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A cube of side $3$ units is formed using a set of smaller cubes of side $1$ unit. Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible.

  1. $ 1: 4$
  2. $ 1: 3$
  3. $ 1: 2$
  4. $ 2: 3$
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Anybody wants to count in real… 

most of us have 3*3 Rubik’s cube try it.

Exactly matches to this problem. 😊

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2 Answers

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Best answer

Option C

Total $27 (3 \times 3 \times 3)$ small cubes of $1$ unit each will be required to form a bigger cube of side $3$ units

Afbeeldingsresultaat voor cube image   

No. of faces per cube $= 6$

Total number of cubes $= 9\times 3 = 27$

Total number of faces $= 27\times 6 = 162$

Total number of non visible faces $= 162-54 = 108$ 

$\frac{\text{No. of  visible  faces}}{\text{No. of non  visible  faces}}= \frac{54}{108}=\frac{1}{2}$

Ref: https://www.quora.com/A-cube-with-sides-of-3-units-is-formed-using-a-set-of-smaller-cubes-with-sides-of-1-unit-How-do-you-find-the-proportion-of-the-number-of-faces-of-the-smaller-cubes-visible-to-those-which-are-not-visible

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might be a silly doubt but how is the faces at the base visible???

it should be only 5 faces i.e 9*5 =45 faces visible cause the base is not visible.

pls clear this
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I kept wondering for a while & then realized that the centre-most piece of the cube is never visible from any side. You can only see it after dismantling the cube.
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Correct Answer = Option C 
 



We need a total of 27 small cubes to make one large cube.

1 small cube has a total of 6 faces, hence for 27 small cubes the total number of faces = 27 * 6 = 162

 

i.e One large cube has a total of 162 faces some of which are visible and some of which are not visible.

 

 


Now we will find the number of visible faces in 1 large cube.

From the figure, we notice that some pieces (like corners) have more faces visible, whereas some pieces (like centre) has only one face visible. Hence we first identify each type and count of visible faces for each type of piece

  1. Corner:

    Now each face has 4 corner pieces. Hence 6 faces would have 6*4 = 24 corners pieces. But we have over-counted here. Notice that, each corner piece is shared among 3 faces, hence the correct number of corner pieces = 24/3 = 8 pieces.

    Each corner piece has 3 faces visible, hence total visible faces for corner piece = 8*3 = 24  ----- (i)
     
  2. Side:

    Each face has 4 side pieces. Hence 6 faces would have 6 * 4 = 24 side pieces, but again we have over-counted. Notice how each side piece is shared among 2 faces, hence the correct number of side pieces = 24/2 = 12 pieces.

    Each side piece has 2 faces visible, hence total visible faces for side piece = 12 * 2 = 24 ----- (ii)
     
  3. Center:

    Each face has 1 centre piece. Hence 6 faces would have 6 centre piece. But are we again overcounting? NO, the centre piece is never shared among any face of a large cube, hence no over-counting. So total centre piece = 6.

    Each centre piece has 1 face visible, hence total visible faces for center piece = 6 ----- (iii)

Adding equation i, ii and iii we get 24 + 24 + 6 = 54 visible faces.
 

Now total invisible faces = Total faces – Total visible faces
                                        =  162 – 54
                                        = 108

Hence the proportion of the number of faces of the smaller cubes visible to those which are NOT visible  
= 54 : 108
= 1 : 2
 

Answer:

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