2015 AMC 12B Problems/Problem 5

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 (ProblemsAnswer KeyResources)
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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