Page 254 - Services Selection Board (SSB) Interviews

250 Cubes
32. (c) We will evaluate the value of ‘K’ in each and Case (v): In this case number of cubes is given by 25
every case: cubes from 1 face and 5 cubes from common edge
Case (i): In this case number of cubes is all 8 corner so number of cubes is 25 × 2 + 5 = 55
ones so number of cubes is 8 Hence option (a) gives the all possible value of ‘K’
Case (ii): In this case number of cubes is 4 corner 35. (c) From the solution of previous question required
ones so number of cubes is 4 number of cubes is 3 × 50 = 150 or 3 × 55 = 165
Case (iii): In this case number of cubes is 4 corner Solution for 36–40:
ones so number of cubes is 4 Here on each face 6 × 6 = 36 cubes that are painted with
Case (iv): In this case number of cubes is 4 corner one colour.
ones so number of cubes is 4 36. (b) Case (i): When red and blue are adjacent to each

Case (v): In this case number of cubes is 2 corner other then from one face we will get 6 × 6 = 36
ones so number of cubes is 2 cubes but out of them 6 cubes from common edge
Hence option (C) gives the all possible value of ‘K’ is common so number of cubes are 2 × 6 × 6 – 6 =
66
33. (b) We will evaluate the value of ‘K’ in each and Case (ii): When red and blue are opposite to each
every case: other then required number of cubes is 2 × 6 × 6 =
Case (i): In this case number of cubes is given by 4 72
common edges so number of cubes is 7 × 4 = 28 37. (a) Case (i): when these three colour are adjacent to
Case (ii): In this case number of cubes is given by each other then from one face we will get 6 × 6 =
4 common edges, out of these 4 edges there are 2 36 cubes but out of them 6 × 3 = 18 cubes from
corner cubes common with these 4 edges so number common edge is common so number of cubes are
of cubes is 7 × 4 – 2 = 26 3 × 6 × 6 – 6 × 3 = 90
Case (iii): In this case number of cubes is given by 2 Case (ii): When red and blue are opposite to each
common edges so number of cubes is 7 × 2 = 14. other (or any two of the given three) then required
Case (iv): In this case number of cubes is given by number of cubes is 3 × 6 × 6 – 2 × 6 = 96
4 common edges, out of these 4 edges there are 2 38. (c) Case (i): When red and blue are opposite to each
corner cubes common with these 4 edges so number other then from one face we will get 6 × 6 = 36
of cubes is 7 × 4 – 2 = 26 cubes but out of them 2 × 6 cubes from common
Case (v): In this case number of cubes is given by edge with green painted face is common so number
3 common edges, out of these 3 edges there are 2 of cubes are 2 × 6 × 6 – 2 × 6 = 60
corner cubes common with these 4 edges so number Case (ii): When red and blue are adjacent to each
of cubes is 7 × 3 – 3 = 18 other then green is either adjacent to these or
st
Hence option (b) gives the all possible value of ‘K’ opposite to any one of red or blue, in 1 condition
number of cubes is 2 × 6 × 6 – 2 × 6 – 11 = 55
34. (a) We will evaluate the value of ‘K’ in each and cubes or in 2 condition 2 × 6 × 6 – 6 – 6 = 60,
nd
every case: required number of cubes is 55 or 60
Case (i): In this case number of cubes is given by 25 39. (b) From solution of previous questions statements
cubes from 1 face so number of cubes is 25 × 2 = (ii) and (iii) are correct.
50 40. (d) None of the cubes can be painted in four faces.
Case (ii): In this case number of cubes is given by 25 Solution for 41–45:
cubes from 1 face so number of cubes is 25 × 2 = Consider the 1 step, initial number of cubes N after
st
3
50
st
removal of 1 set of coloured cubes number of cubes left
Case (iii): In this case number of cubes is given by 25 out is (N – 1) hence number of cubes removed in 1 step
3
st
cubes from 1 face and 5 cubes from common edge (i.e with colour 1) is
so number of cubes is 25 × 2 + 5 = 55 N – (N – 1) = 3N – 3N +1
2
3
3
Case (iv): In this case number of cubes is given by 25 Similarly number of cubes removed in 2 step (i.e with
nd
cubes from 1 face and 5 cubes from common edge colour 2) is
so number of cubes is 25 × 2 + 5 = 55