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Difference between revisions of "2015 AMC 12B Problems/Problem 5"

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==Problem==
 
==Problem==
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The Tigers beat the Sharks 2 out of the 3 times they played. They then played <math>N</math> more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for <math>N</math>?
  
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<math>\textbf{(A)}\; 35 \qquad  \textbf{(B)}\; 37 \qquad \textbf{(C)}\; 39 \qquad \textbf{(D)}\; 41 \qquad \textbf{(E)}\; 43</math>
  
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==Solution==
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The ratio of the Shark's victories to games played is <math>\frac{1}{3}</math>. For <math>N</math> to be at its smallest, the Sharks must win all the subsequent games because <math>\frac{1}{3} < \frac{95}{100}</math>. Then we can write the equation
  
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==Solution==
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<cmath>\frac{1+N}{3+N}=\frac{95}{100}=\frac{19}{20}</cmath>
  
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Cross-multiplying yields <math>20(1+N)=19(3+N)</math>, and we find that <math>N=\fbox{\textbf{(B)}\; 37}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=6|num-b=4}}
 
{{AMC12 box|year=2015|ab=B|num-a=6|num-b=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:06, 4 March 2015

Problem

The Tigers beat the Sharks 2 out of the 3 times they played. They then played $N$ more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for $N$?

$\textbf{(A)}\; 35 \qquad \textbf{(B)}\; 37 \qquad \textbf{(C)}\; 39 \qquad \textbf{(D)}\; 41 \qquad \textbf{(E)}\; 43$

Solution

The ratio of the Shark's victories to games played is $\frac{1}{3}$. For $N$ to be at its smallest, the Sharks must win all the subsequent games because $\frac{1}{3} < \frac{95}{100}$. Then we can write the equation

\[\frac{1+N}{3+N}=\frac{95}{100}=\frac{19}{20}\]

Cross-multiplying yields $20(1+N)=19(3+N)$, and we find that $N=\fbox{\textbf{(B)}\; 37}$.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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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