Page 250 - Services Selection Board (SSB) Interviews

246 Cubes

Answer with Solution

Concept Applicator

1. (b) If total number of cut is 10 then minimum number Solution for 11–15:
of pieces is 11 when cut is made in one plane only. Out of 6 faces of 3 faces are exposed and those were
2. (a) If total number of cut is 10 then for maximum painted.
number of pieces these cuts have to be well Number of vertices with three faces exposed (Painted) is 1
distributed in three planes. For 10 cuts, 3, 3 and 4 is
the distribution of cuts. Number of vertices with 2 faces exposed (Painted) is 3
Hence total number of pieces is Number of vertices with 1 faces exposed (Painted) is 3
(3 + 1)(3 + 1)(4 + 1) = 4 × 4 × 5 = 80 Number of vertices with 0 faces exposed (Painted) is 1
3. (b) For maximum number of pieces cuts has to be 6, Number of sides with 2 sides exposed (Painted) is 3
7 and 7 and maximum number of pieces is (6 + 1) Number of sides with 1 sides exposed (Painted) is 6
(7 + 1)(7 + 1) = 7 × 8 × 8 = 448. Number of sides with no sides exposed (Painted) is 3
Minimum number of pieces is 20 + 1 = 21. From the above observation
Hence required ratio is 448: 21 Number of cubes with 3 faces Painted is 1
4. (b) If 45 = 1 × 1 × 45 then we require only 44 cuts Number of cubes with 2 faces Painted is given by sides
in one plane. which is exposed from two sides and there are 3 such
If 1 × 3 × 15 then we require 2 cuts in one plane sides and from one side we will get 6 such cubes hence
and 14 cuts in other plane so total number of cuts is required number of cubes is 6 × 3 = 18
2 + 14 = 16. Number of cubes with 1 face Painted is given by faces
If 1 × 5 × 9 then we require 4 cuts in one plane and which is exposed from one sides and there are 3 such
8 cuts in other plane so total number of cuts is faces hence required number of cubes is 36 × 3 = 108
4 + 8 = 12 Number of cubes with 0 face Painted is given by difference
If 3 × 3 × 5 then we require 2 cuts in one plane, between total number of cubes – number of cubes with at
rd
nd
2 cuts in 2 plane and 4 cuts in 3 plane so total least 1 face painted = 343 – 1 – 18 – 108 = 216
number of cuts is 2 + 2 + 4 = 8. In other words number of cubes with 0 painted is (7 – 1)
3
5. (a) For maximum number of cuts it has to be in one = 216.
cut only, so number of cuts is 49
6. (c) For minimum number of cuts we will get 50 from 11. (d From the above explanation number of the cubes
with 0 faces painted is 216.
2 × 5 × 5 and cuts is 1 + 4 + 4 = 9
Solution for 7–10: 12. (b) From the above explanation number of the cubes
with 2 faces painted is 18.
Since total number of cubes is hence in the formula we 13. (c) From the above explanation number of the cubes
will substitute n = 6 with at most 2 faces painted is 216 + 108 + 18 =
7. (a) Number of the cubes with 0 faces painted is 342.
3
3
(6 – 2) = 4 = 64 Or else 343 -1 = 342
8. (c) Number of the cubes with 2 faces painted is 14. (a) From the above explanation number of the cubes
2
6(6 – 2) = 6 × 16 = 96 with at least 2 faces painted is 18 + 1 = 19.
9. (a) At most 2 faces painted means number of cubes 15. (d) From the above explanation number of the cubes
with 0 face painted + number of cubes with 1 face with 3 faces painted is 1.
painted + number of cubes with 2 face painted = 64 Solution for 16–20:
+ 48 + 96 = 208
10. (b) At least 2 faces painted means number of cubes Out of 6 faces of 4 faces are exposed and those were
with 2 face painted + number of cubes with 3 face painted.
painted = 96 + 8 = 104.