2002 AMC 12A 第 9 题

先试着解答 2002 AMC 12A 第 9 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 2002 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。

9.

Jamal 想把 3030 个电脑文件存到软盘上,每张软盘容量为 1.441.44 兆字节(MB)。其中三个文件各需要 0.80.8 MB,另有 1212 个各需要 0.70.7 MB,剩下 1515 个各需要 0.40.4 MB。一个文件不能拆分到多张软盘上。装下所有文件最少需要多少张软盘?

Jamal wants to store 3030 computer files on floppy disks, each of which has a capacity of 1.441.44 megabytes (mb). Three of his files require 0.80.8 mb of memory each, 1212 more require 0.70.7 mb each, and the remaining 1515 require 0.40.4 mb each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

1212

1313

1414

1515

1616

答案:B
知识点:最优化极限情形界定
难度评级:1570
解答:

文件总大小为 3(0.8)+12(0.7)3(0.8)+12(0.7) +15(0.4)=16.8+15(0.4) = 16.8 mb,所以仅按容量至少需要 16.81.44=1123\dfrac{16.8}{1.44} = 11\tfrac{2}{3} 张软盘。

含有 0.80.8-mb 文件的软盘只能再放一个 0.40.4-mb 文件,至少留下 0.240.24 mb 空间。三个 0.80.8-mb 文件总共至少浪费 3(0.24)=0.723(0.24) = 0.72 mb,超过半张软盘容量,因此至少需要 1313 张。

十三张也足够:六张各放两个 0.70.7-mb 文件,三张各放一个 0.80.8-mb 文件和一个 0.40.4-mb 文件,四张各放三个 0.40.4-mb 文件。

因此,正确答案是 B

The files need 3(0.8)+12(0.7)3(0.8)+12(0.7) +15(0.4)=16.8+15(0.4) = 16.8 mb, so at least 16.81.44=1123\dfrac{16.8}{1.44} = 11\tfrac{2}{3} disks by volume alone.

A disk containing a 0.80.8-mb file has room for only one more 0.40.4-mb file, leaving at least 0.240.24 mb unused. Across the three 0.80.8-mb files this wastes at least 3(0.24)=0.723(0.24) = 0.72 mb, over half a disk, forcing at least 1313 disks.

Thirteen suffice: six disks each hold two 0.70.7-mb files, three disks each hold one 0.80.8-mb file plus one 0.40.4-mb file, and four disks each hold three 0.40.4-mb files.

Thus, the correct answer is B.

← 第 8 题#8
完整试卷

其他年份的第 9 题