AMC 8 · 2007 · #20

Grade 6 algebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Ratio Shift — Two State

percentageratio-proportionlinear-equations-one-varsystems-of-equations convert-to-algebra ↑ Prerequisites: percentagelinear-equations-one-var

📏 Short solution 💡 2 insights

Problem

Before the district play, the Unicorns had won $45\%$ of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2007 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: Before district play the Unicorns had won $45\%$ of their games. During district play they won $6$ more and lost $2$, ending the season at exactly $50\%$ wins. Find the total number of games played in the whole season.

Givens: Win rate before district play was $45\%$; During district play the Unicorns won $6$ games and lost $2$ games; Final season win rate is $50\%$; Answer choices: (A) $48$, (B) $50$, (C) $52$, (D) $54$, (E) $60$

Unknowns: The total number of games the Unicorns played in the whole season

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities

The problem already hands us five candidate totals, so Tool #6 (Guess and Check) is the most direct route — far simpler than setting up algebra. For each choice we know the games-before count (choice $- 8$), and we just check whether $45\%$ of that is a whole number AND whether adding $6$ wins puts us at exactly half the season total. Tool #3 (Eliminate Possibilities) sharpens the check: $45\% = \tfrac{9}{20}$, so the games-before count must be a multiple of $20$, which knocks out almost every choice instantly.

Execute — Answer: A

#6 Guess and Check 6.RP.A.3 Step 1

Translate the structure once.
If $T$ is the season total, then games before district play is $T - 8$ (district play added $6$ wins $+ 2$ losses).
Wins before are $45\%$ of $T - 8$.
Wins after are half of $T$, and equal (wins before) $+ 6$.

$$\text{games before} = T - 8,\quad \text{wins before} = 0.45(T - 8),\quad \tfrac{T}{2} = 0.45(T - 8) + 6$$

💡 Write everything in terms of the season total $T$ so each guess becomes a single arithmetic check.

#3 Eliminate Possibilities 6.RP.A.3 Step 2

Narrow the field.
$45\% = \tfrac{9}{20}$, so wins-before $= \tfrac{9}{20}(T - 8)$ is a whole number only when $T - 8$ is a multiple of $20$.
Among the choices, $T \in \{48, 50, 52, 54, 60\}$ gives $T - 8 \in \{40, 42, 44, 46, 52\}$.
Only $T - 8 = 40$ is a multiple of $20$.

$$T - 8 \in \{40, 42, 44, 46, 52\}\quad\Rightarrow\quad\text{multiple of }20\text{: only } 40$$

💡 Win counts must be whole numbers, and $45\%$ forces a multiple of $20$ — most choices fail this test before any deeper check.

#6 Guess and Check 6.RP.A.3 Step 3

Verify the survivor $T = 48$.
Games before district play $= 40$; wins before $= 0.45 \times 40 = 18$.
After district play, wins $= 18 + 6 = 24$ out of $48$ games, which is exactly $50\%$.
All conditions check.

$$T = 48:\ \text{wins before} = 0.45 \times 40 = 18,\ \text{wins after} = 18 + 6 = 24,\ \tfrac{24}{48} = \tfrac{1}{2} \checkmark$$

💡 Wins go from $18$ out of $40$ to $24$ out of $48$ — a clean $45\% \to 50\%$ jump powered by $6$ extra wins in $8$ extra games.

#3 Eliminate Possibilities 6.RP.A.3 Step 4

Confirm by ruling out the other choices fast.
For $T = 50, 52, 54$, the value $0.45(T - 8)$ is $18.9, 19.8, 20.7$ — not whole numbers, so those seasons can't have started at exactly $45\%$.
For $T = 60$: wins before $= 0.45 \times 52 = 23.4$, also not whole.
Only (A) survives.

$0.45 \times 42 = 18.9,\ 0.45 \times 44 = 19.8,\ 0.45 \times 46 = 20.7,\ 0.45 \times 52 = 23.4$ — none are whole

💡 Wins are countable objects; a fractional win count means the choice can't match a real season record. Answer: $\textbf{(A)}\ 48$.

[1] #6 6.RP.A.3 Translate the structure once. If $T$ is the season total, then games before dist

[2] #3 6.RP.A.3 Narrow the field. $45\% = \tfrac{9}{20}$, so wins-before $= \tfrac{9}{20}(T - 8)

[3] #6 6.RP.A.3 Verify the survivor $T = 48$. Games before district play $= 40$; wins before $=

[4] #3 6.RP.A.3 Confirm by ruling out the other choices fast. For $T = 50, 52, 54$, the value $0

Review

Reasonableness: Replay the season with $T = 48$: the team starts $18$-$22$ ($18$ wins out of $40$ games, exactly $45\%$), then adds $6$ wins and $2$ losses in district play to finish $24$-$24$ — half wins, half losses. Both percentage conditions land exactly, and all counts are whole numbers. The win rate climbed because the team's district-play rate ($6$ out of $8 = 75\%$) is well above its earlier $45\%$, which is consistent with the rate rising toward $50\%$.

Alternative: Tool #13 (Algebra): set $w/t = 0.45$ and $(w+6)/(t+8) = 0.5$. Substituting $w = 0.45t$ gives $0.45t + 6 = 0.5t + 4$, so $0.05t = 2$ and $t = 40$. Then the season total is $t + 8 = 48$, matching (A). This works, but requires two equations and decimal arithmetic — Guess and Check on the five choices reaches the answer with a one-line divisibility check.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems, including percent (Reading $45\%$ as $\tfrac{9}{20}$ to require games-before to be a multiple of $20$, and checking that $24$ out of $48$ is exactly $50\%$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Setting up the single-step equation $\tfrac{T}{2} = 0.45(T - 8) + 6$ that each candidate $T$ must satisfy.)

⭐ Since $45\% = \tfrac{9}{20}$, the games-before count has to be a multiple of $20$ — that one fact narrows five answer choices to one. The Unicorns played $48$ games in all.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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