abstract-reasoning-s01-q01
The sequence simply alternates between two shapes, so after a circle the square returns.
This is untimed educational practice for one cognitive skill. It is not a clinical, diagnostic, employment, or professionally recognized assessment, and it does not produce an IQ score or credential.
The sequence simply alternates between two shapes, so after a circle the square returns.
The three sizes cycle in a fixed order, so after medium the cycle continues with large.
Each step skips exactly one letter of the alphabet, so G is followed by I.
Each group adds one more star than the last, so three stars are followed by four.
The arrow rotates a quarter turn clockwise each step through a four-direction cycle, so after right it points down.
The repeating block is two black dots followed by one white dot, so the white dot is due next.
Each step moves two letters backwards through the alphabet, so T is followed by R.
The three symbols repeat in a fixed cycle, and the cycle has reached the sun's turn.
The side count climbs 3, 4, 5, so the next shape has six sides, a hexagon.
Each group has twice as many dots as the one before, so 8 doubles to 16.
Three of the shapes are built entirely from straight sides; the circle is the only curved shape.
X, K and T are drawn with straight strokes only, while S is drawn entirely with curves.
A, H and M each have a vertical line of symmetry, their left and right halves mirror each other, but F does not.
Cube, sphere and cylinder are solid three-dimensional forms; the square is a flat two-dimensional shape.
Three of the pairs repeat the same letter twice; only one pairs two different letters.
Three codes place a letter before its matching digit; one reverses the order and puts the digit first.
The triangle, pentagon and heptagon have odd numbers of sides (3, 5, 7); the hexagon's six sides make it the only even one.
B, D and H sit at alphabet positions 2, 4 and 8, each double the last, while K sits at position 11.
Three groups contain exactly three identical symbols; the star group contains only two.
S, N and Z look the same after a half-turn (180° rotation), but E does not share this rotational symmetry.
Track each feature separately: the shape alternates, so a square is due; the colour cycle red-blue-green has reached green.
The cycle is three directions long (up, right, down), and after up the cycle continues with right.
Three hours past 9 wraps around the clock face back to 12, completing the four-step cycle.
The sequence alternates between a fixed A and a letter that advances one step each visit (B, C, D), so E is due.
The two features run at different speeds: the triangle flips every step (so ▼ is due) while the shading holds for two steps (so black continues).
The cycle is four symbols long, so position 10 = two full cycles (8) plus 2 more, the second symbol of the cycle.
Position 12 is exactly four complete cycles of three, so it lands on the final term of a cycle.
The shapes alternate circle/square while the count rises by one each pair, so three squares follow three circles.
The pendulum passes through the centre between every swing to a side, so centre always follows left or right.
Run the two cycles separately: the letter cycle X-Y-Z has reached Y, and the case alternation has reached lower case.
Each group adds two dots to the last, so 7 grows to 9.
The pattern loses two blocks each step, so 4 shrinks to 2.
The amount added grows by one each step (+1, +2, +3, +4), so the next step adds 5 to reach 16.
Each count is twice the one before, so 16 doubles to 32.
Each count is half the one before, so 8 halves to 4.
Two rules alternate: double, then subtract one. After doubling 9 to 18, the next step subtracts one to give 17.
The counts are square grids of side 1, 2, 3, 4, so the next grid is 5 × 5 = 25 dots.
Each triangle adds a new bottom row one dot longer than the last (+2, +3, +4), so the next step adds 5 to reach 15.
The rule alternates +3 then −2. After adding 3 to reach 10, the next step subtracts 2 to give 8.
Adding the last two counts, 5 + 8, gives 13.
The relationship enlarges the shape without changing it, so the square simply becomes large.
b and d are left-right mirror images of each other, and the mirror image of p is q.
The number names how many sides the shape has, and the six-sided shape is the hexagon.
The relationship swaps the shading from black to white while keeping the shape the same.
The relationship reverses the order of the two symbols, so CD becomes DC.
The relationship turns a flat shape into its three-dimensional counterpart, and the solid built from squares is the cube.
The relationship doubles the first number, and 5 doubled is 10.
The relationship moves two places forward in the alphabet, and two letters after M comes O.
The relationship swaps the roles of the two symbols: the outer symbol moves to the middle and vice versa.
The relationship moves the last symbol to the front of the group, so the star leads the two squares.
Reversing writes the symbols in the opposite order, so the last letter comes first.
Only the outer symbols trade places, 2 and 9 swap, while the middle symbols 5 and 8 stay put.
Each symbol is written twice in place before moving to the next, following the worked example exactly.
The operator keeps the 1st, 3rd and 5th symbols and drops the 2nd and 4th, just as in the worked example.
Each letter moves forward exactly one place: C to D, F to G, K to L.
Work from the inside out: swapping first and last turns ABC into CBA, then doubling each symbol gives CCBBAA.
Each rotation moves the front symbol to the back: WXYZ becomes XYZW, then XYZW becomes YZWX.
Follow the cycle one step at a time: the circle first becomes a square, and the square then becomes a triangle.
Reversing 1234 gives 4321, and removing the first symbol of that result leaves 321.
The original string XY is followed by its reverse YX, producing a mirror-symmetric result.
Each row must contain all three shapes exactly once, and the circle is the one still missing from row 3.
Moving one cell to the right adds one dot, so the middle cell holds 4 + 1 = 5 dots.
Each row shifts the previous row one place to the left, so row 3 continues white, black, grey, and grey also completes its column.
Combining the two cells overlays their contents, and a vertical line laid over a horizontal line forms a cross.
Columns 2 and 3 already have their single star, so row 3's star must occupy the only remaining column.
The side count rises 5, 6 along the row, so the final cell holds the seven-sided heptagon.
A quarter turn clockwise carries a left-pointing arrow round to point up.
Applying the stated rule to the bottom-right cell gives 3 × 3 = 9 dots.
The square appears in both cells so it is excluded, leaving the circle and triangle, which each appear only once.
Row 3 already has large and small, so medium completes the row, and it also completes the third column, which holds large and small so far.
Two strands alternate: 1, 2, 3, 4 and 10, 20, 30. The next term belongs to the tens strand.
One strand climbs from the start of the alphabet (A, B, C) while the other descends from the end (Z, Y, X); the climbing strand is due next.
Triangle groups (growing by one) alternate with advancing letters, and it is the triangles' turn with four.
The odd positions climb by two (2, 4, 6, 8) and the even positions climb by three (3, 6, 9); the next term continues the threes with 12.
The letter strand steps back two places each time (Z, X, V, T) while the number strand doubles; the letters are due next with R.
The shapes alternate circle/square while the count rises by one each term, so six squares follow five circles.
The odd positions hold square numbers (1, 4, 9, 16) and the even positions simply count (2, 3, 4, 5); the next square number is 25.
The letters advance two places each step while the digits climb through the odd numbers, giving I paired with 9.
Three strands rotate (a count, a letter, then a dot group that grows by one), and it is the dots' turn with three.
One strand falls by ten (100, 90, 80, 70) while the other doubles (1, 2, 4, 8); the falling strand is due next.
Each example reverses the order of the digits, so 39 becomes 93.
The machine shifts every letter one place forward in the alphabet, so P, Q, R become Q, R, S.
Each output is three times the input plus one, a rule all three examples satisfy, so 9 maps to 28.
The machine repeats only the final symbol, so XY gains a second Y.
Each output is the sum of the input's digits (2+5=7, 6+3=9, 4+8=12), so 5+7 gives 12.
The machine halves the number of dots, so eight dots become four.
Each example removes one side from the shape, so the seven-sided heptagon becomes a six-sided hexagon.
Each output is the pair's product plus two (2×3+2=8, 3×4+2=14, 4×5+2=22), so 5×6+2 gives 32.
Each output is the letter's position in the alphabet, and Q is the 17th letter.
The machine accepts numbers that read the same forwards and backwards, and only 45654 is such a palindrome.
Reversing 58274 gives 47285, and deleting that result's last symbol leaves 4728.
Trace the wrap-around jumps one at a time: 1 to 4, 4 to 7, 7 to 2, 2 to 5, and finally 5 to 8.
Apply the alternating rules in order: 5 doubles to 10, minus 3 is 7, doubles to 14, minus 3 is 11.
Shifting A, C, E forward two places gives C, E, G; reversing that intermediate result gives G, E, C.
Each double flip cancels itself out, so only the three single flips matter: an odd number of flips turns white to black, whatever the order.
Two quarter turns clockwise make a half turn, and a half turn points a left-facing arrow to the right.
The gaps between terms double each step (+1, +2, +4, +8), so the next gap is +16, giving 34.
The gaps grow by two each step (+4, +6, +8, +10), so the next gap is +12, giving 42.
Handle each symbol type with its own rule: D and H step back to C and G, while 3 and 5 double to 6 and 10.
Doubling AB gives AABB, and keeping the 1st and 3rd symbols of that result restores the original AB, the two operators undo each other.