If 7 white cubes and 20 red cubes, all of equal size, are fastened tog

Question

https://gmatclub.com/forum/download/file.php?id=38970 If 7 white cubes and 20 red cubes, all of equal size, are fastened together to form one large cube as shown above, what is the smallest fractional part of the surface area of the large cube that could be white? (A) 1/9 (B) 7/54 (C) 4/27 (D) 1/6 (E) 7/27 2017-12-01_0948_003.png

Sasindran · Answer

A

Let each side of the small cube be 1m. 
The surface area of the large cube is 6*(3*3)=54.

Now one white cube can be placed at the centre of the large cube so that the white cube is invisible. The remaining 6 white cubes can be placed at the center of each face (we have 6 faces) to minimise the white area. Hence the surface area of the large cube that could be white is 6*(1*1)=6.

So the fraction is 6/54=1/9

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arvind910619 · Answer

I think the answer is 1/6
Let the side of each cube be 1 .
Then the surface area of the large cube is 6a^2=6*9=54
Now there are 7 white cubes ans we need minimum area of these cubes so we have to place them in the middle line .
So there are 9 faces of area 1 square units each that are visible so 9/54=1/6

Please correct if i am wrong

Sasindran · Answer

You can reduce the area further by placing one white cube at the core of the big cube. If u place the remaining 6 cubes at the center of each face of the big cube u have only white area as 6 which is lesser than 9

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Archit3110 · Answer

this cube has3 faces with 9 cubes ; total cubes 27
so as to have smallest fractional part of the surface area of cube let there be 1 center piece on all faces be white and on any of the other 2 faces be 2 white
P of 1 white face on one side so that its smallest fraction part ; 1/9
option A

Fdambro294 · Answer

27 cubes: 20 are brown and 7 are white

The 27 cubes would create a 3 by 3 by 3 larger cube (assume each of the 27 cubes are 1 by 1 by 1)

The goal is to cover up as much of the surface area as possible with the brown cubes, starting with placing the brown cubes in the spots at which 3 faces of a single cube show

(1st)cover up the 8 vertices in which 3 faces are showing with smaller 1x1x1 brown cubes

We can place 8 of the brown cubes here ——> 12 smaller brown cubes remain

(2nd) there are 12 edges and on each edge there will be 1 cube that has 2 faces showing (3 cubes on each edge minus the 2 vertices on either side = 1 cube with 2 faces showing)

We can place the remaining 12 brown cubes in these spots at which 2 of the smaller cube faces are showing

(3rd) now the only remaining faces on the larger cube are the smaller cubes placed in the middle of each of the 6 larger cube’s faces

Since we have placed every smaller brown cube, We have to put 6 of the white cubes in these positions such that 1 face of each of the 6 smaller white cubes is showing

(6 white cubes) * (1 face showing for each) = 6 total faces showing that are white

The total number of faces overall will consist of 9 smaller faces on each of the 6 larger faces of the larger cube

9 * 6 = 54 total faces / surface area of the larger cube

(# white faces showing) / (total # of faces showing) = 6 / 54 = 1 / 9

A

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If 7 white cubes and 20 red cubes, all of equal size, are fastened tog

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If 7 white cubes and 20 red cubes, all of equal size, are fastened tog

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