Difference between revisions of "2010 AMC 12A Problems/Problem 5"
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Let <math>k</math> be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes. | Let <math>k</math> be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes. | ||
| − | <cmath>\begin{align*} | + | <cmath>\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ |
| − | & | + | 6n &> 250\end{align*}</cmath> |
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:46, 3 July 2013
Problem
Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next 
shots are bullseyes she will be guaranteed victory. What is the minimum value for ?
Solution
Let 
be the number of points Chelsea currently has. In order to guarantee victory, we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.
The lowest integer value that satisfies the inequality is 
.
See also
| 2010 AMC 12A (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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
