working-memory-s01-q01
First digit 8 + last digit 9 = 17.
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.
First digit 8 + last digit 9 = 17.
With 5 letters, the middle (3rd) position holds P.
Basket and Candle each have 6 letters; Umbrella has 8.
Reading backward: 1 (1st), 5 (2nd), 9 (3rd).
apple (3) + cherry (2) = 5.
Circle appears twice: 1st and 3rd positions.
Add as you go rather than at the end: 5 + 2 = 7, then 7 + 8 = 15. Keeping a running total lightens the memory load.
Rehearsing the list once in order (lamp, river, stone) fixes the positions: River sat in the second slot.
Replaying the sequence in order (T, F, B, X) shows B follows straight after F.
Tally each colour as it passes: green appears in the 2nd and 4th positions, so twice.
The sequence was 7, 1, 8, 4, counting along it puts 8 in the third position. Grouping digits into pairs (71, 84) makes short sequences easier to hold.
The list ended with BRICK. The last item in a studied list is often the freshest in memory, a pattern called the recency effect.
The order was G, M, S, K, V, so the third letter was S. Middle items are the hardest to recall, which makes them worth an extra rehearsal while studying.
The studied list was spoon, ladder, candle, marble, ribbon, carpet never appeared. Checking each option against your mental list one at a time avoids false recognition.
The sequence 3, 9, 9, 2, 6 repeats only the 9, in the second and third positions. Noticing a repeat while studying gives you a strong anchor for later recall.
The order was Priya, Callum, Nadia, Tomas. Callum came directly before Nadia. Replaying the list as a spoken rhythm helps preserve neighbouring pairs.
The instruction specified the second door on the right. Turning directions into a quick mental walk-through makes them far easier to retain than the words alone.
The sequence mixed Q, R, and Z with the digits 4 and 7, three letters in total. Sorting items into categories while studying is itself a working-memory exercise.
Counting along violet, amber, teal, coral, olive, slate places olive fifth. With six items, chunking into two groups of three keeps the positions straight.
The sequence ended ... 1, 9, 3, so the second-to-last number was 9. Seven digits sits near the limit of most people's span: chunking (62, 75, 193) helps.
Reversing 4 - 1 - 7 gives 7 - 1 - 4. A useful shortcut: reversing any sequence leaves the middle element exactly where it was, so it is still 1.
The first letter of a reversed sequence is simply the last letter of the original. Here that is A, with no need to reverse the whole thing.
Reading 2 - 8 - 5 - 3 from the end gives 3, 5, 8, 2, so the first two digits backwards are 3 then 5. Working from the tail end is easier if you rehearse the sequence once forwards first.
Reversed, the sequence reads 7, 2, 6, 4, 9, putting 4 in the fourth position. Equivalently, the fourth-from-the-front of the reversed list is the fourth-from-the-back of the original.
The studied sequence 5, 8, 1, 6 reversed is 6, 1, 8, 5. Doing this from memory combines storage and manipulation: the core of working memory.
Swapping the ends of L - D - Q - J - X moves X to the front and L to the back, so the sequence now starts with X.
Sorted, the digits run 2, 3, 7, 9. The third is 7. Re-sorting held material is harder than reversing because every element changes position.
Moving the 6 to the end gives 1, 8, 4, 5, 6, whose middle (third) digit is 4. Visualising the digits sliding one place left makes the shift concrete.
Reversing the order gives OAK, MINT, PEAR, so the middle word is still MINT, and spelt backwards it reads TNIM. Handling two transformations in a row is a genuine two-step hold.
The reversed sequence starts 4, 6, and 4 + 6 = 10. Notice you only ever needed the last two digits of the original, spotting shortcuts reduces the load.
Update after every step: 5 + 3 = 8, 8 − 2 = 6, 6 + 4 = 10. Discard each old value as you go, only the current total needs holding.
Apple has a and e (2), kiwi has i and i (2), plum has u (1). A running tally reaches 5.
Track the floor after each move: 3 up to 7, down to 5, up to 6. Picturing the lift moving makes each update easier to hold than raw arithmetic.
Only the first letter changes: C becomes P while A-R-T stay put, giving PART. Editing one element of a held word is a small but genuine act of updating.
The evens are 4, 2, 6, and 8, a tally that ticks up to 4. Saying the count in your head at each hit stops it slipping while you scan the next number.
Step by step: 20 halved is 10, plus 6 is 16, minus 9 is 7.
LEFT appears four times and RIGHT twice, so A finishes on 4 and B on 2. Running two counters at once is what makes this harder than a single tally.
Count down in steps: 50 to 43, to 36, to 29. Alternatively, 3 × 7 = 21 subtracted in one go gives the same 29, but the step-by-step version is the memory workout.
Follow each hop: B forward two is D, back one is C, forward three is F. Saying each intermediate letter aloud in your head keeps the chain intact.
Update the stock after every movement: 12 − 5 = 7, 7 + 8 = 15, 15 − 6 = 9, 9 + 2 = 11. Four updates in a row is where unrehearsed totals start to slip.
Doubling 6 gives 12, and DOG has 3 letters, so 12 + 3 = 15. The trick is holding the intermediate 12 while you count the letters.
TRAIN loses its T to become RAIN, then gains an S at the end to become RAINS. Applying edits one at a time, in order, prevents the steps blurring together.
9 is odd, so add 5 to reach 14; 14 is even, so halve it to reach 7. Check each condition against the current value, not the starting one.
North, right turn to east, right again to south, then a left turn from south faces you east. Imagining yourself physically rotating keeps each turn anchored.
TABLE, DESK, and SHELF each contain an E; CHAIR and LAMP do not, so the count is 3. Holding the rule steady while scanning is the real work here.
47 reversed is 74, and 74 + 2 = 76. A common slip is adding 2 before reversing. The order of steps matters.
M moves forward to N. N does not come after P, so the otherwise-branch applies: one more forward lands on O. Conditional steps demand you test before you move.
Swapping the S and T of STONE gives TSONE, with the remaining letters O-N-E untouched. Focus the swap on exactly the two named positions.
14 is over 10, even, and not divisible by 6, so it passes all three tests. 12 and 18 fail the multiple-of-6 condition, and 9 is odd, checking every candidate against every held condition is the exercise.
3 × 4 = 12, then 12 − 5 = 7, then 7 × 7 = 49. The final squaring only works if the 7 was held accurately through the earlier steps.
The studied word was HARBOUR. Doing arithmetic in between creates interference: attaching a quick image (boats in a harbour) protects the word against it.
The studied number was 58. Counting backwards floods the same verbal rehearsal loop the number lives in, which is exactly why distractor tasks make digits slip.
The studied pair was CANDLE and ORANGE. The near-miss alternatives share sounds or meanings with the originals. Precise recall beats gist here.
The sequence was 4, 9, 1, so the middle number was 9. Note how easily the 17 from the distractor task tries to intrude on recall, recognising intrusions is part of the skill.
Of the held letters, T comes later in the alphabet than K. The counting task in between forces you to protect the letters while attention is elsewhere.
The studied note was 3 lemons. Binding the number to the item as a single picture, three lemons in a row, keeps quantity and object from drifting apart.
The studied order was 7, 2, 5. Order errors, recalling the right digits in the wrong sequence, are the most common failure once a distractor intervenes.
The studied phrase was GREEN DOOR. Visualising an actual green door makes the pairing far more robust than rehearsing the two words separately.
34 + 81 = 115. This item stacks interference and computation: the studied numbers must survive the multiplication before you can even start adding them.
The studied allocation was Row F, Seat 12. The wrong options transpose or nudge one detail each. Exactly the errors interference tends to produce with letter-number pairs.
One tea (£2) plus two scones (2 × £3 = £6) totals £8. The juice price is a deliberate extra load. Part of the skill is holding values you end up not needing.
Substitute first, then compute: (5 × 2) + 8 = 10 + 8 = 18. Translating each symbol before starting the arithmetic keeps the two jobs from colliding.
Red (4 kg) plus green (5 kg) makes exactly 9 kg; the other pairs give 11 kg and 12 kg. Testing pairs systematically stops you re-checking combinations you have already ruled out.
From the studied values, 6 + 9 − 1 = 14. The recall and the arithmetic are separate steps: retrieve all three numbers first, then compute.
Doubling each of 5, 3, 8 in turn gives 10, 6, 16. The challenge is keeping the already-transformed digits fixed while you work on the next one.
The distances from 11 are: Anna 1, Ben 2, Chloe 4, so Anna is closest. Converting each held score into a distance is the manipulation step.
Mentally re-ordering the times gives 09:20, 09:35, 09:45. The tram's 09:35 departure is second. The list arrives unsorted on purpose.
The 22 tins split equally as 11 each, so shelf A must pass 3 of its 14 across. Equivalently, move half the difference: (14 − 8) ÷ 2 = 3.
6 × 4 = 24, and the held 27 exceeds it by 3. The held number must survive the multiplication intact before the comparison can happen.
Chain the conversions: 2 tokens = 6 stars, and 6 stars = 12 coins. Two-step unit chains require holding the middle quantity just long enough to convert it again.
Substituting gives 4 + 7 + 4 = 15. Reading the symbols as their numbers in one smooth pass is quicker than translating each one separately.
ALARM contains two As, in the first and third positions, and replacing both gives OLORM. Scanning for every occurrence, not just the first, is the point of the rule.
Applying the mapping digit by digit, 3 becomes F, 1 becomes D, and 2 becomes B, so the decoded sequence is F, D, B, keeping the original order.
Each letter moves forward once: B becomes C and each E becomes F, giving CFF. Apply the shift uniformly: every letter, exactly one place.
Follow the symbols in order: 6 + 10 = 16, 16 − 4 = 12, 12 + 10 = 22. You are holding both the code and a running total at once. A genuine dual load.
One application turns cat into dog; applying the mapping again turns dog into bird. Chained substitutions require re-consulting the held rule at every step.
C is 3, A is 1, and B is 2, so CAB totals 3 + 1 + 2 = 6. Decoding and summing in a single left-to-right pass keeps the load manageable.
2 is even (E), 5 is odd (O), and 8 is even (E), so the sequence becomes E, O, E. Judging each number afresh against the rule avoids drifting into alternation on autopilot.
The chain runs 4 doubled to 8, plus 3 to 11, doubled again to 22. Keeping two named rules and a running value alive together is the heart of rule-based working memory.
Stepping each coded letter back one place turns H into G, B into A, U into T, and F into E, spelling GATE. Decoding letter by letter while holding the partial answer is the exercise.
Jar B receives 2 marbles, then 1 more from jar A, finishing on 3. A transfer changes both jars at once: update the two totals together.
After Beth passes Amir the order is Beth, Amir, Cara; when Cara then passes Amir it becomes Beth, Cara, Amir. Re-stating the full order after every overtake prevents position slips.
Home scores 2 + 1 = 3 goals and Away scores 1 + 2 = 3, so the match ends level at three each. Keeping two independent tallies is the load here.
The kettle becomes 5 + 3 = 8 and the toaster becomes 2 × 2 = 4. Each value gets a different operation, so pairing the right rule with the right item is half the task.
The first swap gives blue, red, green; the second swaps the middle and right cups to give blue, green, red. Visualising actual cups sliding makes each swap easier to hold than the colour names alone.
SALT, STAR, and SOFT start with S (three words); SALT, NET, and SOFT end with T (three words). Note that SALT and SOFT legitimately count towards both tallies.
After the first stop the cart holds 10 − 4 + 2 = 8 books; halving at the second stop leaves 4. The halving must apply to the updated total, not the starting one.
Three steps in total move C3 to D4, then E5, then F6. Letter and number must advance in lockstep: updating one and forgetting the other is the classic error.
Switch 1 is flipped twice, returning it to off, while switches 2 and 3 are each flipped once and stay on. An even number of flips always cancels out.
The transfer leaves P with £30 and lifts Q to £35; Q's £15 payment then drops it to £20, so P holds more with £30. The transfer touches both balances: the payment only one.
From the studied rules: 9 doubled is 18, minus 4 is 14, doubled again is 28. Everything (start value, both rules, and their order) had to be recalled before computing.
Swapping the middle two gives 2, 3, 5, 6, and the first plus third of that new order is 2 + 5 = 7. The reordering must fully settle before the addition starts.
316 reversed is 613, and the smallest digit of the original (1) brings it to 612. Both the reversed number and a fact about the original must be held simultaneously.
PLANET without its last letter is PLANE, which reversed reads ENALP, and its second letter is N. Each transformation output becomes the next step's input.
Monday's 4 plus Wednesday's 5 makes exactly 9; the other pairings give 8, 11, and 6. Four recalled values must all stay available while you test combinations.
8 is even so it becomes 13; 13 is odd so it doubles to 26; 26 exceeds 20 so it drops to 25. Each condition tests the current value, so the chain must be walked strictly in order.
Taking each digit from 10 gives 10 − 9 = 1, 10 − 2 = 8, 10 − 6 = 4, and 10 − 4 = 6, so the sequence becomes 1, 8, 4, 6. Transformed digits must not contaminate the ones still waiting.
After the first swap the seats hold Kofi, Ben, Aisha, Lena; the second swap exchanges Ben and Lena, leaving Lena in seat 2. Re-listing all four seats after each swap keeps the state accurate.
7 days × 3 = 21 and 4 sides × 2 = 8, so 21 − 8 = 13. Retrieving the everyday facts, computing both parts, and holding the first result through the second is the full layered load.
The tokens total 2 + 5 + 7 = 14; halved that is 7, and adding the red token's value of 2 gives 9. The colour mapping must survive right to the final step, because red is needed again at the end.