Cubes and dices deals with the problems of the following types:-
1. Counting the number of cubes or blocks in the given figure.
2. Painting a stack of cubes.
3. Coloring the six faces of a cube.
4. Construction of boxes.
5. Problems on dice.

From the given figure, it consists of 1 column containing 2 blocks and 3 columns containing 1 block each.
Therefore, total number of cubes = (1 x 2) + (3 x 1) = 5.

From the given figure, it consists of 1 column containing 3 cubes and 2 columns containing 2 cubes each and 3 columns containing 1 cubes each.
Number of cubes in columns of 3 cubes = 1 x 3 = 3.
Number of cubes in columns of 2 cubes = 2 x 2 = 4.
Number of cubes in columns of 1 cubes = 3 x 1 = 3.
Therefore, total number of cubes = 3 + 4 + 3 = 10.

From the given figure, it consists of 4 columns containing 1 cube each, 4 columns containing 2 cubes each and 1 column containing 3 cubes.
Therefore, total number of cubes = (4 x 1) + (4 x 2) + (1 x 3) = 15.

Let the upper face and the RHS face of the cube be coloured as shown below:
Then, clearly the rows of smaller cubes (formed by cutting the large cube into 64 parts) which are indicated by dots, have none of their sides coloured.
Since, there are 9 such rows and each row consists of 4 cubes so there are 9 x 4 = 36 cubes which are not coloured at all.
Hence, 36 cubes are not coloured at all.

The big cube can be cut into 27 small cubes as shown below:
Clearly, out of these 27 small cubes, the cubes having only one side painted are those which lie at the centre of each face of the big cube.
Since, there are 6 faces of a cube, required number of cubes is 6.

The original (coloured) cube is divided into 64 smaller cubes as shown in the figure:
The four central cubes on each face of the larger cube, have only one side painted.
There are six faces.
Therefore, required total number of such cubes = 4 x 6 = 24.

From (i), it is clear that red is opposite black i.e. black is on the top i.e. on face 5.
From (iii) and (iv), it is clear that brown and white lie on either sides of blue i.e. brown is on face 4 and white on face 2.
from (v), red is at the bottom i.e. on face 6.
Then, clearly green lies on face 3. This satisfies (ii) also.
Thus the cube will be coloured as indicated in the figure below:
**(1)** From the figure, it is clear that white is opposite brown.
**(2)** Out of the three colours black, blue and white, no two colours lie on opposite faces.
Hence, these three are adjacent colours.
**(3)** From (v) it is clear that, red colour lies at the bottom face and from (i) it is derived that since red is opposite black.
Hence, black lies on the top.

When the piece in figure X is bend to form a box then:
The number 2 will lie inverse 4,
The number 1 will lie inverse 6,
The number 5 will lie inverse 3.
Figure (a) has the numbers 1 and 6 on adjoining faces,
Figure (b) has the numbers 3 and 5 on adjoining faces,
Figure (c) has the numbers 2 and 4 on the adjoining faces.
So, these three alternatives are not possible.
Since, the numbers 1, 3 and 4 can appear on adjoining faces, so figure (d) is possible.
Therefore, only the box shown in figure (d) can be framed by bending figure X.

When the piece in figure X is bend to form a box then dot and the shading must lie inverse to each other.
Hence, figures a, c and d which bear the dot and the shading on adjoining faces cannot possibly be formed by bending the piece in figure X.
Therefore, only box (b) can be formed.

From figures (i), (ii) and (iv), it is clear that 6, 4, 1 and 2 dots appear adjoining to 3 dots.
Therefor, there will be 5 dots on the face inverse the face with 3 dots.

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