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

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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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.

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,

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

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.

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).

Thanks Bunuel Your helps are amazing, this is the greatest GMAT portal on earth.

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.

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.

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.

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

The question should actually read:
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

Very good approach but there is a caveat here, this solution doesn't check if the 96 remainings cubes should be arranged on the edge or in the middle of each face. Because in the smaller cube present on vertices ( of the larger cube)3 faces and those along the edges will have at least 2 faces exposed


The answer still lands correctly because these 56 noncolored cubes will be exhausted filling the 8 vertices and along the 12 edges to minimize the area given to colored faces.