Last visit was: 23 Jul 2024, 10:51 It is currently 23 Jul 2024, 10:51
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Manager
Joined: 27 May 2008
Posts: 93
Own Kudos [?]: 229 [212]
Given Kudos: 4
Concentration: Finance
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643312 [97]
Given Kudos: 86728
Manager
Joined: 10 Jul 2013
Posts: 228
Own Kudos [?]: 1050 [45]
Given Kudos: 102
General Discussion
Manager
Joined: 25 Jun 2009
Posts: 131
Own Kudos [?]: 333 [2]
Given Kudos: 6
Q49  V22 GMAT 2: 700  Q50  V35
1
Kudos
1
Bookmarks
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=6^6 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

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,
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643312 [15]
Given Kudos: 86728
8
Kudos
7
Bookmarks
nitishmahajan 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=6^6 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

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,

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...
Intern
Joined: 11 May 2010
Posts: 1
Own Kudos [?]: 4 [4]
Given Kudos: 7
2
Kudos
2
Bookmarks
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=6^6 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643312 [3]
Given Kudos: 86728
1
Kudos
2
Bookmarks
zsokhan 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=6^6 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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643312 [8]
Given Kudos: 86728
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
4
Kudos
4
Bookmarks
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 to the coloured is :

A. 25.9%
B. 44.44%
C. 35%
D. 61%
E. None of these

Similar questions to practice:
if-a-4-cm-cube-is-cut-into-1-cm-cubes-then-what-is-the-107843.html
a-large-cube-consists-of-125-identical-small-cubes-how-110256.html
64-small-identical-cubes-are-used-to-form-a-large-cube-151009.html
the-entire-exterior-of-a-large-wooden-cube-is-painted-red-155955.html
Senior Manager
Joined: 17 Apr 2013
Status:Verbal Forum Moderator
Posts: 360
Own Kudos [?]: 2235 [0]
Given Kudos: 298
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
Thanks Bunuel Your helps are amazing, this is the greatest GMAT portal on earth.
Intern
Joined: 07 Oct 2013
Posts: 20
Own Kudos [?]: 33 [2]
Given Kudos: 266
GMAT 1: 770 Q50 V45 (Online)
GRE 1: Q167 V170
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
2
Bookmarks
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
Intern
Joined: 07 Mar 2014
Posts: 7
Own Kudos [?]: 2 [1]
Given Kudos: 2
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
1
Kudos
jps245 wrote:
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

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).
Intern
Joined: 15 Jul 2012
Posts: 24
Own Kudos [?]: 45 [0]
Given Kudos: 245
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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?
Math Expert
Joined: 02 Sep 2009
Posts: 94592
Own Kudos [?]: 643312 [0]
Given Kudos: 86728
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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.
Intern
Joined: 14 Jan 2015
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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.
Tutor
Joined: 16 Oct 2010
Posts: 15140
Own Kudos [?]: 66810 [2]
Given Kudos: 436
Location: Pune, India
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
2
Kudos
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.
Intern
Joined: 28 Jan 2013
Posts: 16
Own Kudos [?]: 7 [0]
Given Kudos: 184
Location: India
GPA: 3.6
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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.
Tutor
Joined: 16 Oct 2010
Posts: 15140
Own Kudos [?]: 66810 [0]
Given Kudos: 436
Location: Pune, India
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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.
Intern
Joined: 16 May 2016
Posts: 7
Own Kudos [?]: 1 [0]
Given Kudos: 6
Location: United States (CA)
GMAT 1: 760 Q49 V45
GPA: 3.55
A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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
Tutor
Joined: 16 Oct 2010
Posts: 15140
Own Kudos [?]: 66810 [0]
Given Kudos: 436
Location: Pune, India
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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
Intern
Joined: 26 Dec 2022
Posts: 34
Own Kudos [?]: 6 [0]
Given Kudos: 422
Re: A big cube is formed by rearranging the 160 coloured and 56 [#permalink]
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.

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 --> $$\frac{96}{216}\approx(0.4444)=44.44%$$.