Non-Verbal Reasoning — Mirror Images, Folding, Dice & Figure Counting

A water image is the reflection you would see in a still pond. The reflecting surface now lies on the horizontal axis, below the object. So it swaps top and bottom and leaves left and right exactly as they were. Crucially, the order of the letters does NOT reverse — the word still reads left to right.

Why does the water list differ from the mirror list? Because symmetry is about a specific axis. M has a perfect left–right balance (its left half mirrors its right half), so it survives a mirror — but flip M top-to-bottom and it becomes a W-shape, so it fails in water. E is the opposite: its upper half mirrors its lower half, so E survives water, but flipped left-to-right it becomes a backwards-E, so it fails in a mirror. Only characters balanced about both axes — H, I, O, X and 0, 1, 8 — survive either reflection.

1. CHOKED → the order stays C-H-O-K-E-D. Check each letter against the water list: C ✓, H ✓, O ✓, K ✓, E ✓, D ✓ — all six are symmetric about the horizontal axis. So the water image reads CHOKED, completely unchanged. Its mirror image, by contrast, reads D-E-K-O-H-C with C, K, E, D all laterally inverted.
2. 1308 → all four digits (1, 3, 0, 8) are on the water list, so the water image reads 1308, unchanged. But the mirror image reverses the order to 8031, with the 3 flipped (1, 0 and 8 keep their shape).
3. MOM → the reverse case. In a mirror MOM is unchanged; in water M is not symmetric about the horizontal axis, so each M becomes a W-like shape and the water image looks like WOW. One word, two opposite answers.
4. HIT → H, I, T. H ✓ and I ✓ are on the water list, but T is not (flip T top-to-bottom and the crossbar moves to the bottom, giving a ⊥ shape). So HIT is not unchanged in water, even though it is unchanged in shape in a mirror (where it reads TIH).

The whole family of paper questions rests on one idea: a punch cuts every layer it passes through, so the number of holes equals the number of layers pierced. Never try to picture the finished sheet in one jump. Work backwards, unfolding one step at a time, and at each unfolding double the holes by reflecting them across that fold line.

Because each fold doubles the layers, n folds turn a single punch into 2ⁿ holes — provided the punch misses every fold line. So 1 fold → 2 holes, 2 folds → 4 holes, 3 folds → 8 holes, 4 folds → 16 holes. Two independent punches after 2 folds give 4 + 4 = 8 holes.

Read the rule out of that table: after the two folds, the hole set is symmetric about both centre lines. Take the punched cell, mirror its column (C1↔C4, C2↔C3), mirror its row (R1↔R4, R2↔R3), and take all four combinations. With only one fold you take only two combinations — mirror the column and stop. This mechanical recipe never fails and needs no drawing.

Paper cutting follows the same logic, with one extra rule about where the cut is made. A cut that touches the fold line unfolds into a single symmetric shape straddling the fold. A cut made on a free edge (away from the fold) unfolds into two mirror-image notches. That is exactly how children make paper snowflakes, and it is exactly what the option-setter tests.

1. Cut on the crease. Fold the sheet's right half onto its left half; the packet's right-hand edge is the crease. Halfway up that crease, snip off a small triangle whose base lies along the crease. Unfold: the triangle is doubled across the crease and the two halves join into one diamond-shaped (rhombus) hole sitting exactly on the vertical centre line, at mid-height. One cut, one hole — not two.
2. Cut on the free edge. Same fold, but now cut the identical triangle from the packet's left-hand edge. Unfolding gives two separate triangular notches — one on the sheet's left edge and one on its right edge, at the same height, mirror images of each other. Only the position of the cut changed, and the answer changed completely.
3. Two folds, one cut at the centre corner. Fold vertically, then horizontally. The packet corner where the two creases meet is the centre of the sheet. Snip that corner off and unfold: you get a single hole at the dead centre, symmetric about both centre lines — four quarters of the cut joined into one shape.
4. Three folds, one punch. Fold right onto left, then bottom onto top, then the resulting quarter in half again along a new vertical line. Layers = 2 × 2 × 2 = 8, so one clear punch yields 8 holes (2³), and the pattern is symmetric about all three fold lines. Count layers first, positions second — the layer count alone eliminates two options.

A cube has 6 faces, 12 edges and 8 corners. Any single view of a cube can show at most three faces, and those three meet at one corner — so they are mutually adjacent. From this comes the golden rule of every dice question: any two numbers seen together in the same view can never be opposite each other. Each face has exactly four neighbours and one opposite. Solve by ruling out neighbours until one candidate is left.

Worked case — Rule 1. Position I of a cube shows the faces 4, 1, 2; Position II of the same cube shows 4, 1, 5. The common faces are 4 and 1, so the odd faces 2 and 5 are opposite each other. Reason it out rather than memorising: faces 4 and 1 share one edge, and that edge has two ends. At one corner the third face is 2; at the other corner it must be the face opposite to 2 — and Position II shows 5 there. Hence 2 is opposite 5, and 4 and 1 are both adjacent to 2 and to 5.

Verify that answer against every view before you tick it — a habit worth ten marks over a paper. Pairs 1–5, 2–6, 3–4. Position I shows 2, 1, 3: no two of them form a pair ✓. Position II shows 2, 3, 5: none is a pair ✓. Position III shows 2, 5, 4: none is a pair ✓. The die is therefore consistent — and note that 2 + 6 = 8, not 7, so this is not an ordinary die and the sum-to-7 shortcut would have given the wrong answer.

A cube of side n units is painted on all six faces and then cut into n³ unit cubes. How many little cubes have 3, 2, 1 or 0 painted faces? Do not count — think about where a small cube sits inside the big one. Its position alone decides how many of its faces were on the outside.

• Always run the total check. The four counts must add to n³, because every small cube is a corner, an edge-middle, a face-middle or an interior cube — there is no fifth possibility. For n = 4: 8 + 24 + 24 + 8 = 64 = 4³ ✓. For n = 5: 8 + 36 + 54 + 27 = 125 = 5³ ✓. For n = 3: 8 + 12 + 6 + 1 = 27 = 3³ ✓. If your four numbers do not sum to n³ you have slipped in arithmetic — fix it before choosing an option.
• Read "exactly" versus "at least" carefully. Exactly one painted face is 6(n−2)², a much smaller number than at least one, which is n³ − (n−2)³. For n = 4 the two answers are 24 and 56 — and both usually appear among the options. This single word is the commonest trap in the topic.
• The degenerate cases. If n = 2 there is no interior and no edge-middle: all 8 small cubes are corners with 3 painted faces (8 + 0 + 0 + 0 = 8 = 2³ ✓). If n = 1 the single cube has all 6 faces painted, which none of the four formulae covers — the formulae assume n ≥ 2.
• Number of cuts. To slice a cube of side n into n³ unit cubes you need (n−1) cuts along each of the three directions, so total cuts = 3(n−1). For n = 4: 3 × 3 = 9 cuts. For n = 5: 3 × 4 = 12 cuts. In general, m cuts along one direction produce (m+1) slices, so k cuts split into k+1 parts — and if the cuts are distributed as a, b, c along the three axes, the number of pieces is (a+1)(b+1)(c+1). With 9 cuts split 3-3-3 that is 4 × 4 × 4 = 64 ✓.

Counting questions are lost to carelessness, not difficulty. Never scan a figure and guess. Use one fixed order every single time: count the smallest units first, then figures made of two units, then three, and so on, and finally the whole figure. Add the groups at the end. A second, foolproof check for triangles: list every set of three labelled points, and keep only those where all three joining lines are actually drawn. Two independent methods agreeing is what lets you tick and move on.

1. Case A — one cevian. Triangle ABC, with D on BC and the line AD drawn. Smallest triangles: ABD, ACD = 2. Made of two units: ABC = 1. Total = 2 + 1 = 3 triangles.
2. Case B — two cevians from the same vertex. Triangle ABC with D and E both on BC (in the order B, D, E, C), and both AD and AE drawn. Smallest: ABD, ADE, AEC = 3. Two units: ABE, ADC = 2. Three units: ABC = 1. Total = 3 + 2 + 1 = 6 triangles. General rule: with k cevians from one vertex the base is split into (k+1) parts and the count is (k+1)(k+2)/2. Check: k = 1 gives 2 × 3 / 2 = 3 ✓; k = 2 gives 3 × 4 / 2 = 6 ✓; k = 3 gives 4 × 5 / 2 = 10 ✓.
3. Case C — two medians. In triangle ABC let D be the midpoint of BC and E the midpoint of AC; draw AD and BE, crossing at the centroid G. The two lines cut the triangle into four regions: triangles ABG, BDG, AEG and quadrilateral GDCE. Count in groups — 1 region: ABG, BDG, AEG = 3; 2 regions: ABD (ABG+BDG), ABE (ABG+AEG), ACD (AEG+GDCE), BCE (BDG+GDCE) = 4; 4 regions: ABC = 1. Nothing uses exactly 3 regions. Total = 3 + 4 + 1 = 8 triangles.
4. Verify Case C by the point method. The eight triangles are exactly ABC, ABD, ABE, ABG, ACD, AEG, BCE, BDG. Sets like A-C-E or B-D-C fail because those three points are collinear; sets like A-D-E or C-D-G fail because the joining line (DE, or CG) was never drawn. Two methods, one answer.
5. Case D — rectangle with both diagonals. Rectangle ABCD with diagonals AC and BD meeting at O. Smallest triangles: AOB, BOC, COD, DOA = 4. Made of two regions: ABC, BCD, CDA, DAB = 4. No triangle uses three regions, and the whole rectangle is not a triangle. Total = 4 + 4 = 8 triangles, plus exactly 1 rectangle (ABCD itself).

The square total is the sum of the first n square numbers, n(n+1)(2n+1)/6. Check each row of the table: n = 3 gives 3 × 4 × 7 / 6 = 84 / 6 = 14 ✓; n = 4 gives 4 × 5 × 9 / 6 = 180 / 6 = 30 ✓; n = 8 gives 8 × 9 × 17 / 6 = 1224 / 6 = 204 ✓. The rectangle total counts all choices of two vertical grid lines and two horizontal grid lines out of (n+1) each, which is C(n+1, 2)² = [n(n+1)/2]²; for n = 2 that is 3² = 9, and counting by hand gives 4 + 2 + 2 + 1 = 9 ✓. Note that squares are a subset of rectangles, so the rectangle number is always the larger one.

These are the types that genuinely need pictures, so be honest about what a text page can do: it cannot draw the series for you, but it can hand you the examiner's toolbox. Every figure series and figure analogy is built from a small, fixed set of transformations. Your job is never to "see" the answer — it is to name the rule that turns one figure into the next, and then apply that same rule once more. Test the transformations in a set order and the rule surfaces in seconds.

1. Test in this order. (1) Same shape but turned? → rotation; find the angle. (2) Turned, but no angle works? → reflection. (3) Has the number of parts changed? → addition or removal. (4) Has a small mark moved? → track only the mark. (5) Has shading moved? (6) Has the number of sides changed? Stop at the first rule that explains every step, not just the first step.
2. The rule must fit every step. Beginners find a rule that explains figure 1 → figure 2 and jump to the options. Always test it on figure 2 → figure 3 as well, and on 3 → 4 if given. If it fails anywhere, the rule is wrong however elegant it looked.
3. Analogy is the same skill, shortened. In A : B :: C : ?, isolate the single change from A to B and apply exactly that to C. It works identically on letters, which you can verify in text: AZ : BY :: CX : ? — the first letter moves one forward, the second one backward, so the answer is DW. Derive the rule from the given pair; never reverse-engineer it from the options.
4. Count before you look. Most series answers can be narrowed to two options purely by counting elements (lines, dots, sides, shaded cells) in the given figures and extending the count. Do the counting first; only then compare orientations. This order halves your time.

In an embedded (hidden) figure question you are given a small figure X and four larger options, and exactly one option contains X complete and unchanged. It may be rotated, but it is never stretched, shrunk or bent. Blind searching wastes a minute; a fixed checking order finds it in fifteen seconds.

1. Find the most distinctive component first — an unusual angle, the one curve among straight lines, a sharp spike, a very short segment. Scan the four options only for that component. Options lacking it are eliminated at once, with no tracing at all. This is component-matching, and it is the whole method.
2. Count the line segments of X. The correct option must contain at least that many segments arranged the same way. If a candidate has fewer lines in the relevant region than X needs, it is out — a count is faster and safer than an impression.
3. Verify proportions. X keeps its exact size and shape. If a candidate would need X stretched longer, squashed shorter or its angle widened, reject it — the examiner never rescales the hidden figure.
4. Trace it with your pencil point. Follow X's outline over the surviving option. If you must lift the pencil, or invent a line that is not printed, X is not there. One clean unbroken trace = the answer.

Figure completion (pattern completion) gives you a large design with one piece cut out, usually a corner or a quarter, plus four candidate pieces. The design is almost always symmetric, and that symmetry is your answer key. Identify the symmetry first: is the design symmetric about the vertical centre line (MV), the horizontal centre line (MH), both, or under a 90°/180° rotation? Then read off what the missing piece must contain by reflecting or rotating the part of the design that is still printed.

1. Worked reasoning. Suppose the design fills a 3 × 3 grid, the piece missing is cell A3 (top-right), and you can see that the printed part is symmetric about the vertical centre line. MV maps A3 ↔ A1. So whatever is drawn in A1 — say a quarter-circle opening downwards-right — the missing piece must be its left-right mirror: a quarter-circle opening downwards-left. Read A1, flip it, pick that option.
2. If the symmetry is 180°-rotational instead, the partner of A3 is C1 (bottom-left), not A1. Getting the partner cell right is the whole question; the rotation table earlier in this chapter gives every partner for MV, MH, 180°, CW 90° and ACW 90°.
3. Check continuity at the edges. Lines of the design that run into the cut edge must continue at the same height and the same angle in the missing piece. Two options often satisfy the symmetry but only one lines up at the boundary — the edge check is the tie-breaker.

Grouping of identical figures gives you five to nine small figures with labels (often numbered 1 to 9 or lettered) and asks you to sort them into groups that share a property. Do not stare at the set as a whole. Instead run down a fixed checklist of properties and stop at the first one that splits the set cleanly, because the paper always intends exactly one such property.

• Number of sides or lines — triangles together, quadrilaterals together, curves together. This alone settles most grouping questions and takes five seconds.
• Open versus closed figures — does the outline enclose an area, or does it have loose ends? An open figure never groups with a closed one.
• Number of enclosed regions — one region, two regions, three. Count the compartments, not the lines.
• Shading — fully shaded, half shaded, unshaded; and where the shading sits.
• Symmetry — figures with a line of symmetry versus figures with none. This is the property candidates forget, and it is a favourite of the paper-setter.
• Rotation-only identity — two figures that are the same shape turned differently belong together; a figure that is a mirror of the others does not. This is the rotate-vs-reflect distinction again, now used as a sorting key.