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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.

  1. If number of cubes painted on exactly one face with colour 1 and colour 2 is 150, then how many cubes are painted with only colour 4?








  1. View Hint View Answer Discuss in Forum

    Consider the 1st step, initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N - 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is
    N3 - (N - 1)3 = 3N2 - 3N + 1
    Similarly number of cubes removed in 2nd step (i.e with colour 2) is
    Similarly number of cubes removed in 3rd step is (i.e with colour 3) and so on.
    = 3(N - 1)2 - 3(N - 1) + 1
    Number of cubes remaining after 1st step is (N - 1)3
    Number of cubes remaining after 2nd 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 1st 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.

    Correct Option: C

    Consider the 1st step, initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N - 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is
    N3 - (N - 1)3 = 3N2 - 3N + 1
    Similarly number of cubes removed in 2nd step (i.e with colour 2) is
    Similarly number of cubes removed in 3rd step is (i.e with colour 3) and so on.
    = 3(N - 1)2 - 3(N - 1) + 1
    Number of cubes remaining after 1st step is (N - 1)3
    Number of cubes remaining after 2nd 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 1st 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.

  1. Of all the removed cubes which one of the following could be the number of cubes with exactly 2 face painted after 3 steps?








  1. View Hint View Answer Discuss in Forum

    Consider the 1st step, initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N - 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is
    N3 - (N - 1)3 = 3N2 - 3N + 1
    Similarly number of cubes removed in 2nd step (i.e with colour 2) is
    Similarly number of cubes removed in 3rd step is (i.e with colour 3) and so on.
    = 3(N - 1)2 - 3(N - 1) + 1
    Number of cubes remaining after 1st step is (N - 1)3
    Number of cubes remaining after 2nd 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 2nd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 11
    And after 3rd 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.

    Correct Option: B

    Consider the 1st step, initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N - 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is
    N3 - (N - 1)3 = 3N2 - 3N + 1
    Similarly number of cubes removed in 2nd step (i.e with colour 2) is
    Similarly number of cubes removed in 3rd step is (i.e with colour 3) and so on.
    = 3(N - 1)2 - 3(N - 1) + 1
    Number of cubes remaining after 1st step is (N - 1)3
    Number of cubes remaining after 2nd 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 2nd step number of cubes with exactly 2 face painted is 5(N - 2) - 6 = 5N - 11
    And after 3rd 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.