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Manager  Joined: 27 May 2008
Posts: 97
A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 33% (02:54) correct 67% (03:02) wrong based on 1122 sessions

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A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area to the coloured is :

A. 25.9%
B. 44.44%
C. 35%
D. 61%
E. None of these
Math Expert V
Joined: 02 Sep 2009
Posts: 60625

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mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

Good question, +1. Though I think there is a little typo, the question should be: "The percentage of exposed area THAT is coloured is:"

We have 160+56=216=6^3 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of exposed area THAT is coloured is --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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A big cube is formed by rearranging the 160 colored and 56 non-colored similar cubes in such a way that the exposure of the colored cubes to the outside is minimum. The percentage of exposed area that is colored:

A. 25.9%
B. 44.44%
C. 35%
D. 61%
E. None of these
......
My explanation and diagram:
Attachments cube 44.jpg [ 217.18 KiB | Viewed 31321 times ]

##### General Discussion
Manager  Joined: 25 Jun 2009
Posts: 216

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mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area to the coloured is :

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

200 cubes will be exposed outside and 16 cubes will be inside which will not be exposed at all.

To minimise the coloured area 16 coloured cubes will be used inside and rest outside.

So total 144 coloured cubes + 56 non-coloured cubes will be used outside.

The percentage of exposed area to the coloured$$= \frac{200}{144} *100=139 % ( approx)$$

Can you please confirm the OA for this one ?

Cheers,
Manager  Joined: 25 Jun 2009
Posts: 216

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1
Bunuel wrote:
mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

Good question, +1. Though I think there is a little typo, the question should be: "The percentage of exposed area THAT is coloured is:"

We have 160+56=216=6^6 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of exposed area THAT is coloured is --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Bunuel,

Can you please explain how did you get that non exposed cubes will be 4 ^3 cubes? I always tend to get confused with questions like these and try to use the brute force to calculate the number of exposed or non exposed cubes.

Cheers,
Math Expert V
Joined: 02 Sep 2009
Posts: 60625

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nitishmahajan wrote:
Bunuel wrote:
mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

Good question, +1. Though I think there is a little typo, the question should be: "The percentage of exposed area THAT is coloured is:"

We have 160+56=216=6^6 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of exposed area THAT is coloured is --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Bunuel,

Can you please explain how did you get that non exposed cubes will be 4 ^3 cubes? I always tend to get confused with questions like these and try to use the brute force to calculate the number of exposed or non exposed cubes.

Cheers,

Well the big cube which has the side built with 6 little cubes (6^3) will have "inside" the cube which has the side built with 4 little cubes (6-2 edges), so 4^3... Similarly, the cube which has the side built with 4 little cubes (4^3) will have "inside" the cube which has the side built with 2 little cubes, (4-2 edges), so 2^3...
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Manager  Joined: 25 Jun 2009
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Quote:
Well the big cube which has the side built with 6 little cubes (6^3) will have "inside" the cube which has the side built with 4 little cubes (6-2 edges), so 4^3... Similarly, the cube which has the side built with 4 little cubes (4^3) will have "inside" the cube which has the side built with 2 little cubes, (4-2 edges), so 2^3...

Thanks a lot Bunuel...!
Manager  Joined: 27 May 2008
Posts: 97

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Well the big cube which has the side built with 6 little cubes (6^3) will have "inside" the cube which has the side built with 4 little cubes (6-2 edges), so 4^3... Similarly, the cube which has the side built with 4 little cubes (4^3) will have "inside" the cube which has the side built with 2 little cubes, (4-2 edges), so 2^3...

Nice explanation "Bunuel", you got it absolutely correct. It took me lot of time to get the concept and the answer.
Intern  Joined: 11 May 2010
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2
Bunuel wrote:
mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

Good question, +1. Though I think there is a little typo, the question should be: "The percentage of exposed area THAT is coloured is:"

We have 160+56=216=6^6 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of exposed area THAT is coloured is --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Thanks for an explanation of the "cube" concept but here is a question. Why did you divide the number of outside colored cubes by the number of all small cubes? The questions asks about relationship of exposed area and the coloured cubes.
Math Expert V
Joined: 02 Sep 2009
Posts: 60625

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zsokhan wrote:
Bunuel wrote:
mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area THAT is coloured is:

1) 25.9%
2) 44.44%
3) 35%
4) 61%
5) None of these

Good question, +1. Though I think there is a little typo, the question should be: "The percentage of exposed area THAT is coloured is:"

We have 160+56=216=6^6 big cube.

Non exposed will be 4^3 little cubes, which will be "inside" the big cube, so exposed will be 216-4^3=152 little cubes out of which there will be 152-56=96 colored cube (as colored cubes are minimized on the surface).

Now surface area of big cube is 6*36=216 (exposed area) out of which the area of 96 is colored, so the percentage of exposed area THAT is coloured is --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Thanks for an explanation of the "cube" concept but here is a question. Why did you divide the number of outside colored cubes by the number of all small cubes? The questions asks about relationship of exposed area and the coloured cubes.

We are asked to find the ratio of colored area to whole surface area. Colored area is 96, as 96 cubes are exposed.Total surface area is: 6 faces * area of one face =6*(6*6)=216.

Hope it's clear.
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Thanks for answer but it seems I haven't got something. I thought we have 64 cubes inside ( the smaller inside cube, 4x4x4). As I understood they aren't exposed and we have to exlude them from 216.

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Nice Explaination Bunuel
Intern  Joined: 23 Jul 2013
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Just a thought... But if the number of non colored cubes was to be 30 and the number of colored cubes 186, then it would have made the problem a whole lot tougher. Assuming that all the sides of the colored cubes are indeed colored, they would need to be placed at the edges and corners first to increase the surface are that is colored.

Please correct me if I am wrong Math Expert V
Joined: 02 Sep 2009
Posts: 60625
Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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mission2009 wrote:
A big cube is formed by rearranging the 160 coloured and 56 non-coloured similar cubes in such a way that the exposure of the coloured cubes to the outside is minimum. The percentage of exposed area to the coloured is :

A. 25.9%
B. 44.44%
C. 35%
D. 61%
E. None of these

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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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Thanks Bunuel Your helps are amazing, this is the greatest GMAT portal on earth.
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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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Bunuel,

The percentage of exposed area that is colored will be:
Number of colored cubes exposed (96)/Total number of cubes which are exposed(152=216-64)

Please clarify this. I think the question needs to be worded more specifically such as,
The %tage of colored cubes which are exposed or something on similar lines.
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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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vabhs192003 wrote:
Bunuel,

The percentage of exposed area that is colored will be:
Number of colored cubes exposed (96)/Total number of cubes which are exposed(152=216-64)

Please clarify this. I think the question needs to be worded more specifically such as,
The %tage of colored cubes which are exposed or something on similar lines.

I think this question arose because one small thing in Bunuel's answer is missing.
96 is not only a number of colored сubes exposed, but also the exposed surface area of these сubes. only 1 side of each of 96 colored сubes will be exposed.
216 is the total exposed surface area of all 216 cubes (hope no questions with that).
We are asked for "The percentage of exposed area that is coloured", thus, 96/216 = 0.4444.

And I totally agree with earlier note of ashivapu, in case the number of colored cubes were more than 160, the questions would be even tougher!
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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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160+56=216=6^3, total number of cubes.

So first inside cube has 4^3=64 and that cube has 2^3=8, leading to 64+8=72 cubes inside, giving 216-72=144 cubes will be exposed.
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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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2
Good question

Out of the 216 cubes:

64 (4^3) can be used to create the middle 4by4 cube - use all color cubes for this

152 (216 - 96) will have at least one face showing on the outside. out of these

96 will have one side showing - use all color cubes (6 sides * 4^2 on each side)
48 will have two sides showing - use all non-color cubes (12 edges with 4 cubes each)
8 will have three sides showing at the corners - use all non-color cubes

so the percentage is 96 / (6^3). you could also count the faces for the area 96*1 / (96*1 + 48*2 + 8*3).

Great solution Bunuel
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Re: A big cube is formed by rearranging the 160 coloured and 56  [#permalink]

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jps245 wrote:
Good question

Out of the 216 cubes:

64 (4^3) can be used to create the middle 4by4 cube - use all color cubes for this

152 (216 - 96) will have at least one face showing on the outside. out of these

96 will have one side showing - use all color cubes (6 sides * 4^2 on each side)
48 will have two sides showing - use all non-color cubes (12 edges with 4 cubes each)
8 will have three sides showing at the corners - use all non-color cubes

so the percentage is 96 / (6^3). you could also count the faces for the area 96*1 / (96*1 + 48*2 + 8*3).

Great solution Bunuel

This is key information that Bunuel leaves out. He simply says that 96 colored cubes (i.e. the remaining 96 after accounting for the 4^3 cube not exposed at all) are on the outside, and that 96 out of a total of 216 cubes yields the correct answer. He's not factoring in that some cubes have two or even three sides exposed. It's very important to know that exactly 96 cubes will only have one side exposed, AND that there happens to be 216 total squares exposed (in addition to there being 216 total cubes exposed - this happens because there are six faces). Re: A big cube is formed by rearranging the 160 coloured and 56   [#permalink] 07 Mar 2014, 14:40

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