2002 AMC 10A Problem 11

Attempt Problem 11 of the 2002 AMC 10A below, then check your answer against the professionally curated solution 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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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

Answer: B
Concepts:optimizationbounding to limit cases
Difficulty rating: 1420
Solution:

The files need 3(0.8)+12(0.7)+15(0.4)3(0.8)+12(0.7)+15(0.4) =16.8=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.

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