Figure transformations — SSC CGL Reasoning Notes

Concepts and rules

A square sheet is folded one or more times, a hole is punched through all the layers, and you must find the hole pattern when the sheet is fully opened.

Grid used in these notes: the sheet is a 4×4 grid, rows R1 (top) to R4 (bottom), columns C1 (left) to C4 (right); R2C3 means row 2, column 3.

Every fold is a reflection. When you open it, each hole that went through the folded flap gets a twin at the mirror position across the fold line, at the same perpendicular distance from the line.

Vertical fold (fold line runs up and down): the row stays the same and the column is mirrored. Horizontal fold: the column stays the same and the row is mirrored.

Middle vertical fold (between C2 and C3): column c goes to 5 − c, so C1↔C4 and C2↔C3. Middle horizontal fold (between R2 and R3): row r goes to 5 − r.

General rule: a fold line between column k and column k + 1 sends column c to 2k + 1 − c (the same rule works for rows).

Diagonal fold along the diagonal from the top-left corner to the bottom-right corner: cell RrCc and cell RcCr are twins (row and column swap).

Diagonal fold along the other diagonal (top-right corner to bottom-left corner): cell RrCc and cell R(5 − c)C(5 − r) are twins.

Counting rule: if a punch lies away from every fold line and every fold covers it fully, the holes double with each fold: 1 fold gives 2 holes, 2 folds give 4, 3 folds give 8.

Exception 1: a hole punched exactly on a fold line has no separate twin for that fold; it opens as one hole on the line (a half-circle notch at the folded edge opens into one full circle).

Exception 2: in an uneven fold the flap is narrower than the part it lies on; a hole outside the area covered by the flap passes through fewer layers and gets no twin from that fold.

Shape of the hole: a circle or square looks the same after reflection, but an asymmetric cut is flipped; a triangle pointing right becomes a triangle pointing left across a vertical fold.

Method and shortcuts

Step 1: write the folds in order as Fold 1, Fold 2, Fold 3 and note each fold line and which part moves.

Step 2: mark the punch on the final folded shape using the cell name of the original grid.

Step 3: open the last fold first. Reflect every current hole across that fold line and add the twins, skipping holes on the line and holes outside the flap.

Step 4: repeat for the earlier folds until the sheet is fully open, then count the holes and match the pattern with the options.

Why reverse order: the last fold is on top, so physically it is the first one you can open; opening Fold 1 first would reflect holes across a line that is still hidden inside the folded stack.

Shortcut 1 (symmetry test): the final pattern must be symmetric about every fold line the punch passed through, so first strike out every option that is not symmetric about those lines.

Shortcut 2 (quarter fold): after the middle vertical and middle horizontal folds, a punch at RrCc opens into RrCc, RrC(5 − c), R(5 − r)Cc and R(5 − r)C(5 − c).

Shortcut 3 (count first): when options differ in the number of holes, count with the doubling rule and the 2 exceptions before checking positions.

When shortcuts fail: uneven folds, punches on a fold line and diagonal folds mixed with straight folds must be unfolded step by step.

Example 1: one vertical fold

Question: A 4×4 sheet is folded along the middle vertical line so that the left half lies on the right half. A hole is punched at R2C3. Where are the holes after opening?

Solution: The folded sheet occupies C3 and C4. Opening the fold reflects R2C3 to column 5 − 3 = 2, that is R2C2.

Answer: 2 holes, at R2C2 and R2C3.

Example 2: quarter fold

Question: A 4×4 sheet is folded left half onto right half (Fold 1) and then top half onto bottom half (Fold 2). A hole is punched at R4C3. Find the holes after opening.

Solution: The folded sheet is the block R3–R4, C3–C4. Open Fold 2 first: R4C3 gets the twin R1C3 (row 5 − 4 = 1).

Open Fold 1: R4C3 gets R4C2 and R1C3 gets R1C2.

Answer: 4 holes, at R1C2, R1C3, R4C2 and R4C3.

Example 3: punch on the fold line

Question: A 4×4 sheet is folded left half onto right half. One hole is punched exactly on the folded edge in row R1, and another hole at R3C4. How many holes appear after opening?

Solution: The edge hole lies on the fold line (between C2 and C3), so it stays one hole centred on that line. The hole at R3C4 is away from the line and gets the twin R3C1.

Answer: 3 holes: one on the middle line in row R1, plus R3C1 and R3C4.

Example 4: diagonal fold

Question: A 4×4 square sheet is folded along the diagonal from the top-left corner to the bottom-right corner, so that the lower-left triangle lies on the upper-right triangle. Holes are punched at R1C3 and R2C4. Which cells have holes after opening? (a) R1C3, R2C4, R1C2, R2C1 (b) R1C3, R2C4, R4C3, R3C4 (c) R1C3, R2C4, R3C1, R4C2 (d) R1C3, R2C4, R1C1, R4C4

Solution: For this diagonal, twins swap row and column. R1C3 gives R3C1 and R2C4 gives R4C2. Option (a) is the vertical-fold twin and option (b) the horizontal-fold twin, both wrong axes.

Answer: (c) R1C3, R2C4, R3C1, R4C2.

Example 5: three folds

Question: A 4×4 sheet: Fold 1 puts the left half on the right half, Fold 2 puts the top half on the bottom half, and Fold 3 folds column C4 onto column C3 of the remaining 2×2 block. One hole is punched at R4C3. What is the pattern after opening? (a) 4 holes at the 4 corners (b) 8 holes filling rows R1 and R4 (c) 8 holes filling columns C1 and C4 (d) 8 holes filling rows R3 and R4

Solution: Open Fold 3 (line between C3 and C4, so c goes to 7 − c): R4C3 gets R4C4. Open Fold 2 (r goes to 5 − r): add R1C3 and R1C4.

Open Fold 1 (c goes to 5 − c): C3 gives C2 and C4 gives C1, adding R1C1, R1C2, R4C1 and R4C2. All 8 cells of rows R1 and R4 have holes; 3 folds, 8 holes, as the doubling rule predicts.

Answer: (b) 8 holes filling rows R1 and R4.

Example 6: uneven fold

Question: A 4×4 sheet: Fold 1 turns column C1 over onto column C2 (fold line between C1 and C2). Fold 2 puts the top half on the bottom half. Holes are punched at R3C2 and R4C4. How many holes are there after opening? (a) 8 (b) 4 (c) 7 (d) 6

Solution: Open Fold 2: R3C2 gets R2C2 and R4C4 gets R1C4, so 4 holes so far.

Open Fold 1: the flap was only column C1 and it covered only column C2. R3C2 and R2C2 get twins R3C1 and R2C1, but R4C4 and R1C4 were never under the flap and get no twins.

Answer: (d) 6 holes, at R1C4, R2C1, R2C2, R3C1, R3C2 and R4C4. Doubling blindly would give 8, which is option (a), the trap.

Common mistakes

Opening Fold 1 first. The last fold must be opened first; otherwise twins land across the wrong line.

Doubling a hole that lies on the fold line. A hole on the line stays one hole for that fold.

Reflecting a hole that the flap never covered in an uneven fold. Check whether the flap reached that cell before adding a twin.

Mirroring the wrong coordinate. A vertical fold changes the column and keeps the row; a horizontal fold changes the row and keeps the column.

Treating a diagonal fold like a vertical or horizontal fold. For a diagonal fold, swap row and column (or use R(5 − c)C(5 − r) for the other diagonal).

Forgetting that asymmetric cuts flip. A right-pointing triangle opens as a left-pointing twin across a vertical fold.

Practice questions

Q1. A 4×4 sheet is folded top half onto bottom half and then left half onto right half. A hole is punched at R3C4. Where are the holes after opening? (a) R1C1, R1C4, R4C1, R4C4 (b) R2C1, R2C4, R3C1, R3C4 (c) R2C2, R2C3, R3C2, R3C3 (d) R3C1 and R3C4 only

Q2. A 4×4 sheet is folded along the diagonal from the top-right corner to the bottom-left corner, so that the upper-left triangle lies on the lower-right triangle. Holes are punched at R4C4 and at R3C2 (a cell centred on the fold line). How many holes are there after opening? (a) 4 (b) 2 (c) 3 (d) 6

Q3. A 4×4 sheet: Fold 1 puts the left half on the right half, Fold 2 turns column C3 over onto column C4, and Fold 3 puts the top half on the bottom half. A hole is punched at R3C4. What is the pattern after opening? (a) 8 holes filling rows R2 and R3 (b) 8 holes filling columns C2 and C3 (c) 4 holes at R2C3, R2C4, R3C3 and R3C4 (d) 6 holes

Answers

Q1. (b) Open the vertical fold (last): R3C4 gives R3C1. Open the horizontal fold: R3C1 and R3C4 give R2C1 and R2C4.

Q2. (c) For this diagonal the twin of RrCc is R(5 − c)C(5 − r). R4C4 gives R1C1; R3C2 is on the line (3 + 2 = 5) and stays single. Total 3.

Q3. (a) Fold 3: R3C4 gives R2C4. Fold 2 (c goes to 7 − c): add R3C3 and R2C3. Fold 1 (c goes to 5 − c): add R3C1, R3C2, R2C1 and R2C2. All 8 cells of rows R2 and R3.

Quick revision

Open the last fold first and reflect every hole across that fold line.

Middle folds on a 4×4 grid: column c goes to 5 − c, row r goes to 5 − r; a fold between k and k + 1 sends c to 2k + 1 − c.

Diagonal folds: swap row and column, or use R(5 − c)C(5 − r) for the other diagonal.

Holes double per fold, except holes on a fold line and holes outside a narrow flap.

Use symmetry about every fold line to strike out wrong options quickly.

Concepts and rules

Figure completion: a figure has one portion missing, and you choose the option that completes it so that every line, shading and symmetry of the figure continues correctly.

Embedded figure: a small target shape is hidden inside a larger figure, and you find which option contains it (or does not contain it).

Text model used here: a 4×4 grid with rows R1 to R4 (top to bottom) and columns C1 to C4 (left to right); a pattern is the list of shaded cells.

Mirror about the vertical middle line: RrCc pairs with RrC(5 − c). Mirror about the horizontal middle line: RrCc pairs with R(5 − r)Cc.

Mirror about the main diagonal (top-left to bottom-right): RrCc pairs with RcCr.

Rotation by 90° clockwise about the centre: RrCc goes to RcC(5 − r). Rotation by 180°: RrCc goes to R(5 − r)C(5 − c). Rotation by 90° anticlockwise: RrCc goes to R(5 − c)Cr.

Symmetry about both middle lines gives 180° symmetry as well, but 90° rotational symmetry does not imply any mirror symmetry; a pinwheel pattern turns onto itself yet has no mirror line.

An embedded shape counts as present only if every one of its parts lies on parts that actually exist in the big figure; extra lines or extra shaded cells around it do not cancel it.

Unless the question says otherwise, the hidden shape may be turned (rotated); it may be flipped (reflected) only if the question allows it. If the question says the shape may be turned but not flipped, a mirror image does not count.

The size of the hidden shape cannot change, and you cannot invent a missing edge or bridge a gap.

Method and shortcuts

Completion step 1: from the given parts, test which symmetry they obey: vertical mirror, horizontal mirror, diagonal mirror, 90° rotation or 180° rotation.

Completion step 2: map each given shaded cell to the missing region with that symmetry rule and list the cells you get.

Completion step 3: check the edges where the missing part meets the figure; every line or shaded strip that reaches the gap must continue in the option.

Shortcut: if 3 quadrants are given, compare the 2 quadrants beside the gap. If each is the mirror of the next, the rule is mirror; if each is a turned copy, the rule is rotation.

Embedded step 1: pick the most distinctive part of the target, such as a cell with 3 shaded neighbours (a T-junction) or a step (a zig-zag).

Embedded step 2: search the option for that feature; if it is missing, reject the option at once. If present, trace the rest of the target around it in every allowed orientation.

Why this works: a feature like a T-junction survives every rotation and reflection, so its absence proves the shape cannot be there in any orientation.

Example 1: mirror completion

Question: A 4×4 pattern is symmetric about its vertical middle line. The left half has shaded cells R1C1, R2C2, R3C1 and R4C2. Which cells of the right half are shaded?

Solution: Keep the row and send column c to 5 − c: R1C1 to R1C4, R2C2 to R2C3, R3C1 to R3C4, R4C2 to R4C3.

Answer: R1C4, R2C3, R3C4 and R4C3.

Example 2: symmetry about both lines

Question: A 4×4 pattern is symmetric about both middle lines. In the top-left quadrant (R1–R2, C1–C2) the shaded cells are R1C2 and R2C1. Which cells of the bottom-right quadrant are shaded?

Solution: Two mirrors in a row change both the row and the column: RrCc goes to R(5 − r)C(5 − c). R1C2 goes to R4C3 and R2C1 goes to R3C4.

Answer: R3C4 and R4C3.

Example 3: rotation completion

Question: A 4×4 pattern looks the same after a 90° clockwise turn. Its top-left quadrant has R1C1 and R2C1 shaded. Which cells of the top-right quadrant are shaded?

Solution: A 90° clockwise turn carries the top-left quadrant onto the top-right quadrant, with RrCc going to RcC(5 − r). R1C1 goes to R1C4 and R2C1 goes to R1C3.

Answer: R1C3 and R1C4 (a vertical mirror would wrongly give R1C4 and R2C4).

Example 4: hidden T-shape

Question: The target is a T-shape of 4 cells: 3 cells in a line with a 4th cell attached to the middle one (it may be turned or flipped). In which shaded pattern is it hidden? (a) R1C1, R1C2, R1C3, R2C1, R3C1 (b) R1C1, R1C2, R2C1, R2C2, R3C3 (c) R1C1, R1C2, R2C2, R2C3, R3C3 (d) R1C1, R2C1, R2C2, R3C1, R3C3

Solution: A T needs one cell with 3 shaded side-neighbours. In (a) every cell has at most 2, in (b) the square block gives at most 2, in (c) the staircase gives at most 2. In (d), R2C1 touches R1C1, R3C1 and R2C2, which is a T standing upright with its arm to the right; R3C3 is just an extra cell.

Answer: (d)

Example 5: rotation, not mirror

Question: A 4×4 pattern has these shaded cells in 3 quadrants: top-left R1C1, R1C2, R2C2; top-right R1C4, R2C4, R2C3; bottom-right R4C4, R4C3, R3C3. Which cells complete the bottom-left quadrant? (a) R3C2, R4C1, R4C2 (b) R3C1, R3C2, R4C2 (c) R3C1, R3C2, R4C1 (d) R3C1, R4C1, R4C2

Solution: Test the rule. Mirroring the top-left quadrant across the vertical line would give R1C4, R1C3, R2C3, but the top-right quadrant has R2C4 instead of R1C3, so it is not a mirror. Turning top-left by 90° clockwise (RrCc to RcC(5 − r)) gives R1C4, R2C4, R2C3, which matches; turning again gives R4C4, R4C3, R3C3, which also matches.

So the pattern is a pinwheel. One more turn, from bottom-right to bottom-left, sends R4C4 to R4C1, R4C3 to R3C1 and R3C3 to R3C2. Option (a) is the mirror image of the bottom-right quadrant, the trap.

Answer: (c) R3C1, R3C2, R4C1.

Example 6: turned but not flipped

Question: The target is a Z-shape: R1C1, R1C2, R2C2, R2C3. It may be turned but not flipped. In which pattern is it hidden? (a) R1C2, R2C1, R2C2, R3C1, R4C4 (b) R1C2, R1C3, R2C1, R2C2, R4C4 (c) R1C1, R2C1, R2C2, R3C2, R1C4 (d) R1C1, R1C2, R1C3, R2C3, R3C3

Solution: The Z has 2 orientations when turned: lying (top pair to the left, bottom pair to the right) and standing (R1C2, R2C1, R2C2, R3C1, that is right column on top, left column below). Its flipped form is the S-shape, which is not allowed.

Option (b) has a lying S (R1C2, R1C3, R2C1, R2C2) and option (c) has a standing S (R1C1, R2C1, R2C2, R3C2); both are flips. Option (d) is a straight line with a bend and has no step at all. Option (a) has exactly the standing Z.

Answer: (a)

Common mistakes

Using a mirror when the figure has rotational symmetry. Always test the rule on 2 given quadrants before applying it to the gap.

Copying a half without flipping it. A mirror changes column c to 5 − c; sliding the left half to the right (c + 2) is not a mirror.

Rejecting an embedded shape because extra lines or cells surround it. Only the target's own parts need to exist.

Accepting a flipped copy when the question allows only turning. Check whether the step or arm goes the same way as in the target.

Matching only the outline. Inner lines, dots and shading must also continue across the missing part.

Practice questions

Q1. A 4×4 pattern is symmetric about its vertical middle line. The left half has R1C2, R2C1, R3C1, R3C2 and R4C2 shaded. Which cells of the right half are shaded? (a) R1C4, R2C3, R3C3, R3C4, R4C4 (b) R1C3, R2C4, R3C3, R3C4, R4C3 (c) R1C3, R2C4, R3C3, R4C3 (d) R1C3, R2C3, R3C3, R3C4, R4C4

Q2. A 4×4 pattern is symmetric about the diagonal from the top-left corner to the bottom-right corner. Above the diagonal the shaded cells are R1C2, R1C4 and R2C3. Which cells below the diagonal are shaded? (a) R3C1, R4C1, R4C2 (b) R2C1, R4C1, R4C3 (c) R2C1, R3C1, R3C2 (d) R2C1, R4C1, R3C2

Q3. The target is an L-shape: 3 cells in a line with a 4th cell attached beside one end (it may be turned or flipped). Which pattern does NOT contain it? (a) R1C1, R2C1, R3C1, R3C2, R3C3 (b) R1C1, R1C2, R1C3, R1C4, R2C4 (c) R1C1, R1C2, R2C2, R2C3, R3C3 (d) R1C4, R2C2, R3C2, R3C3, R3C4

Answers

Q1. (b) Send c to 5 − c and keep the row: R1C3, R2C4, R3C3, R3C4, R4C3. Option (a) slides the half without flipping.

Q2. (d) Swap row and column: R1C2 to R2C1, R1C4 to R4C1, R2C3 to R3C2.

Q3. (c) An L needs 3 cells in a straight line; the staircase in (c) never has 3 in a line. (a) has R1C1, R2C1, R3C1, R3C2; (b) has R1C2, R1C3, R1C4, R2C4; (d) has R2C2, R3C2, R3C3, R3C4.

Quick revision

Vertical mirror: c to 5 − c. Horizontal mirror: r to 5 − r. Diagonal mirror: swap r and c.

90° clockwise: RrCc to RcC(5 − r); 180°: RrCc to R(5 − r)C(5 − c).

Test the rule on given parts first: mirror and rotation give different answers for pinwheel patterns.

For hidden shapes, look for the feature that never changes with turning: a T-junction, a step, 3 in a line.

Turned but not flipped means a mirror copy is wrong.