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# If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain

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If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain  [#permalink]

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Updated on: 01 Aug 2016, 10:56
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If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after painting it by red color then which of the following is/are true.

I) Ratio of number of painted surfaces of resultant cubes to total number of surfaces is 1/4
II) Number of cubes having no faces pointed is equal to number of cubes having three faces painted.
III) Number of cubes having 2 faces painted is same as number of cubes having one face painted

a) None
b) II and III
c) II only
d) I, II & III
e) I & III only

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Originally posted by GMATinsight on 01 Aug 2016, 10:50.
Last edited by GMATinsight on 01 Aug 2016, 10:56, edited 1 time in total.
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If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain  [#permalink]

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01 Aug 2016, 12:28
1
GMATinsight wrote:
If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after painting it by red color then which of the following is/are true.

Long explanation, probably too detailed

Hypercube facts ($$n=3$$ for a cube):
$$2^n\text{ vertices} \\ 2^{n−1}n\text{ edges}\\ 2 + E - V \text{ faces}$$

For a cube: 8 vertices, 12 edges and 6 faces

Let $$G = 4$$ be the size of the original cube.

I) Ratio of number of painted surfaces of resultant cubes to total number of surfaces is 1/4

There is 1 large cube with 6 sides of $$G^2$$ painted faces.

There is $$G^3$$ 1x1 cubes with 6 sides of surface area 1.

$$\frac{1 \times 6 \times G^2}{G^3 \times 6 \times 1} = \frac{G^2}{G^3} = G^{-1} = \frac{1}{4}$$

True

II) Number of cubes having no faces pointed is equal to number of cubes having three faces painted.

There are always 8 vertices on a cube, each vertex is the only 1x1 block which has 3 coloured faces.

Inside the cube, there is a secondary unpainted cube. Each side of this unpainted cube is equal to the size of the original cube minus two (on each axis, there will be two 1x1 cubes on the exterior of the larger cube). The size of this cube is $$(G-2)^3$$.

There are $$(G-2)^3$$ cubes with 0 sides painted. $$= 2^3 = 8$$
There are $$V = 2^3$$ cubes with 3 sides painted $$= 8$$

True

III) Number of cubes having 2 faces painted is same as number of cubes having one face painted

There are 12 edges on a cube. Each edge consists of G cubes, two of which are the vertices of the cube (and have 3 colours), the rest have two colours.

On each exterior face, there is a square of $$(G-2)$$ cubes, surrounded by 4 vertex cubes and $$(G-2) \times 4$$ edge cubes.
Each square consists of $$(G-2)^2$$ 1-coloured faces. There is 1 square per face.

There are $$(G-2)\times E$$ cubes with 2 sides painted. $$= 2 \times 12 = 24$$
There are $$(G-2)^2 \times F$$ cubes with 1 side painted. $$= 2^2 \times 6 = 24$$

True

d) I, II & III

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Re: If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain  [#permalink]

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01 Aug 2016, 10:54
GMATinsight wrote:
If a cube of dimension 4 by 4 by 4 cuts into 1 by 1 by 1 after painting it by red color then which of the following is/are true.

I) Ratio of number of painted surfaces of resultant cubes to total number of surfaces is 1/4
II) Number of cubes having no faces pointed is equal to number of cubes having three faces painted.
III) Number of cubes having 2 faces painted is same as number of cubes having one face painted

a) None
b) II and III
c) II only
d) I, II & III
e) I & III only

Source: http://www.GMATinsight.com

Please check the detailed of calculating everything about the painted cubes as shown in attachment

Attachments

File comment: www.GMATinsight.com

Answer 4.jpg [ 144.82 KiB | Viewed 1990 times ]

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Re: If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain  [#permalink]

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29 Jul 2019, 17:06
DAllison2016 wrote:
GMATinsight wrote:
If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after painting it by red color then which of the following is/are true.

Long explanation, probably too detailed

Hypercube facts ($$n=3$$ for a cube):
$$2^n\text{ vertices} \\ 2^{n−1}n\text{ edges}\\ 2 + E - V \text{ faces}$$

For a cube: 8 vertices, 12 edges and 6 faces

Let $$G = 4$$ be the size of the original cube.

I) Ratio of number of painted surfaces of resultant cubes to total number of surfaces is 1/4

There is 1 large cube with 6 sides of $$G^2$$ painted faces.

There is $$G^3$$ 1x1 cubes with 6 sides of surface area 1.

$$\frac{1 \times 6 \times G^2}{G^3 \times 6 \times 1} = \frac{G^2}{G^3} = G^{-1} = \frac{1}{4}$$

True

II) Number of cubes having no faces pointed is equal to number of cubes having three faces painted.

There are always 8 vertices on a cube, each vertex is the only 1x1 block which has 3 coloured faces.

Inside the cube, there is a secondary unpainted cube. Each side of this unpainted cube is equal to the size of the original cube minus two (on each axis, there will be two 1x1 cubes on the exterior of the larger cube). The size of this cube is $$(G-2)^3$$.

There are $$(G-2)^3$$ cubes with 0 sides painted. $$= 2^3 = 8$$
There are $$V = 2^3$$ cubes with 3 sides painted $$= 8$$

True

III) Number of cubes having 2 faces painted is same as number of cubes having one face painted

There are 12 edges on a cube. Each edge consists of G cubes, two of which are the vertices of the cube (and have 3 colours), the rest have two colours.

On each exterior face, there is a square of $$(G-2)$$ cubes, surrounded by 4 vertex cubes and $$(G-2) \times 4$$ edge cubes.
Each square consists of $$(G-2)^2$$ 1-coloured faces. There is 1 square per face.

There are $$(G-2)\times E$$ cubes with 2 sides painted. $$= 2 \times 12 = 24$$
There are $$(G-2)^2 \times F$$ cubes with 1 side painted. $$= 2^2 \times 6 = 24$$

True

d) I, II & III

HI! Can you please explain how did you get G3 in the first statement?

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Re: If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain  [#permalink]

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06 Aug 2019, 06:36
GMATinsight wrote:
If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after painting it by red color then which of the following is/are true.

I) Ratio of number of painted surfaces of resultant cubes to total number of surfaces is 1/4
II) Number of cubes having no faces pointed is equal to number of cubes having three faces painted.
III) Number of cubes having 2 faces painted is same as number of cubes having one face painted

a) None
b) II and III
c) II only
d) I, II & III
e) I & III only

Source: http://www.GMATinsight.com

given: large cube with side=4 has its faces painted in red, then its cut into small cubes with side=1;

num of smaller cubes is 4^3=64;
num of faces not red is (4-2)^3=8;
num of 1 red faces: (cubes in the center)(faces)=4•6=24
num of 2 red faces: (cubes in the edge that are not vertices)(edges)=2•12=24
num of 3 red faces: (cubes in the vertices)=8

Re: If a cube of dimension 4 by 4 by 4 is cut into 1 by 1 by 1 after pain   [#permalink] 06 Aug 2019, 06:36
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