Find out the next series from the Answer Figure that follows the sequence of the Question Figure:

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.

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.
The required number of cubes must be equal to difference between two positive integer
Since 64 - 27 = 37
125 - 27 = 98
125 - 64 = 61

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.
Total number of Cubes left after 7^thstep in (N - 7)³ in the form of (N - 7) x (N - 7) x (N - 7) cubes.
And out of these number of cubes whose two sides are painted is given by three edges with each edge has (N - 8) so total number of cubes is 3 x (N - 8)
From the given information 3(N - 8) = 21 or N = 15
Number of cubes removed in 3^rd step (i.e with colour 3) is = 3(N - 2)² - 3(N - 2) + 1 = 469

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.
From the given condition
Number of cubes removed in 3^rd step (i.e with colour 3) is = 3(N - 2)² - 3(N - 2) + 1 = 217 hence N = 11 So number of cubes with colour 5 is = 3(N - 4)² - 3(N - 4) + 1 = 127

The circle at the corner of the square moves clock wise, where as the cross inside the square moves anti clock wise.

In one step, the arrow and the small line segment turn to other side of the more line and in the next figure.

Every time the inner figure enlarges and become the outer figure and the row inner figure changes to new element.

The number of projections doubles in each step.

The tail of the figured moves 45 degree clock wise.