2002 AMC 10A Problem 11

The files need $3(0.8)+12(0.7)+15(0.4)$ $=16.8$ mb. On any disk holding a $0.8$ mb file, only one $0.4$ mb file fits alongside it (since $0.8+0.7\gt 1.44$), leaving at least $0.24$ mb wasted. Across the three such disks that is at least $0.72$ mb, so the effective demand is at least $16.8+0.72=17.52$ mb, requiring at least $\left\lceil\dfrac{17.52}{1.44}\right\rceil=13$ disks. This is achievable: $3$ disks each hold one $0.8$ file and one $0.4$ file, $6$ disks each hold two $0.7$ files, and $4$ disks each hold three $0.4$ files. Thus, the correct answer is B.