Question

$A$ 队和 $B$ 队在一个篮球联赛中比赛，每场比赛必有一队胜、一队负。 $A$ 队赢了自己所有比赛的 $\tfrac23$ ， $B$ 队赢了自己所有比赛的 $\tfrac58$ 。此外， $B$ 队比 $A$ 队多赢 $7$ 场，也多输 $7$ 场。 $A$ 队一共打了多少场比赛？

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?

$21$

$27$

$42$

$48$

$63$

答案：C

知识点：方程组分数

难度评级：1290

解答：

设 $A$ 队打了 $a$ 场， $B$ 队打了 $b$ 场。 $A$ 队的胜场数和负场数分别为 $\tfrac23 a$ 和 $\tfrac13 a$ $B$ 队的胜场数和负场数分别为 $\tfrac58 b$ 和 $\tfrac38 b$ 所以且 $\tfrac58 b = \tfrac23 a + 7$ $\tfrac38 b = \tfrac13 a + 7.$

两式相减得 $\tfrac14 b = \tfrac13 a$ 所以 $b = \tfrac43 a$ 代回负场方程，得 $\tfrac38 \cdot \tfrac43 a = \tfrac13 a + 7$ 即 $\tfrac12 a = \tfrac13 a + 7$ 因此 $\tfrac16 a = 7$ ， $a = 42$ 。

所以正确答案是 C。

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 $\tfrac58 b = \tfrac23 a + 7$ and $\tfrac38 b = \tfrac13 a + 7.$

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 第 5 题

答案：C

难度评级：1290

解答：

其他年份的第 5 题