2010 AMC 12A Problems/Problem 5

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

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 12 Problems and Solutions

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2010 AMC 12A Problems/Problem 5

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