Cubes and Dice problems

Consider the 1^st step, initial number of cubes N³ after removal of 1^st set of coloured cubes number of cubes left out is (N - 1)³ hence number of cubes removed in 1^st step (i.e with colour 1) is
N³ - (N - 1)³ = 3N² - 3N + 1
Similarly number of cubes removed in 2^nd step (i.e with colour 2) is
Similarly number of cubes removed in 3^rd step is (i.e with colour 3) and so on.
= 3(N - 1)² - 3(N - 1) + 1
Number of cubes remaining after 1^st step is (N - 1)³
Number of cubes remaining after 2^nd step is (N - 2)³ and so on.
Number of cubes with only face is painted with colour 1 is 3(N - 2)(N - 1) = 3N² - 9N + 6
Number of cubes with only face is painted with colour 2 is 3 (N - 3)(N - 2) = 3N² - 15 N + 18
From the given condition (3N² - 9N + 6) + (3N² - 15 N + 18) = 6N² - 24N + 24 = 150 from this we will get N = 7.
Number of cubes left after step 3 is 4 x 4 x 4 = 64
When all the exposed faces are painted with colour 4 then number of cubes with only one face painted is 3 x 2 x 3 = 18
From the observation for 1^st step we have seen that number of cubes is 3(N - 2)(N - 1) or in other words 3 times the product of 2 consecutive integer that is satisfied only by 18 which is 3 times of 2 x 3.

Consider the 1^st step, initial number of cubes N³ after removal of 1^st set of coloured cubes number of cubes left out is (N - 1)³ hence number of cubes removed in 1^st step (i.e with colour 1) is
N³ - (N - 1)³ = 3N² - 3N + 1
Similarly number of cubes removed in 2^nd step (i.e with colour 2) is
Similarly number of cubes removed in 3^rd step is (i.e with colour 3) and so on.
= 3(N - 1)² - 3(N - 1) + 1
Number of cubes remaining after 1^st step is (N - 1)³
Number of cubes remaining after 2^nd step is (N - 2)³ and so on.
After step 1 number of cubes with exactly 2 face painted is 4(N - 1) + (N - 2) = 5N - 6
Similarly after 2^nd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 11
And after 3^rd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 16
So total number of such cubes is 15N - 33 out of the given options only option B satisfy the given condition.

There is 6 columns with 2 cubes each.
Total boxes = 6 × 2 = 12

There are total 4 rows and 5 boxes in each row,
then 5 × 4 = 20 boxes.

4 columns with 1 cubes = 4 cubes
1 columns with 3 cubes = 3
1 columns with 4 cubes = 4
Total cubes = 4 + 3 + 4 = 11

From both the figures we find that numbers 1,2 and 4 dots appear adjacent to 6. thus, the number 5 dots will appear opposite to 6. Therefore when 6 is at the bottom, then 5 will be at the top.

as we can observe the position of 6 is same in both the figures and position of 4 moves from Right hand side to Left hand side (clock wise direction).

as both the figure contains one dot and 5 dots, fixing the 5 dots face and moving one dot face position to fig (i). we find 4 dots is opposite to 2 dots.

For the better understanding take a dice and apply the same steps which is mentioned above.

as 1, 3 5 and 6 are adjacent to 2 in all the three positions, if 3 is opposite to 5 that we came across from the figures. So it is obvious that one will be opposite to 6.

For the better understanding take a dice ans apply the same steps which is mentioned above.

as the dot face cannot be adjacent to both the dark shaded faces. Hence option c cannot be possible.