Question

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 $\tfrac23$ of its games and team $B$ has won $\tfrac58$ 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?

$21$

$27$

$42$

$48$

$63$

Solution:

Let $a$ be the number of games team $A$ played and $b$ the number team $B$ played. Team $A$ wins $\tfrac23 a$ and loses $\tfrac13 a;$ team $B$ wins $\tfrac58 b$ and loses $\tfrac38 b.$ The conditions give $\tfrac58 b = \tfrac23 a + 7 \quad\text{and}\quad \tfrac38 b = \tfrac13 a + 7.$

Subtracting the equations gives $\tfrac14 b = \tfrac13 a,$ so $b = \tfrac43 a.$ Substituting into the loss equation: $\tfrac38 \cdot \tfrac43 a = \tfrac13 a + 7,$ i.e. $\tfrac12 a = \tfrac13 a + 7,$ so $\tfrac16 a = 7$ and $a = 42.$

Thus, the correct answer is C.

2020 AMC 12B Problem 5

Solution:

Problem 5 in Other Years