1997 AMC 12 第 11 题

Question

在本赛季的第六、七、八、九场篮球比赛中，一名球员分别得到

23

、

14

、

11

和

20

分。打完九场后的场均得分高于前五场的场均得分。若打完十场后的场均得分大于

18

，那么她在第十场最少可能得到多少分？

In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored

23,

14,

11,

and

20

points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than

18,

what is the least number of points she could have scored in the tenth game?

26

27

28

29

30

答案：D

知识点：averages inequalities

难度评级：1440

小提示：

设她前五场比赛的总得分为

S

Let

S

be her point total in the first five games

大提示：

利用九场比赛后的平均分条件求出

S

的最大值，再应用十场比赛后的严格平均分下界

Use the nine-game comparison to maximize

S

, then apply the strict ten-game average bound

解答：

第

6

至第

9

场的总得分为

68

。由条件

\frac{S+68}{9}>\frac{S}{5}

可得

S<85

，所以整数

S

的最大值为

84

。十场后的场均得分大于

18

，意味着总得分至少为

181

，因此第十场至少得到

181-(84+68)=29

分。这个分数可以达到，所以正确答案是 D。

Games

6

–

9

total

68.

The condition

\frac{S+68}{9}>\frac{S}{5}

gives

S<85,

so the greatest integral

S

is

84.

A ten-game average above

18

requires a total at least

181,

hence the tenth score is at least

181-(84+68)=29.

This is attainable, so D is correct.

1997 AMC 12 第 11 题

答案：D

难度评级：1440

小提示：

大提示：

解答：

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