Input Output – MBA CET questions with solutions

Approach: Input–output word sorting — fix one word per step, left to right.

Solution:
Rule from the illustration: longer (5-letter) words come first in alphabetical order, then the shorter words in alphabetical order; words are placed one by one from the left.
Target for the new input: cough, fight, sough (5 letters, a→z) then acts, cats, idea, more (4 letters, a→z).
Input: more fight cats cough sough acts idea
Step I: cough more fight cats sough acts idea
Step II: cough fight more cats sough acts idea
Step III: cough fight sough more cats acts idea
Step IV: cough fight sough acts more cats idea
Step V: cough fight sough acts cats more idea
Step VI: cough fight sough acts cats idea more — final arrangement.
No word is already in its final place, so all six placements are needed.

Why the other options are wrong:
A, B, C: stopping at III, IV or V leaves the 4-letter words unsorted.
E: VII would mean placing the last word too, but it falls into place automatically.

Cetking shortcut: Count how many of the first (n − 1) positions hold the wrong word in the input — here all 6 → 6 steps.

Answer: D

Approach: Word arrangement — alternately place the alphabetically last word and the alphabetically first word of the remaining words, from the left.

Solution:
Illustration: take (last) → are (first) → station (2nd last) → brought (2nd first); a word already in place costs no step.
Order: car, demand, fat, on, quick, star.
Input: car on star quick demand fat
Step I: star car on quick demand fat (star to position 1; car is now already 2nd, so no separate step)
Step II: star car quick on demand fat (quick to position 3)
Step III: star car quick demand on fat (demand to position 4)
Now 'on' and 'fat' are already in place → Step III is the last step.

Why the other options are wrong:
A), B), C), E) — none follows the last/first alternation (e.g. C puts 'on' second instead of 'car').

Cetking shortcut: Write the final output first (star car quick demand on fat) — if it appears in the options for an early step, check whether some words were already in place.

Answer: D

Approach: Each step brings the next largest number to the next position from the left (descending order).

Solution:
Input: 03 31 43 22 11 09
Step I: 43 03 31 22 11 09
Step II: 43 31 03 22 11 09
Step III: 43 31 22 03 11 09
Step IV: 43 31 22 11 03 09
Step V: 43 31 22 11 09 03 → fully descending
Last step = Step V

Cetking shortcut: Steps = numbers that are not already in their final position when their turn comes; here only the last number (03) falls into place automatically → 6 − 1 = 5.

Answer: D

Approach: Find the position mapping between Batch 1 and Batch 3, then reverse it.

Solution:
Batch 1: is(1) not(2) ready(3) cloth(4) simple(5) harmony(6) burning(7)
Batch 3: cloth is not ready simple harmony burning = positions 4, 1, 2, 3, 5, 6, 7 of Batch 1
So Batch 1 = Batch 3's words in positions 2, 3, 4, 1, 5, 6, 7.
Batch 3 (11–12): soap shy miss pen yet the she
Batch 1 = shy, miss, pen, soap, yet, the, she → shy miss pen soap yet the she

Why the other options are wrong:
B) has only six words and a wrong order.
C), D) reorder the last three words, but positions 5–7 stay unchanged between Batch 1 and Batch 3.
E) is just the given Batch 3 password.

Answer: A

Approach: Each step brings the next largest number to the next position from the left.

Solution:
Input: 09 25 16 30 32 18 17 06
Step I: 32 09 25 16 30 18 17 06
Step II: 32 30 09 25 16 18 17 06
Step III: 32 30 25 09 16 18 17 06

Why the other options are wrong:
B) repeats 18 and drops 17.
C) contains 26, which is not in the input.
D) does not move 25 and swaps 17/18.
E) is only Step I with 30 not moved.

Answer: A

Approach: Each step moves one number (ascending, to the front) and one word (alphabetical, to the end).

Solution:
Illustration: Step I puts 22 in front and 'defeat' at the end; Step II puts 34 in front and 'ink' at the end, and so on — one number + one word per step; 6 pairs → 6 steps.
New input numbers: 67, 54, 23, 45, 90, 88 (six); words: thankful, message, listing, known, booklist, quit (six).
Final output: 90 88 67 54 45 23 booklist known listing message quit thankful.
No pair is already in place at the start, so 6 pairs need 6 steps.

Answer: C

Approach: Input–output: words in reverse alphabetical order alternate with numbers in ascending order, one element placed per step.

Solution:
Words (Z→A): White, Sky, Night, Lotus, Bear. Numbers (ascending): 19, 36, 55, 90.
Input: 36 Sky 19 Night 90 55 Bear Lotus White
Step I: White 36 Sky 19 Night 90 55 Bear Lotus
Step II: White 19 36 Sky Night 90 55 Bear Lotus
Step III: White 19 Sky 36 Night 90 55 Bear Lotus → 3rd from right = 55
Steps IV–V only rearrange from position 4 onward, so the 2nd element from left stays 19.
Answer: 55 – 19.

Why the other options are wrong: B and E swap the elements; C and D come from sorting words alphabetically (A→Z) or moving all numbers at once.

Cetking shortcut: Once positions 1–2 are fixed (Step II), the second element from the left never changes.

Answer: A

Approach: Input–output machine that places ONE element per step, alternating word (reverse alphabetical) and number (ascending).

Solution:
Example rule: yellow (step I), 15 (II), war (III), 19 (IV), then quick and 28 … final: yellow 15 war 19 quick 28 fire 39.
Words in reverse alphabetical order: hold, goal, die, age. Numbers ascending: 26, 35, 53, 72.
Input: age die 72 53 35 hold goal 26
Step I: hold age die 72 53 35 goal 26
Step II: hold 26 age die 72 53 35 goal
Step III: hold 26 goal age die 72 53 35
Step IV: hold 26 goal 35 age die 72 53
Step V: hold 26 goal 35 die age 72 53
Step VI: hold 26 goal 35 die 53 age 72 — final arrangement.

Why the other options are wrong:
Three/Four/Five: stop before 53 and 72 are in place. Two: far too few — one element moves per step.

Cetking shortcut: Count elements that are NOT already in their final position; here 6 of 8 need a move (age and 72 fall into place automatically), so 6 steps.

Answer: D

Approach: Input–output: words placed two at a time alphabetically, numbers two at a time in descending order, alternating.

Solution:
Rule from the illustration: odd steps bring the next TWO words in alphabetical order to the front block; even steps bring the next TWO numbers in descending order; everything else keeps its original order.
New input: Oral 11 Micy 25 Pink 47 81 Pick 49 Vineet 16 Neenu
Words (A→Z): Micy, Neenu, Oral, Pick, Pink, Vineet. Numbers (high→low): 81, 49, 47, 25, 16, 11.
Step I: Micy Neenu Oral 11 25 Pink 47 81 Pick 49 Vineet 16
Step II: Micy Neenu 81 49 Oral 11 25 Pink 47 Pick Vineet 16
Step III: Micy Neenu 81 49 Oral Pick 11 25 Pink 47 Vineet 16
Step IV: Micy Neenu 81 49 Oral Pick 47 25 11 Pink Vineet 16
Step V: Micy Neenu 81 49 Oral Pick 47 25 Pink Vineet 11 16
Step VI: Micy Neenu 81 49 Oral Pick 47 25 Pink Vineet 16 11
In Step III: … 81 49 Oral Pick … ⇒ Oral is between 49 and Pick.

Why the other options are wrong:
Pink: comes after Pick, near 47. 11: after Pick. Vineet: still at the far end in Step III. 49: one of the two named terms itself.

Cetking shortcut: Only Steps I–III are needed: after Step II (81 49) the next words are Oral and Pick, so Oral lands right after 49.

Answer: D

Approach: Input–output: replay the same pair-wise rule up to Step V.

Solution:
Rule from the illustration: odd steps bring the next TWO words in alphabetical order to the front block; even steps bring the next TWO numbers in descending order; everything else keeps its original order.
New input: Oral 11 Micy 25 Pink 47 81 Pick 49 Vineet 16 Neenu
Words (A→Z): Micy, Neenu, Oral, Pick, Pink, Vineet. Numbers (high→low): 81, 49, 47, 25, 16, 11.
Step I: Micy Neenu Oral 11 25 Pink 47 81 Pick 49 Vineet 16
Step II: Micy Neenu 81 49 Oral 11 25 Pink 47 Pick Vineet 16
Step III: Micy Neenu 81 49 Oral Pick 11 25 Pink 47 Vineet 16
Step IV: Micy Neenu 81 49 Oral Pick 47 25 11 Pink Vineet 16
Step V: Micy Neenu 81 49 Oral Pick 47 25 Pink Vineet 11 16
Step VI: Micy Neenu 81 49 Oral Pick 47 25 Pink Vineet 16 11
In Step V: … Oral Pick 47 25 Pink … ⇒ 47 and 25 sit next to each other ⇒ zero elements between them.

Why the other options are wrong:
Two, Three, Four, Five: all assume 25 and 47 are still at their input positions; by Step IV they were placed together as the second number pair.

Cetking shortcut: 47 and 25 are the 3rd and 4th largest numbers — they move together in Step IV and stay adjacent afterwards.

Answer: A

Approach: Find what decides the last step in the illustration, then apply it to the new input.

Solution:
In the illustration, Step 1 drops the last letter of every word and moves the smallest number to the end; Step 2 only shifts that number behind the first word; Steps 3, 4 and 5 each drop one more letter from every word. The machine stops when the shortest word, Pacer (5 letters), is down to 1 letter: Pace (Step 1), Pace (Step 2), Pac, Pa, P (Step 5).
New input: 20 Quicker towel Parker 41 29 87 Harness. The shortest word is towel (5 letters), the same length as Pacer; the others (Parker 6, Quicker 7, Harness 7) are longer.
So towel reaches 1 letter in Step 5, and the arrangement is complete there: Har 20 Pa 29 Qui 41 t 87.

Why the other options are wrong:
A (3), B (4): the shortest word still has 2 or more letters at these steps.
D (6), E (7): the process has already stopped at Step 5.

Cetking shortcut: Here the last step number equals the length of the shortest word (5 letters, Step 5); compare only the shortest word with the illustration.

Answer: C

Approach: Machine places one element per step: numbers in descending order alternating with words in alphabetical order (number, word, number, word …).

Solution:
Rule from illustration: 65 box 47 cat 9 solve 2 when → largest number, 1st word, next number, 2nd word …; one element moved per step.
New input: denizen 41 denial 18 28 denitrify 5 63 denim denigrate.
Numbers descending: 63, 41, 28, 18, 5. Words alphabetical (all start "deni"): denial, denigrate, denim, denitrify, denizen.
Step 1: 63 denizen 41 denial 18 28 denitrify 5 denim denigrate
Step 2: 63 denial denizen 41 18 28 denitrify 5 denim denigrate
Step 3: 63 denial 41 denizen 18 28 denitrify 5 denim denigrate
Step 4: 63 denial 41 denigrate denizen 18 28 denitrify 5 denim
Step 5: 63 denial 41 denigrate 28 denizen 18 denitrify 5 denim
From the right: denim (1), 5 (2), denitrify (3), 18 (4). Fourth from right = 18.
Required difference = 80 − 18 = 62.

Why the other options are wrong:
A (67) = 80 − 13, B (77) = 80 − 3, C (65) = 80 − 15, D (80) = 80 − 0: none of 13, 3, 15, 0 sits 4th from right in Step 5.

Cetking shortcut: Words sharing a long prefix (deni-) are ranked by the 5th letter: a < g < m < t < z.

Answer: E

Approach: Input–output: words in reverse alphabetical order and numbers in descending order, arranged in blocks of three words + three numbers, one word and one number per step.

Solution:
Rule from illustration: each step brings the next word in reverse alphabetical order into the word block at the front and the next largest number into the number block; elements are placed in groups of three words followed by three numbers (sand rock monk 99 80 79, then jam hole fog 73 60 34 …). The rest keep their order.
New input: 23 mat belt 50 car 36 73 hay 66 owl 40 sell fool 47.
Words (reverse alphabetical): sell, owl, mat, hay, fool, car, belt. Numbers (descending): 73, 66, 50, 47, 40, 36, 23.
Step I: sell 73 23 mat belt 50 car 36 hay 66 owl 40 fool 47
Step II: sell owl 73 66 23 mat belt 50 car 36 hay 40 fool 47
Step III: sell owl mat 73 66 50 23 belt car 36 hay 40 fool 47
Step IV: sell owl mat 73 66 50 hay 47 23 belt car 36 40 fool
Step V: sell owl mat 73 66 50 hay fool 47 40 23 belt car 36
Step VI: sell owl mat 73 66 50 hay fool car 47 40 36 23 belt
In Step VI, between 50 and 36 lie: hay, fool, car, 47, 40 → 5 elements.

Why the other options are wrong:
A (Three), B (Four), C (Two), E (one): miscounts; exactly five elements (hay, fool, car, 47, 40) sit between 50 and 36.

Cetking shortcut: Write only the finished front part of Step VI (sell owl mat 73 66 50 hay fool car 47 40 36) — the question needs nothing else.

Answer: D

Approach: Same machine as the previous question: build steps until the string "47 23 belt car" appears.

Solution:
Rule from illustration: each step brings the next word in reverse alphabetical order into the word block at the front and the next largest number into the number block; elements are placed in groups of three words followed by three numbers (sand rock monk 99 80 79, then jam hole fog 73 60 34 …). The rest keep their order.
New input: 23 mat belt 50 car 36 73 hay 66 owl 40 sell fool 47.
Words (reverse alphabetical): sell, owl, mat, hay, fool, car, belt. Numbers (descending): 73, 66, 50, 47, 40, 36, 23.
Step I: sell 73 23 mat belt 50 car 36 hay 66 owl 40 fool 47
Step II: sell owl 73 66 23 mat belt 50 car 36 hay 40 fool 47
Step III: sell owl mat 73 66 50 23 belt car 36 hay 40 fool 47
Step IV: sell owl mat 73 66 50 hay 47 23 belt car 36 40 fool
Step V: sell owl mat 73 66 50 hay fool 47 40 23 belt car 36
Step VI: sell owl mat 73 66 50 hay fool car 47 40 36 23 belt
"47 23 belt car" appears consecutively only in Step IV (… hay 47 23 belt car 36 40 fool).

Why the other options are wrong:
A (2), B (3): 47 has not yet moved next to 23 (Step III: … 50 23 belt car 36 hay 40 fool 47).
D (5), E (6): fool and 40 are inserted, breaking the run (Step V: … 47 40 23 belt car 36).

Cetking shortcut: The remaining elements keep their input order (23 belt car …); just find the step where 47 is placed right before 23 — that is the step in which 47 enters the number block (Step IV).

Answer: C

Approach: Step I of the machine removes every word with an even number of letters; the remaining words keep their order.

Solution:
Illustration: COVER(5) LAYER(5) MINIMUM(7) CONTROL(7) AMAZE(5) UNIQUE(6) INITIAL(7) ENTER(5) FINAL(5) EXIT(4).
Step I drops UNIQUE(6) and EXIT(4) — the only even-length words — and keeps the rest in order. ✔
New input: FILES(5) GRADE(5) INSIDE(6) AWARE(5) ADVANCE(7) PAIRS(5) FOREIGN(7) ALTER(5) OVER(4) TOWARDS(7).
Remove INSIDE and OVER → Step I: FILES GRADE AWARE ADVANCE PAIRS FOREIGN ALTER TOWARDS.
Word between FOREIGN and TOWARDS = ALTER.

Why the other options are wrong:
B (ADVANCE), C (AWARE), D (GRADE): these lie before FOREIGN.
E (CONTROL): not in the new input at all.

Cetking shortcut: Count letters first whenever words disappear in an input-output step — odd/even length or vowel-start rules are the usual filters.

Answer: A

Approach: Decode the rule from the example — alternately place the alphabetically last and first remaining words from the left.

Solution:
Example: take (last) → are (first) → station (2nd last) → brought (2nd first); gone, enough already fall in place.
New input: car on star quick demand fat. Alphabetical order: car, demand, fat, on, quick, star.
Step 1: star to front → star car on quick demand fat (car is now already 2nd).
Step 2: next largest quick to 3rd → star car quick on demand fat.
Step 3: next smallest demand to 4th → star car quick demand on fat.
(on, fat are then already in order, so Step 3 is also the last step.)

Why the other options are wrong:
A – demand placed before car/quick.
C – quick placed second; car is the smallest and must be second.
D – repeats 'on' and drops star.
E – car first; the largest word (star) must lead.

Answer: B

Approach: From the example, each step moves the next largest number to the next position from the left; a step is skipped if the number is already in place.

Solution:
Input: 03 31 43 22 11 09
Step I: 43 03 31 22 11 09
Step II: 43 31 03 22 11 09
Step III: 43 31 22 03 11 09
Step IV: 43 31 22 11 03 09
Step V: 43 31 22 11 09 03 → fully descending.
Last step = Step V.

Why the other options are wrong: After Steps I–IV, 03 and 09 are still out of order, so the machine needs Step V.

Cetking shortcut: Steps = number of placements needed. Here no number is already in its final place before it is moved, so 6 numbers need 5 placements (the last one falls in place).

Answer: C

Approach: Compare batches two apart — Batch 1 → Batch 3 → Batch 5 follow one fixed rearrangement.

Solution:
Label Batch 1 words 1–7: is(1) not(2) ready(3) cloth(4) simple(5) harmony(6) burning(7).
Batch 3 = cloth is not ready simple harmony burning = 4 1 2 3 5 6 7.
Check Batch 3 → Batch 5: ready cloth is not simple harmony burning = again 4 1 2 3 5 6 7 ✓.
So Batch 3 = (4th word, 1st, 2nd, 3rd, 5th, 6th, 7th) of Batch 1. Reverse it: Batch 1 = (2nd, 3rd, 4th, 1st, 5th, 6th, 7th) of Batch 3.
Batch 3 given: soap shy miss pen yet the she.
Batch 1 = shy miss pen soap yet the she.

Why the other options are wrong: B, C, E reverse the last three words (that happens only in the even batches); D moves the last three words to the front.

Cetking shortcut: The last three words stay fixed between Batch 1 and Batch 3 — keep only options ending 'yet the she' (A only).

Answer: A

Approach: Each step moves the next largest number to the next place from the left; numbers already in place are skipped.

Solution:
Input: 16 09 24 28 15 04
Step 1: 28 to front → 28 16 09 24 15 04
Step 2: 24 to 2nd → 28 24 16 09 15 04 (16 is now already 3rd)
Step 3: 15 to 4th → 28 24 16 15 09 04 (final)

Why the other options are wrong:
A – 04 before 09: neither has been moved yet, so they keep input order (09, 04).
B – 15 before 16 is not descending.
C – 09 cannot come before 16 after 24 is placed.
E – 24 before 28: the largest number goes first.

Answer: D

Approach: Read the final step of the example — words end in descending order of length, and alphabetically within the same length.

Solution:
Example final: crept going light (5 letters, A→Z) | but for sir (3 letters, A→Z) | te (2 letters).
New input: more fight cats cough sough acts idea.
5-letter words: cough, fight, sough (A→Z).
4-letter words: acts, cats, idea, more (A→Z).
Last step: cough fight sough acts cats idea more.

Why the other options are wrong:
B – 'more' placed among 5-letter words.
C – starts with a 4-letter word.
D – 'more' before 'idea' breaks alphabetical order.
E – 'more' before 'cats' and 'idea'.

Cetking shortcut: For a 'last step' question you only need the final ordering rule, not the intermediate steps.

Answer: A

Approach: Decode the machine rule from the example, then run the new input.

Solution:
Rule: Step I – digits of each number in ascending order; Step II – numbers in ascending order; Step III – keep the middle two digits; Step IV – add those two digits.
Step I: 2579 4578 2358 2568 1568 2467 2356 2456
Step II: 1568 2356 2358 2456 2467 2568 2579 4578
Step III: 56 35 35 45 46 56 57 57
Step IV: 11 8 8 9 10 11 12 12
2nd + 4th + 6th + 8th = 8 + 9 + 11 + 12 = 40.

Cetking shortcut: After Step I you only need the middle two digits of each number, and Step II order can be read off the sorted 4-digit values directly.

Answer: D

Approach: Input–output: selection-sort by digit sum, one swap per step.

Solution: Example digit sums: 245(11), 316(10), 436(13), 519(15), 868(22), 710(8), 689(23). Each step swaps the next-smallest-digit-sum number into the next position; a number already in place costs no step.
New input with digit sums: 544(13), 653(14), 325(10), 688(22), 461(11), 231(6), 857(20).
Step I: 231 ↔ 544 → 231, 653, 325, 688, 461, 544, 857
Step II: 325 ↔ 653 → 231, 325, 653, 688, 461, 544, 857
Step III: 461 ↔ 653 → 231, 325, 461, 688, 653, 544, 857
Step IV: 544 ↔ 688 → 231, 325, 461, 544, 653, 688, 857
653 already in place; Step V: 857 ↔ 688 → 231, 325, 461, 544, 653, 857, 688 (final).
Total = 5 steps.

Cetking shortcut: Replace each number by its digit sum (13, 14, 10, 22, 11, 6, 20) and run the swaps on these small numbers — far faster than copying full rows.

Answer: B

Approach: Input–output: alternate word (A→Z) and number (high→low), one placement per step.

Solution: Rule (from the example): words in alphabetical order and numbers in descending order alternate, starting with a word. Each step moves the next required element into the next position, shifting the rest right; if the required element is already in place, no step is used.
Input: 98 11 64 22 but will an it
Step I: an 98 11 64 22 but will it (98 is already in 2nd place, so nothing moves for it)
Step II: an 98 but 11 64 22 will it
Step III: an 98 but 64 11 22 will it
Step IV: an 98 but 64 it 11 22 will
Step V: an 98 but 64 it 22 11 will
Step VI: an 98 but 64 it 22 will 11 (final).

Why the other options are wrong:
A — Step VI exists (it is the last step).
B — that is Step V.
D, E — numbers in ascending order; the rule is descending.

Cetking shortcut: Write the final output first (an 98 but 64 it 22 will 11); the last step is always the final output, so check whether the asked step is the last one.

Answer: C

Approach: Same machine rule.

Solution: Rule (from the example): words in alphabetical order and numbers in descending order alternate, starting with a word. Each step moves the next required element into the next position, shifting the rest right; if the required element is already in place, no step is used.
Input: 32 now 20 gift 53 box 62 at
Step I: at 32 now 20 gift 53 box 62
Step II: at 62 32 now 20 gift 53 box
Step III: at 62 box 32 now 20 gift 53
Step IV: at 62 box 53 32 now 20 gift
(Step V: at 62 box 53 gift 32 now 20 — final.)

Why the other options are wrong:
B — that is Step V (final output).
C, D — numbers not in descending order of placement.
E — Step III is not the last step.

Cetking shortcut: Step IV must have exactly four elements placed (at 62 box 53) followed by the rest in their original relative order: 32 now 20 gift.

Answer: A

Approach: Same machine rule — count steps till the final output.

Solution: Rule (from the example): words in alphabetical order and numbers in descending order alternate, starting with a word. Each step moves the next required element into the next position, shifting the rest right; if the required element is already in place, no step is used.
Input: pay by 18 36 nose ear 72 54
Step I: by pay 18 36 nose ear 72 54
Step II: by 72 pay 18 36 nose ear 54
Step III: by 72 ear pay 18 36 nose 54
Step IV: by 72 ear 54 pay 18 36 nose
Step V: by 72 ear 54 nose pay 18 36
Step VI: by 72 ear 54 nose 36 pay 18 (final — pay and 18 are already in place).
Last step = 6.

Cetking shortcut: Final output has 8 places; count how many fall into place by themselves (here the last two) — 8 − 2 = 6 steps.

Answer: D

Approach: Net count: every 1/2/3 adds one fruit, every 4 removes two fruits.

Solution: Sequence: 1 2 3 3 2 1 4 | 2 3 1 4 | 2 2 3 3 1 4 | 1 1 3 2 3 4
Count: '1' = 6, '2' = 6, '3' = 7, '4' = 4 (total 23 orders).
Fruits added = 6 + 6 + 7 = 19.
Fruits removed = 4 × 2 = 8 (each '4' removes one apple and one guava).
Fruits left = 19 − 8 = 11.
(Check: before every '4' there is at least one apple and one guava in the basket, so every removal is possible.)

Cetking shortcut: Total left = (number of orders) − 3 × (number of 4s) = 23 − 12 = 11.

Answer: B

Approach: Track guavas only: '2' adds one guava, '4' removes one guava.

Solution: Sequence: 1 2 3 3 2 1 4 | 2 3 1 4 | 2 2 3 3 1 4 | 1 1 3 2 3 4
Count: '1' = 6, '2' = 6, '3' = 7, '4' = 4 (total 23 orders).
Guavas added = number of 2s = 6.
Guavas removed = number of 4s = 4.
Running check: 1, 2 → 4 → 1; 2 → 2 → 4 → 1; 2, 2 → 3 → 4 → 2; 2 → 3 → 4 → 2.
Guavas left = 6 − 4 = 2.

Cetking shortcut: Guavas = (#2) − (#4).

Answer: B

Approach: Track apples only: '1' adds one apple, '4' removes one apple.

Solution: Sequence: 1 2 3 3 2 1 4 | 2 3 1 4 | 2 2 3 3 1 4 | 1 1 3 2 3 4
Count: '1' = 6, '2' = 6, '3' = 7, '4' = 4 (total 23 orders).
Apples added = number of 1s = 6.
Apples removed = number of 4s = 4.
Running check: after '1 2 3 3 2 1' apples = 2, '4' → 1; '1' → 2, '4' → 1; '1' → 2, '4' → 1; '1 1' → 3, '4' → 2.
Apples left = 6 − 4 = 2.

Cetking shortcut: Apples = (#1) − (#4).

Answer: B

Approach: Input–output (word sorting): one word is fixed per step in alphabetical order.

Solution:
Rule used: in each step the machine fixes the next word in dictionary (alphabetical) order at the left, after the words already fixed.
Input: in this volume provides the. Alphabetical order: in, provides, the, this, volume.
Step I: "in" is fixed first (already in place) — in this volume provides the.
Step II: "provides" is fixed — in provides this volume the.
Step III: "the" is fixed — in provides the this volume.
Now "this" and "volume" are already in order, so the arrangement is complete at Step III.

Cetking shortcut: Write the sorted order first; the last step is the step that fixes the last word that is out of place.

Answer: C

Approach: Input–output (word sorting): fix one word per step in alphabetical order.

Solution:
Rule used: in each step the machine fixes the next word in dictionary (alphabetical) order at the left, after the words already fixed.
Input: these many other clinicians and. Alphabetical order: and, clinicians, many, other, these.
Step I: and these many other clinicians.
Step II: and clinicians these many other.
Step III: and clinicians many these other.
Step III matches the given line (the paper prints "others" for "other").

Cetking shortcut: Count how many words from the left are already in sorted order in the target line: and, clinicians, many = 3 ⇒ Step III.

Answer: C

Approach: Input–output (reverse): the machine moves one word per step into its alphabetical place at the left; the input that needs exactly three such moves to reach this line is the answer.

Solution:
Step III: in play role schooling vital (the first three words in, play, role are fixed).
Option B: schooling play vital role in → Step I: in schooling play vital role → Step II: in play schooling vital role → Step III: in play role schooling vital ✔ (three moves needed).
A (in play schooling vital role): only role is out of place — one move.
C (in schooling play vital role), D (in schooling play role vital), E (in schooling vital role play): in is already first, so only two moves are needed.

Why the other options are wrong:
A, C, D, E all reach the line in fewer than three steps, so it would not be Step III.

Cetking shortcut: An input that reaches this line only at Step III must have its first three sorted words (in, play, role) all out of place.

Answer: B

Approach: Input–output (word sorting): fix one word per step in alphabetical order.

Solution:
Rule used: in each step the machine fixes the next word in dictionary (alphabetical) order at the left, after the words already fixed.
Input: modern man is becoming a chronic victim. Alphabetical order: a, becoming, chronic, is, man, modern, victim.
Step I: a modern man is becoming chronic victim.
Step II: a becoming modern man is chronic victim.

Why the other options are wrong:
B: this is Step III (chronic also fixed).
C: "becoming" is missing from its place — not a valid step.
D, E: words appear out of the machine's order.

Cetking shortcut: At Step II exactly two words (a, becoming) are fixed at the left; the rest keep input order.

Answer: A