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filipembribeiro
Joined: 20 Jul 2020
Last visit: 06 Mar 2026
Posts: 37
Given Kudos: 52
Location: Brazil
Concentration: Leadership, Technology
Schools: Stanford '26 Kellogg '26 INSEAD Jan '26 IESE '26 HBS '26
GMAT Focus 1: 645 Q80 V83 DI83 
GMAT Focus 2: 675 Q84 V84 DI83
GPA: 3.0
WE:Consulting (Consulting)
Schools: Stanford '26 Kellogg '26 INSEAD Jan '26 IESE '26 HBS '26
GMAT Focus 2: 675 Q84 V84 DI83
Posts: 37
12 Jul 2021, 03:46
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B
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Difficulty:
95%
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Question Stats:
27% (02:21) correct
73%
(02:49)
wrong
based on 44
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There are 152 black cubies and 64 white cubies, they were assembled to make one bigger cube and arranged in a way that the exposure of black cubies to the outside is minimum. What is the estimated percentage of the exposed area that is covered by black cubies?
A) 33%
B) 41%
C) 44%
D) 56%
E) None
A) 33%
B) 41%
C) 44%
D) 56%
E) None
12 Jul 2021, 09:42
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filipembribeiro
Let each cube be of unit 1, so the volume = 152+64=216, that is \(s^3=216……s=6\)
That is each side has 6 smaller cubes.
Now, the 6*6*6 cube will have following sides of small cubes exposed to outside.
1) 3 sides : The corner ones, so 8 of them.
2) 2 sides: The edges minus the corner ones, so 4 on each edge. As number of edges = 12, cubes on the edges = 12*4=48
3) 1 side: All the faces less the edges. 6*6-(2*6+2*4)=16. As number of faces are 6, cubes =16*6=96
We will first fix white cubes with max exposure.
So 8+48+8=64.
Thus black cubes with one face exposed = 96-8=88, meaning area = 88*1*1=88
Total surface area of cube = \(6s^2=216\)
Minimum % of black = \(\frac{88}{216}*100\)~41%
B
filipembribeiro
Joined: 20 Jul 2020
Last visit: 06 Mar 2026
Posts: 37
Given Kudos: 52
Location: Brazil
Concentration: Leadership, Technology
Schools: Stanford '26 Kellogg '26 INSEAD Jan '26 IESE '26 HBS '26
GMAT Focus 1: 645 Q80 V83 DI83
GMAT Focus 2: 675 Q84 V84 DI83
GPA: 3.0
WE:Consulting (Consulting)
Schools: Stanford '26 Kellogg '26 INSEAD Jan '26 IESE '26 HBS '26
GMAT Focus 2: 675 Q84 V84 DI83
Posts: 37
12 Jul 2021, 09:47
Kudos
Bookmarks
To discover what figure we are building, let's add the cubes
152 + 64 = 216 => 6x6x6 cube
This cube has a "mini cube" inside with size 4x4x4 plus the external part. As we want as few black cubes on the outside as possible, we have to put the black ones on the inner cube first.
So there are 64 inner black cubes (4x4x4) and 152-64 = 88 outer black cubes.
Among the external parts, we have the central cubes that have only one side exposed. There are 4x4 = 16 of these per face, thus 96 in total.
That is, we can put all 88 black cubes that are left in these centers.
Total external area = 6x6x6 = 216 faces
Black outer area = 88 faces
88/216 = 40.7% = 41%
Ans. B
152 + 64 = 216 => 6x6x6 cube
This cube has a "mini cube" inside with size 4x4x4 plus the external part. As we want as few black cubes on the outside as possible, we have to put the black ones on the inner cube first.
So there are 64 inner black cubes (4x4x4) and 152-64 = 88 outer black cubes.
Among the external parts, we have the central cubes that have only one side exposed. There are 4x4 = 16 of these per face, thus 96 in total.
That is, we can put all 88 black cubes that are left in these centers.
Total external area = 6x6x6 = 216 faces
Black outer area = 88 faces
88/216 = 40.7% = 41%
Ans. B
