2009 AMC 12B Problem 11

Problem 11 of 25IntermediateAlgebra

On Monday, Millie puts a quart of seeds,

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\%

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?

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Solution

Each day the birds leave

\dfrac{3}{4}

of the millet and Millie adds

\dfrac{1}{4}

quart of new millet, so after

n

days the millet is

\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

\dfrac{3}{4}

quart, so millet exceeds half when

1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4},

that is,

\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}.

Since

\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4}

and

\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4},

this first happens on day

5,

which is Friday. Thus, the correct answer is D.

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