diagrammatic-reasoning-s01-q01
An 'Add 4' box outputs its input plus 4, so 9 becomes 9 + 4 = 13.
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.
An 'Add 4' box outputs its input plus 4, so 9 becomes 9 + 4 = 13.
Doubling multiplies the input by 2, so 7 becomes 14.
Reversing writes the sequence back to front, so P-Q-R becomes R-Q-P. The middle item stays in place for a three-item sequence.
The box deletes only the final item (plum), leaving the rest in their original order.
A quarter turn clockwise moves an upward arrow to point right, imagine turning a clock hand from 12 to 3.
Only the two end symbols trade places (★ and ■); the middle symbols stay exactly where they were.
Halving divides the input by 2, so 26 becomes 13.
The box changes shading only, so the triangle keeps its shape but flips from white to black.
Repeating the whole sequence writes the complete run again after itself, giving A-B followed by A-B, not each symbol doubled in place.
6 − 6 = 0, a valid output; an operator can produce zero.
Work left to right: 5 + 3 = 8, then 8 × 2 = 16.
12 ÷ 2 = 6, then 6 − 2 = 4.
Reversing gives T-A-C; removing the last letter of that result leaves T-A. Apply each box to the previous box's output, not to the original input.
North turned 90° clockwise points east; a further 180° turn points it the opposite way, west.
Removing the first item leaves 4, 6, 8; reversing that gives 8, 6, 4.
3 squared is 9, and 9 + 1 = 10.
Toggling makes the circle black, and rotating a circle produces no visible change. Some operators have no effect on symmetric figures.
Swapping the end letters of STAR gives RTAS; reversing RTAS gives SATR. Track the intermediate result carefully at each stage.
18 − 4 = 14, then 14 ÷ 2 = 7.
Order matters in a pipeline: doubling first gives 12, then adding 2 gives 14, a different result from the original wiring.
Test each candidate against every pair: adding 3 fits 3→6 but fails 5→10, and squaring would turn 9 into 81. Only doubling fits all three examples.
Halving fits 6→3 but fails 10→7; subtracting 3 is the only rule that fits every pair.
For a three-letter word several rules look identical, so use the longer example to separate them: swapping only the end letters of STOP would give PTOS, not POTS. Full reversal fits both.
Multiplying by 4 fits 4→16 but fails 5→25, and adding 12 fails 5→25 too. Each output is the input multiplied by itself, so the box squares.
In both examples the output is the original sequence with its opening symbol appended at the end, nothing else moves.
Adding 8 fits 7→15 but fails 10→21, and tripling-minus-6 also fails 10→21. Doubling then adding 1 fits all three pairs. Always check every example, not just the first.
Both examples keep their shape while black and white exchange, so the rule acts on shading alone.
Subtracting 6 fits 12→6 but fails 20→10; each output is exactly half its input, so the box halves.
Every output is the input with just its final letter missing: removing vowels would have turned RAIN into RN.
All four candidate rules turn 2 into 8, but only cubing also turns 3 into 27: a single example rarely pins down a rule, so seek a second one before deciding.
The count rises by one dot per frame, so the next frame shows 5 dots.
The shading alternates every frame, and the last square shown was white, so black follows.
Each frame turns the arrow 90° clockwise (up → right → down), so the next quarter turn points it left.
Two features change independently: the count rises by one each frame while the shading alternates. After 4 white comes 5 black.
Each shape has one more side than the last (3, 4, 5), so a six-sided hexagon comes next.
The star moves one corner clockwise each frame, so from bottom-right it reaches bottom-left.
Two rules alternate: double, then subtract 1 (2→4→3→6→5→10→9). The next step doubles again: 9 × 2 = 18.
The gaps grow by one each step: +2, +3, +4, so the next gap is +5. Five letters past J is O.
Track each feature separately: the size cycle returns to small, while the rotation continues from down to left.
Each row rises by 1 from left to right (and each row starts one higher than the row above), so the final cell holds 5.
7 is odd, so the No branch applies: 7 + 3 = 10.
12 is greater than 10, so it is halved to 6.
9 is odd, so it becomes 10; 10 is greater than 8, so subtracting 5 gives 5. Re-test the condition on the updated value, not the original.
ORANGE has 6 letters, so it becomes ORANG; ORANG starts with the vowel O, so it is reversed to GNARO.
15 is a multiple of 3, giving 5; 5 is odd, so adding 1 gives 6.
At 1.5 kg the parcel passes the weight test, so the fragile question decides: fragile parcels go to Depot C.
Weight is tested first, so the 2.5 kg fragile parcel still goes to Depot B. Only a parcel at 2 kg or under that is not fragile reaches Depot A.
24 halves to 12; 12 divides by 3 to give 4; 4 is not greater than 5, so adding 10 gives 14.
Check each backwards path for consistency: an even input would be 9, but 9 is odd, contradiction. An odd input of 13 gives 13 + 5 = 18 and stays consistent.
6 is even, so the pair becomes (9, 6); the sum 15 exceeds 14, so the sum itself is output. Note that swapping never changes a sum.
From Amber, one change reaches Red and a second completes the cycle to Green.
Turning the key moves Locked to Closed, and pushing then opens the door.
The coin moves the machine to Ready and the button press moves it on to Printing.
The first press does nothing in Idle; the coin then moves it to Ready and the second press starts Printing. Signals with no listed transition are simply ignored.
Trace floor by floor: 2 → 3 → 4 → 3 → 4.
Fill advances to Wash, then Rinse; the hold keeps it at Rinse; the final advance reaches Spin.
The first coin unlocks it, the second coin is ignored, the first push relocks it, and the second push is ignored, ending Locked.
Every crate passes through Station 1 first; crate 8 is even, so it is routed on to Station 3 and never sees Station 2.
Two tasks drop the status to Tired; the third task only counts one of two; the rest restores Fresh and clears the count; the final task leaves the status still Fresh.
The second 20p lifts credit to 40p, triggering a vend that leaves 10p; the final 10p brings credit to 20p, below the vend threshold.
Both examples double their input, so ◆ turns 6 into 12.
Both examples reverse the sequence, so R-S-T becomes T-S-R.
● adds 3 in both examples, so the pipeline gives 4 × 2 = 8, then 8 + 3 = 11.
In both examples only the first item disappears, so ▲ removes the opening item and X-Y-Z becomes Y-Z.
Squaring fits 3→9 but fails 5→15, and adding 6 fails 5→15 too, only multiplying by 3 fits both examples.
P turns 2 into 4 and 5 into 10, so Q must turn 4 into 9 and 10 into 15, adding 5 fits both runs.
The rotation turns the arrow to point right, and the toggle then flips it from white to black.
Test both wirings: halving first gives 3 then 5, while adding 2 first gives 8 then 4, only halve-then-add matches the observed output.
Compute the intermediate stage first: reversing gives 2-7-4 and 5-1, so ◇ must drop the final item to leave 2-7 and 5.
Work backwards: the final Double must have received 7, so X turned 10 into 7: it subtracts 3, turning 12 into 9.
3 → 6 → 12 → 24; 24 is the first value over 20, and the loop stops there rather than doubling again.
2 → 4 → 8 → 16 → 32, four doublings bring the number past 30.
100 → 85 → 70 → 55 → 40; the condition '40 or less' is met exactly at 40, so no further pass runs.
1 → 4 → 7 → 10 → 13: four additions reach the threshold exactly.
DIAGRAMS has 8 letters, so four removals from the end leave the first four: DIAG.
9 is odd → 14; 14 is even → 7; 7 is odd → 12. Re-test the even/odd condition on each new value.
Track the pass number separately from the value: 5 → 7 → 6 → 8 → 7 after passes 1 to 4.
48 → 24 → 12 → 6 → 3; the loop stops at 3 because the condition 'while even' finally fails.
7 → 10 → 5 → 8 → 4. Note that 5 is not below 5, so the loop must continue past it. Read stopping conditions literally.
Adding 2 to an odd number always gives another odd number, so the stopping condition can never be met, checking whether a loop can terminate is part of reading a process diagram.
Simulate each fault: with Add 5 broken the output would be 6, and with both broken it would be 3. Only a broken Double explains 8, since 3 + 5 = 8 passed straight through.
If Reverse did nothing, the Remove box would trim A-B-C-D to A-B-C. Exactly what was observed. A broken Remove box would instead have produced D-C-B-A.
With Subtract 2 broken, the Triple box receives 6 unchanged and outputs 18, matching the observation. A broken Triple box would have given 4.
The rotation clearly happened (the arrow points down), but the shading never flipped, so the toggle box passed its input through untouched.
Simulate each single fault: a broken Add 1 gives 4 → 8 → 5; a broken Double gives 2; a broken Subtract 3 gives 10. Only the first matches the actual output of 5.
A broken first Double gives 5 → 1 → 2; a broken second Double gives 6; a broken Subtract 4 gives 10 → 10 → 20, which matches the observation.
Correct behaviour would give 17 and doing nothing would give 7; the observed 27 equals 7 + 10 + 10, the rule applied twice over.
With Halve broken the sum is 12 + 4 = 16, matching the observation; a broken Subtract 8 would give 6 + 12 = 18 and both broken would give 24.
With the first Remove broken, the full list is reversed to 4-3-2-1 and then trimmed to 3-2-1: the observed result. A broken Reverse would give 3-4, and a broken second Remove would give 4-3-2.
Judge each run separately against its expected output: 6 matches 3 doubled, but 8 does not match 16, an intermittent fault that only some test runs reveal.
Stage by stage: the arrow turns east, is toggled to black, then turns south. The two quarter turns combine into a half turn regardless of the toggle between them.
From the examples, ● adds 3 (12 → 15 and 20 → 23). Reversing the order changes the result: 8 + 3 = 11, then doubled to 22.
3 × 5 = 15, and the difference between 15 and 4 is 11.
First pass: 9 is odd, so (9 + 7) ÷ 2 = 8. Second pass: 8 is even, so it is squared to 64. Re-test the routing condition on the new value.
The first pair shows a 90° clockwise rotation with shading preserved; rotating a left-pointing arrow 90° clockwise points it up, still black.
Only a broken Add 2 explains the test: 5 → 10 → 9 (a broken Double gives 6 and a broken Subtract 1 gives 14). Through the same fault, 10 doubles to 20 and then loses 1, giving 19.
Each full pass maps n to 2n + 1: 2 → 5 → 11 → 23, and 23 is the first value past 20.
The machine pairs arrivals: 3 is stored, 4 triggers the output 7; then 6 is stored and 2 triggers the output 8.
Black dots rise by 1 while white dots fall by 2 each frame, so the fifth figure has 5 black and 0 white, 5 dots in total.
★ triples and ✦ halves, so the chain runs 4 → 12 → 6 → 18 → 9. Deduce each operator from its examples first, then trace the chain one step at a time.