2009 AMC 10B Problem 14

Attempt Problem 14 of the 2009 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

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
Answer: D
Concepts:geometric sequencerecursioninequality
Small hint:
Each day the standing millet is multiplied by 34,\dfrac34, then 14\dfrac14 quart is added
Big hint:
Other seeds always total 34\dfrac34 quart; find when millet first exceeds that
Solution:
Each quart adds 14\dfrac14 quart of millet, and the birds leave 34\dfrac34 of the standing millet. On day nn the millet present is 14(1+34+⋯+(34)n−1)=1−(34)n. \begin{gathered} \dfrac14\left(1+\dfrac34+\cdots+\Big(\dfrac34\Big)^{n-1}\right) \\ = 1-\Big(\dfrac34\Big)^{n}. \end{gathered}
The other seeds always total 34\dfrac34 quart. Millet exceeds half when 1−(34)n>34,1-\Big(\dfrac34\Big)^n\gt\dfrac34, i.e. (34)n<14.\Big(\dfrac34\Big)^n\lt\dfrac14.
Since (34)4=81256>14\Big(\dfrac34\Big)^4=\dfrac{81}{256}\gt\dfrac14 but (34)5=2431024<14,\Big(\dfrac34\Big)^5=\dfrac{243}{1024}\lt\dfrac14, this first happens on day 5,5, which is Friday.
Thus, the correct answer is D.
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