2002 AMC 12A Problem 9

Below is the professionally curated solution for Problem 9 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:optimizationbounding to limit cases

Difficulty rating: 1570

9.

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

Solution:

The files need 3(0.8)+12(0.7)+15(0.4)=16.83(0.8)+12(0.7)+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.

Problem 9 in Other Years