Page 255 - Services Selection Board (SSB) Interviews

Cubes 251
rd
st
Similarly number of cubes removed in 3 step is (i.e with From the observation for 1 step we have seen that
colour 3) and so on. number of cubes is 3(N – 2)(N – 1) or in other words
2
= 3(N – 1) – 3(N – 1) + 1 3 times the product of 2 consecutive integer that is
st
Number of cubes remaining after 1 step is (N – 1) 3 satisfied only by 18 which is 3 times of 2 × 3.
3
Number of cubes remaining after 2 nd step is (N – 2) and 45. (b) After step 1 number of cubes with exactly 2 face
so on. painted is 4(N – 1) + (N – 2) = 5N – 6
nd
41. (b) From the given condition Similarly after 2 step number of cubes with exactly
2 face painted is 5(N – 1) – 6 = 5N – 11
rd
Number of cubes removed in 3 step (i.e with colour rd
2
3) is = 3(N – 2) – 3(N – 2) +1 = 217 hence N = 11 And after 3 step number of cubes with exactly 2
face painted is 5(N – 2) – 6 = 5N – 16
2
So number of cubes with colour 5 is = 3(N – 4) –
3(N – 4) + 1 = 127 So total number of such cubes is 15N – 33 out of
the given options only option B satisfy the given
th
42. (a) Total number of Cubes left after 7 step in (N – condition.
3
7) in the form of (N – 7) × (N – 7) × (N – 7) cubes.
And out of these number of cubes whose two sides Solution for 46–50:
are painted is given by three edges with each edge If C3 means number of cuboids with 3 face painted, C2
has (N – 8) so total number of cubes is 3 × (N – 8) means number of cubes with 2 face painted and so on:
From the given information 3(N – 8) = 21 or N = 15 Case P = Q = R = C3 C2 C1 C0
rd
Number of cubes removed in 3 step (i.e with colour (i) 1 × 1 × 105 0 0 104 2 102 0 0
2
3) is = 3(N – 2) – 3(N – 2) +1 = 469 (ii) 1 × 3 × 35 0 2 34 4 68 33 0
43. (c) The required number of cubes must be equal to (iii) 1 × 5 × 21 0 4 20 4 44 57 0
difference between two positive integer (iv) 1 × 7 × 15 0 6 14 4 36 65 0
Since 64 – 27 = 37 (v) 3 × 5 × 7 2 4 6 8 36 46 15
125 – 27 = 98 46. (b) From the table condition is same as that of case
125 – 64 = 61 (iii) then number of cubes painted with 1 face is 57
44. (c) Number of cubes with only face is painted with 47. (c) From the table condition is same as that of case
2
colour 1 is 3(N – 2)(N – 1) = 3N – 9N + 6 (ii), (iii) or (iv) then number of cubes painted with 1
Number of cubes with only face is painted with face is either 33 or 57 or 65
2
colour 2 is 3 (N – 3)(N – 2) = 3N – 15N +18 48. (a) From the table condition is same as that of case
2
2
From the given condition (3N – 9N + 6) + (3N – (iv) or (v) then number of cubes painted with 1 face
2
15N +18) = 6N – 24N + 24 = 150 from this we is either 46 or 65
will get N = 7. 49. (c) From the table condition is same as that of case
Number of cubes left after step 3 is 4 × 4 × 4 = 64 (iii) then number of cubes painted with 1 face is
either 46 or 65
When all the exposed faces are painted with colour
4 then number of cubes with only one face painted 50. (d) From the table we can say that no such exists so
is 3 × 2 × 3 = 18 data is inconsistence.