2009 AMC 12B 第 11 题

先试着解答 2009 AMC 12B 第 11 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 2009 AMC 12B 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。

11.

星期一,Millie 把一夸脱鸟食倒入喂鸟器,其中 25%25\% 是小米。此后每天她都再加入一夸脱同样混合比例的鸟食,而且不取出剩下的鸟食。每天鸟只吃掉喂鸟器中小米的 25%25\%,但会吃掉所有其他种子。在星期几,Millie 刚放入鸟食后,鸟会发现喂鸟器中超过一半的种子是小米?
On Monday, Millie puts a quart of seeds, 25%25\% of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only 25%25\% of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?
星期二
Tuesday
星期三
Wednesday
星期四
Thursday
星期五
Friday
星期六
Saturday
答案:D
知识点:等比数列递推不等式
小提示:
每天早上,之前小米的 34\tfrac{3}{4} 留下,又加入 14\tfrac{1}{4} 夸脱新的小米;其他种子总是 34\tfrac{3}{4} 夸脱。
Each morning 34\tfrac{3}{4} of the previous millet remains and 14\tfrac{1}{4} quart of new millet is added, while the other seeds always total 34\tfrac{3}{4} quart
大提示:
nn 天后小米量为 1−(34)n1 - \left(\tfrac{3}{4}\right)^n 夸脱;要求它超过 34\tfrac{3}{4}。
After nn days the millet is 1−(34)n1 - \left(\tfrac{3}{4}\right)^n quart; require this to exceed 34\tfrac{3}{4}
解答:
每天鸟留下小米的 34\dfrac{3}{4},而 Millie 又加入 14\dfrac{1}{4} 夸脱新小米,所以 nn 天后小米量为 14(1+34+⋯+(34)n−1)=1−(34)n \begin{aligned} &\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) \\ &= 1 - \left(\dfrac{3}{4}\right)^n \end{aligned} 夸脱。
非小米种子总量始终是 34\dfrac{3}{4} 夸脱,所以小米超过一半等价于 1−(34)n>341 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4},即 (34)n<14\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}。
因为 (34)4=81256>14\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4},而 (34)5=2431024<14\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4},所以首次发生在第 55 天,也就是星期五。
所以正确答案是 D。
Each day the birds leave 34\dfrac{3}{4} of the millet and Millie adds 14\dfrac{1}{4} quart of new millet, so after nn days the millet is 14(1+34+⋯+(34)n−1)=1−(34)n \begin{aligned} &\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) \\ &= 1 - \left(\dfrac{3}{4}\right)^n \end{aligned} quart.
The non-millet seeds always total 34\dfrac{3}{4} quart, so millet exceeds half when 1−(34)n>34,1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4}, that is, (34)n<14.\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}.
Since (34)4=81256>14\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4} and (34)5=2431024<14,\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4}, this first happens on day 5,5, which is Friday.
Thus, the correct answer is D.
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