2010 AMC 12A 第 5 题

先试着解答 2010 AMC 12A 第 5 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2010 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

在一场 100100 箭的射箭比赛进行到一半时,Chelsea 领先 5050 分。每箭射中靶心得 1010 分,其他可能得分为 88、44、22 和 00 分。Chelsea 每箭至少得 44 分。如果 Chelsea 接下来 nn 箭都射中靶心,她就能保证获胜。nn 的最小值是多少?
Halfway through a 100100-shot archery tournament, Chelsea leads by 5050 points. For each shot a bullseye scores 1010 points, with other possible scores being 8,8, 4,4, 2,2, and 00 points. Chelsea always scores at least 44 points on each shot. If Chelsea’s next nn shots are bullseyes she will be guaranteed victory. What is the minimum value for n?n?
3838
4040
4242
4444
4646
答案:C
知识点:不等式极限情形界定
小提示:
Chelsea 的对手在剩下的 5050 箭中每箭最多得 1010 分。
Chelsea’s opponent can score at most 1010 points on each of the remaining 5050 shots
大提示:
若有 nn 箭靶心,其他 50−n50-n 箭至少 44 分,则需要 10n+4(50−n)>500−5010n+4(50-n)\gt 500-50。
With nn bullseyes and at least 44 points on the other 50−n50-n shots, require 10n+4(50−n)>500−5010n+4(50-n)\gt 500-50
解答:
对手在最后 5050 箭中最多可得 50⋅10=50050\cdot10=500 分。Chelsea 已领先 5050 分,所以她还需要超过 500−50=450500-50=450 分。
设 Chelsea 有 nn 箭靶心,得 10n10n 分,其余 50−n50-n 箭至少得 4(50−n)4(50-n) 分。要保证获胜,需要 10n+4(50−n)>450。10n+4(50-n)\gt450\text{。} 化简得 6n>2506n\gt250,即 n>4123n\gt41\tfrac23。
因此至少需要 4242 箭靶心。
所以正确答案是 C。
The opponent can score at most 50⋅10=50050\cdot10=500 on the last 5050 shots. Since Chelsea leads by 50,50, she must score more than 500−50=450500-50=450 points on her remaining shots to guarantee victory.
Her nn bullseyes give 10n10n points, and her other 50−n50-n shots give at least 4(50−n)4(50-n) points, so 10n+4(50−n)>450.10n+4(50-n)\gt450. This simplifies to 6n>250,6n\gt250, i.e. n>4123.n\gt41\tfrac23.
Therefore Chelsea needs at least 4242 bullseyes.
Thus, C is the correct answer.
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