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Difference between revisions of "2002 AMC 12A Problems/Problem 9"

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==Solution==
 
==Solution==
  
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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.
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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.
  
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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.
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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.
  
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Thus our answer is <math>3+10 = \boxed{ \text{(B)}\ 13 }</math>.
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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>.
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~quantumpsiinverted
  
 
==See Also==
 
==See Also==

Latest revision as of 20:42, 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.


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?

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$

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.

We are left with $12$ files of $0.7$ MB each, and $12$ files of $0.4$ MB each. Their total size is $12\cdot (0.7 + 0.4) = 13.2$ MB. The total capacity of $9$ disks is $9\cdot 1.44 = 12.96$ 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.

Thus our answer is $3+10 = \boxed{\textbf{(B) }13 }$.

Solution 2

Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ 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 $3$ files per disks for the 0.4 MB files and hence would require 4 disks.

We are left with $12$ files($0.7$ MB), where the optimal number of files per disks is $2$, so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$.

~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

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