AMC12 2010 A

AMC12 2010 A · Q5

AMC12 2010 A · Q5. It mainly tests Linear inequalities, Arithmetic misc.

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 $n$?

在一个 100 箭的射箭比赛进行到一半时，Chelsea 领先 50 分。每箭射中靶心得 10 分，其他可能得分为 8、4、2 和 0 分。Chelsea 每箭至少得 4 分。若 Chelsea 接下来的 $n$ 箭都是靶心，则她将确保获胜。求 $n$ 的最小值。

(A) 38 38

(B) 40 40

(D) 44 44

(E) 46 46

Answer

Correct choice: (C)

正确答案：（C）

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 $\boxed{42\ \textbf{(C)}}$.

设 $k$ 为 Chelsea 当前的得分。为了保证获胜，我们必须考虑对手每箭都射中靶心从而得到最大得分的情况。 \begin{align*}k+ 10n + 4(50-n) &> (k-50) + 50\cdot{10}\\ 6n &> 250\end{align*} 满足该不等式的最小整数为 $\boxed{42\ \textbf{(C)}}$。

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