Question

Jamal 想把 $30$ 个电脑文件存到软盘上，每张软盘容量为 $1.44$ 兆字节（MB）。其中三个文件各需要 $0.8$ MB，另有 $12$ 个各需要 $0.7$ MB，剩下 $15$ 个各需要 $0.4$ 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 $30$ computer files on floppy disks, each of which has a capacity of $1.44$ megabytes (mb). Three of his files require $0.8$ mb of memory each, $12$ more require $0.7$ mb each, and the remaining $15$ require $0.4$ mb each. No file can be split between floppy disks. What is the minimal number of floppy disks that will hold all the files?

$12$

$13$

$14$

$15$

$16$

答案：B

知识点：最优化极限情形界定

难度评级：1570

小提示：

所需总容量为 $3(0.8)+12(0.7)+15(0.4)$ mb。

The total memory needed is $3(0.8)+12(0.7)+15(0.4)$ mb

大提示：

放有一个 $0.8$ 文件的软盘最多只能再放一个 $0.4$ 文件，会浪费 $0.24$ mb；把这些浪费加起来。

A disk holding a $0.8$ file has room for only one $0.4$ file, wasting $0.24$ mb; add up this waste

解答：

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

含有 $0.8$ -mb 文件的软盘只能再放一个 $0.4$ -mb 文件，至少留下 $0.24$ mb 空间。三个 $0.8$ -mb 文件总共至少浪费 $3(0.24) = 0.72$ mb，超过半张软盘容量，因此至少需要 $13$ 张。

十三张也足够：六张各放两个 $0.7$ -mb 文件，三张各放一个 $0.8$ -mb 文件和一个 $0.4$ -mb 文件，四张各放三个 $0.4$ -mb 文件。

因此，正确答案是 B。

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

A disk containing a $0.8$ -mb file has room for only one more $0.4$ -mb file, leaving at least $0.24$ mb unused. Across the three $0.8$ -mb files this wastes at least $3(0.24) = 0.72$ mb, over half a disk, forcing at least $13$ disks.

Thirteen suffice: six disks each hold two $0.7$ -mb files, three disks each hold one $0.8$ -mb file plus one $0.4$ -mb file, and four disks each hold three $0.4$ -mb files.

Thus, the correct answer is B.

2002 AMC 12A 第 9 题

答案：B

难度评级：1570

小提示：

大提示：

解答：

其他年份的第 9 题