2009 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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Concepts:geometric sequencerecursioninequality

Difficulty rating: 1610

11.

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?

Tuesday

Wednesday

Thursday

Friday

Saturday

Solution:

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)n1)=1(34)n \dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) = 1 - \left(\dfrac{3}{4}\right)^n 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.

Problem 11 in Other Years