2004 AMC 8 第 16 题

先试着解答 2004 AMC 8 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2004 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

两个 600600 毫升的水壶中装有橙汁。一个水壶装了 13\frac 13 满,另一个装了 25\frac 25 满。向每个水壶中加水直到装满,然后把两个水壶都倒入一个大容器。大容器中混合液的橙汁占几分之几?
Two 600600 mL pitchers contain orange juice. One pitcher is 13\frac 13 full and the other pitcher is 25\frac 25 full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?
18\dfrac{1}{8}
316\dfrac{3}{16}
1130\dfrac{11}{30}
1119\dfrac{11}{19}
1115\dfrac{11}{15}
答案:C
知识点:混合问题分数
小提示:
先计算加水前每个水壶中的橙汁量
Compute the orange juice in each pitcher before water is added.
大提示:
两个水壶装满后,混合液总量为 12001200 毫升
After filling both pitchers, the total mixture volume is 12001200 mL.
解答:
第一个水壶含有 600⋅13=200600 \cdot \dfrac{1}{3} = 200 毫升橙汁。第二个含有 600⋅25=240600 \cdot \dfrac{2}{5} = 240 毫升。
大容器中共有 200+240=440200 + 240 = 440 毫升橙汁。混合液总量为 2⋅600=12002 \cdot 600 = 1200 毫升。
因此橙汁所占比例为 4401200=1130\dfrac{440}{1200} = \dfrac{11}{30}。
所以正确答案是 C。
The first pitcher contains 600⋅13=200600 \cdot \dfrac{1}{3} = 200 mL of orange juice. The second one has 600⋅25=240600 \cdot \dfrac{2}{5} = 240 mL.
The large container then has 200+240=440200 + 240 = 440 mL of orange juice. The total amount of mixture is 2⋅600=12002 \cdot 600 = 1200 mL.
Then the fraction of orange juice is 4401200=1130.\dfrac{440}{1200} = \dfrac{11}{30}.
Thus, C is the correct answer.
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