Non-verbal reasoning and dice together are 17 of the 692 reasoning questions the Punjab Police SI has asked — the smallest topic in the section, and worth exactly the hour it takes to read this page. Eight are dice, seven are figure series, and two are one-off puzzles about mirror images and counting squares. The dice are the valuable half, because they have a method that always works and never varies. The figure series are mostly settled by counting rather than by looking. If you are short of time, learn the dice rule and skim the rest. Eight marks for one reliable technique is a good trade; seven marks for an eye for rotation is not.
The teaching part above and below — the explanations, the figures and the worked examples on screen — is about 8 minutes. The collapsed question banks hold 17 solved past questions in all; working through every one of them takes roughly 48 minutes. Do that across several sittings, not in one go.
Part 1 · Dice — 8 questions, one method
List what you can see together, then eliminate
A cube has six faces in three opposite pairs, and the question shows you two views of the same die and asks which face is opposite which. The method never changes. Any two faces visible in the same view are adjacent, so they cannot be opposite each other. Write down every pair you can see together across both views. Each face has four neighbours and exactly one opposite, so once four faces are ruled out as neighbours of a given face, the fifth is its opposite by elimination. It helps to find a face that appears in both views and use it as the pivot: it is adjacent to everything shown beside it in either view, which usually rules out four faces at once. If the die is a standard one, opposite faces add to 7, and the question will say so when that applies.
Six faces, three pairs. Once you have named one pair, the remaining four faces form the other two pairs, and a single further adjacency usually fixes both.
Every other dice question (7)Tap to open
Six letters A, B, C, D, E and F are written on the different faces of a die. Two positions of this die are shown in the figure : in the first position the top face carries E, the front face carries F and the right face carries A; in the second position the top face carries F, the front face carries D and the right face carries B. Find the letter on the face opposite B. (SI 20 Jun 2023)
- 1F
- 2E
- 3C
- 4A
Show answer
Answer: 2 — E
F is visible in both views, so F touches E and A (view 1) and D and B (view 2); a face cannot be adjacent to its own opposite, so F is opposite C, and the remaining four letters A, B, D, E form the other two pairs. Now rotate view 1 backwards about the left-right axis so that F moves from the front to the top: A stays on the right, and the face opposite E comes to the front. A further quarter-turn about the vertical axis that brings the left face to the front leaves top = F, front = opposite(A), right = opposite(E). View 2 shows top F, front D, right B, so opposite(A) = D and opposite(E) = B. Hence B lies opposite E. The trap is to pair B with A or D merely because they share a view, but letters seen together are adjacent, never opposite.
Six letters J, K, L, M, N and O are written on the different faces of a dice. Two positions of this dice are shown in the figure below. Find the letter on the face opposite to K. (Figure : two corner views of the same cube. View 1 shows M on the top face, J on the front-left face and L on the front-right face. View 2 shows O on the top face, K on the front-left face and M on the front-right face.) (SI 1 Jul 2023)
- 1O
- 2L
- 3J
- 4N
Show answer
Answer: 2 — L
M appears in both views, so use it as the pivot. - View 1: M is adjacent to J and L. - View 2: M is adjacent to O and K. So M touches J, L, O and K — four faces — which leaves N as the face opposite M. The other four (J, L, O, K) therefore form the two remaining opposite pairs. J and L are seen together in view 1, so they are adjacent, not opposite; O and K are seen together in view 2, so they are adjacent too. That leaves only two possible pairings, and the turning direction settles it: rotate view 1 so that M moves from the top to the right-hand face and it becomes exactly view 2, carrying L to the position opposite K. So K is opposite L. Option 3 (J) is the trap — J and K are both left-hand faces in their own views, which tempts you to pair them, but they are adjacent through M.
Six letters A, B, C, D, E and F are written on different faces of a dice. Two positions of this dice are shown in the figure. In the first view the top face carries A, the front-left face carries B and the front-right face carries C. In the second view the top face again carries A, the front-left face carries D and the front-right face carries E. Find the letter on the face opposite to C. (SI 3 Jul 2023)
- 1E
- 2F
- 3A
- 4D
Show answer
Answer: 1 — E
A sits on top in both views, so A is fixed and the face hidden underneath is the same in both — it must be F, the only letter never shown. That leaves B, C, D and E as the four side faces, which form a single ring around the vertical axis. In an isometric view the front-right face follows the front-left one clockwise when seen from above, so view 1 gives the clockwise step B → C and view 2 gives D → E. The only 4-cycle containing both steps is B → C → D → E → B. In a ring of four, opposite faces are two steps apart, so B faces D and C faces E. The letter opposite C is therefore E. F is the trap answer because it is the unseen letter, but F is the bottom face, opposite A.
Six letters A, B, C, D, E and F are written on different faces of a dice. Two positions of this dice are shown. In the first position the top face shows A, the front face shows E and the right face shows B. In the second position the top face shows F, the front face shows A and the right face shows C. Find the letter on the face opposite to F. (SI 9 Jul 2023)
- 1B
- 2E
- 3D
- 4A
Show answer
Answer: 1 — B
Face A is visible in both views, so use it as the anchor. In view one A is adjacent to E and B; in view two A is adjacent to F and C. That makes four distinct neighbours of A — E, B, F, C — so the remaining letter D must be opposite A. Now line the two views up. Tilt the first die forward so its top face A comes to the front: A is now at the front, E moves to the bottom and B stays on the right, with the hidden back face on top. The second die also has A at the front, but with C on the right and F on top. Rotating the tilted first die a quarter turn clockwise about the front-back axis sends right to bottom, bottom to left and top to right — this puts the old back face on the right and the old left face on top. Matching that to view two gives back = C and left = F, while B stays on the right in the untilted labelling. So in one consistent labelling the die reads top A, bottom D, front E, back C, right B, left F, and the pair left–right is F–B. Therefore F is opposite B. E is ruled out because E and F are adjacent in that labelling, and D is already taken by A.
Six letters A, B, C, D, E and F are written on different faces of a dice. Two positions of this dice are shown in the figure : in the first position the top face shows F, the front face shows E and the right face shows C; in the second position the top face shows B, the front face shows F and the right face shows A. Find the letter on the face opposite to A. (SI 15 Jul 2023)
- 1C
- 2F
- 3E
- 4D
Show answer
Answer: 3 — E
Step 1 — collect adjacencies. Both views show F. View 1 (top F, front E, right C) makes E and C adjacent to F; view 2 (top B, front F, right A) makes B and A adjacent to F. So F touches A, B, C and E, and the only letter left for the face opposite F is D. Step 2 — pair the other four. Roll view 1 forward so the top face comes to the front: now front = F, top = the face opposite E, right = C (rolling about the left-right axis leaves the right face alone). View 2 has front = F as well, so it is this position turned about the front-back axis. Turning about that axis cycles (top, right) through (opp E, C), (C, E), (E, opp C) and (opp C, opp E). Only the last pair can give top = B and right = A with B and A distinct from C and E. Hence B is opposite C and A is opposite E. Option 2 (F) is the trap for anyone who stops after noticing F appears in both views.
Six letters A, B, C, D, E and F are written on the different faces of a dice. Two positions of this dice are shown in the figure below : in the first position the top face shows B, the front face shows C and the right face shows A; in the second position the top face shows D, the front face shows F and the right face shows B. Find the letter on the face opposite C. (SI 29 Jul 2023)
- 1D
- 2B
- 3E
- 4F
Show answer
Answer: 1 — D
B is visible in both positions, so every letter sharing a view with B is adjacent to it: C and A from the first position, D and F from the second. That accounts for four neighbours, so the sixth letter E lies opposite B. Now rotate the first dice so that B moves from the top to the right, which is where it stands in the second position; C then moves to the front and A to the bottom. Tilting that cube forward to bring F to the front puts the face opposite C on top — and the second position shows D on top. So C and D are opposite, and A and F are the remaining pair. Option 4 (F) is the trap, since F merely shares a corner with C through the B face.
Six letters A, B, C, D, E and F are written on different faces of a dice. Two positions of this dice are shown below. In the first position the top face shows B, the front face shows D and the right-hand face shows C. In the second position the top face shows D, the front face shows F and the right-hand face shows E. Find the letter on the face opposite to E. (SI 31 Jul 2023)
- 1C
- 2B
- 3D
- 4A
Show answer
Answer: 2 — B
D appears in both views. In the first view D is adjacent to B and C; in the second it is adjacent to F and E. So D touches B, C, E and F — four faces — and the only face left for it to be opposite is A. That leaves B, C, E and F to form the other two opposite pairs, and since B–C are adjacent (first view) and E–F are adjacent (second view), B must pair with E or with F. Fix it by rotating: tip the first die backwards so its front face D comes to the top; it then reads top D, back B, right C. The second die reads top D, front F, right E. Turning the first die a quarter turn anticlockwise about the vertical axis brings its left face to the front and its front face to the right, giving front = opposite of C and right = opposite of B. Matching against the second die, opposite of C is F and opposite of B is E. So E is opposite B, and A is opposite D, C is opposite F. Option 3 (D) is the trap of pairing the shared face with the new one.
Part 2 · Figure series — 7 questions
Count first; rotation is the third thing to look at
A figure series looks like a test of visual intuition and is mostly a test of counting. Work through three questions in order. How many sides? If the shapes are polygons, write their side counts in a row — 5, 6, 7 — and the pattern usually appears at once. How many elements? Count dots, lines, arms, shaded cells. A growing or shrinking count settles the answer without any need to judge orientation. Which way has it turned? Only now look at rotation, and pick one distinctive feature — a single arrow, a shaded corner — and track that one feature rather than the whole figure. Two of the SI's seven are alternating series, where the odd-numbered figures form one sequence and the even-numbered figures another, exactly as in a number series. If the figures seem to change erratically, split them before anything else.
All seven figure series questions (7)Tap to open
A square is divided into four equal quarters by its vertical and horizontal mid-lines. Three of the quarters — top-left, top-right and bottom-left — carry the same element : a straight line runs from that quarter's outer corner of the square in to the centre of the square, and a small circle is drawn centred on that line a short way out from the centre, so that the circle bulges into its own quarter and passes close to the centre point. The bottom-right quarter is blank. Which figure completes the pattern in the bottom-right quarter? (SI 2016)
- 1A straight line from the centre of the square out to the bottom-right corner, with a small circle centred on that line a short way out from the centre, bulging into the quarter
- 2A straight line from the centre of the square out to the bottom-right corner, with a small circle centred on that line close to the bottom-right corner
- 3A straight line from the centre of the square out to the bottom-right corner, with no circle on it
- 4A straight line from the mid-point of the right edge to the mid-point of the bottom edge, cutting the corner off, with a small circle centred on that line
Show answer
Answer: 1 — A straight line from the centre of the square out to the bottom-right corner, with a small circle centred on that line a short way out from the centre, bulging into the quarter
The figure is built by one rule repeated four times round the centre. Each quarter gets a straight line joining the square's outer corner of that quarter to the centre, and a small circle centred on that same line a short way out from the centre, so the circle sits inside its own quarter and just touches the centre point. Rotating that element through 90 degrees produces the next quarter, and the three drawn quarters agree exactly. The blank bottom-right quarter must therefore carry the same element in its own orientation: a line from the centre out to the bottom-right corner, with the circle on that line near the centre. Option 2 keeps the line but slides the circle out to the corner, which breaks the ring of circles round the centre; option 3 drops the circle altogether; option 4 replaces the corner-to-centre line with an edge-to-edge line that cuts the corner off, which appears nowhere in the given figure.
Each of the following questions contain four problem figures and four answer figures. Select a figure from amongst the answer figures which will continue the same series as given in the problem figures. (SI 18 Aug 2021)
- 1A hexagon drawn as 4 nested hexagons
- 2A hexagon drawn as 5 nested hexagons
- 3A hexagon drawn as 3 nested hexagons
- 4A hexagon drawn as 4 nested hexagons, smaller in the frame
Show answer
Answer: 2 — A hexagon drawn as 5 nested hexagons
Two things change together along the series. The number of sides grows by one at each step: 2 line segments in the angle, then 3 in the triangle, 4 in the square, 5 in the pentagon — so the fifth figure must be a hexagon, 6 sides. At the same time the figure is drawn as one more nested copy each step: 1 outline, then 2, then 3, then 4. The count of copies is always one less than the number of sides. A hexagon therefore needs 6 − 1 = 5 nested outlines, which is the figure at option 2. Option 3 has only 3 outlines and options 1 and 4 have 4, so each of them repeats a count already used earlier in the series instead of continuing it.
Each of the following questions, contain four problem figures and four answer figures. Select a figure from amongst the answer figures which will continue the same series as given in the problem figures. Each problem figure is a box holding a large outline arrow and, inside it, a small arrow. In order the four figures are : (1) large arrow pointing right, small arrow pointing left; (2) large arrow pointing left, small arrow pointing up; (3) large arrow pointing right, small arrow pointing right; (4) large arrow pointing left, small arrow pointing down. (SI 18 Aug 2021)
- 1Large arrow pointing left, small arrow pointing left
- 2Large arrow pointing right, small arrow pointing down
- 3Large arrow pointing right, small arrow pointing left
- 4Large arrow pointing left, small arrow pointing right
Show answer
Answer: 3 — Large arrow pointing right, small arrow pointing left
Track the two arrows separately. The large outer arrow simply alternates: right, left, right, left — so the fifth figure must have it pointing right again. The small inner arrow turns 90° clockwise each step: left, then up, then right, then down — so the next turn brings it back to left. The figure required is therefore a large arrow pointing right with a small arrow pointing left inside it, which is option 3. Option 2 keeps the outer arrow right but leaves the inner arrow at down, repeating the fourth figure's inner direction instead of advancing it; options 1 and 4 both point the outer arrow left, which breaks the alternation.
In the question given below, there are four problem figures and four answer figures. Select the one answer figure that will continue the series given in the problem figures. Each figure is a square with a circle drawn inside it, and the circle holds plus signs. The four problem figures, in order, are : (1) two rows of two plus signs, four in all; (2) a single row of two plus signs; (3) a row of three plus signs above a row of two, five in all; (4) a single row of three plus signs. (SI 19 Aug 2021)
- 1A circle holding a row of four plus signs above a row of three - seven in all
- 2A circle holding a row of three plus signs above a row of three - six in all
- 3A circle holding a row of four plus signs above a row of four - eight in all
- 4A circle holding a row of three plus signs above a row of two - five in all
Show answer
Answer: 2 — A circle holding a row of three plus signs above a row of three - six in all
Separate the series into its two alternating strands. The odd-numbered figures are the two-row ones and their totals run 4 (2 over 2), then 5 (3 over 2) — going up by one. The even-numbered figures are the single-row ones and their totals run 2, then 3 — also going up by one. The fifth figure is odd-numbered, so it belongs to the two-row strand and must carry the next total after 5, namely 6, laid out as two rows. Six plus signs in two rows is three over three, which is option 2. Option 4 repeats the five of problem figure 3 and so adds nothing to the series; option 1 (seven) and option 3 (eight) skip a step.
Which answer figure will complete the pattern in the question figure? The question figure is a large square divided into four equal cells by a vertical and a horizontal centre line. In the top-left cell a straight line runs from the cell's bottom-left corner to its top-right corner (that is, from the mid-point of the big square's left edge to the mid-point of its top edge), and a small black dot inside a little box sits in the triangle on the centre side of that line. In the top-right cell a line runs from its top-left corner to its bottom-right corner (mid-point of the top edge to mid-point of the right edge), again with the dot-in-a-box on the centre side. In the bottom-left cell a line runs from its top-left corner to its bottom-right corner (mid-point of the left edge to mid-point of the bottom edge), with the dot-in-a-box on the centre side. The bottom-right cell is blank and carries the question mark. (SI 16 Oct 2022)
- 1A square with a straight line from its top-left corner down to its bottom-right corner, and the small dot-in-a-box placed in the upper-right triangle
- 2A square with a straight line from its bottom-left corner up to its top-right corner, and the small dot-in-a-box placed in the lower-right triangle
- 3A square with a straight line from its bottom-left corner up to its top-right corner, and the small dot-in-a-box placed in the upper-left triangle
- 4A square with a straight line from its top-left corner down to its bottom-right corner, and the small dot-in-a-box placed in the upper-right triangle, close to the top edge
Show answer
Answer: 3 — A square with a straight line from its bottom-left corner up to its top-right corner, and the small dot-in-a-box placed in the upper-left triangle
Look at what the three drawn lines do together. Their endpoints are the mid-points of the big square's edges: Cell · Line joins · Dot sits Top-left · left mid-point → top mid-point · on the centre side Top-right · top mid-point → right mid-point · on the centre side Bottom-left · left mid-point → bottom mid-point · on the centre side Those three lines are three sides of a diamond (a square standing on its corner) whose vertices are the four edge mid-points. The fourth side is missing, and it must join the bottom mid-point to the right mid-point. Inside the bottom-right cell on its own, that line runs from the cell's bottom-left corner up to its top-right corner. The dot is always inside the diamond, that is, on the side of the line nearer the centre of the big square. For the bottom-right cell the centre of the big square is its top-left corner, so the dot goes in the upper-left triangle. Required figure: diagonal from bottom-left to top-right, dot in the upper-left triangle — option 3. Why the others fail: options 1 and 4 draw the diagonal the other way (top-left to bottom-right), which would point the fourth side away from the diamond and leave the shape open; option 2 has the correct diagonal but puts the dot in the lower-right triangle, outside the diamond, breaking the rule the other three cells follow.
In the following question, select a figure from amongst the four alternatives, which when placed in the question mark of figure would complete the pattern. The question figure is a large square divided into four equal cells by a vertical and a horizontal centre line. In the top-left cell a straight line runs from the mid-point of the big square's left edge to the mid-point of its top edge, with two small outline triangles in the part of the cell on the centre side of that line. In the top-right cell the line runs from the mid-point of the top edge to the mid-point of the right edge, again with two small triangles on the centre side. In the bottom-right cell the line runs from the mid-point of the bottom edge to the mid-point of the right edge, with two small triangles on the centre side. The bottom-left cell is blank and carries the question mark. (SI 16 Oct 2022)
- 1A square with a straight line from its top-left corner down to its bottom-right corner, and two small outline triangles in the upper-right triangle
- 2A square with a straight line from its bottom-left corner up to its top-right corner, and two small outline triangles in the upper-left triangle
- 3A square with a straight line from its bottom-left corner up to its top-right corner, and two small outline triangles in the lower-right triangle
- 4A square with a straight line from its top-left corner down to its bottom-right corner, and two small outline triangles in the lower-left triangle
Show answer
Answer: 1 — A square with a straight line from its top-left corner down to its bottom-right corner, and two small outline triangles in the upper-right triangle
This is the same construction as the previous figure question, with the blank in a different corner and pairs of small triangles instead of a dot. Cell · Line joins · Triangles sit Top-left · left mid-point → top mid-point · on the centre side Top-right · top mid-point → right mid-point · on the centre side Bottom-right · bottom mid-point → right mid-point · on the centre side Bottom-left · ? · ? The three drawn lines are three sides of a diamond whose corners are the mid-points of the big square's four edges. The missing fourth side joins the left mid-point to the bottom mid-point. Seen inside the bottom-left cell alone, that line runs from the cell's top-left corner down to its bottom-right corner. The two small triangles always lie inside the diamond, that is, on the side of the line facing the centre of the big square. For the bottom-left cell the centre is its top-right corner, so the triangles go in the upper-right triangle. Required figure: diagonal from top-left to bottom-right, two triangles in the upper-right part — option 1. Why the others fail: option 4 has the right diagonal but puts the triangles in the lower-left part, outside the diamond; options 2 and 3 use the opposite diagonal (bottom-left to top-right), which would not meet the ends of the other three lines at all.
Select the correct figure : In the first problem figure a square frame contains a triangle, and inside that triangle is a small circle. In the second problem figure the square frame contains a circle, and inside that circle is a small triangle. In the third problem figure the square frame contains a hexagon, and inside that hexagon is a small square. Which answer figure completes the analogy? The answer figures are : A — a square frame containing a square, with a small hexagon inside it; B — a square frame containing a circle, with a small square inside it; C — a square frame containing a triangle, with a small square inside it; D — a square frame containing a hexagon, with a small square inside it. (SI 16 Oct 2022)
- 1B
- 2C
- 3D
- 4A
Show answer
Answer: 4 — A
Compare the first pair of problem figures: · Outer frame · Middle shape · Inner shape Figure 1 · square · triangle · circle Figure 2 · square · circle · triangle The frame is unchanged; the middle shape and the inner shape have swapped places. That is the whole rule. Figure 3 is frame square, middle hexagon, inner square. Swapping the middle and the inner gives frame square, middle square, inner hexagon. That is answer figure A. Since the options list the answer figures by letter, and (D) is the option whose text is "A", the correct choice is option 4. Why the others fail: answer figure D is figure 3 copied unchanged — no swap at all; answer figures B and C change the middle shape to a circle or a triangle, shapes that do not occur in figure 3 and so cannot come from a swap; only A keeps exactly the same two shapes and exchanges them.
One mirror-image question gives a key — a figure and its image — and asks you to apply the same transformation to a new figure. Work out the rule from the key before touching the options; it is usually a left-right flip, which reverses horizontal directions and leaves vertical ones alone. One counting question asks how many squares there are in a grid. In an n × n grid the answer is the sum of the squares 1² + 2² + … + n², because a k × k square can sit in (n − k + 1)² positions. For a 3 × 3 grid that is 9 + 4 + 1 = 14.
Both one-off puzzles, solved (2)Tap to open
Find the figure from the answer figure that will replace the question mark (?) from the problem figure. The first problem figure is a key-like shape : an upright stem with a ring at the bottom and, near the top, two short horizontal arms sticking out to the right of the stem, one at the very top and one a little lower, so the top of the key reads like the letter F. A horizontal bar crosses the stem between those two arms and ends on the left in an arrowhead pointing left. The second problem figure is the same key with the two arms sticking out to the left instead and the arrowhead pointing right, the ring still at the bottom. The third problem figure is a square divided into four small cells : top-left a solid black diamond, top-right an outline star, bottom-left an arrow pointing right, bottom-right an arrow pointing up. The fourth figure is the question mark. (SI 16 Oct 2022)
- 1A four-cell square with an outline star top-left, a solid black diamond top-right, an arrow pointing up bottom-left and an arrow pointing left bottom-right
- 2A four-cell square with an arrow pointing right top-left, an arrow pointing down top-right, a solid black diamond bottom-left and an upside-down outline star bottom-right
- 3A four-cell square with an arrow pointing down top-left, an arrow pointing left top-right, an upside-down outline star bottom-left and a solid black diamond bottom-right
- 4A four-cell square with a solid black diamond top-left, an outline star top-right, an arrow pointing right bottom-left and an arrow pointing up bottom-right
Show answer
Answer: 1 — A four-cell square with an outline star top-left, a solid black diamond top-right, an arrow pointing up bottom-left and an arrow pointing left bottom-right
First work out the rule from the key. In figure 1 the two arms point right and the arrow points left; in figure 2 the arms point left and the arrow points right. Crucially the ring stays at the bottom in both. A 180-degree rotation would have carried the ring to the top, and a top-to-bottom flip would have done the same. Only a left-to-right mirror (reflection in a vertical axis) swaps left and right while leaving top and bottom alone. So the rule is: mirror the figure about a vertical axis. Now apply that to figure 3. A vertical mirror swaps the two cells in each row and also mirrors each symbol: Cell in figure 3 · Goes to · Symbol after mirroring top-left, solid diamond · top-right · diamond (symmetric, unchanged) top-right, outline star · top-left · star (near-symmetric, unchanged) bottom-left, right arrow · bottom-right · left arrow bottom-right, up arrow · bottom-left · up arrow (symmetric, unchanged) Required figure: star top-left, diamond top-right, up arrow bottom-left, left arrow bottom-right — that is option 1. Why the others fail: option 2 is the top-to-bottom mirror (rows swapped, star turned upside down, up arrow turned into a down arrow); option 3 is a 180-degree rotation (diagonally opposite cells swapped, star inverted); option 4 is figure 3 unchanged.
Count the number of squares in the following figure : a square divided by two horizontal and two vertical lines into a 3 × 3 arrangement of nine equal smaller squares. (SI 2016)
- 118
- 214
- 310
- 49
Show answer
Answer: 2 — 14
In an n × n grid the number of squares is the sum of k × k squares for k = 1 to n, and each size k can be placed in (n − k + 1)² positions. For n = 3: Size · Positions · Count 1 × 1 · 3 × 3 · 9 2 × 2 · 2 × 2 · 4 3 × 3 · 1 × 1 · 1 Total = 9 + 4 + 1 = 14. The tempting wrong answer is option 4 (9), which counts only the smallest cells and forgets the four 2 × 2 blocks and the whole outer square; option 1 (18) over-counts by treating rectangles as squares.
- Assuming opposite faces add to 7 when the die is lettered. That rule is for standard numbered dice only; with letters, use adjacency and elimination.
- Treating two faces seen in one view as opposite. Anything visible together is adjacent, which is exactly what rules it out.
- Judging rotation before counting. Side counts and element counts are objective; an impression of turning is not.
- Reading an alternating figure series as one sequence. Split the odd and even positions, exactly as in a number series.
- Counting only the small squares in a grid. A 3 × 3 grid has 14 squares, not 9 — the 2 × 2 and 3 × 3 squares count too.
- Faces seen together are adjacent, never opposite.
- Each face has four neighbours and one opposite.
- Use the face common to both views as the pivot.
- Standard numbered die: opposites add to 7.
- Figure series: count sides, then elements, then turn.
- Erratic figures → split odd and even positions.
- Track one feature, not the whole figure.
- Mirror: horizontal reverses, vertical does not.
- Squares in an n × n grid: 1² + 2² + … + n².
- Smallest topic in the section — learn the dice first.