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# A big cube is formed by rearranging the 160 coloured and 56

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08 Mar 2014, 09:49
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=6^3 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Hi there,

Thank you for the explanation and for valiant posts.
Quick question though. I don't understand why the exposed part 4^3. In a cube, only 1 side is covered (the one at the bottom), so shouldn't the exposed part be 5 faces, therefore 5^3?
As well, I don't understand the second part, after: 216-4^3=152.

Thanks,

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09 Mar 2014, 09:44
Yela 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=6^3 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

Hi there,

Thank you for the explanation and for valiant posts.
Quick question though. I don't understand why the exposed part 4^3. In a cube, only 1 side is covered (the one at the bottom), so shouldn't the exposed part be 5 faces, therefore 5^3?
As well, I don't understand the second part, after: 216-4^3=152.

Thanks,

Yela - the nonexposed part is 4^3. Don't picture the cube lying on a surface; rather, picture the cube floating (i.e. so that all six faces are "exposed"). Now, consider the fact that the cube is 6x6x6. What are the dimensions of the smaller cube that lies immediately inside that 6x6x6 (i.e. 6^3) cube?

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

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06 Oct 2014, 20:37
Another way of looking at this problem:

Minimising the surface area of coloured cubes means maximising the surface area of uncoloured cubes in the big cube.
Since all cubes are similar, total cubes is 160+56=216 and if we take the length of cube as 1 unit,then total surface area will be 6x (length)xlength=216 units

Now in any big cube, maximum surface area exposed will be at 8 corners where 3 surfaces of small cubes will be exposed and then 2 surfaces of all small cubes along all the edges of the big cube will be exposed thereafter. this gives total surface area as

8 corners of big cube x 3 unit area (surfaces) of small cube at each corner + 2 unit area (surfaces) of small cube x 4 such small cubes at each edge of big cube
= 8 X 3 + 2 x 4 x 12 (total edges of a cube is 12)
=120 (max surface area of uncoloured cubes)

So total surface area of coloured cubes will be 216-120= 96
and hence the percentage will be 96/216= 44.44%

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

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08 Oct 2014, 02:15
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=6^3 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

hey Bunuel,
i have a doubt with the colored part
in the corner portion of the outer surface part of the cubes every cube will be shown in 2 sides.
so how will you account for that?

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

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08 Oct 2014, 02:22
saggii27 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=6^3 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

hey Bunuel,
i have a doubt with the colored part
in the corner portion of the outer surface part of the cubes every cube will be shown in 2 sides.
so how will you account for that?

By placing colored cubes in the center of each face so that only one face is exposed.
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Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]

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21 Nov 2014, 23:09
Bunuel

Great Solution ! However the way it has been worded by mission 2009 the answer comes out to be 10%. As you suggested it has to be reworded to obtain 44.44% !

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

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21 May 2015, 20:40
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=6^3 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

I may be over thinking this but, going with the "that" modification ...
Big cube. 6^3 = 216, hidden/interior cube 4^3, minimizing color exposed interior cube gets 64 colored leaving 96 on the outside.
The exterior surface is 6*36 = 216 but only 152 blocks ... Interesting.
There are 8 corners each accounting for 3 of the 216, use 8 non-colored for the corners.
There are 12 edges, each 4 long, each covering 2 of the 216, use the remaining non-colored for these positions.
Since everything else is colored, the needed info is available.
There are 8*3 + 12*4*2 = 24+96 = 120 non-colored of 216, factoring 24 out leaves 5/9 which is the complement of the desired percentage. 44.44% is correct.

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

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21 May 2015, 22:09
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MarkAF wrote:
I may be over thinking this but, going with the "that" modification ...
Big cube. 6^3 = 216, hidden/interior cube 4^3, minimizing color exposed interior cube gets 64 colored leaving 96 on the outside.
The exterior surface is 6*36 = 216 but only 152 blocks ... Interesting.
There are 8 corners each accounting for 3 of the 216, use 8 non-colored for the corners.
There are 12 edges, each 4 long, each covering 2 of the 216, use the remaining non-colored for these positions.
Since everything else is colored, the needed info is available.
There are 8*3 + 12*4*2 = 24+96 = 120 non-colored of 216, factoring 24 out leaves 5/9 which is the complement of the desired percentage. 44.44% is correct.

Yes Mark, your logic is sound. You are taking the long way but focus on the surface of the big cube is warranted (in this question, it doesn't matter).
The exterior of the big cube has 216 small cube surfaces - 36 surfaces on each face. On each face, there are 4*4 = 16 cubes which have only one face exposed. So of all the cubes that make the exterior, there are exactly 16*6 = 96 cubes which have only a single surface exposed. We should use our leftover coloured cubes for these 96 cubes to minimise colour exposure.
Hence, the percentage of coloured surface = 96/216 = 44.44%

This would come in very handy if the question were a little different:

A big cube is formed by rearranging the 180 coloured and 36 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:

Big cube. 6^3 = 216, hidden/interior cube 4^3, minimizing color exposed interior cube gets 64 colored leaving 116 on the outside. Use 96 such that only one surface is exposed. You have 20 leftover coloured cubes.
How many cubes have 2 surfaces exposed? The edges but not the corners. 4 cubes on each edge have only 2 surfaces exposed. So total 12*4 = 48 cubes have 2 surfaces exposed. Use 20 leftover cubes here so another 40 surfaces are coloured.
Total 96 + 40 of the 216 outside surfaces are coloured.
So the answer here will not be 116/216 but 136/216.

Alternatively, place the non coloured cubes on 8 vertices and leftover 36 - 8 = 28 on edges. So non coloured cubes make 8*3 + 28*2 = 24 + 56 = 80 surfaces. So coloured cubes will make 216 - 80 = 136 surfaces.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17842 [1], given: 235 Intern Joined: 28 Jan 2013 Posts: 20 Kudos [?]: 4 [0], given: 184 Location: India GPA: 3.6 Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink] ### Show Tags 25 Jun 2015, 00:15 Great question! Though, I believe the question can be better phrased to clarify what needs to be calculated. I worked out the math correctly - but failed to understand what the question intended us to calculate. Even with Bunuel 's revised text in the post above - "The percentage of exposed area THAT is colored is:" leads me to calculate the proportion of colored cubes among the exposed cubes. So 96/152. Since we determined 64 cubes are 'inside' and are hidden (not exposed). The question for the answer that we've arrived to ~ 96/216 can be stated as - find the proportion of exposed cubes that are colored to the total number of cubes? I just want to confirm if the verbiage of the question is a little misleading or am I the only one misunderstanding this? Would hate to do such a mistake on the official exam. Kudos [?]: 4 [0], given: 184 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7745 Kudos [?]: 17842 [0], given: 235 Location: Pune, India Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink] ### Show Tags 25 Jun 2015, 21:11 sehej wrote: Great question! Though, I believe the question can be better phrased to clarify what needs to be calculated. I worked out the math correctly - but failed to understand what the question intended us to calculate. Even with Bunuel 's revised text in the post above - "The percentage of exposed area THAT is colored is:" leads me to calculate the proportion of colored cubes among the exposed cubes. So 96/152. Since we determined 64 cubes are 'inside' and are hidden (not exposed). The question for the answer that we've arrived to ~ 96/216 can be stated as - find the proportion of exposed cubes that are colored to the total number of cubes? I just want to confirm if the verbiage of the question is a little misleading or am I the only one misunderstanding this? Would hate to do such a mistake on the official exam. The verbiage of the question is spot on. You need to find the percentage of EXPOSED AREA that is coloured. How much is the exposed area? It is 216 small cube faces. Each big cube face is made up of 6*6 small cube faces. The big cube has 6 faces so total exposed AREA = 6*6*6 = 216 (This is not the number of cubes which are exposed. Small cubes on vertices contribute 3 faces to the exposed area) 96 small cube faces are coloured and exposed. Each of these cubes have only one face exposed and hence they contribute 96 small cube faces to the total exposed area. That is how you get 96/216 = Exposed coloured area/Total exposed area If the number of coloured cubes were much higher, we would have to account for their multiple exposed color faces but we don't need to do that in this question. I have discussed this hypothetical situation in my post above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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04 Oct 2015, 02:37
First of all, you can figure out that since there are a total of 216 small cubes, the big cube is 6x6x6 small cubes. Also, there are 152 small cubes visible on the outside of the big cube and 64 small cubes hidden inside. If you want minimal color exposure, then all 64 inside cubes should be colored. That leaves 96 colored cubes that must be visible. You can position these so that the middle 16 cubes on each of the 6 sides are colored. (That way only 1 side of each colored cube is exposed.)

So the percentage of exposed area that is colored is 96/216 = 4/9 = 44.44%
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Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]

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20 Jan 2016, 06:15
convinced with the answer . But while trying to solve it, I was looking at the problem in a different manner and got stuck

Let the smaller cubes be 1*1*1 cubes.
so we have 160 coloured and 56 non-coloured equals 216 cubes .
216 when arranged in one big cube becomes 6*6*6 cube.

Now I was looking at the surface area instead of volume.
Surface area of big cube = 6â^2= 216 (turns out volume and surface are same for this cube)
This surface area is occupied by 1*1 small non coloured cubes 56 in number , thus surface area covered by non coloured is 56. remaining surface area = 216-56= 160

160/216 = 74% is clearly not the solution.
but I am unable to understand the flaw in my logic.

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

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20 Jan 2016, 06:58
shreyashid wrote:
convinced with the answer . But while trying to solve it, I was looking at the problem in a different manner and got stuck

Let the smaller cubes be 1*1*1 cubes.
so we have 160 coloured and 56 non-coloured equals 216 cubes .
216 when arranged in one big cube becomes 6*6*6 cube.

Now I was looking at the surface area instead of volume.
Surface area of big cube = 6â^2= 216 (turns out volume and surface are same for this cube)
This surface area is occupied by 1*1 small non coloured cubes 56 in number , thus surface area covered by non coloured is 56. remaining surface area = 216-56= 160

160/216 = 74% is clearly not the solution.
but I am unable to understand the flaw in my logic.

Hi,
where you are going wrong is that you are taking that only one side of each non coloured is open/visible, while it is written that we have to make the coloured one least..

Remember the cubes in the vertices will have 3 faces open and along the edge will have two faces open..
place the noncolured ones at these places and then check..
8 vertices will have 3 faces each=8*3=24..

each edge, we are left with 4 places as the corner two are already taken care of by vertices..
12 edges with four cubes each=48, and 2 faces open for each=48*2=96..

we realize here the noncoloured are over 8+48=56..

so total noncolured faces open=24+96=120..
so coloured faces=216-120=96..
%=96/216=44.44%..
hope you are clear now where you went wrong..
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A big cube is formed by rearranging the 160 coloured and 56 [#permalink]

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25 Oct 2017, 18:48
Is it just me or is this question poorly worded?

The percentage of exposed area = 216-64 = 152 <- exposed cubes

to the coloured: 160 <-- total colored cubes

(colored exterior): 96 <- total colored exterior

is: 96/152 or 160/152, neither of which is the answer

Percentage of exposed colored pieces to the total number of cubes = 96 exposed colored cubes /216 total cubes

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

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26 Oct 2017, 05:08
supracuhz wrote:
Is it just me or is this question poorly worded?

The percentage of exposed area = 216-64 = 152 <- exposed cubes

to the coloured: 160 <-- total colored cubes

(colored exterior): 96 <- total colored exterior

is: 96/152 or 160/152, neither of which is the answer

Percentage of exposed colored pieces to the total number of cubes = 96 exposed colored cubes /216 total cubes

I have explained this here:
https://gmatclub.com/forum/a-big-cube-i ... l#p1542298
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Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]

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15 Nov 2017, 02:11
Thank Bunuel for the diagram. Really helped me understand.

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Re: A big cube is formed by rearranging the 160 coloured and 56   [#permalink] 15 Nov 2017, 02:11

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