Question

There are $20$ until cubes all of whose faces are white, and $44$ until cubes all of whose faces are red They are put together to form a bigger cube $(4\times 4\times 4)$. What is the minimum number of white faces visible on this larger cube?

• A. $20$
• B. $14$
• C. $12$
• D. $8$

Answer 1

The correct option is C $12$

$20$ unit cubes are white $44$ are red. Now of the inner cube formed, there will be a core inside which want be visible. That shall contain $2\times 2\times 2=8$ cubes. For minimum white cubes outside let these all $8$ be white.

Now remaining $12$ cubes will be visible.

Hence, the answer is $12.$