Difference between revisions of "2015 AMC 12B Problems/Problem 5"
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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>? | 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>? | ||
| − | <math>\textbf{(A)}\; | + | <math>\textbf{(A)}\; 35 \qquad \textbf{(B)}\; 37 \qquad \textbf{(C)}\; 39 \qquad \textbf{(D)}\; 41 \qquad \textbf{(E)}\; 43</math> |
==Solution== | ==Solution== | ||
| + | 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 | ||
| + | <cmath>\frac{1+N}{3+N}=\frac{95}{100}=\frac{19}{20}</cmath> | ||
| + | |||
| + | 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 
more times, and the Sharks ended up winning at least 95% of all the games played. What is the minimum possible value for ?
Solution
The ratio of the Shark's victories to games played is 
. For to be at its smallest, the Sharks must win all the subsequent games because 
. Then we can write the equation
![\[\frac{1+N}{3+N}=\frac{95}{100}=\frac{19}{20}\]](http://latex.artofproblemsolving.com/d/7/d/d7da7b103568399e687d9760bcbf7caa5ffea4e8.png)
Cross-multiplying yields 
, and we find that 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
