2020 AMC 12B Problems/Problem 5

Problem 5

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 $A_g=3(14)=\boxed{\textbf{(C) } 42}$ .

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2020 AMC 12B Problems/Problem 5

Problem 5

Solution