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How do we know what are the leght and width among 3 given

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Director
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How do we know what are the leght and width among 3 given [#permalink]

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New post 03 Sep 2007, 00:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How do we know what are the leght and width among 3 given numbers, if need to know, of course.

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24
30
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New post 03 Sep 2007, 01:08
What is the maximum number fo 4*4*4 cubes that can fit in a rectangular box measuring 10*12*16

If the rectangular box has a base 10*12, then it can fit only 3*2*4 = 24 cubes.
If the rectangular box has a base 10*16, then it can fit only 4*2*3 = 24 cubes.
If the rectangular box has a base 12*16, then it can fit only 4*3*2 = 24 cubes.

So the maximum is 24 cubes.
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New post 03 Sep 2007, 01:10
Note: Common GMAT Volume Trick:
When fitting 3D objects into other 3D objects, knowing the respective volume is not enough. We must know the specific dimensions (length, width, height) of each object to determine if it can fit
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New post 03 Sep 2007, 09:57
ywilfred wrote:
Note: Common GMAT Volume Trick:
When fitting 3D objects into other 3D objects, knowing the respective volume is not enough. We must know the specific dimensions (length, width, height) of each object to determine if it can fit


Right, I got it, now.

Thank you.
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New post 23 Oct 2007, 08:49
ywilfred wrote:
Note: Common GMAT Volume Trick:
When fitting 3D objects into other 3D objects, knowing the respective volume is not enough. We must know the specific dimensions (length, width, height) of each object to determine if it can fit


Yup. This is very relevant in regard to the cylinder inside a rectangular box question. It depends on the base of the box.
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Re: Cube and box [#permalink]

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New post 22 Dec 2007, 22:37
IrinaOK wrote:
How do we know what are the leght and width among 3 given numbers, if need to know, of course.

12
18
20
24
30


asked an engineering friend. he said it doesnt matter what the base is in this case. we divide out the dimensions by 4. there is no need to rearrange the dimensions because the divisor is the same (all are divided by 4)

16/4 * 10/4 * 12/4
4* 2.5 * 3
we cannot have half a cube. we can only have 2 cubes. we cannot round up a cube. it is impossible.

4*2*3 = 24 max
Re: Cube and box   [#permalink] 22 Dec 2007, 22:37
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