In this question, the sets of numbers given in the alternatives are represented. The columns and rows of Matrix I are numbered from 0 to 4 and that of Matrix II are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, e.g., D can be represented by 04,12, etc., and 'I' can be represented by 65, 79, etc. Similarly you have to identify the set for the word 'BANK'.

Initial total number of cubes = 343,
Number of cubes removed = 27
Smaller 27 cubes painted blue
Exposed faces of original big cube (3 faces with 9 cube on each face i.e total 27 cubes) painted with black
Out of 27 small cubes from 3 x 3 x 3, outer 26 cubes are 1st painted with blue and then it is kept back with original cube and painted with yellow so out of 26 cubes only 5 edges will give us cubes with both the colours and number of such cubes are 12

In any plane, leave 4 sides cube and select (3 x 3 x 3) inter section. But
the cubes 2 x 2 x 1 give 2 less cube because that part we are already removed.
No. of cubes = (3 x 3 x 3) - 2 = 25.

For maximum number of pieces cuts has to be 6, 7 and 7 and maximum number of pieces is (6 + 1)(7 + 1)(7 + 1) = 7 x 8 x 8 = 448.
Minimum number of pieces is 20 + 1 = 21.
Hence required ratio is 448:21

Number of cubes with no face painted is
105 - 34 - 24 - 20 = 27
Or else all the 3 x 3 x 3 inner cubes will remain coloured.

Out of 6 faces of 5 faces are exposed and those were painted.
Number of vertices with three faces exposed (Painted) is 4
Number of vertices with 2 faces exposed (Painted) is 4
Number of vertices with 1 faces exposed (Painted) is 0
Number of vertices with 0 faces exposed (Painted) is 0
Number of sides with 2 sides exposed (Painted) is 8
Number of sides with 1 sides exposed (Painted) is 4
Number of sides with no sides exposed (Painted) is 0
From the above observation:
Number of cubes with 3 faces Painted is 4
Number of cubes with 2 faces Painted is given by sides which is exposed from two sides, out of 8 such edges 4 vertical edges will give us 6 cubes per edge and 4 edges from top surface will give us 5 such cubes from each edge and required number of cubes is 6 x 4 + 4 x 5 = 44.
Number of cubes with 1 face Painted is given by faces which is exposed from one sides four vertical faces will give us 6 x 5 = 30 cubes per face and top face will give us 5 x 5 = 25 and required number of cubes is 30 x 4 + 25 x 1 = 145
Number of cubes with 0 face Painted is given by difference between total number of cubes - number of cubes with at least 1 face painted = 343 - 4 - 44 - 145 = 150
In other words number of cubes with 0 painted is 6 x 5 x 5 = 150
From the above explanation number of the cubes with 0 faces painted is 150.
From the above explanation number of the cubes with 2 faces painted is 44.

For least number of cuts 120 = 4 x 5 x 6 i.e number of cuts must be 3, 4 and 5 in three planes in this case number of cubes on a face is either 6 x 5 = 30 or 6 x 4 = 24 or 4 x 5 = 20 cubes . And number of cuboids on an edge is 4 or 5 or 6
To satisfy this case all the cuboids on the edges and corners must have more than one colour on them. And in that case opposite face must have painted in the same colour.
In that case number of cuboids with 3 colours on them = 8
In that case number of cuboids with 2 colours on them = 4 x (2 + 3 + 4 ) = 36
Hence number of cuboids with at least 1 colour on them is 120 - 36 - 8 = 76

Since total number of cubes is hence in the formula we will substitute n = 6
Number of the cubes with 0 faces painted is (6 - 2)³ = 4³ = 64

Number of cubes with three coloured face on the top side = 4
Number of cubes with three coloured face on the 2^nd from top side = 4
Number of cubes with three coloured face on the bottom side = 12
Total number of such cubes = 12 + 8 = 20

Here on each face 6 x 6 = 36 cubes that are painted with one colour.
None of the cubes can be painted in four faces.

From top face (out of 3 x 3 square face) only one cubes is with one face painted.
From 4 vertical faces each face will give us 6 cubes hence total number of cubes from vertical faces is 6 x 4 = 24.
From bottom face we will get 3 x 3 = 9 cubes
So total number of cubes with one face painted is 1 + 24 + 9 = 34