# 2020 AMC 12B Problems/Problem 5

## Problem

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?

$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

## Solution

First, let us assign some variables. Let

$$A_w=2x, A_l=x, A_g=3x,$$ $$B_w=5y, B_l=3y, B_g=8y,$$

where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$. Using the given information, we can set up the following two equations:

$$B_w=A_w+7\implies 5y=2x+7,$$ $$B_l=A_l+7\implies 3y=x+7.$$

We can solve through substitution, as the second equation can be written as $x=3y-7$, and plugging this into the first equation gives $5y=6y-7\implies y=7$, which means $x=3(7)-7=14$. Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{\textbf{(C) } 42}$.

~Argonauts16

## Solution 2

Using the information from the problem, we can note that team A has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can find the following list of possible win-lose scenarios for $A$, represented in the form $(w, l)$ for convenience:

$$A \implies (14, 7)$$ $$B \implies (18, 9)$$ $$C \implies (28, 14)$$ $$D \implies (32, 16)$$ $$E \implies (42, 21)$$

Thus, we have 5 matching $B$ scenarios, simply adding 7 to $w$ and $l$. We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting 7 from $w$ and $l$ gives us the point $(28, 14)$, making the answer $\boxed{\textbf{C}}$.

## Solution 3 (Answer Choices)

Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is $$\frac{2}{3} \implies n \equiv 0 \pmod{3}$$ Similarly, since the ratio of $B$ is $$\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}$$ Now, we can go through the answer choices and see which ones work:

$$\textbf{(A) } 21 \implies 21 + 14 = 35 \not \equiv \pmod{8}$$ $$\textbf{(B) } 27 \implies 27 + 14 = 41 \not \equiv \pmod{8}$$ $$\textbf{(C) } 42 \implies 42 + 14 = 56 \equiv \pmod{8}$$ $$\textbf{(D) } 48 \implies 48 + 14 = 62 \not \equiv \pmod{8}$$ $$\textbf{(E) } 63 \implies 63 + 14 = 77 \not \equiv \pmod{8}$$

So we can see $\boxed{\textbf{(C) } 42.0} \text{ \tiny nice}$ is the only valid answer.

~herobrine-india

~IceMatrix

## See Also

 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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