Non-Verbal Reasoning - MCQ Practice Questions

Understanding:

A solid cube is divided into $3\times 3\times 3=27$ smaller cubes by making 2 cuts along each axis. We need to find how many smaller cubes have no exposed face.

Step 1: Identify interior cubes.

A smaller cube is completely interior only if it is not on any face of the original cube. Along each axis, only the middle slice position (position 2 out of 3) is not on an outer face.

Step 2: Count interior positions.

There is exactly one cube at the very center of the $3\times 3\times 3$ arrangement.

Step 3: Verify.

In a $3\times 3\times 3$ grid, along each dimension the two end positions are on outer faces. Only the middle position (1 per axis) is interior. So $1\times 1\times 1=1$ cube is fully hidden.

Answer:

Only 1 smaller cube is completely interior with no face exposed.

Quick Tip:

For an $n\times n\times n$ cube cut into unit cubes, the number of fully interior cubes is $(n-2{)}^3$. For $n=3$: $(3-2{)}^3=1$.

Understanding:

We cut squares of side 2 cm from each corner of a $10\text{ cm}\times 8\text{ cm}$ sheet and fold up the flaps to form an open box.

Formula:

Step 1: Find new dimensions after cutting.

Step 2: Calculate volume.

Answer:

The volume of the open box is 96 cubic centimetres.

Quick Tip:

Always subtract twice the cut size from each dimension (both ends are cut). The height of the box equals the side of the square cut.

Understanding:

We start with a cube where P is on top and Q faces the observer. We rotate the cube 90° to the right (clockwise when seen from above) and observe that R now faces us.

Step 1: Identify the rotation axis.

Rotating 90° to the right about the vertical (top-bottom) axis means the top and bottom faces do NOT change. Only the four side faces cycle.

Step 2: Determine what changes.

Before rotation: Q faces front. After rotating 90° clockwise (viewed from top), the face that was on the right now faces front. The top and bottom remain the same.

Step 3: Identify the top face.

Since the rotation is about the vertical axis (top to bottom), the face on top before the rotation remains on top after the rotation. P was on top before the rotation, so P is still on top.

Step 4: Confirm consistency.

Q was facing front; after a 90° clockwise rotation (from above), Q moves to the left side. R, which was on the right, now faces front — consistent with the information given.

Answer:

The face on top remains P, since a rotation about the vertical axis does not change the top face.

Quick Tip:

A rotation about the vertical axis only cycles the four lateral faces (front, right, back, left). The top and bottom faces always stay fixed during such a rotation.

Understanding:

The sheet is rolled along the longer side (22 cm), so the circumference of the base circle equals 22 cm and the height of the cylinder equals 10 cm.

Formula:

Step 1: Find the radius.

Step 2: Calculate the volume.

Answer:

The volume of the hollow cylinder is 385 cubic centimetres.

Quick Tip:

When a rectangle is rolled into a cylinder, the longer side becomes the circumference and the shorter side becomes the height (or vice versa depending on which edge meets). Always identify which edge forms the circumference before computing the radius.

Understanding:

A paper is folded three times along a vertical axis. A single triangular notch is cut from the folded right edge (which is a multi-layer edge). We need to find the number of cut-outs after unfolding.

Step 1: Count the layers after three folds.

Each vertical fold doubles the number of layers.

Step 2: Effect of cutting through all layers.

Cutting a notch through all 8 layers produces 8 separate triangular cut-outs when the paper is unfolded. The cut-out appears once for each layer.

Step 3: Why they appear on the edge.

Each fold brings more paper to the folded edge. Since the notch is cut from the folded right edge, upon unfolding, each layer reveals one triangular hole positioned along that edge.

Answer:

When the paper is unfolded, 8 triangular cut-outs will appear.

Quick Tip:

For edge cuts (not interior punches), the number of cut-outs equals the number of layers, which is $2^n$ for $n$ folds. For interior punches that do not touch any fold line, each layer also gives one hole, totalling $2^n$.

Understanding:

We need to find how many cubes of side 3 cm fit into a box of dimensions $12\text{ cm}\times 9\text{ cm}\times 6\text{ cm}$.

Formula:

where $a$ is the side of the small cube.

Step 1: Find how many cubes fit along each dimension.

Step 2: Multiply to get total.

Verification:

Answer:

24 cubes of side 3 cm can be packed into the box.

Understanding:

A square sheet is folded twice — first diagonally and then in half — and a hole is punched through all layers. We need to determine how many holes appear on unfolding.

Step 1: Count layers after each fold.

After the first diagonal fold: 2 layers.

After the second fold (in half): 4 layers.

Step 2: Count holes produced.

Punching once through 4 layers creates 4 holes when fully unfolded, one per layer.

Step 3: Position of holes.

The first diagonal fold creates reflection symmetry along one diagonal of the square. The second fold creates another line of symmetry. The punch point is replicated by each unfolding across these two axes, producing 4 symmetrically placed holes.

Step 4: Confirm no holes coincide on a fold line.

Since the punch is through the center of the smallest triangle (not on a fold edge), all 4 holes are distinct and separate on the fully opened sheet.

Answer:

When the paper is fully unfolded, 4 circular holes will appear.

Quick Tip:

Each fold that does not pass through the punch location doubles the number of holes. Two such folds → $2^2=4$ holes. If the punch lands exactly on a fold line, it counts as a half-hole on that edge, giving fewer distinct holes.

Understanding:

We need to identify which complex figure contains a right-angled triangle hidden (embedded) within it.

Step 1: Analyse each option

A rectangle divided by one of its diagonals produces exactly two right-angled triangles — each having the right angle at a corner of the rectangle. The right angle sits at the bottom-left corner of the lower triangle, matching the given figure precisely.

Step 2: Eliminate other options

A pentagon with all acute interior angles cannot contain a right-angled triangle as a recognisable sub-figure because no internal right angle is formed.

A regular hexagon with only horizontal interior lines produces parallelogram-like strips, not right-angled triangles.

A circle with only a vertical diameter produces two semicircles, not triangles.

Answer:

The rectangle divided by a diagonal contains the given right-angled triangle embedded within it.

Quick Tip:

In embedded figures, always look for the shape that naturally produces the given figure as one of its parts — a diagonal of a rectangle always creates two congruent right-angled triangles.

Understanding:

We need to identify which complex figure does NOT contain an equilateral triangle as an embedded part.

Step 1: Check Option A

A regular hexagon divided from its centre to all six vertices creates six equilateral triangles. The given figure is clearly embedded.

Step 2: Check Option B

A large equilateral triangle divided by joining midpoints of its sides creates four smaller equilateral triangles. The given figure is embedded.

Step 3: Check Option C

A rhombus formed by joining two equilateral triangles along one side directly contains two equilateral triangles. The given figure is embedded.

Step 4: Check Option D

A rectangle divided only by a horizontal midline produces two smaller rectangles. No equilateral triangle (with all equal sides and 60° angles) can be identified within rectangles divided this way, since all angles in a rectangle are 90° and no 60° angle is formed.

Answer:

A rectangle divided by a horizontal midline does not contain an equilateral triangle.

Quick Tip:

Whenever all angles in a complex figure are 90°, no equilateral triangle (which requires 60° angles) can be embedded in it.

Understanding:

We need to find which complex figure contains a perfect square (equal sides, 90° angles) embedded within it.

Step 1: Analyse Option B

When a rectangle is divided by both its diagonals, the two diagonals bisect each other at the centre. The four triangles formed share the centre point. Within the rectangle, a square can be traced by connecting the midpoints of all four sides — this is a standard construction that always produces a square embedded inside any rectangle.

Step 2: Eliminate other options

A regular pentagon has interior angles of 108°; no arrangement of its sides or diagonals produces four equal sides meeting at 90° angles.

A rhombus with angles of 60° and 120° has no 90° angles, so a square cannot be embedded.

An equilateral triangle divided into four smaller equilateral triangles contains only 60° angles — no square is possible.

Answer:

The rectangle divided by both its diagonals contains the square embedded within it.

Quick Tip:

Joining the midpoints of any rectangle always produces a rhombus; if that rectangle is itself a square, the midpoint figure is also a square — a key embedded-figure fact for exams.

Understanding:

We need to find where an isosceles triangle (horizontal base, two equal sides, apex pointing upward) is embedded.

Step 1: Analyse Option A

A regular hexagon divided by lines from each vertex to the centre creates six equilateral triangles. Each of these triangles is isosceles (in fact equilateral, which is a special case of isosceles) with a horizontal base and equal sides. The given isoscopic triangle shape is clearly present.

Step 2: Eliminate other options

Three equally spaced radii in a circle create three sectors but not enclosed triangles — no triangle is fully formed.

Horizontal strips inside a large isosceles triangle create trapezoids (except the topmost piece, which is a smaller triangle), but the whole embedded triangle matching the given shape requires the full figure, not a strip.

A square with both diagonals creates four right-angled triangles, each with a right angle at the corner. These are not isosceles with a horizontal base and apex pointing upward in the required orientation.

Answer:

The regular hexagon divided by lines to the centre contains the given isosceles triangle embedded within it.

Quick Tip:

A regular hexagon divided from its centre always hides six equilateral (and therefore isosceles) triangles — one of the most frequently tested embedded-figure combinations.

Understanding:

We need to identify the complex figure that contains a parallelogram (no right angles, opposite sides parallel and equal) as an embedded shape.

Step 1: Analyse Option A

A regular hexagon with the three main diagonals (connecting opposite vertices) drawn divides the hexagon into six equilateral triangles. Adjacent pairs of these triangles form parallelograms. Specifically, any two neighbouring triangles sharing a long diagonal create a rhombus-shaped parallelogram. The given parallelogram is clearly embedded.

Step 2: Eliminate other options

A circle with two perpendicular diameters creates four quarter-circle sectors — no straight-sided parallelogram is formed.

A medial construction in an equilateral triangle (three medians) creates six smaller triangles meeting at the centroid — the enclosed shapes are triangles, not parallelograms.

A square with one diagonal produces two right-angled isosceles triangles. A triangle is not a parallelogram.

Answer:

The regular hexagon with its three main diagonals contains the parallelogram embedded within it.

Quick Tip:

In a regular hexagon, every pair of adjacent equilateral triangles (formed by the three main diagonals) creates a rhombus — a special parallelogram. Always check hexagon-based figures for hidden parallelograms.

Understanding:

The given figure is a plus/cross shape made of five unit squares: one centre and four arms (top, bottom, left, right). We need to find which complex figure contains this embedded.

Step 1: Understand the cross shape

A plus sign of five equal squares occupies the centre cell and four adjacent cells of a 3×3 grid — everything except the four corner cells.

Step 2: Match with the options

A 3×3 grid with the four corner squares removed (unshaded) leaves exactly the five central and edge-middle squares, forming the cross/plus shape perfectly. The given figure is embedded as the remaining visible area.

Step 3: Eliminate other options

A 3×3 grid with corners shaded highlights the corners, not the cross — the cross is the unshaded region but the figure presented is the full grid with shading, making the cross less directly identifiable as the embedded figure.

A 2×2 grid has only four squares total; a five-square cross cannot fit inside it.

A single square divided by a vertical midline gives only two rectangles — no cross shape.

Answer:

The 3×3 grid with the four corner squares removed directly reveals the embedded cross shape.

Quick Tip:

For cross/plus embedded figures, always visualise a 3×3 grid and mark the five required cells. Any complex figure that naturally isolates those five cells contains the cross.

Understanding:

We need to find the complex figure that contains a right-angled isosceles triangle (90°-45°-45°, two equal legs) embedded in it.

Step 1: Analyse Option A

A square with one diagonal drawn splits the square into exactly two congruent right-angled isosceles triangles. Each triangle has a 90° angle at a corner of the square and two 45° angles at the ends of the diagonal. The two legs are the equal sides of the square. This matches the given figure perfectly.

Step 2: Eliminate other options

A regular hexagon has interior angles of 120°; its diagonals create triangles with angles of 60° and 30°/120° — not 45°.

An equilateral triangle divided into four smaller equilateral triangles contains only 60° angles.

A rectangle with length twice its width and no internal lines is a single closed shape; no triangle is embedded without internal lines.

Answer:

A square divided by one diagonal embeds the right-angled isosceles triangle.

Quick Tip:

A diagonal of any square always produces two 45°-45°-90° triangles — one of the most direct and commonly tested embedded-figure facts.

Understanding:

We need to identify the complex figure that embeds a trapezium — a quadrilateral with exactly one pair of parallel sides.

Step 1: Analyse Option C

When a line is drawn parallel to the base of a triangle and intersects the other two sides, the lower portion (between the base and the new line) is a trapezium: the base of the triangle and the new line are the two parallel sides, while the two cut portions of the triangle's other sides are the non-parallel sides. This directly embeds the given trapezium.

Step 2: Eliminate other options

A circle with two parallel chords forms a trapezium-like region but the boundary includes two arcs, not straight lines — it is not a true straight-sided trapezium.

A regular pentagon with no internal lines is a single closed five-sided figure; no quadrilateral sub-figure is accessible without internal lines.

A rectangle with a vertical midline creates two smaller rectangles — these have two pairs of parallel sides, making them parallelograms, not trapezoids.

Answer:

A triangle with a line drawn parallel to its base embeds a trapezium in its lower portion.

Quick Tip:

Any line drawn parallel to the base of a triangle and intersecting the other two sides always creates a trapezium in the lower region — a fundamental property used repeatedly in embedded-figure and mensuration questions.

Understanding:

We need to find which complex figure contains a downward-pointing equilateral triangle embedded within it.

Step 1: Analyse Option A

When the midpoints of all three sides of an upward-pointing equilateral triangle are joined, the large triangle is divided into exactly four smaller equilateral triangles. Three of them point upward (occupying the corners) and one central triangle points downward. The downward-pointing equilateral triangle is directly embedded at the centre.

Step 2: Eliminate other options

A square with both diagonals creates four right-angled isosceles triangles — all with 90° angles, not equilateral.

A regular pentagon with all diagonals creates a smaller regular pentagon inside with acute triangles (36°-72°-72°), not equilateral triangles.

Three equally spaced radii in a circle without rim-connecting chords create three sectors — no enclosed triangle is formed because the arc boundaries are curves.

Answer:

A large equilateral triangle divided by midpoint-joining lines embeds a downward-pointing equilateral triangle at its centre.

Quick Tip:

Joining the midpoints of an equilateral triangle always creates one central downward-pointing triangle — this is a classic embedded-figure pattern that appears in many competitive exams.

Understanding:

We need to identify the complex figure that contains a semicircle — half a circle bounded by a diameter — as an embedded part.

Step 1: Analyse Option A

A rectangle with a semicircle drawn on top of one of its shorter sides forms a composite shape (like a classic door or window arch). The upper curved portion is precisely a semicircle with the diameter along the top edge of the rectangle. The semicircle is directly and clearly embedded in this figure.

Step 2: Eliminate other options

A triangle with an inscribed circle contains a full circle, not a semicircle. You cannot identify a standalone semicircle (half-circle with a visible diameter) within it.

A square with a circumscribed circle again contains a full circle surrounding the square. No semicircle sub-figure is identifiable as a distinct embedded shape.

Two equal circles touching at a point form a figure-eight-like arrangement. Each circle is complete; no diameter line divides either into a semicircle sub-figure.

Answer:

The rectangle with a semicircle on top of one shorter side directly embeds the given semicircle.

Quick Tip:

In embedded-figure questions involving curved shapes, look for composite figures (shapes built by attaching curved parts to straight-sided figures). The curved component is almost always the embedded figure being tested.