Difference between revisions of "2002 AMC 12A Problems/Problem 9"
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Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files? | Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files? | ||
| − | <math>\ | + | <math>\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16</math> |
==Solution== | ==Solution== | ||
| − | A 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is better to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file. | + | A <math>0.8</math> MB file can either be on its own disk, or share it with a <math>0.4</math> MB. Clearly it is better to pick the second possibility. Thus we will have <math>3</math> disks, each with one <math>0.8</math> MB file and one <math>0.4</math> MB file. |
| − | We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each. Their total size is <math>12\cdot (0.7 + 0.4) = 13.2</math> MB. The total capacity of 9 disks is <math>9\cdot 1.44 = 12.96</math> MB, hence we need at least 10 more disks. And we can easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each. | + | We are left with <math>12</math> files of <math>0.7</math> MB each, and <math>12</math> files of <math>0.4</math> MB each. Their total size is <math>12\cdot (0.7 + 0.4) = 13.2</math> MB. The total capacity of <math>9</math> disks is <math>9\cdot 1.44 = 12.96</math> MB, hence we need at least <math>10</math> more disks. And we can easily verify that <math>10</math> disks are indeed enough: six of them will carry two <math>0.7</math> MB files each, and four will carry three <math>0.4</math> MB files each. |
| − | Thus our answer is <math>3+10 = \boxed{ \ | + | Thus our answer is <math>3+10 = \boxed{\textbf{(B) }13 }</math>. |
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| + | ==Solution 2== | ||
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| + | Similarly to Solution 1, we see that there must be <math>3</math> disks to account for the <math>0.8</math> MB file. Secondly, since there are <math>[30-(3+12)]-3 = 12</math> files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be <math>3</math> files per disks for the 0.4 MB files and hence would require 4 disks. | ||
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| + | We are left with <math>12</math> files(<math>0.7</math> MB), where the optimal number of files per disks is <math>2</math>, so the optimal number of disks for this type of file would be <math>6</math> disks. Therefore, the answer is <math>3+4+6=\boxed{13}</math>. | ||
| + | |||
| + | ~quantumpsiinverted | ||
==See Also== | ==See Also== | ||
Revision as of 19:42, 2 October 2023
- The following problem is from both the 2002 AMC 12A #9 and 2002 AMC 10A #11, so both problems redirect to this page.
Contents
Problem
Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
Solution
A 
MB file can either be on its own disk, or share it with a 
MB. Clearly it is better to pick the second possibility. Thus we will have 
disks, each with one MB file and one MB file.
We are left with 
files of 
MB each, and files of MB each. Their total size is 
MB. The total capacity of 
disks is 
MB, hence we need at least 
more disks. And we can easily verify that disks are indeed enough: six of them will carry two MB files each, and four will carry three MB files each.
Thus our answer is 
.
Solution 2
Similarly to Solution 1, we see that there must be disks to account for the MB file. Secondly, since there are ![$[30-(3+12)]-3 = 12$](http://latex.artofproblemsolving.com/a/8/c/a8c13ceae0a0eb198a99c24abd73d715d9719cd1.png)
files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with files( MB), where the optimal number of files per disks is 
, so the optimal number of disks for this type of file would be 
disks. Therefore, the answer is 
.
~quantumpsiinverted
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
