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>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/jj3eCwD7Bms | ||
| + | |||
| + | ~savannahsolver | ||
== See also == | == See also == | ||
Revision as of 12:58, 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.
Contents
Problem
On Monday, Millie puts a quart of seeds, 
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 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?
Solution
On Monday, day 1, the birds find 
quart of millet in the feeder. On Tuesday they find
![\[\frac 14 + \frac 34 \cdot \frac 14\]](http://latex.artofproblemsolving.com/c/4/0/c4050b4283bef8e8479cea9c214ea95a608e378a.png)
quarts of millet. On Wednesday, day 3, they find
![\[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14\]](http://latex.artofproblemsolving.com/6/c/4/6c4238653fc34a7f23ef0e4efcd926c173ee0c21.png)
quarts of millet. The number of quarts of millet they find on day 
is
![\[\frac 14 + \frac 34 \cdot \frac 14 + \left(\frac34\right)^2 \cdot \frac 14 + \cdots + \left(\frac 34\right)^{n-1} \cdot \frac 14 = \frac {(\frac 14)(1 - (\frac 34)^n)}{1 - \frac 34} = 1 - \left(\frac 34\right)^n .\]](http://latex.artofproblemsolving.com/d/7/4/d743236100c927b7f7d555e4cb9fbd2deaa41520.png)
The birds always find 
quart of other seeds, so more than half the seeds are millet if 
, that is, when 
. Because 
and 
, this will first occur on day 
which is 
. The answer is 
.
Video Solution
~savannahsolver
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 (Problems • Answer Key • Resources) | |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
