AMC10 2013 A

AMC10 2013 A · Q4

AMC10 2013 A · Q4. It mainly tests Fractions, Casework.

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?

垒球队打了十场比赛，得分为1, 2, 3, 4, 5, 6, 7, 8, 9和10分。他们恰好在五场比赛中以一分的劣势输掉。在其他比赛中，他们得分的两次是对手得分。他们的对手总共得了多少分？

(A) 35 35

(B) 40 40

(D) 50 50

(E) 55 55

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The softball team could only have scored twice as many runs as their opponent when they scored an even number of runs. In those games their opponents scored $\frac{2}{2}+\frac{4}{2}+\frac{6}{2}+\frac{8}{2}+\frac{10}{2}=15$ runs. In the games the softball team lost, their opponents scored $(1+1)+(3+1)+(5+1)+(7+1)+(9+1)=30$ runs. The total number of runs scored by their opponents was $15+30=45$ runs.

答案（C）：垒球队只有在他们得分为偶数时，才可能得分是对手的两倍。在这些比赛中，他们的对手得分为 $\frac{2}{2}+\frac{4}{2}+\frac{6}{2}+\frac{8}{2}+\frac{10}{2}=15$ 分。在垒球队输掉的比赛中，他们的对手得分为 $(1+1)+(3+1)+(5+1)+(7+1)+(9+1)=30$ 分。他们对手的总得分为 $15+30=45$ 分。

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