Analogy Practice Test Questions - Non Verbal Reasoning

Figure B is figure A rotated 135 degrees clockwise, with the pivot dot at the tail and the small barb on the shaft both carried along unchanged. Applying the same 135-degree clockwise rotation to figure C (pointing down-left) brings it all the way back around to pointing straight up.
Option 1 shows the arrow pointing straight down, only 180 degrees short of a full 360-degree turn from C rather than exactly 135 degrees further.
Option 2 shows a diagonal up-left arrow, 45 degrees short of the correct position.
Option 4 shows a horizontal right-pointing arrow, only 90 degrees of rotation from C instead of 135.
Option 5 shows a diagonal up-right arrow, 45 degrees past the correct position.
Option 3 correctly shows the arrow pointing straight up, exactly 135 degrees clockwise from figure C, so it is the right answer.

Rule: A → B is a 135° clockwise turn (pivot dot and barb move with it) → apply the same turn to C (down-left, 225°) → 225+135=360=0° (up)
∴ Option 3 is correct ✓

Two things must both carry over from A to B: the shape becomes solid black, and the small red reference dot stays fixed in the top-left corner throughout. Applying both to figure C (a hollow square with the dot top-left) gives a solid black square with the dot still in the top-left corner.
Option 2 correctly fills the square solid black but drops the reference dot entirely.
Option 3 keeps the dot in place but leaves the square hollow, not applying the fill toggle.
Option 4 correctly fills the shape solid black and keeps a dot in the top-left corner, but changes the shape itself to a circle instead of keeping it a square.
Option 5 correctly fills the square solid black but moves the dot to the top-right corner instead of leaving it top-left.
Option 1 correctly shows a solid black square with the red dot still in the top-left corner, so it is the right answer.

Rule: A → B fills the shape solid black AND keeps the corner dot fixed → apply both to C (square) → solid black square, dot still top-left
∴ Option 1 is correct ✓

Figure B has one more dot than figure A (2 dots becomes 3 dots), the shape itself unchanged. Applying the same "count increases by 1" rule to figure C (2 stars) gives 3 stars, evenly spaced.
Option 1 simply repeats figure C unchanged, with no increase in count.
Option 2 shows 3 stars but unevenly spaced, with two stars bunched close together and the third set apart -- the count is right but the spacing is wrong, so look carefully rather than just counting.
Option 3 shows 4 stars, two more than C rather than one more.
Option 5 shows 3 evenly spaced stars plus an extra small dot beneath them, an unrelated addition rather than a true count increase.
Option 4 correctly shows exactly 3 evenly spaced stars, one more than C, so it is the right answer.

Rule: A → B adds one more of the same item → apply to C (2 stars) → 3 stars
∴ Option 4 is correct ✓

Figure B swaps the outer and inner shapes of figure A: the circle (previously outer) becomes the small inner shape, and the square (previously inner) becomes the outer shape. Applying the same swap to figure C (triangle outside, circle inside) gives a circle outside with a small solid triangle inside.
Option 1 simply repeats figure C unchanged, with no swap.
Option 2 swaps the outer shape to a square instead of a circle.
Option 4 keeps the circle outside but leaves a small solid square inside instead of a triangle.
Option 5 keeps the triangle outside and only makes the inner circle hollow, without swapping the shapes at all.
Option 3 correctly shows a circle outside with a small solid triangle inside, so it is the right answer.

Rule: A → B swaps which shape is outside and which is the small inner shape → apply to C (triangle-out, circle-in) → circle-out, triangle-in
∴ Option 3 is correct ✓

Two marks are added from A to B together: a small tick on top and a small dot at the bottom-right, with nothing else changed. Applying both additions to figure C (a plain square) gives a square with a tick on top and a dot at the bottom-right.
Option 1 simply repeats figure C unchanged, with neither mark added.
Option 3 correctly adds both the tick and the dot but also adds an extra, unwanted tick on the left side.
Option 4 adds only the top tick and leaves out the bottom-right dot.
Option 5 correctly adds both marks but changes the shape to a circle instead of keeping it a square.
Option 2 correctly shows the square with exactly the top tick and the bottom-right dot added, nothing more and nothing less, so it is the right answer.

Rule: A → B adds a top tick AND a bottom-right dot together → apply both to C (square) → square with tick on top and dot at bottom-right, nothing extra
∴ Option 2 is correct ✓

From A to B, the inner star stays exactly the same while only the outer shape changes (square becomes circle). Continuing figure C (star inside a triangle), the outer shape should change again while the star stays the same; the outer shape sequence square, circle, triangle continues to a pentagon.
Option 1 repeats the same triangle outline as figure C, with no change to the outer shape.
Option 2 reverts the outer shape back to a square instead of advancing it.
Option 3 uses a pentagon outline but replaces the inner star with a plain dot.
Option 4 uses a circle outline, repeating figure B's outer shape instead of advancing further.
Option 5 correctly keeps the star unchanged inside a new pentagon outer shape, so it is the right answer.

Rule: inner star never changes, only the outer shape advances (square → circle → triangle → pentagon) → apply to C (triangle) → pentagon with same star inside
∴ Option 5 is correct ✓

Figure B replaces the solid black fill of figure A with diagonal line hatching, the outline unchanged. Applying the same rule to figure C (a solid black square) gives a square filled with diagonal hatching.
Option 2 keeps the shape a circle instead of changing it to a square.
Option 3 uses a dotted fill pattern instead of diagonal line hatching.
Option 4 leaves the square completely hollow instead of hatched.
Option 5 simply repeats figure C unchanged, still solid black.
Option 1 correctly shows the square filled with diagonal hatching, so it is the right answer.

Rule: A → B replaces solid fill with diagonal hatching, outline unchanged → apply to C (solid square) → hatched square
∴ Option 1 is correct ✓

Figure B is figure A mirrored left-to-right, with nothing else changed. Applying the same left-right mirror to figure C (an arrow pointing up-right) gives an arrow pointing up-left.
Option 1 simply repeats figure C unchanged, with no mirroring.
Option 2 rotates the arrow 180 degrees instead of mirroring it, pointing down-left.
Option 3 flips the arrow top-to-bottom instead of left-to-right, pointing down-right.
Option 5 shows a mirrored arrow that is noticeably shorter than the original, an inconsistent change in length.
Option 4 correctly shows the arrow mirrored left-to-right, pointing up-left at the same length, so it is the right answer.

Rule: A → B is a left-right mirror flip → apply to C (up-right arrow) → up-left arrow
∴ Option 4 is correct ✓

Figure B has exactly double the number of spokes of figure A (2 spokes becomes 4 spokes). Applying the same doubling rule to figure C (3 spokes) gives 6 spokes.
Option 1 shows 5 spokes, not double the 3 spokes in C.
Option 3 shows 8 spokes, more than double.
Option 4 shows only 3 spokes, simply repeating figure C unchanged.
Option 5 shows 4 spokes, repeating figure B's count instead of doubling C.
Option 2 correctly shows exactly 6 spokes, double the count of figure C, so it is the right answer.

Rule: A → B doubles the spoke count (2 → 4) → apply to C (3 spokes) → 6 spokes
∴ Option 2 is correct ✓

Two changes happen from A to B together: the shape shrinks to exactly half its size, and its fill changes from hollow to diagonal hatching. Applying both changes to figure C (a large hollow hexagon) gives a half-size hatched hexagon.
Option 1 correctly applies the hatched fill but keeps the hexagon at full size, not shrinking it.
Option 2 correctly shrinks the hexagon to half size but leaves it hollow instead of hatching it.
Option 4 correctly shrinks the shape to half size and hatches it, but the shape is a square instead of a hexagon.
Option 5 correctly shrinks the hexagon to half size but fills it fully solid black instead of with diagonal hatching.
Option 3 correctly shows a half-size hexagon with diagonal hatching, applying both changes together, so it is the right answer.

Rule: A → B shrinks to half size AND hollow becomes hatched → apply both to C (large hexagon) → half-size hatched hexagon
∴ Option 3 is correct ✓

Figure B is figure A with exactly one corner cut off (replaced by a short diagonal edge), everything else unchanged. Applying the same rule to figure C (a plain pentagon) gives a pentagon with one corner cut off.
Option 1 simply repeats figure C unchanged, with no corner cut.
Option 2 keeps the shape a square instead of a pentagon.
Option 3 changes the shape to a hexagon instead of cutting a corner off the pentagon.
Option 4 correctly cuts one corner but also adds an unrelated dot near the top, an extra change beyond the established rule.
Option 5 correctly shows the pentagon with exactly one corner cut off and nothing else changed, so it is the right answer.

Rule: A → B cuts exactly one corner off the shape → apply to C (pentagon) → pentagon with one corner cut, no other change
∴ Option 5 is correct ✓

Two changes happen from A to B together: the arrow rotates 90 degrees clockwise (up becomes right), and a red dot is added at its tip. Applying both changes to figure C (a plain arrow pointing down) gives an arrow pointing left with a red dot at its tip.
Option 2 keeps the arrow pointing right and adds the dot, but does not apply the further rotation needed from C.
Option 3 rotates the arrow to point up and adds the dot, which is a 180-degree rotation from the correct direction rather than 90 degrees.
Option 4 correctly rotates the arrow to point left but leaves out the red dot entirely.
Option 5 rotates the arrow to a diagonal up-left direction with a dot, overshooting the correct 90-degree step.
Option 1 correctly shows the arrow pointing left with a red dot at its tip, applying both changes together, so it is the right answer.

Rule: A → B does two things at once -- rotate 90° clockwise AND add a dot at the tip → apply both to C (down arrow) → left arrow with dot
∴ Option 1 is correct ✓

Two changes happen from A to B together: the count increases from one shape to two, and the fill toggles from solid black to hollow. Applying both changes to figure C (one solid black square) gives two hollow squares side by side.
Option 1 shows two squares but they remain solid black instead of toggling to hollow.
Option 2 correctly toggles to hollow but keeps the count at one square instead of two.
Option 3 correctly shows two hollow shapes but they are triangles, not squares, ignoring figure C's shape.
Option 5 shows three hollow squares, one more than the correct count of two.
Option 4 correctly shows exactly two hollow squares side by side, applying both changes together, so it is the right answer.

Rule: A → B does two things at once -- count goes from one to two AND fill toggles solid-to-hollow → apply both to C (one solid square) → two hollow squares
∴ Option 4 is correct ✓

Figure B swaps the left and right positions of the two elements in figure A, with the elements themselves unchanged. Applying the same left-right swap to figure C (triangle on the left, circle on the right) gives the circle on the left and the triangle on the right.
Option 1 simply repeats figure C unchanged, with no positions swapped.
Option 3 keeps the circle on the right but replaces the triangle with a star, changing the elements instead of just swapping them.
Option 4 swaps the elements' vertical arrangement (circle above, triangle below) rather than swapping left and right.
Option 5 correctly places the circle on the left but replaces the triangle with a dot on the right, an unrelated element change.
Option 2 correctly shows the circle on the left and the triangle on the right, the same two elements simply swapped, so it is the right answer.

Rule: A → B swaps the left and right elements, elements unchanged → apply to C (triangle-left, circle-right) → circle-left, triangle-right
∴ Option 2 is correct ✓

Figure B has one fewer dot than figure A (4 dots becomes 3 dots), the shape itself unchanged. Applying the same "count decreases by 1" rule to figure C (5 stars) gives 4 stars.
Option 1 simply repeats figure C unchanged, with no decrease in count.
Option 2 changes the shape to dots instead of keeping it stars.
Option 3 shows 3 stars, two fewer than C rather than one fewer.
Option 4 shows 4 stars but with an extra small dot beneath them, an unrelated addition rather than a clean count decrease.
Option 5 correctly shows exactly 4 stars with nothing else added, one fewer than C, so it is the right answer.

Rule: A → B removes one item from the row → apply to C (5 stars) → 4 stars
∴ Option 5 is correct ✓