2002 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:optimizationbounding to limit cases

Difficulty rating: 1420

11.

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. On any disk holding a 0.80.8 mb file, only one 0.40.4 mb file fits alongside it (since 0.8+0.7>1.440.8+0.7\gt 1.44), leaving at least 0.240.24 mb wasted. Across the three such disks that is at least 0.720.72 mb, so the effective demand is at least 16.8+0.72=17.5216.8+0.72=17.52 mb, requiring at least 17.521.44=13\left\lceil\dfrac{17.52}{1.44}\right\rceil=13 disks.

This is achievable: 33 disks each hold one 0.80.8 file and one 0.40.4 file, 66 disks each hold two 0.70.7 files, and 44 disks each hold three 0.40.4 files.

Thus, the correct answer is B.

Problem 11 in Other Years