Non-Verbal & Visual Reasoning

A water image is the reflection you see in a still pond. The mirror 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.

1. BOOK → order stays B-O-O-K. Now check each letter against the water list: B ✓, O ✓, O ✓, K ✓ — all four are symmetric about the horizontal axis. So the water image of BOOK reads BOOK, completely unchanged.
2. HIDE → H, I, D, E are all on the water list, order unchanged, so the water image reads HIDE unchanged. Its mirror image, however, reads EDIH with the E and D flipped.
3. 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 1 and 3 flipped.
4. MOM → the reverse case. In a mirror MOM is unchanged, but in water M is not symmetric about the horizontal axis, so each M turns into a W-like shape: the water image looks like WOW, not MOM.

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. To find the answer, do not try to imagine 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. Two punches with 2 folds give 4 + 4 = 8 holes.

1. Worked case 1 — one fold. Take a square sheet and think of it as four quadrants: top-left, top-right, bottom-left, bottom-right. Fold the right half exactly onto the left half, so the crease is the vertical centre line and the packet is two layers thick. Now punch once, high up in the middle of the packet, clearly away from the crease.
2. Answer to case 1: 2 layers pierced → 2 holes (= 2¹). Unfold and reflect across the vertical centre line: one hole sits in the top-left quadrant and its twin in the top-right quadrant, both at the same height from the top edge and at the same distance from the centre line.
3. Worked case 2 — two folds. Start again with the square. Fold 1: right half onto left half (vertical crease at the centre). Fold 2: bottom half onto top half (horizontal crease at the centre). The packet is now one quadrant — say the top-left — and is four layers thick. Punch once near the middle of that quadrant, on neither crease.
4. Answer to case 2: 4 layers pierced → 4 holes (= 2²). Unfold the horizontal crease first: the hole in the top-left quadrant gains a twin in the bottom-left quadrant. Now unfold the vertical crease: those two are reflected into the top-right and bottom-right quadrants. Final picture: one hole in each of the four quadrants, all equally far from the vertical centre line and equally far from the horizontal centre line — the four holes are the corners of a rectangle centred on the middle of the sheet.

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. This is exactly how children make paper snowflakes.

1. Worked cutting case. Fold the square's right half onto its left half. The packet's right-hand edge is the crease; its left-hand edge holds the sheet's original left and right edges lying on top of each other. Halfway up the crease, snip off a small triangle whose base lies along the crease.
2. Unfold it: the triangle is doubled across the crease, so the two halves join into one diamond-shaped (rhombus) hole sitting exactly on the vertical centre line, at mid-height of the sheet. One cut, one hole — not two.
3. Now change only the position. Cut the same triangle from the packet's left-hand edge instead. 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.
4. Two folds, one cut at the centre corner. Fold the square twice (vertical, then horizontal). The corner of the packet 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.

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 faces meet at one corner — so they are mutually adjacent. From this comes the golden rule: any two numbers seen together in the same view can never be opposite each other. Every dice question is solved by ruling out neighbours until only one candidate is left.

Here 2 and 3 are common to both views, and only the third face changes (5, then 6). Reason it out: the faces 2 and 3 share one edge, and that edge has two ends. At one end the third face is 5; at the other end it must be the face opposite to 5. Since the second view shows 6 at that corner, 6 is opposite 5. General result: when two views share two common faces, the two odd faces are opposite each other — and the two common faces are adjacent to both.

The number 1 has now been seen with 2, 3, 4 and 5 — that is four different neighbours, and a face has exactly four neighbours. So the only number left, 6, is opposite 1. Finish the die: 3 has been seen with 2 and with 4, so 3 cannot be opposite either — therefore 3 is opposite 5, and the last pair is 2 opposite 4. Full answer: 1–6, 2–4, 3–5. Notice the method: find the number with four known neighbours first; the rest falls out by elimination.

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 in the big one. Its position decides how many of its faces were on the outside.

• Always run the total check. The four counts must add up to n³, because every small cube is either a corner, an edge-middle, a face-middle or an interior cube. For n = 4: 8 + 24 + 24 + 8 = 64 = 4³ ✓. For n = 5: 8 + 36 + 54 + 27 = 125 = 5³ ✓. If your four numbers do not sum to n³, you have made an arithmetic slip — fix it before choosing an option.
• "At least one face painted" = n³ − (n−2)³ (everything except the hidden interior). For n = 4: 64 − 8 = 56. For n = 5: 125 − 27 = 98. Read the question carefully: exactly one face is 6(n−2)², while at least one is this much bigger number.
• Smallest case, n = 3 (27 cubes): 8 corners with 3 faces, 12 × 1 = 12 with 2 faces, 6 × 1 = 6 with 1 face, and 1³ = 1 with no face — the single cube at the very heart. 8 + 12 + 6 + 1 = 27 ✓. If n = 2, there is no interior at all: all 8 small cubes have 3 painted faces.

Counting questions are lost to carelessness, not difficulty. Never scan the figure and guess. Use a fixed order every 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.

1. Case A — one median. Take triangle ABC. Let D be the midpoint of BC, and draw the median AD. Smallest triangles: ABD and ACD = 2. Made of two units: the whole triangle ABC = 1. Total = 2 + 1 = 3 triangles.
2. Case B — two medians. In the same triangle ABC, let D be the midpoint of BC and E the midpoint of AC. Draw both medians AD and BE; they cross at the centroid G. The two medians cut the triangle into four regions: three triangles (ABG, BDG, AEG) and one quadrilateral (GDCE).
3. Count Case B in groups. Made of 1 region: ABG, BDG, AEG = 3. Made of 2 regions: ABD (=ABG+BDG), ABE (=ABG+AEG), ACD (=AEG+GDCE), BCE (=BDG+GDCE) = 4. Made of 4 regions: ABC = 1. Nothing is made of exactly 3 regions. Total = 3 + 4 + 1 = 8 triangles.
4. Verify by the point method. The eight triangles are exactly ABC, ABD, ABE, ABG, ACD, AEG, BCE, BDG. Sets like A-C-E or B-G-E 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, same answer — that is how you gain confidence in the exam.

The pattern generalises. An n × n grid contains n² + (n−1)² + … + 2² + 1² squares, which is the sum of the first n square numbers, n(n+1)(2n+1) / 6. Check it: for n = 3, 3 × 4 × 7 / 6 = 84 / 6 = 14 ✓, matching the table. For n = 4: 16 + 9 + 4 + 1 = 30, and 4 × 5 × 9 / 6 = 180 / 6 = 30 ✓. For a chessboard, n = 8: 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204, and 8 × 9 × 17 / 6 = 1224 / 6 = 204 ✓.

These are the question types that genuinely need pictures, so let us be honest: this page cannot draw the series for you. What it can do is give 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 identify 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 you will find the rule in seconds.

1. Test in this order. (1) Is the outline the same shape but turned? → rotation. (2) Is it turned but the turning does not work? → 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. If it fails there, the rule is wrong, however pretty it looked.
3. Analogy is the same skill, shortened. In A : B :: C : ?, find the single change from A to B and apply exactly that to C. It works identically for letters: AZ : BY :: CX : ? — the first letter moves one forward, the second one backward, so the answer is DW. Find the rule from the given pair; never guess from the options.

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 checking order finds it in fifteen seconds.

1. Find the most distinctive feature first — an unusual angle, the one curve among straight lines, a sharp spike, a very short segment. Scan the four options only for that feature. Options that lack it are eliminated at once, without any tracing.
2. Count the line segments of X. The correct option must contain at least that many segments arranged in the same way. If a candidate has fewer lines in that region than X needs, it is out.
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.