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Each of 27 white 1-centimeter cubes will have exactly one face painted

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Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 26 Apr 2019, 06:35
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Each of 27 white 1-centimeter cubes will have exactly one face painted red. If these 27 cubes are joined together to form one large cube, as shown above, what is the greatest possible fraction of the surface area that could be red?

A. 11/27
B. 13/27
C. 1/2
D. 5/9
E. 19/27

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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 21 May 2019, 12:27
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fraction exposed = colored area / total area
total area = all the exposed sides of exposed cubes
= 9 * 6 ( total 6 faces of the big cube and 9 on each face )

colored area = one for each cube for all exposed cubes
total number of cubes in the diagram = 3 ^ 3 = 27
number of cubes exposed = 27-1 = 26
(only one cube at the center is not exposed)

fraction = 26/54
= 13/27
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Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post Updated on: 17 Nov 2019, 16:37
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Number of cubes having 3 faces outside- 8 (corner ones)

Number of cubes having 2 faces outside- 12*(n-2)= 12*(3-2)= 12

Number of cubes having 1 face outside- \(6*(n-2)^2\) =\( 6*{(3-2)^2}\) = 6

Number of cubes having 0 face outside- \((n-2)^2\) = \({(3-2)^2}\)=1

Maximum number of colored faces outside=8+12+6=26

greatest possible fraction of the surface area that could be red= \(\frac{26}{54} \)= \(\frac{13}{27}\)

Originally posted by nick1816 on 26 Apr 2019, 07:33.
Last edited by nick1816 on 17 Nov 2019, 16:37, edited 1 time in total.
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 28 Apr 2019, 03:34
2
total faces ; 27
centers painted= 3
2 faces; 6
edges; 1
single faces ; 3
sum ; 13
13/27
IMO B

Bunuel wrote:
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Each of 27 white 1-centimeter cubes will have exactly one face painted red. If these 27 cubes are joined together to form one large cube, as shown above, what is the greatest possible fraction of the surface area that could be red?

A. 11/27
B. 13/27
C. 1/2
D. 5/9
E. 19/27

PS93602.01
Quantitative Review 2020 NEW QUESTION


Attachment:
2019-04-26_1734.png
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 23 May 2019, 05:13
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Solution


Given
In this question, we are given
    • Each of 27 white 1-centimeter cubes will have exactly one face painted red.
    • These 27 cubes are joined together to form one large cube, as shown in the given diagram.

To Find
We need to determine
    • The greatest possible fraction of the surface area that could be red.

Approach & Working
To maximise the surface area, count as red, except the cube which is at the centre, all the other cubes must have their red surface area exposed.
    • Number of cubes with red surface area exposed = 27 – 1 = 26

However, in each surface of the bigger cube, there will be 9 smaller cube surfaces, and there will be total 6 surfaces for the bigger cube.
    • Hence, total number of cube surfaces exposed = 9 * 6 = 54
    • Therefore, the greatest fraction = 26/54 = 13/27

Hence, the correct answer is option B.

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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 27 May 2019, 19:05
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Bunuel wrote:
Image
Each of 27 white 1-centimeter cubes will have exactly one face painted red. If these 27 cubes are joined together to form one large cube, as shown above, what is the greatest possible fraction of the surface area that could be red?

A. 11/27
B. 13/27
C. 1/2
D. 5/9
E. 19/27

PS93602.01
Quantitative Review 2020 NEW QUESTION


Attachment:
2019-04-26_1734.png


The total surface area of the big cube is 9 x 6 = 54.

The maximum number of red faces that can be exposed on the surface of the big cube is 26 (since one of the unit cubes is inside of the big cube and can’t been seen).

Thus, the greatest possible fraction of the surface area that could be red is 26/54 =13/27.

Answer: B
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 29 May 2019, 20:43
EgmatQuantExpert ScottTargetTestPrep Bhanupriya05

I'm sorry, I don't understand the solution(s) provided, especially when you say all the 'exposed' sides, why do we consider the exposed sides (as per the diagram) when even the ones on the bottom could also have been painted red? [Bunuel could you please clarify on this?]

As per the 'exposed' sides, the most bottom right cube would be considered as two red faces, but the question stem clearly says, "Each of 27 white 1-centimeter cubes will have exactly one face painted red"

I'm not sure where am I getting confused in here, and would sincerely appreciate any help on this.

Thanks in advance.
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 30 May 2019, 01:05
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Hi rishabhjain13

I am not considering exposed sides as per the diagram.
even the bottom ones are considered.
lets calculate colored (exposed) area again,

Now how many cubes do we have?
27

how many are exposed?
26. (all except the one at the center)

for each exposed cube, how many faces are colored?
1 (given that Each of 27 white cubes will have exactly one face painted red)

since we are calculating greatest possible surface area that could be red,
Maximum red faces can be 27 * 1 = 27 but one red face for the cube at the center can never be exposed.
so maximum red surface exposed will be 26.
(imagine arrangement where in colored side of each cube is exposed/facing outward).

To calculate fraction, we need to divide this with total surface exposed in a cube (painted or not painted)

total surface area exposed in a cube = number of faces in big cube * no of faces of 1-centimeter cubes in each face of big cube
= 6 * 9 ( you can understand this from the diagram)
= 54
ans = 26/54
=13/27

I hope its clear. Maybe experts can explain it in a better manner.

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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 01 Jun 2019, 06:15
rishabhjain13 wrote:
EgmatQuantExpert ScottTargetTestPrep Bhanupriya05

I'm sorry, I don't understand the solution(s) provided, especially when you say all the 'exposed' sides, why do we consider the exposed sides (as per the diagram) when even the ones on the bottom could also have been painted red? [Bunuel could you please clarify on this?]

As per the 'exposed' sides, the most bottom right cube would be considered as two red faces, but the question stem clearly says, "Each of 27 white 1-centimeter cubes will have exactly one face painted red"

I'm not sure where am I getting confused in here, and would sincerely appreciate any help on this.

Thanks in advance.


Let’s rephrase the question to make sure we clearly understand what the question is asking: We have 27 small cubes with a side length of 1. We paint exactly one face of each cube red and then stack these smaller cubes to form the shape depicted on the diagram. Though that’s not exactly what the question is asking, we actually need to figure out how many red faces we will see at most.

Having that said, I think you are getting confused about the number of visible faces (which is 56) versus the number of small cubes (which is 27). Let me provide an alternative explanation to overcome that. Let’s classify the smaller cubes into four groups: a) The cubes with 3 exposed faces (the cubes which are on the corner of the big cube, there are 8 of them) b) The cubes with 2 exposed faces (the cubes which are on the edge but not on the corner of the big cube, there are 12 of them) c) The cubes with 1 exposed face (the cubes which are at the center of each face, there are 6 of them) d) The cube with no exposed face (the only cube which is at the core of the larger cube)

Let’s pick one face from each group besides the no visible face group and paint it red. Then, we will have 8 + 12 + 6 = 26 of them. That’s why you can have at most 26 faces painted red.
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Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 20 Jun 2019, 14:30
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Easiest way for me is just to draw the unfolded cube and do it in steps counting the red faces:

Image

Total = 9 faces per side * 6 sides = 54

1) There are 2 All red sides that are opposite each other (e.g. top and bottom face): 9 faces * 2 sides = 18

Next, there are 4 adjacent (side) faces, all of them must not touch any red, and a pair of those adjacent faces must not touch the other pair's red edges. This is my shorthand for saying that since the bottom and top 6 cubes are "exposed" their adjacent faces on the sides won't be red. So all that's left to be red is the cubes in the middle row that weren't affected by the top and bottom face, and among these they also cannot touch.

2) First pair of adjacents: 3*2 = 6

3) Second pair: 1*2 = 2

18+6+2 = 26 / 54 --> 13/27
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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New post 30 Nov 2019, 10:41
This must be super simple but I am not getting the part where we subtract 1 from 27. ie: 26 (all except the one at the center). why is that one not exposed?
Why? Pls explain:=)))
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted   [#permalink] 30 Nov 2019, 10:41
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