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Difference between revisions of "2014 AIME I Problems/Problem 2"

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==Problem==
 
 
== Problem 2 ==
 
== Problem 2 ==
  
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== Solution ==
 
== Solution ==
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First, we find the probability both are green, then the probability both are blue, and add the two probabilities which equals <math>0.58</math>. 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, we get <math>N=\boxed{144}</math>.
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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  
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<cmath> \frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}</cmath>
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Solving this equation,
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<cmath>20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29</cmath>
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Multiplying both sides by <math>16+N</math>, we get
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<cmath>\begin{align*}
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20\cdot16+30\cdot N&=29(16+N)\\
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320+30N&=464+29N\\
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N&=\boxed{144}
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\end{align*}</cmath>
  
 
== See also ==
 
== See also ==

Latest revision as of 19:12, 15 January 2022

Problem 2

An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$. Find $N$.

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 $0.58$.

The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$, and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$, so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\] Solving this equation, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$, we get

\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144} \end{align*}

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

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