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# What is the maximum number of 4x4x4 cubes that can fit in a

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What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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03 Nov 2006, 23:01
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What is the maximum number of 4x4x4 cubes that can fit in a rectangular box measuring 10x12x16 ?

A. 12
B. 18
C. 20
D. 24
E. 30

M23-06
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04 Nov 2006, 00:54
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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04 Nov 2006, 03:34
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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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04 Nov 2006, 05:50
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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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04 Nov 2006, 06:23
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

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04 Nov 2006, 06:50
Nope. Check out ezo's fine solution. You are figuring out volume and squishing your boxes.
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04 Nov 2006, 09:34
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

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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16 Sep 2011, 20:54
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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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18 Sep 2011, 00:14
Its 24.
(10/4) * (12/4) * ( 16/4) = 24.
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http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
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Re: What is the maximum number of 4X4X4 cubes that can fit in a [#permalink]

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29 Oct 2013, 02:39
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 13:41
Bunuel, I still believe that this Could some how be solved by LCM and HCF logic.

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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 15:08
honchos wrote:
Bunuel, I still believe that this Could some how be solved by LCM and HCF logic.

That honestly seems like overkill. Just picture a box. You'd try to fill it as efficiently as you could. So you'd start packing on the 12x16 side, so 3x4 = 12 boxes. How many stacks of 12? 10/4 = 2.5, so there are 2 full stacks. 12x2 = 24.

12 x 16 x 10 --> 3*4*2 = 24 is the fastest way to do this problem.
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 19:32
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$$\frac{(10 * 12 * 16)}{(4 * 4 * 4)}$$

$$\frac{12}{4} = 3$$

$$\frac{16}{4} = 4$$

$$\frac{10}{4} = 2$$ max (We cant do a decimal calculation here; cannot adjust dimensions of the boxes; so max quotient is 2)

Answer = 4 x 2 x 3 = 24 = D
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 22:01
PareshGmat wrote:
$$\frac{(10 * 12 * 16)}{(4 * 4 * 4)}$$

$$\frac{12}{4} = 3$$

$$\frac{16}{4} = 4$$

$$\frac{10}{4} = 2$$ max (We cant do a decimal calculation here; cannot adjust dimensions of the boxes; so max quotient is 2)

Answer = 4 x 2 x 3 = 24 = D

Yes that is a far better approach, I think you have used that LCM/HCF concept, Isn't it.

Do you remember the relationship between HCF/LCM
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 22:49
honchos wrote:
CCMBA wrote:
honchos wrote:
Bunuel, I still believe that this Could some how be solved by LCM and HCF logic.

That honestly seems like overkill. Just picture a box. You'd try to fill it as efficiently as you could. So you'd start packing on the 12x16 side, so 3x4 = 12 boxes. How many stacks of 12? 10/4 = 2.5, so there are 2 full stacks. 12x2 = 24.

12 x 16 x 10 --> 3*4*2 = 24 is the fastest way to do this problem.
Go and sit in actual exam, exam will tell you what honestly kills any one. Better and time saving approach's are thumbs up in the real Examination.

Your answer doesn't make sense, Just a rat race marathon. Every can and has solved it by that method.

But this question can certainly be solve through LCM/HCF.

This is not a "rat race marathon" as you suggest. I'm pretty sure you're the one who's trying to look for a harder, more time-consuming solution. My visual was meant to concretize the problem, which you seemed to have trouble understanding. When I did this, I used the exact approach many are advocating: Divide each dimension by 4. Multiply the integer parts of the quotient.

So 12/4 * 16/4 * 10/4 becomes 3*4*2. I did this in 30 seconds. On my paper, I had 7 numbers written down: first row (12, 16, 10), second row (3, 4, 2, 24). Just because someone takes the time to explain her logic does not mean she doesn't understand how to get the answer. I think about how to do the problem. There was really no need to insult me. If you cannot tell this is the exact same method others have proposed, it speaks to an inability to recognize and apply concepts when they are presented to you in a novel format. That appears to be your problem.
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 22:56
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I just drew it out on paper. 12 cubes per level (I made the base 12x16 bc both are div by 4) and 2 levels of cubes.

So 24
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 23:04
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honchos wrote:
PareshGmat wrote:
$$\frac{(10 * 12 * 16)}{(4 * 4 * 4)}$$

$$\frac{12}{4} = 3$$

$$\frac{16}{4} = 4$$

$$\frac{10}{4} = 2$$ max (We cant do a decimal calculation here; cannot adjust dimensions of the boxes; so max quotient is 2)

Answer = 4 x 2 x 3 = 24 = D

Yes that is a far better approach, I think you have used that LCM/HCF concept, Isn't it.

Do you remember the relationship between HCF/LCM

Not used HCF/LCM. Cant recall the relation

In this type of question, the learning lesson for me is that we cannot "adjust" the dimensions given. So 10/4 max can yield 2 as quotient; cannot "borrow" from the adjacent nos. at make it a perfect division.

Its a nice problem.
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 23:07
CCMBA,

What you are trying to preach me even I could solve by that method or rather any one who is preparing for GMAT can solve by this method.

But GMAT is far beyond you comprehension as rumor mongers always rumor's that quantitative questions are quite easy.

I have given the GMAT and scored 710. Q50 V36

As soon as you keep on doing all the questions correctly the difficulty level continue to rise and one questions pops up, which were even far difficult then CAT one's. One question, I will not post it as it against policy, could be solved only by the use of Binomial theorem.

You are continuously arguing for a method , which was already discussed on this threads and ON many threads before. There seems no logic you are TRYING TO PREACH ME THE the same method what was discussed before in the thread.

My idea of posting is to enrich the Post not to clutter it with the same methods, which are already in place.

Neither do I post just for the sake of posting.

All those who believes GMAT quant is easy, make sure that GMAT writers are smartest people on earth and they will check down to bottom before the give you Q51.

Life is like Boomerang you send insult, it will come back to you as an insult. you TRIED TO PLAY OVER-SMART BY THESE WORDS-
That honestly seems like overkill.

If you are too possessive in your comfort Zone and do not want to diversify the learning process, it seems you are infatuated with your frog well.

I simply requested Bunuel to intervene, instead you did, infact you can, But you just repeat the same method, which every one on thos Forum Knows and initiated a insulting and sarcastic language-
That honestly seems like overkill.

If you cant handle sarcasm, better you should drop it application on others.
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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06 Mar 2014, 23:11
PareshGmat wrote:
honchos wrote:
PareshGmat wrote:
$$\frac{(10 * 12 * 16)}{(4 * 4 * 4)}$$

$$\frac{12}{4} = 3$$

$$\frac{16}{4} = 4$$

$$\frac{10}{4} = 2$$ max (We cant do a decimal calculation here; cannot adjust dimensions of the boxes; so max quotient is 2)

Answer = 4 x 2 x 3 = 24 = D

Yes that is a far better approach, I think you have used that LCM/HCF concept, Isn't it.

Do you remember the relationship between HCF/LCM

Not used HCF/LCM. Cant recall the relation

In this type of question, the learning lesson for me is that we cannot "adjust" the dimensions given. So 10/4 max can yield 2 as quotient; cannot "borrow" from the adjacent nos. at make it a perfect division.

Its a nice problem.

Thanks Paresh, we all know that trick, but my point was to learn some arithmetic approach that may have broad application in other questions also.
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Re: What is the maximum number of 4x4x4 cubes that can fit in a [#permalink]

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08 Mar 2014, 09:56
honchos wrote:
CCMBA,

What you are trying to preach me even I could solve by that method or rather any one who is preparing for GMAT can solve by this method.

But GMAT is far beyond you comprehension as rumor mongers always rumor's that quantitative questions are quite easy.

I have given the GMAT and scored 710. Q50 V36

As soon as you keep on doing all the questions correctly the difficulty level continue to rise and one questions pops up, which were even far difficult then CAT one's. One question, I will not post it as it against policy, could be solved only by the use of Binomial theorem.

You are continuously arguing for a method , which was already discussed on this threads and ON many threads before. There seems no logic you are TRYING TO PREACH ME THE the same method what was discussed before in the thread.

My idea of posting is to enrich the Post not to clutter it with the same methods, which are already in place.

Neither do I post just for the sake of posting.

All those who believes GMAT quant is easy, make sure that GMAT writers are smartest people on earth and they will check down to bottom before the give you Q51.

Life is like Boomerang you send insult, it will come back to you as an insult. you TRIED TO PLAY OVER-SMART BY THESE WORDS-
That honestly seems like overkill.

If you are too possessive in your comfort Zone and do not want to diversify the learning process, it seems you are infatuated with your frog well.

I simply requested Bunuel to intervene, instead you did, infact you can, But you just repeat the same method, which every one on thos Forum Knows and initiated a insulting and sarcastic language-
That honestly seems like overkill.

If you cant handle sarcasm, better you should drop it application on others.

The GMAT is in part a reasoning test. You have to pick the best method in a giving situation. I only explained why this was the best method. Overkill = too much work. That wasn't meant to be an insult. Your response, however, was deliberately insulting. Both times.

Furthermore, you could simply look up the LCM/HCF method, which no one seems to remember. That way people might have a better idea of how to answer your question. And don't have to risk being attacked while doing it.

Stop being rude.
Re: What is the maximum number of 4x4x4 cubes that can fit in a   [#permalink] 08 Mar 2014, 09:56

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