Difference between revisions of "2009 AMC 10B Problems/Problem 14"

 
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<cmath>\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .</cmath>
 
<cmath>\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .</cmath>
 
The birds always find <math>\frac 34</math> quart of other seeds, so more than half the seeds are millet if <math>1 - \left(\frac 34\right)^n > \frac 34</math>, that is, when <math>\left(\frac 34\right)^n < \frac 14</math>.  Because <math>\left(\frac 34\right)^4 = \frac {81}{256} > \frac 14</math> and <math>\left(\frac 34\right)^5 = \frac {243}{1024} < \frac 14</math>, this will first occur on day <math>5</math> which is <math>\boxed {\text{Friday}}</math>. The answer is <math>\mathrm{(D)}</math>.
 
The birds always find <math>\frac 34</math> quart of other seeds, so more than half the seeds are millet if <math>1 - \left(\frac 34\right)^n > \frac 34</math>, that is, when <math>\left(\frac 34\right)^n < \frac 14</math>.  Because <math>\left(\frac 34\right)^4 = \frac {81}{256} > \frac 14</math> and <math>\left(\frac 34\right)^5 = \frac {243}{1024} < \frac 14</math>, this will first occur on day <math>5</math> which is <math>\boxed {\text{Friday}}</math>. The answer is <math>\mathrm{(D)}</math>.
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==Video Solution==
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https://youtu.be/jj3eCwD7Bms
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~savannahsolver
  
 
== See also ==
 
== See also ==

Latest revision as of 13:58, 16 January 2021

The following problem is from both the 2009 AMC 10B #14 and 2009 AMC 12B #11, so both problems redirect to this page.

Problem

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?

$\textbf{(A)}\ \text{Tuesday}\qquad \textbf{(B)}\ \text{Wednesday}\qquad \textbf{(C)}\ \text{Thursday}\qquad \textbf{(D)}\ \text{Friday}\qquad \textbf{(E)}\ \text{Saturday}$

Solution

On Monday, day 1, the birds find $\frac 14$ quart of millet in the feeder. On Tuesday they find \[\frac 14 + \frac 34 \cdot \frac 14\] quarts of millet. On Wednesday, day 3, they find \[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14\] quarts of millet. The number of quarts of millet they find on day $n$ is \[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .\] The birds always find $\frac 34$ quart of other seeds, so more than half the seeds are millet if $1 - \left(\frac 34\right)^n > \frac 34$, that is, when $\left(\frac 34\right)^n < \frac 14$. Because $\left(\frac 34\right)^4 = \frac {81}{256} > \frac 14$ and $\left(\frac 34\right)^5 = \frac {243}{1024} < \frac 14$, this will first occur on day $5$ which is $\boxed {\text{Friday}}$. The answer is $\mathrm{(D)}$.

Video Solution

https://youtu.be/jj3eCwD7Bms

~savannahsolver

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2009 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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