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

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Question Stats: 53% (02:36) correct 47% (02:34) wrong based on 788 sessions

### HideShow timer Statistics 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 [ 10.48 KiB | Viewed 9961 times ]

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

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5
1

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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9
7
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
##### General Discussion
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Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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1
5
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, 06:33.
Last edited by nick1816 on 17 Nov 2019, 15:37, 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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2
total faces ; 27
centers painted= 3
2 faces; 6
edges; 1
single faces ; 3
sum ; 13
13/27
IMO B

Bunuel wrote: 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
Target Test Prep Representative V
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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Bunuel wrote: 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.

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

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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.

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

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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.

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

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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.

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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# Scott Woodbury-Stewart

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

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Easiest way for me is just to draw the unfolded cube and do it in steps counting the red faces: 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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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]

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A77777 wrote:
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:=)))

Hey mate,
Because it is surrounded by other cubes, on top of it, bottom, left and right there are other cubes so this specific cube does not affect the exterior part of the larger cube.
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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ScottTargetTestPrep wrote:
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.

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.

Hello. This is a great explanation, however, I think I still don´t understand the question.
If they´re telling you that only one of the faces of each cube is painted, why are you considering the rest?
My apologies, i am sure it is super simple, but i can´t see it! Thank you!
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Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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EgmatQuantExpert wrote:

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.

Hello ,

I am little bit confused regarding the approach ,
Here are my doubts :
•The question already mentions that there are 27 cubes and exactly one face is painted red => if that’s the case then why we are considering the cubes having more than 1 faces coloured as red ?

• the question asks about the greatest possible fraction of the surface area that could be Red ,

Then the setup should be :
surface Area for red (max case)/ total surface area

Then why you took red coloured faces number , instead of their surface area !

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

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ScottTargetTestPrep wrote:
Bunuel wrote: 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.

While visualising , there are 2 cubes at the centre which are hidden The question states only one face of the each of the 27 cubes are red ,

Should we consider only one face or to maximise the fraction ( mentioned in the last line of the question ) , should we consider more than 1 coloured face of a single cube !

Posted from my mobile device
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Joined: 31 Mar 2019
Posts: 99
Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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LeenaSai wrote:
EgmatQuantExpert wrote:

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.

Hello ,

I am little bit confused regarding the approach ,
Here are my doubts :
•The question already mentions that there are 27 cubes and exactly one face is painted red => if that’s the case then why we are considering the cubes having more than 1 faces coloured as red ?

• the question asks about the greatest possible fraction of the surface area that could be Red ,

Then the setup should be :
surface Area for red (max case)/ total surface area

Then why you took red coloured faces number , instead of their surface area !

Posted from my mobile device

chetan2u plz guide
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Posts: 8741
Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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Bunuel wrote: 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:
The attachment 2019-04-26_1734.png is no longer available

A very simple and less confusing way is..

STEP I
There are 27 cubes joined to make a bigger cube. As you can see the figure too, 26 of these cubes have at least one of their faces open to air, so we have 26 of $$1*1 cm^2$$ faces colored red.
OR there is 1 cube right in the center which does not have any face open to air, so remaining 26 have at least 1 face open to air.
To maximize the red color, we take that the red colored face is surely open to air.

STEP II
Total surface area in $$cm^2=6*(3*3)=54 cm^2$$

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

If you look at the sketch attached.
There are 13 faces of smaller cube colored in THREE sides of the bigger cube, and similarly for the other 3 sides.. smaller cube faces in these THREE faces of bigger cube = 3*3*3
Fraction = $$\frac{13}{3*(3*3)}$$

B
Attachments 2019-04-26_1734.png [ 10.93 KiB | Viewed 1611 times ]

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

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chetan2u wrote:
Bunuel wrote: 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

A very simple and less confusing way is..

STEP I
There are 27 cubes joined to make a bigger cube. As you can see the figure too, 26 of these cubes have at least one of their faces open to air, so we have 26 of $$1*1 cm^2$$ faces colored red.
OR there is 1 cube right in the center which does not have any face open to air, so remaining 26 have at least 1 face open to air.
To maximize the red color, we take that the red colored face is surely open to air.

STEP II
Total surface area in $$cm^2=6*(3*3)=54 cm^2$$

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

If you look at the sketch attached.
There are 13 faces of smaller cube colored in THREE sides of the bigger cube, and similarly for the other 3 sides.. smaller cube faces in these THREE faces of bigger cube = 3*3*3
Fraction = $$\frac{13}{3*(3*3)}$$

B

Thankuuu so much for your prompt reply with in dept explanation chetan2u Posted from my mobile device

I was assuming that the cubes are fixed on the ground , so the lower portion’s exposed cubes will be 8 (3+3+2) , thereby the answer has to be 25/54 ! But now I understood my mistake !

It’s no where written that the cube has to be fixed on the ground

My bad , for this assumption
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Re: Each of 27 white 1-centimeter cubes will have exactly one face painted  [#permalink]

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LeenaSai wrote:
ScottTargetTestPrep wrote:
Bunuel wrote: 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.

While visualising , there are 2 cubes at the centre which are hidden The question states only one face of the each of the 27 cubes are red ,

Should we consider only one face or to maximise the fraction ( mentioned in the last line of the question ) , should we consider more than 1 coloured face of a single cube !

Posted from my mobile device

Response:

There’s only one smaller cube at the center of the big cube which does not have any exposed faces. The bottom layer is 3 x 3, so are the middle and top layers. The only hidden cube is the middle cube on the middle 3 x 3 layer.

It could be the case that none of the red painted faces are visible; however, as we want to maximize the number of visible red painted faces, we have to assume that the red painted face of every cube is visible. On the other hand, no matter what we do, we cannot have any faces of the “hidden” cube visible. Thus, the maximum number of red painted faces is 27 - 1 = 26; i.e. all red painted faces minus the red painted face of the hidden cube. That’s how we get the number 26.

The total number of faces is simply the number of faces of the big cube, which is 6, multiplied by the number of smaller faces on each face of the big cube, which is 9. Thus, there are a total of 6 x 9 = 54 faces (red or not).

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

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews Re: Each of 27 white 1-centimeter cubes will have exactly one face painted   [#permalink] 28 May 2020, 16:01

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