2020 AMC 12B 第 5 题

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

小提示：

设

A

队打了

a

场，

B

队打了

b

场，分别写出胜场和负场的条件。

Let

A

play

a

games and

B

play

b

games, then write both the win and loss conditions

大提示：

胜场给出

\tfrac58 b = \tfrac23 a + 7

；负场给出

\tfrac38 b = \tfrac13 a + 7

。

Wins:

\tfrac58 b = \tfrac23 a + 7;

losses:

\tfrac38 b = \tfrac13 a + 7

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

。题意给出

\tfrac58 b = \tfrac23 a + 7

以及

\tfrac38 b = \tfrac13 a + 7\text{。}

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

，所以

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

小提示：

大提示：

解答：

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