### Discussion :: Cube and Cuboid - Cube and Cuboid 5 (Q.No.5)

- Cube and Cuboid - Introduction
- Cube and Cuboid - Cube and Cuboid 1
- Cube and Cuboid - Cube and Cuboid 2
- Cube and Cuboid - Cube and Cuboid 3
- Cube and Cuboid - Cube and Cuboid 4
- «« Cube and Cuboid - Cube and Cuboid 5
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- Cube and Cuboid - Cube and Cuboid 9

The following questions are based on the information given below:

All the opposite faces of a big cube are coloured with red, black and green colours. After that is cut into 64 small equal cubes.

Shelbin said: (Nov 12, 2012) | |

This is a good one. At the most two faces coloured = max 2 faces coloured! (So minimum faces to be coloured is 0). Hence 24(2 faces coloured)+ 24(1 face coloured)+ 8(0 face coloured) = 56. |

Karthik.P said: (Nov 1, 2013) | |

Maximum faces colored for any cube is '3' in the given scenario. It's better to subtract '3' faces colored cubes from total cubes then we will get the number of cubes that are at most two faces colored. i.e. (4^3)-8 = 64-8= 56. |

Nitin said: (Aug 3, 2018) | |

Why we need to count the cubes whose faces have no colour? When it's mentioned that at the most two sides "coloured". |

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