Verbal Reasoning - Cube and Cuboid - Discussion
Discussion Forum : Cube and Cuboid - Cube and Cuboid 5 (Q.No. 5)
Directions to Solve
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.
5.
How many small cubes are there whose at the most two faces are coloured ?
Answer: Option
Explanation:
Number of small cubes having two faces coloured = 8 + 8 + 4 + 4 = 24
and Number of small cubes having only one face coloured = 4 x 6 = 24
and Number of small cubes having no face coloured = 4 + 4 = 8
Therefore, total number of small cubes whose at the most two faces are coloured = 24 + 24 + 8 = 56.
Discussion:
4 comments Page 1 of 1.
Arun said:
2 months ago
Why do they include one-face coloured cubes?
Anyone, please explain to me.
Anyone, please explain to me.
Nitin said:
6 years ago
Why we need to count the cubes whose faces have no colour? When it's mentioned that at the most two sides "coloured".
Karthik.P said:
1 decade ago
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.
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.
(1)
Shelbin said:
1 decade ago
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.
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.
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