Difference between revisions of "2014 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | First, we find the probability both are green, then the probability both are blue, and add the two probabilities | + | First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to <math>0.58</math>. |
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| + | The probability both are green is <math>\frac{4}{10}\cdot\frac{16}{16+N}</math>, and the probability both are blue is <math>\frac{6}{10}\cdot\frac{N}{16+N}</math>, so | ||
| + | <cmath> \frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}</cmath> | ||
| + | Solving this equation, | ||
| + | <cmath>20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29</cmath> | ||
| + | Multiplying both sides by <math>16+N</math>, we get | ||
| + | <cmath>20\cdot 16 + 30\cdot N = 29(16+n)\Rightarrow 320+30N=464+29N \Rightarrow N = \boxed{144}</cmath> | ||
== See also == | == See also == | ||
Revision as of 14:09, 28 November 2019
Problem 2
An urn contains 
green balls and 
blue balls. A second urn contains 
green balls and 
blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is 
. Find .
Solution
First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to .
The probability both are green is 
, and the probability both are blue is 
, so
![\[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\]](http://latex.artofproblemsolving.com/3/f/8/3f803fe9c2a3ba3f05533a2c351610dda78c116c.png)
Solving this equation,
![\[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\]](http://latex.artofproblemsolving.com/5/c/0/5c0c506554a3219b40d67f33153c47b2c5ac6836.png)
Multiplying both sides by 
, we get
![\[20\cdot 16 + 30\cdot N = 29(16+n)\Rightarrow 320+30N=464+29N \Rightarrow N = \boxed{144}\]](http://latex.artofproblemsolving.com/a/4/a/a4a4771596af38005ed0f29a15a8322fc1c38aeb.png)
See also
| 2014 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
