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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.
  
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<cmath>\begin{align*}+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\
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<cmath>\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\
 
  6n &> 250\end{align*}</cmath>
 
  6n &> 250\end{align*}</cmath>
  
  
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The lowest integer value that satisfies the inequality is <math>\boxed{42\ \textbf{(C)}}</math>.
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The lowest integer value that satisfies the inequality is gyat <math>\boxed{42\ \textbf{(C)}}</math>.
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By ~ Skibidi Ohio Babygronk Rizzler
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:59, 1 December 2024

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 $n$ shots are bullseyes she will be guaranteed victory. What is the minimum value for $n$?

$\textbf{(A)}\ 38 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 46$

Solution

Let $k$ 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.

\begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ 6n &> 250\end{align*}


The lowest integer value that satisfies the inequality is gyat $\boxed{42\ \textbf{(C)}}$. By ~ Skibidi Ohio Babygronk Rizzler

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

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