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What is the maximum number of 4X4X4 cubes that can fit in a [#permalink]
 03 Nov 2006, 23:01
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0% (00:00) correct
100% (00:11) wrong based on 0 sessions
What is the maximum number of 4X4X4 cubes that can fit in a rectangular box measuring 10X12X16?
1) 12
2) 18
3) 20
4) 24
5) 30
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Manager
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I go for 30. I multiplied 4 x 4 x 4 = 64 and 10 x 12 x 16 = 1920. 64 can only go 30 times into 1920.
Last edited by lfox2 on 06 Nov 2006, 00:46, edited 1 time in total.
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Manager
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How would you omplete this arrangement, I would be interested in seeing other ways to solve this problem.
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Intern
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I go for 24
the 10x12x16 rectangular box fits for 24 4x4x4 cubs.
Dimensions of the rectangular box are not all perfectly divisible by 4(cube Dimensions) so there must be an empty part in the box, since 10 is not divisible by 4.
==> Therefore we pretend that the box rectangle is 8x12x16, with a total area of 1536. ===>1536/64=24
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Senior Manager
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I went with 16/4 (4) * 12/4 (3) * 10/4 (apparently 2 cubes) = 24
(or say how many cubes fit on each side)
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SVP
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assuming that the base of th recatngle = 12*16 = 192
the maximum one level cubes = 12 ie: (192/16)
hight = 10 ( 4*2+2) so it can take two levels of the cubes = 12*2 = 24
assuming base of rectangle = 16*10 = 160
the maximum one level cubes = 160/16 = 10
hight is 12 = 4*3 so it can take three levels of cubes with hight 4
maximum number of cubes = 30
my answer is 30
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Manager
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Nope. Check out ezo's fine solution. You are figuring out volume and squishing your boxes.
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Director
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yezz wrote: assuming that the base of th recatngle = 12*16 = 192
the maximum one level cubes = 12 ie: (192/16)
hight = 10 ( 4*2+2) so it can take two levels of the cubes = 12*2 = 24
assuming base of rectangle = 16*10 = 160
the maximum one level cubes = 160/16 = 10
hight is 12 = 4*3 so it can take three levels of cubes with hight 4
maximum number of cubes = 30
my answer is 30
Here in second case we can not have more than 8 cubes in each row.
You can only 4 along the length i.e16 and 2 along the width that is 10.
Thus no will be 8. So max will be 24.
Please correct me if I am wrong.
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Senior Manager
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Sumithra wrote: I went with 16/4 (4) * 12/4 (3) * 10/4 (apparently 2 cubes) = 24
(or say how many cubes fit on each side)
This looks good.
Thanks a lot EZO,Sumithra and others 
OA is 24
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Senior Manager
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Completely tricked by this one.  Agree with 24. GMAT won't simply ask you to divide one number by another. Now would they?
_________________
My dad once said to me: Son, nothing succeeds like success.
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Director
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side with length 16
number of segments of length 4 we can fit on this side <= 16/4
=> maximum = 4
side with length 12
number of segments of length 4 we can fit on this side <= 12/4 = 3
=> maximum = 3
side with length 10
number of segments of length 4 we can fit on this side <= 10/4 = 2
=> maximum = 2
=> maximum number of cubes with length 4 that can be fit in a rectangular box of dimensions 16,12,10
= 4*3*2 = 24
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Intern
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completely tricked by this one.........good tricky question...initially i also went for 30.......
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