Cubes
Direction: 343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i.e., 7 x 7 x 7) and kept on the surface of a room, all the exposed surfaces( in this case there are 5) are painted.
 How many of the cubes have at most faces painted?

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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 at most 2 faces painted is 150 + 145 + 44 = 339.
Or else 343  4 = 339Correct Option: A
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 at most 2 faces painted is 150 + 145 + 44 = 339.
Or else 343  4 = 339
 How many of the cubes have 3 faces painted?

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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 3 faces painted is 4.Correct Option: D
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 3 faces painted is 4.
Direction: N ^ 3 number of cubes of similar size are arranged in the from of a bigger cube (N cubes on each side, i. e., N x N x N ) and kept at the corner of a room, all the exposed surfaces are painted with colour 1, then all the coloured smaller cubes are removed and all the exposed surfaces are painted with colour 2, then all the coloured smaller cubes are removed and all the exposed surfaces are painted with colour 3, this process is repeated 'K' number of times.
 If number of cubes painted with colour 3 is 217 then how many cubes are painted with colour5

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Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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 = 127Correct Option: B
Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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
 If after 7^{th} step number of cubes painted in exactly 2 faces with colour 7 is 21, then what is the number of cubes removed in 3^{rd} step.

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Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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^{th}step 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 = 469Correct Option: A
Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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^{th}step 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
 Which of the following can be the number of cubes removed from the original N^{3} number of cubes.
(i) 37
(ii) 61
(iii) 98

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Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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 = 61Correct Option: C
Consider the 1^{st} step, initial number of cubes N^{3} 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
N^{3}  (N  1)^{3} = 3N^{2}  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