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 1/4There 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.
\(\frac{1 \times 6 \times G^2}{G^3 \times 6 \times 1} = \frac{G^2}{G^3} = G^{-1} = \frac{1}{4}\)
TrueII) 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 \(= 8\)
TrueIII) Number of cubes having 2 faces painted is same as number of cubes having one face paintedThere 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
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