AMC10 2002 A

AMC10 2002 A · Q11

AMC10 2002 A · Q11. It mainly tests Rounding & estimation, Money / coins.

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?

Jamal 想要将 30 个计算机文件存储在软盘上，每个软盘容量为 1.44 兆字节 (mb)。其中 3 个文件每个需要 0.8 mb 内存，另外 12 个每个需要 0.7 mb，剩余 15 个每个需要 0.4 mb。文件不能分割存储在软盘之间。需要的最少软盘数量是多少？

(A) 12 12

(B) 13 13

(D) 15 15

(E) 16 16

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) First note that the amount of memory needed to store the 30 files is $3(0.8)+12(0.7)+15(0.4)=16.8\ \text{mb},$ so the number of disks is at least $\dfrac{16.8}{1.44}=11+\dfrac{2}{3}.$ However, a disk that contains a 0.8-mb file can, in addition, hold only one 0.4-mb file, so on each of these disks at least 0.24 mb must remain unused. Hence, there is at least $3(0.24)=0.72\ \text{mb}$ of unused memory, which is equivalent to half a disk. Since $\left(11+\dfrac{2}{3}\right)+\dfrac{1}{2}>12,$ at least 13 disks are needed. To see that 13 disks suffice, note that: Six disks could be used to store the 12 files containing 0.7 mb; Three disks could be used to store the three 0.8-mb files together with three of the 0.4-mb files; Four disks could be used to store the remaining twelve 0.4-mb files.

（B）首先注意到存储这 30 个文件所需的内存量为 $3(0.8)+12(0.7)+15(0.4)=16.8\ \text{mb},$ 因此磁盘数量至少为 $\dfrac{16.8}{1.44}=11+\dfrac{2}{3}.$ 然而，包含一个 0.8 mb 文件的磁盘，此外只能再容纳一个 0.4 mb 文件，因此在每一块这样的磁盘上至少有 0.24 mb 必须空置不用。于是，至少有 $3(0.24)=0.72\ \text{mb}$ 的未使用内存，这相当于半块磁盘。由于 $\left(11+\dfrac{2}{3}\right)+\dfrac{1}{2}>12,$ 所以至少需要 13 块磁盘。为了说明 13 块磁盘足够，注意：可以用 6 块磁盘存储 12 个容量为 0.7 mb 的文件；可以用 3 块磁盘存储 3 个 0.8 mb 的文件，并在每块上再放入 1 个 0.4 mb 的文件（共 3 个 0.4 mb 文件）；可以用 4 块磁盘存储剩下的 12 个 0.4 mb 文件。

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