IAS 2026 — CSAT Reasoning

The correct option is (a) - 16.

Solution

The problem states that after 3 cuts, the pieces are identical and unmoved. For a cube to be divided into identical pieces without moving them, the cuts must be made along the three axes (X, Y, and Z) or parallel to each other.

Step 1: Analyze the state after 3 cuts (x₃)
To get identical pieces after 3 cuts without moving them, there are two primary configurations:
1. Parallel cuts: All 3 cuts are in the same direction. This results in $3 + 1 = 4$3+1=4 identical rectangular slabs. (x₃ = 4)
2. Mutually perpendicular cuts: 1 cut along the X-axis, 1 along the Y-axis, and 1 along the Z-axis. This results in $2 \times 2 \times 2 = 8$2×2×2=8 identical smaller cubes. (x₃ = 8)

Step 2: Calculate possible values for x₄ (after the 4th cut)
The 4th cut is applied to the existing configuration.
Case A: If x₃ = 4 (4 slabs)
- If the 4th cut is parallel to the first three, it divides all 4 slabs, resulting in $4 + 1 = 5$4+1=5 pieces.
- If the 4th cut is perpendicular to the first three, it intersects all 4 slabs, resulting in $4 \times 2 = 8$4×2=8 pieces.
Case B: If x₃ = 8 (8 small cubes)
- If the 4th cut is parallel to any of the previous cuts (e.g., a second cut on the X-axis), it will pass through all existing pieces. Since we have a 2x2x2 grid, a new plane parallel to one axis will intersect 4 pieces. However, the question states "one single cut can be used to cut more than one object at a time." A single plane cut through the 2x2x2 arrangement will bisect 4 pieces, adding 4 new pieces. Total = $8 + 4 = 12$8+4=12 pieces.

Conclusion: The possible values for x₄ are 5, 8, and 12. Therefore, 16 is not possible.

Why the other options are incorrect

• Option (b) - 12: This is a possible value. If the first 3 cuts are mutually perpendicular (creating 8 pieces in a 2x2x2 grid), a 4th cut parallel to any of the existing planes will slice through 4 of the small cubes, adding 4 more pieces ($8 + 4 = 12$8+4=12).
• Option (c) - 8: This is a possible value. If the first 3 cuts are parallel (creating 4 slabs), a 4th cut perpendicular to them will bisect all 4 slabs, resulting in $4 \times 2 = 8$4×2=8 pieces.
• Option (d) - 5: This is a possible value. If all 4 cuts are made parallel to each other in the same direction, the number of pieces is simply n + 1, which is $4 + 1 = 5$4+1=5.

Key Concept

The maximum number of pieces produced by n cuts in a cube (where pieces are not rearranged) is determined by the distribution of cuts across the three axes ($p + q + r = n$p+q+r=n), but the constraint of "identical pieces after 3 cuts" limits the starting configuration for the 4th cut.