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 that of those which are NOT visible.

Consider the cube of side 1 unit shown below.

Now, a set of similar cubes as shown above is joined to form a cube of side 3 units, and we have to determine the ratio of faces of smaller cubes visible to those which are not visible. Complete surface area of the cube of side 3 units is determined as: A = 6 ^_x (3)² = 54 Again, the area of one face of cube of side 1 unit is: A1 = (1)² = 1 So, the total number of visible faces is: N_visible = ^{$\frac{A}{A_1} = 54$} Again, total number of smaller cubes required to form the big cube is: Number of smaller cubes = ^{$\frac{\text{volume of cube of side unit}}{\text{volume of cube of side out}}$}                                     = ^{$\frac{(3)^3}{(1)^3} = 27$}

So, the total number of faces of smaller cubes is given as N_total = 6 ^_x (number of smaller cubes)             = 6 ^_x 27 = 162 Therefore, number of invisible faces is N _invisible = N_total - N_visible             = 162 - 54 = 108 Hence, the desired ratio is ^{$\frac{N_{visible}}{N_{visible}} = \frac{54}{108} = \frac{1}{2}$} or 1 : 2