that involve a Middle School Eight team

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #16 Count Carefully

The season has two clearly different kinds of games — conference games (both teams are inside the league) and non-conference games (only one team is inside). Tool #7 (Identify Subproblems) says to split the total into these two independent pieces, count each correctly, and add. Tool #16 (Count Carefully) handles the trap inside the conference subproblem: a game between Team A and Team B is the same game whether we list it as "A vs B" or "B vs A", so we must avoid double-counting the pair. The fact that home and away are distinct games is just a $\times 2$ on top of the unordered pair count.

Execute — Answer: B

#16 Count Carefully 7.SP.C.8 Step 1

Subproblem 1: count conference games.
First count unordered pairs of teams.
Choosing $2$ teams from $8$ gives $\binom{8}{2}$ pairs.

$$\binom{8}{2} = \dfrac{8 \times 7}{2} = 28 \text{ pairs}$$

💡 $8$ choices for the first team, $7$ for the second, divided by $2$ because order does not matter — this is the Grade 7 compound-event counting move.

#7 Identify Subproblems 3.OA.A.3 Step 2

Each pair plays $2$ games (home and away), so multiply the number of pairs by $2$.

$$28 \times 2 = 56 \text{ conference games}$$

💡 Equal groups of $2$ — Grade 3 multiplication as repeated counting.

#7 Identify Subproblems 3.OA.A.3 Step 3

Subproblem 2: count non-conference games.
Each of the $8$ teams plays $4$ such games, and the other team is outside the league, so no pair is shared between two Middle School Eight teams.
Multiply teams by games per team.

$$8 \times 4 = 32 \text{ non-conference games}$$

💡 $8$ equal groups of $4$ — straight Grade 3 multiplication, with no double-count to worry about because the opponent is outside.

#7 Identify Subproblems 4.OA.A.3 Step 4

Combine the two subproblems by adding.

$$56 + 32 = 88 \;\Rightarrow\; \textbf{(B)}$$

💡 Combining results of two sub-counts into a final total is the Grade 4 multi-step word-problem habit.

[1] #16 7.SP.C.8 Subproblem 1: count conference games. First count unordered pairs of teams. Choo

[2] #7 3.OA.A.3 Each pair plays $2$ games (home and away), so multiply the number of pairs by $2

[3] #7 3.OA.A.3 Subproblem 2: count non-conference games. Each of the $8$ teams plays $4$ such g

[4] #7 4.OA.A.3 Combine the two subproblems by adding.

Review

Reasonableness: Sanity-check the conference count from a single team's view: each team plays the other $7$ teams twice, so each team is in $7 \times 2 = 14$ conference games. Summing over $8$ teams gives $8 \times 14 = 112$, but every conference game has $2$ Middle School Eight teams in it, so the actual count is $112 / 2 = 56$ — matches Step 2. Adding $32$ non-conference games gives $88$, which is choice (B). The other choices fail simple checks: $160 = 8 \times 20$ over-counts by treating each team's $20$ games independently; $144$ doubles conference games without halving; $60$ and $96$ don't match any clean split.

Alternative: Tool #10 (Find a Pattern) per-team: each team plays $14$ conference games $+$ $4$ non-conference games $= 18$ games. Across $8$ teams that is $8 \times 18 = 144$ team-game appearances. Conference games contribute $2$ appearances each, non-conference games contribute $1$ each. Let $C = 56$ (already known) and let $N$ be non-conference games; then $2(56) + N = 144$, so $N = 32$, and the total game count is $C + N = 56 + 32 = 88$.

CCSS standards used (min grade 7)

3.OA.A.3 Use multiplication and division within 100 to solve word problems involving equal groups (Computing $28 \times 2 = 56$ conference games and $8 \times 4 = 32$ non-conference games as equal-groups multiplications.)
4.OA.A.3 Solve multistep word problems using the four operations (Combining the two subtotals $56 + 32 = 88$ into a final answer at the end of a multi-step word problem.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting the $\binom{8}{2} = 28$ unordered pairs of conference teams, the same compound-event counting reasoning used in Grade 7 probability.)

⭐ Split the season into conference games and non-conference games, count each pair only once, then add — basic multiplication plus a careful pair count gets you to $88$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 7)

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