Quick Summary: 3D visualisation (dice, cube nets, painted cubes, cube counting, assembling and grouping of solids) gives 1 to 3 marks in GATE General Aptitude 2027 across every paper (DA, RA, CS, ME, EE, CE, EC). These questions look like they need imagination; they actually need four rules. This guide gives you the two-views rule for dice, the 11 nets of a cube and the skip-one rule for opposite faces, the 8 / 12(n − 2) / 6(n − 2)2 / (n − 2)3 painted-cube formulas, the layer method for stacked blocks, Euler’s formula, four diagrams, 8 solved previous-year and GATE-style questions, a 5-question practice set and a one-page rule sheet.
📋 Table of Contents
Why this topic matters in GATE GA
Every GATE paper now carries a Spatial Aptitude question or two, and the 3D ones are the ones students fear most. A die in two positions, a net with a shaded face, a painted cube cut into small cubes, a pile of blocks to count, a solid to identify from its views: none of these needs artistic talent. Each has a rule, and the rule is faster than imagination. My students who learnt the four rules here stopped skipping the question and started picking up the mark in under a minute.
The questions are usually 1-mark MCQs, occasionally 2-mark, sometimes NAT. GA is 10 questions for 15 marks (5 × 1-mark and 5 × 2-mark) with negative marking of 1/3 and 2/3 for MCQ only, so a wrong guess on a 1-mark dice question costs 1/3. Learning the rule converts a guess into a certain mark.
Here is roughly how much the cluster has contributed. Candidates targeting 15/15 in GA should finish any 3D question here in 60 to 90 seconds by rule, not by mental rotation.
| Year | Marks from this topic (approximate, across papers) | Typical question type |
|---|---|---|
| 2019 | 0 to 1 | Cube with coloured faces, which view is possible |
| 2020 | 1 to 2 | Net that folds into the given cube |
| 2021 | 1 to 3 | Painted cube cut into small cubes, pieces that assemble a solid |
| 2022 | 1 to 2 | Dice in two positions, opposite face |
| 2023 | 1 to 3 | Counting stacked cubes, surface area of a stack |
| 2024 | 1 to 2 | Solid from top, front and side views |
| 2025 | 1 to 3 | Cube net with pattern, grouping identical solids |
| 2026 | 1 to 2 | Painted cuboid, number of unpainted cubes |
What the official GA syllabus says
Dice, nets, painted cubes and block counting are the “3 dimensions” half of “patterns in 2 and 3 dimensions”, and the words “assembling and grouping” point directly at questions where pieces combine into a solid or identical solids are grouped. The previous post covered the 2D half, 2D Transformations, Rotation, Reflection, Mirror and Water Images, and the next post finishes the spatial section with Paper Folding, Cutting and Pattern Completion. The complete syllabus and weightage are in the GATE General Aptitude 2027 hub.
Dice and cubes: opposite faces
A cube has 6 faces, 12 edges and 8 vertices. The faces come in three opposite pairs, and almost every dice question is really asking “which face is opposite which”. Three facts settle it:
- Standard die: opposite faces add to 7, so the pairs are 1-6, 2-5 and 3-4. GATE will say if the die is standard; if two visible faces add to 7, the die is not standard, because two opposite faces can never be seen together.
- Adjacent faces are never opposite. If you can see faces 2, 3 and 6 together in one view, then 2 is not opposite 3 or 6, and 3 is not opposite 6. One view kills three pairs.
- The two-views rule: if two views share two common faces, the two remaining faces are opposite (the shared edge has two ends, and the third face at each end completes a corner; those two faces are always opposite). If the views share one common face in the same position, the die was rotated about that face: see which face moved where, and the four side faces form a ring whose opposite positions are opposite faces.
Solution: Both views contain the edge where 1 and 2 meet; the third faces at its two ends are 4 and 3, so 3 is opposite 4. Face 2 is adjacent to 1, 3 and 4, so its opposite is 5 or 6; the same holds for face 1. The two views alone cannot separate the last two pairs, so the question must intend a standard die, where opposite faces sum to 7.
Answer: 5 (standard die). The views alone prove only that 3 is opposite 4.
Open dice versus standard dice. An open (non-standard) die is one whose opposite faces do not sum to 7; only adjacency and the two-views rule work for it. The giveaway is a view showing two faces that sum to 7, such as 3 beside 4, which is impossible on a standard die.
Nets of a cube
A net is the cube unfolded flat. There are exactly 11 distinct nets of a cube (ignoring rotations and reflections). You do not need to memorise all eleven as pictures; you need the pattern of rows and the three quick rejections.
- 1-4-1 family (6 nets): a strip of four squares with one square attached above and one below. The classic cross is one of these.
- 2-3-1 family (3 nets): a strip of three with a domino attached on one side and a single square on the other.
- 2-2-2 (1 net): a staircase of three dominoes.
- 3-3 (1 net): two strips of three joined end to end with an overlap of one column.
- Instant rejections: a straight line of 5 or 6 squares; any 2 × 2 block (so any 2 × 3 rectangle); and a strip of four with both extra squares on the same side of the strip. These never fold into a cube because two squares would land on the same face or one face would be left uncovered.
Which faces become opposite. In any straight row of the net, squares separated by exactly one square are opposite (positions 1 and 3, 2 and 4 in a strip of four). In a 1-4-1 net the two attached squares are opposite each other. Squares that share an edge in the net share an edge in the cube, so they are never opposite.
How to check whether a net folds into a given cube. The cube in the question shows three faces meeting at a corner. (1) Find the opposite pairs in the net with the skip-one rule. (2) If any two of the three visible faces are an opposite pair, reject the net. (3) If two options survive, check the clockwise order of the three faces around the corner; the mirror-image arrangement is a different cube and is the usual trap.
Solution: Along the strip, skip one: P is opposite R and Q is opposite S. T and U are the two attached squares in a 1-4-1 net, so T is opposite U.
Answer: T is opposite U; Q is opposite S.
Solution: A view can only show three mutually adjacent faces, so any view containing an opposite pair is impossible. Opposite pairs: P-R, Q-S, T-U. Option (C) contains P and R.
Answer: (C).
Painted cube and cube counting
A large cube of side n is painted on all six faces, then cut into n3 unit cubes. Each unit cube is a corner, an edge piece, a face-centre piece or an interior piece, and the four counts are fixed by geometry:
| Painted faces | Where the cube sits | Count for side n | n = 3 | n = 4 | n = 5 |
|---|---|---|---|---|---|
| 3 | Corner | 8 (always) | 8 | 8 | 8 |
| 2 | Edge, not corner | 12(n − 2) | 12 | 24 | 36 |
| 1 | Face, not edge | 6(n − 2)2 | 6 | 24 | 54 |
| 0 | Interior | (n − 2)3 | 1 | 8 | 27 |
The four counts always add to n3, the check I ask my students to do every time: for n = 4, 8 + 24 + 24 + 8 = 64. The 12 is the number of edges, the 6 the number of faces, and (n − 2) is what remains of each edge after removing its two corners: “8 corners, 12 edges, 6 faces, 1 core”.
Cuboids. For an a × b × c block apply the same logic edge by edge: corners = 8; two faces = 4[(a − 2) + (b − 2) + (c − 2)]; one face = 2[(a − 2)(b − 2) + (b − 2)(c − 2) + (c − 2)(a − 2)]; unpainted = (a − 2)(b − 2)(c − 2). A dimension of 2 makes the unpainted count zero, a common trick question.
Partially painted cubes. If only some faces are painted, count the unit cubes touching no painted face. One face painted: unpainted = n3 − n2. Two opposite faces: n2(n − 2). Two adjacent faces: n(n − 1)2, because the n cubes along the shared edge are counted twice if you simply add.
Solution: n = 5. One face: 6(5 − 2)2 = 6 × 9 = 54. No face: (5 − 2)3 = 27. Check: 8 + 36 + 54 + 27 = 125 = 53.
Answer: 54 with one red face, 27 with none.
Counting cubes in a stacked figure. GATE draws a pile of unit cubes in isometric view and asks for the total, the minimum, or the number of hidden cubes. Use the layer method: count the top layer, then each layer below, remembering that every cube must be supported by a cube directly beneath it unless the question says otherwise, so no layer has fewer cubes than the one above. Assume hidden positions are filled when asked for “the number of cubes” and empty when asked for “the minimum number”.
Solution: Count faces by direction. Top and bottom: 6 each (one per column). Front and back: the columns have heights 3, 2, 1, so 6 each. Left and right: the rows have heights 3 and 2, so 5 each. Total = 6 + 6 + 6 + 6 + 5 + 5 = 34. For any stack with no overhangs, opposite views are equal, so count three views and double.
Answer: 34 square units.
Assembling, grouping and views of solids
The syllabus words “assembling and grouping” cover four question styles.
1. Which pieces combine to form the target solid? Pieces are shown and you must say which two make a given cube or cuboid. Do not rotate anything yet: count unit cubes. A 2 × 2 × 2 cube has 8, so pieces of 3 and 5 or 4 and 4 are candidates and 6 with 3 is not. Only then check whether the holes in one piece are the bumps of the other.
2. Counting faces, edges and vertices. For any convex polyhedron, V − E + F = 2 (Euler). Cube: 8 − 12 + 6 = 2. Triangular prism: 6, 9, 5. Square pyramid: 5, 8, 5. Tetrahedron: 4, 6, 4. Given two numbers, Euler gives the third. For a modified cube, recount: slicing off one corner adds a triangular face, replaces one vertex with three and adds three edges, so V = 10, E = 15, F = 7, and the formula still holds.
3. Identifying the solid from top, front and side views. The top view gives the footprint, the front view the maximum height in each column (left to right), the side view the maximum height in each row (front to back). Write the footprint as a grid of heights: column maxima must match the front view, row maxima the side view. The fewest cubes consistent with all views is the “minimum”, the most is the “maximum”. With options, test each against the views and eliminate.
4. Grouping identical 3D shapes. Solids are shown in different orientations and you must say which are identical. Rotation keeps a solid identical; reflection does not, just as a left shoe is not a right shoe. Pick a distinctive feature (a notch, a protruding cube, a shaded face) and check whether its position relative to two other features runs clockwise or anticlockwise. Opposite handedness means a mirror image, the 3D version of the rule in the 2D transformations post.
Solution: Grid of heights. The only cell that can be 3 in both its column (left) and its row (front) is left-front = 3. The back row needs a 2 and the right column allows at most 1, so left-back = 2. The right column needs a 1: right-front = 1, right-back empty. Minimum = 3 + 2 + 1 = 6.
Answer: 6 cubes. (Maximum is 7, with right-back = 1.)
Shortcuts and traps
❌ Treating adjacent squares in a net as opposite faces.
❌ Swapping the 12 (edges) and the 6 (faces) in the painted-cube formulas.
❌ Applying the cube formulas to a cuboid without separate (a − 2), (b − 2), (c − 2) terms.
❌ Counting only the visible cubes in a stack and forgetting the hidden supporting cubes.
❌ Accepting a 2 × 2 block or a strip of five as a valid net.
❌ Grouping a mirror image with its original as “identical”.
Solved previous-year and GATE-style questions
Years are given only where I am confident the question appeared; the rest are built to the same pattern. Many more are solved in the PiyushAI Aptitude Test Series with video explanations.
Solution: Two painted faces means an edge cube that is not a corner. Each of the 12 edges has 4 − 2 = 2 such cubes. Count = 12 × 2 = 24. Check: 8 + 24 + 6 × 4 + 8 = 8 + 24 + 24 + 8 = 64.
Answer: 24.
Solution: Face 1 stays in front, so the die was rotated about the front-back axis. In view 1 the ring around face 1 has top 3, right 2. In view 2 face 3 has moved from top to right, a quarter turn, so the face that was on the left has moved to the top: that is 5. Ring: top 3, right 2, bottom X, left 5. Opposite positions in the ring are opposite faces, so 2 is opposite 5. In position 2 the top is 5, so the bottom is 2.
Answer: (A) 2.
Solution: 1-4-1 net. Along the strip: circle opposite square symbol, triangle opposite star. The two attached squares, cross and dot, are opposite each other.
Answer: (B) Cross.
Solution: Unpainted = (3 − 2)(4 − 2)(5 − 2) = 1 × 2 × 3 = 6. One face: 2[(1)(2) + (2)(3) + (3)(1)] = 2[2 + 6 + 3] = 22. Check the whole: corners 8; two faces 4[1 + 2 + 3] = 24; total 8 + 24 + 22 + 6 = 60 = 3 × 4 × 5.
Answer: 6 unpainted, 22 with exactly one painted face.
Solution: Total = 2 + 1 + 1 + 3 + 2 + 0 + 1 + 0 + 0 = 10. Front view = column maxima: max(2, 3, 1) = 3, max(1, 2, 0) = 2, max(1, 0, 0) = 1.
Answer: 10 cubes; the front view is three columns of heights 3, 2, 1.
Solution: Euler: V − E + F = 2, so F = 2 − 8 + 12 = 6. A cube (or any cuboid) has 8 vertices, 12 edges, 6 faces. The hexagonal pyramid has 7 vertices and 12 edges; the octahedron has 6 vertices, 12 edges, 8 faces.
Answer: (B) 6 faces, a cube.
Solution: Count first: 8 = 3 + 5 or 4 + 4, so (P, R) or (Q, S). A straight line of 4 cannot fit inside a 2 × 2 × 2 cube at all, so (Q, S) fails. R fills the bottom layer and one cell of the top layer; the L of 3 fills the other three cells of the top layer.
Answer: P and R.
Solution: A unit cube is painted if it touches either painted face. Touching the first face: 9 cubes. Touching the second (adjacent) face: 9 cubes, but the 3 cubes along the shared edge are counted twice. Painted = 9 + 9 − 3 = 15. Unpainted = 27 − 15 = 12. Formula check: n(n − 1)2 = 3 × 4 = 12.
Answer: 12.
Watch: Mission GATE 2027 Aptitude Test Series, Score 15/15 (PYQ Analysis)
In this session I walk through how the 15-mark General Aptitude section is set for every GATE branch, analyse previous-year questions, and show exactly how the Best GATE Aptitude Test Series 2027 is structured to get you to 15/15.
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General Aptitude is 15 marks in every GATE paper and the easiest 15 marks to lose to silly mistakes. The PiyushAI Aptitude Test Series by Piyush Wairale (IIT Madras) gives you topic-wise tests on exactly this chapter, full-length GA mocks in the real GATE interface, detailed video and text solutions, and All-India rank analysis.
Practice set
Attempt these in 8 minutes before opening the key.
- A cube of side 6 cm is painted on all faces and cut into 1 cm cubes. How many small cubes have exactly one painted face? (A) 96 (B) 64 (C) 48 (D) 150
- Two views of a die show the faces {1, 2, 3} and {2, 3, 6} respectively (each view shows three faces meeting at a corner). Which pair of faces is definitely opposite? (A) 1 and 6 (B) 2 and 5 (C) 3 and 4 (D) 2 and 6
- Which of the following arrangements of six squares is a valid net of a cube? (A) Six squares in a straight line (B) A 2 × 3 rectangle (C) A strip of four with one square above the first and one square below the fourth (D) A strip of four with one square above the first and one square above the second
- A pile of unit cubes has a 2 × 3 base. Seen from the front, the three columns have heights 2, 3, 1; seen from the right side, the two rows have heights 3, 2. What is the maximum number of cubes in the pile? (A) 6 (B) 9 (C) 11 (D) 12
- A convex solid has 6 faces and 9 edges. The number of vertices is ___. (A) 5 (B) 6 (C) 7 (D) 8
Show answer key
- (A). 6(n − 2)2 = 6 × 16 = 96.
- (A). The views share two faces (2 and 3), so the two remaining faces, 1 and 6, are opposite each other. The other pairs cannot be fixed from these views alone.
- (C). It is a 1-4-1 net. A straight line of six and a 2 × 3 rectangle (which contains 2 × 2 blocks) never fold; option (D) also contains a 2 × 2 block, because the two extra squares sit side by side above two adjacent strip squares.
- (C). Grid 2 rows × 3 columns; each cell takes the smaller of its column max (2, 3, 1) and row max (3, 2). Row 1: 2, 3, 1; row 2: 2, 2, 1. Maximum = 6 + 5 = 11.
- (A). Euler: V = 2 + E − F = 2 + 9 − 6 = 5. The solid is a triangular bipyramid (two tetrahedra glued on a face): 6 triangular faces, 9 edges, 5 vertices. Do not confuse it with the triangular prism, which has 5 faces, 9 edges and 6 vertices.
One-page rule sheet
2. Standard die: opposite faces sum to 7 (1-6, 2-5, 3-4). Two faces summing to 7 seen together means a non-standard die.
3. Faces seen together are adjacent, never opposite. Eliminate before rotating.
4. Two views with two common faces: the two remaining faces are opposite. One common face in the same position: walk the ring of four side faces; opposite positions are opposite faces.
5. A cube has exactly 11 nets: six 1-4-1, three 2-3-1, one 2-2-2, one 3-3.
6. Invalid nets: a line of 5 or 6, any 2 × 2 block, both extras on the same side of a strip of four.
7. In a net, skip-one in a straight row = opposite; the two squares attached to a strip of four are opposite; edge-sharing squares are adjacent.
8. Painted cube of side n: 8, 12(n − 2), 6(n − 2)2, (n − 2)3 for 3, 2, 1, 0 painted faces; total n3; at least one face = n3 − (n − 2)3.
9. Painted cuboid a × b × c: unpainted = (a − 2)(b − 2)(c − 2); replace each (n − 2) in the cube formulas by the matching edge term.
10. Stacked cubes: grid of heights; total = sum; front view = column maxima; side view = row maxima; every cube needs support.
11. Euler: V − E + F = 2. Cube 8-12-6, tetrahedron 4-6-4, square pyramid 5-8-5, triangular prism 6-9-5.
12. Assembling: match unit-cube counts first, then shapes. Grouping: rotations are identical, mirror images are not.
Common mistakes
❌ Applying the sum-to-7 rule to an open die.
❌ Reading opposite faces from a net by looking at squares that share an edge.
❌ Writing 12(n − 1) or 12n for edge cubes; (n − 2) removes the two corners from each edge.
❌ Reporting visible cubes in a stack instead of the total including hidden supports.
❌ Choosing the mirror-image option in “which cube is formed from this net” questions.
❌ Leaving a NAT question blank: NAT has no negative marking, so always compute and enter a value.
FAQs
How many 3D visualisation questions appear in GATE GA?
Usually 1 question, sometimes 2, across dice, nets, painted cubes, block counting and views of solids. The cluster contributes 1 to 3 marks a year, most often as a 1-mark MCQ.
Do I need to memorise all 11 nets of a cube?
No. Remember the four families (1-4-1, 2-3-1, 2-2-2, 3-3) and the three instant rejections (a line of five or six, a 2 × 2 block, both extras on the same side of a strip of four). Then use the skip-one rule to find opposite faces in any net GATE gives you.
What is the fastest way to solve a painted-cube question?
Write 8, 12(n − 2), 6(n − 2)2, (n − 2)3 for three, two, one and zero painted faces, add them to confirm n3, and read off the one the question asks for. For a cuboid replace n − 2 by the three separate terms (a − 2), (b − 2), (c − 2).
Is the PiyushAI Aptitude Test Series enough for GA?
Yes. The Best GATE Aptitude Test Series 2027 has dedicated tests on puzzles, arrangements, blood relations and direction sense, tests on every other GA chapter, full-length GA mocks in the GATE interface, solved PYQs from 2010 to 2026 and All-India rank analysis. With the free guides on this site it covers the whole GA section.
How do I count cubes when some are hidden in the figure?
Count layer by layer from the top, and assume every cube is supported by a cube directly below it. The hidden cubes are the supporting ones at the back and bottom; if the question asks for the minimum number, fill only the positions that must be occupied for support, and if it asks for the maximum, fill every hidden position consistent with the views.
Stop losing easy marks. Practise this topic under exam pressure.
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