2013 AMC 10A 第 4 题

Question

一支垒球队打了十场比赛，得分分别为

1

、

2

、

3

、

4

、

5

、

6

、

7

、

8

、

9

和

10

分。他们恰好有五场输一分。在其余每场比赛中，他们的得分是对手的两倍。对手总共得了多少分？

A softball team played ten games, scoring

1,

2,

3,

4,

5,

6,

7,

8,

9,

and

10

runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?

35

40

45

50

55

答案：C

知识点：奇偶性分类讨论

难度评级：1070

小提示：

得分是对手两倍的比赛中，本队得分必须是偶数。

The doubled-score games must be the even scoring games

大提示：

偶数得分对应对手得一半；奇数得分对应输一分。

Pair odd scores with one-run losses and even scores with halved opponent scores

解答：

若本队得分是对手两倍，则本队得分必须为偶数。

因此得

2, 4, 6, 8,

和

10

分的比赛中，对手分别得

1, 2, 3, 4,

和

5

分。

这些分数的和为

1 + 2 + 3 + 4 + 5 = 15\text{。}

其余比赛中，对手分别得

2, 4, 6, 8,

和

10

分。

这些分数的和为

2 + 4 + 6 + 8 + 10 = 30\text{。}

对手总得分为

15 + 30 = 45\text{。}

所以正确答案是 C。

Note that if they scored twice as many runs as their opponents, then they scored an even number of runs.

This means in the games where they scored

2, 4, 6, 8,

and

10

runs, their opponents scored

1, 2, 3, 4,

and

5

runs respectively.

This sums to

1 + 2 + 3 + 4 + 5 = 15.

In the other games, their opponents scored

2, 4, 6, 8,

and

10

runs.

This sums to

2 + 4 + 6 + 8 + 10 = 30.

The total number of runs is then

15 + 30 = 45.

Thus, C is the correct answer.

2013 AMC 10A 第 4 题

答案：C

难度评级：1070

小提示：

大提示：

解答：

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