Cubes, Games, and Tournaments: Aptitude Formulas Derived

Cube problems, knockout brackets, and round-robin schedules each test a single logical invariant. Find the invariant, derive the formula once, and every variation in the question set becomes arithmetic.

This article covers three problem families: cube cutting, painted cubes, and tournament formats. All three appear on the TCS NQT, AMCAT Quantitative Aptitude, Wipro NLTH, and CoCubes tests. All formulas are derived from scratch rather than stated as rules to memorise.

What Aptitude Tests Actually Test With Cubes

A cube has 6 square faces, 12 edges, and 8 corners. That much is geometry. Aptitude questions extend this into three problem types:

• Cube cutting: the number of smaller pieces that result from a given count of cuts.
• Painted cubes: after painting all six faces and cutting, the count of small pieces with 0, 1, 2, or 3 painted faces.
• Hollow cubes: when the inner core is absent, the count of unit cubes remaining in the shell.

The painted-cube family appears most frequently in placement tests. Understanding the structure makes all three types accessible without separate memorisation.

Cube-Cutting: Deriving the Formula

Start with a single straight cut through a cube: 1 cut along one dimension produces 2 slices. Two cuts produce 3 slices. In general:

• k cuts along one dimension → k+1 slices along that dimension.

A cube has three independent dimensions (length, width, height). Applying cuts along all three:

• p cuts along length → p+1 slices
• q cuts along width → q+1 slices
• r cuts along height → r+1 slices
• Total small cubes = (p+1)(q+1)(r+1)

For equal cuts in all three directions (p = q = r = k):

• Total small cubes = (k+1)³

Worked Example — Cube Cutting

• Q1: A cube is cut into 4 equal parts along each dimension. Find the number of smaller cubes.

• Answer: 4 equal parts means 3 cuts per dimension (4-1=3). So (3+1)³ = 4³ = 64 smaller cubes.

• Q2: A cube is cut into 125 equal smaller cubes. Find the number of cuts per dimension.

• Answer: 125 = 5³, so there are 5 slices per dimension, which means 4 cuts per dimension.

Painted Cubes: Four Categories

Paint all six faces of a large cube, then cut it into an n×n×n grid of identical smaller cubes. Each smaller cube can have 0, 1, 2, or 3 painted faces depending on its position.

The positions map directly to cube geometry:

Position	Number of painted faces	Count
Corner	3 (sits at intersection of 3 faces)	8
Edge (non-corner)	2 (sits on intersection of 2 faces)	12(n-2)
Face centre (non-edge)	1 (sits on exactly 1 face)	6(n-2)²
Interior	0 (completely inside)	(n-2)³

Derivation

A cube has 8 corners. Each corner sits at the intersection of 3 faces, so there are always 8 three-painted-face pieces regardless of n.

Each edge has (n-2) non-corner pieces. There are 12 edges, giving 12(n-2) two-painted-face pieces.

Each face has (n-2)² non-edge pieces. There are 6 faces, giving 6(n-2)² one-painted-face pieces.

The fully interior core is a smaller cube of side (n-2), giving (n-2)³ unpainted pieces.

Verification: sum all four to confirm they equal n³: 8 + 12(n-2) + 6(n-2)² + (n-2)³ = [(n-2) + 2]³ = n³. This follows from the binomial expansion of (m+2)³ with m = n-2.

For this to give non-negative counts, n must be at least 2. For n=2, only corner pieces exist (all 8 pieces have 3 painted faces; the edge, face, and interior counts reduce to 0). For n=1, the cube is not cut at all; it is itself the piece, with all 6 faces painted.

Worked Example — 3×3×3 Painted Cube

Cut a fully-painted cube into 3×3×3 = 27 smaller cubes (n=3):

• 3 faces painted: 8
• 2 faces painted: 12(3-2) = 12(1) = 12
• 1 face painted: 6(3-2)² = 6(1) = 6
• 0 faces painted: (3-2)³ = 1
• Total: 8 + 12 + 6 + 1 = 27 ✓

Worked Example — 4×4×4 Painted Cube

Cut a fully-painted cube into 4×4×4 = 64 smaller cubes (n=4):

• 3 faces painted: 8
• 2 faces painted: 12(4-2) = 12(2) = 24
• 1 face painted: 6(4-2)² = 6(4) = 24
• 0 faces painted: (4-2)³ = 8
• Total: 8 + 24 + 24 + 8 = 64 ✓

Hollow Cubes

A hollow cube is a shell with the inner core removed. Starting from an n×n×n arrangement, the interior is (n-2)³ unit cubes. The shell count is:

• Shell = n³ - (n-2)³

For n=4: 64 - 8 = 56 unit cubes in the shell.

For n=3: 27 - 1 = 26 unit cubes in the shell (only the single centre piece is missing).

Note: hollow-cube questions occasionally specify that only some faces are open. In those cases, apply the face-counting logic per open face rather than the simple subtraction formula.

Games and Tournaments

Tournament problems test logical reasoning rather than calculation. The key is identifying the invariant that constrains the answer, not reverse-engineering the bracket.

Knockout Tournaments

In a knockout (single-elimination) tournament, each match eliminates exactly one team or player. To reduce n participants to a single champion, exactly n-1 participants must be eliminated. Each match eliminates one participant, so:

• Total matches in a knockout tournament = n-1

For 16 teams: 16 - 1 = 15 matches. For 64 teams: 64 - 1 = 63 matches.

This result is independent of how the bracket is drawn, whether byes are used, or whether matches can end in a draw, as long as each match produces exactly one loser.

Worked Examples — Knockout

• Q3: A knockout cricket tournament has 32 teams. Find the total match count.

• Answer: 32 - 1 = 31 matches.

• Q4: A knockout tournament ends after 27 matches. Find the number of teams that participated.

• Answer: 27 matches eliminate 27 teams, leaving 1 winner. So 27 + 1 = 28 teams participated.

Round-Robin Tournaments

In a round-robin tournament, every participant plays every other participant exactly once. The number of matches equals the number of distinct pairs:

• Total matches = n(n-1)/2

For 4 teams: 4 × 3 / 2 = 6 matches. For 8 teams: 8 × 7 / 2 = 28 matches. For 10 teams: 10 × 9 / 2 = 45 matches.

This is the combination formula C(n, 2): choosing 2 participants from n to form a match.

Worked Examples — Round-Robin

• Q5: 6 players enter a round-robin chess tournament. Find the total game count.

• Answer: 6 × (6-1) / 2 = 6 × 5 / 2 = 15 games.

• Q6: A round-robin tournament produces 21 total matches. Find the number of teams.

• Answer: n(n-1)/2 = 21, so n(n-1) = 42. Since 7×6=42, the answer is n=7. Seven teams.

Hybrid and Points-Based Formats

Some questions combine knockout and round-robin stages, or ask about points accrued. The approach:

• Identify the format of each stage separately.
• For points questions, the question usually states the point allocation (e.g., 2 for win, 1 for draw, 0 for loss). Apply it directly to the match results given.

For additional problem variations, IndiaBix cube and cuboid problems covers the full range of question types that appear on Indian placement tests.

Practice Strategy for Test Day

Placement aptitude sections give 60 to 90 seconds per question. Cube and tournament problems reward preparation in three ways.

First, the formulas are compact. The four painted-cube counts, the knockout invariant, and the round-robin combination formula fit on a single sheet. Derive them once before the exam, not during it.

Second, the question variations are predictable. Painted cubes almost always ask for exactly one category (3-face, or 0-face), not all four. Knowing which formula to call without deriving it on the spot saves 30 seconds.

Third, these problems do not require approximation. Unlike time-speed-distance questions that occasionally need estimation under time pressure, cube and tournament answers are exact integers. Verify by checking the total sums back to n³ (for cubes) or n-1 / n(n-1)/2 (for tournaments).

The same aptitude papers that include cube problems also include clock problems and calendar problems. All three reward the derivation approach over rote memorisation. Coding and decoding questions appear in the same test sections and follow the same pattern-recognition structure.

The formula-derivation habit (understanding why the formula works rather than what the output is) builds the systematic thinking that AI tasks reward too. If placement prep is your current focus and you want to extend that reasoning into building AI-assisted tools, TinkerLLM starts at ₹299 and focuses on practical AI application rather than abstract certification.