Question 1

Classification (family relations – role type): Three terms denote sibling/extended relations; one denotes a direct parent. Identify the odd one out.

Accepted Answer

Introduction / Context: Family-term classification often hinges on role type. Brother and cousin are lateral relations (sibling/extended), and nephew is a next-generation collateral relation. Mother, however, is an ascendant (direct parent). The split between lateral/collateral kin and direct ascendant establishes a crisp odd-one-out criterion familiar from basic reasoning tests. Given Data / Assumptions: Brother → sibling (same generation). Cousin → extended lateral relation (same generation across branches). Nephew → collateral next generation (child of sibling). Mother → direct ascendant (parent). Concept / Approach: Group by genealogical direction: lateral/collateral versus ascendant. The only ascendant in the list is mother, so that term becomes the outlier while the remaining three can be grouped as non-ascendant relatives. Step-by-Step Solution: Tag brother, cousin, nephew as non-ascendant kin.Tag mother as ascendant.Select mother as the unique ascendant item. Verification / Alternative check: Check dependency language: “child of” applies only to mother among the options; the others depend on sibling/extended links. Why Other Options Are Wrong: They do not represent a direct parent-child relation. Common Pitfalls: Confusing “nephew” with “son.” Nephew is the child of a sibling, not a direct descendant. Final Answer: Mother

Question 2

Classification (plant parts we eat): Three options are underground storage/organ parts; one option is primarily a leaf/seed herb used above ground. Identify the odd one out.

Accepted Answer

Introduction / Context: Botany-based classification often uses the edible plant part. Onion (bulb), beetroot (storage root), and ginger (rhizome) are underground plant parts commonly used as vegetables/spices. Coriander, however, is chiefly used as above-ground leaves (cilantro) and seeds (a spice), not as an underground storage organ. This contrast between subterranean edible parts and an aerial herb yields a clear odd one out. Given Data / Assumptions: Onion → bulb (modified stem/leaf base underground). Beetroot → storage root. Ginger → rhizome (underground stem). Coriander → leaves and seeds used; not an underground storage organ. Concept / Approach: Categorize by plant anatomy: underground storage organs versus aerial herb parts. The unique aerial herb in this set is coriander, establishing the odd entry without ambiguity. Step-by-Step Solution: Group onion, beetroot, ginger as underground edible structures.Identify coriander as aerial herb/seeds.Select coriander as the outlier. Verification / Alternative check: Market perspective: onion/beetroot/ginger are sold from below-ground harvest; coriander is sold as fresh leaves/bunches and dry seeds. Why Other Options Are Wrong: They are all underground storage or stem/leaf base structures. Common Pitfalls: Assuming “root vegetables” include coriander due to culinary familiarity. Distinguish usage from plant anatomy. Final Answer: Coriander

Question 3

Classification (tableware – function): Three items are drinking vessels; one item is a flat dish for serving/eating. Identify the odd one out.

Accepted Answer

Introduction / Context: Household-object classification commonly distinguishes items by function. Cup, tumbler, and jug are used to hold/pour liquids for drinking; a plate is used to serve/eat solid food and is not a vessel. The function boundary (container for liquids vs flat dish) creates a simple and widely understood odd-one-out scenario appropriate for quick reasoning checks. Given Data / Assumptions: Cup → small drinking vessel. Tumbler → drinking glass. Jug → vessel for holding/pouring liquids. Plate → flat dish for food. Concept / Approach: Group by “designed to contain and/or pour liquids for drinking.” The three vessels form the majority class; the plate does not meet the containment function and is therefore the outlier. Step-by-Step Solution: Tag cup/tumbler/jug as vessels.Tag plate as flat serving dish.Select plate as odd one out. Verification / Alternative check: Attempt the sentence “Fill the ____ with water.” Works for cup, tumbler, jug; not for a plate in normal use. Why Other Options Are Wrong: They satisfy the vessel function and are meant to contain liquids. Common Pitfalls: None significant; this is a straightforward functional classification. Final Answer: Plate

Question 4

Classification (natural vs man-made phenomena/objects): Three are natural entities/phenomena; one is a human-made machine. Identify the odd one out.

Accepted Answer

Introduction / Context: World-knowledge classification frequently separates natural objects and phenomena from artificial constructs. River (natural watercourse), Earth (natural celestial body), and breeze (natural movement of air) are all natural. An aeroplane is a human-engineered machine designed for flight. The natural vs man-made boundary provides the intended odd-one-out contrast. Given Data / Assumptions: River → natural geographic formation. Earth → natural planet. Breeze → natural atmospheric phenomenon. Aeroplane → man-made vehicle/machine. Concept / Approach: Partition by origin: natural vs artificial. Exactly one entry is artificial, so it becomes the outlier under the standard 3:1 structure used in reasoning questions. Step-by-Step Solution: Mark river/Earth/breeze as natural.Mark aeroplane as human-made.Select aeroplane as the odd one. Verification / Alternative check: Consider whether the item exists without human intervention. Aeroplanes require design and manufacture; the other three exist independently of humans. Why Other Options Are Wrong: They are not artifacts; they occur naturally. Common Pitfalls: None; this is a direct natural/artifact split. Final Answer: Aeroplane

Question 5

Classification (number properties): Three numbers are not perfect numbers; one number is a perfect number (equal to the sum of its proper divisors). Identify the odd one out.

Accepted Answer

Introduction / Context: Number-property classification often focuses on special categories such as primes, squares, and perfect numbers. A perfect number equals the sum of its proper divisors (divisors less than the number itself). Among the options, 28 is the classic small perfect number; the others are not perfect. Recognizing perfect numbers is a standard competitive-exam micro-fact that can quickly unlock such items. Given Data / Assumptions: 12 → proper divisors: 1, 2, 3, 4, 6 (sum = 16 ≠ 12). 28 → proper divisors: 1, 2, 4, 7, 14 (sum = 28 → perfect). 52 → proper divisors: 1, 2, 4, 13, 26 (sum = 46 ≠ 52). 96 → proper divisors sum is much larger than 96 (abundant), but not equal to 96. Concept / Approach: Compute the sum of proper divisors or recall the well-known early perfect numbers (6, 28, 496, 8128). If exactly one option matches, it is the outlier by property. This method exploits memorized landmarks to avoid time-consuming divisor calculations in the exam setting. Step-by-Step Solution: Recall perfect numbers list: 6, 28, 496, 8128...Check provided options against this list or verify by summing divisors.28 fits; others do not. Verification / Alternative check: Explicitly sum divisors for 28 as shown to confirm equality. Why Other Options Are Wrong: 12, 52, 96 are not perfect; they are either abundant or deficient. Common Pitfalls: Confusing abundant numbers with perfect ones due to large divisor sums; equality is required. Final Answer: 28

Question 6

Classification (divisibility feature): Exactly one number is divisible by 11; the others are not. Identify the odd one out.

Accepted Answer

Introduction / Context: Quick divisibility checks are common in classification problems. A routine trick is to test for divisibility by notable small primes such as 11. Among the options, one number is a clean multiple of 11, while the others are not. Spotting this property gives a fast and reliable route to the answer without heavy computation. Given Data / Assumptions: 143 = 11 * 13. 217 = 7 * 31 (not divisible by 11). 241 = prime (not divisible by 11). 157 = prime (not divisible by 11). Concept / Approach: Use the 11-divisibility test or factor recognition. Only one number among the set is a multiple of 11, giving the required 3-to-1 split for an odd-one-out item. Step-by-Step Solution: Apply divisibility checks/factoring.Find 143 = 11*13 → divisible by 11.Verify others are not multiples of 11. Verification / Alternative check: Perform the alternating-sum test for 11: for 143, (1 − 4 + 3) = 0 → divisible by 11. Why Other Options Are Wrong: 241 and 157 are primes; 217 factors as 7*31. None equal 11*k. Common Pitfalls: Assuming 217 = 11*19 (it is not); 11*19 = 209. Always verify. Final Answer: 143

Question 7

Classification (materials – origin): Three are synthetic polymers; one is a natural protein fibre. Identify the odd one out.

Accepted Answer

Introduction / Context: Materials classification often hinges on source and composition. Nylon, polythene (polyethylene), and generic plastics are man-made synthetic polymers derived from petrochemical processes. Silk, in contrast, is a natural protein fibre produced by silkworms. The natural vs synthetic origin creates the standard odd-one-out boundary used in many reasoning exams. Given Data / Assumptions: Nylon → synthetic polymer. Polythene → synthetic polymer (polyethylene). Plastic → umbrella term generally referring to synthetic polymers. Silk → natural animal fibre (protein). Concept / Approach: Group items by origin (natural vs synthetic). With exactly one natural item present, the selection is straightforward. Step-by-Step Solution: Tag nylon/polythene/plastic as synthetic.Tag silk as natural.Pick silk as the odd one. Verification / Alternative check: Ask which item is produced biologically rather than industrially; silk alone fits this description. Why Other Options Are Wrong: They are all synthetic polymers made via chemical processing. Common Pitfalls: Equating “plastic” with a specific polymer. The generic label does not change its synthetic nature. Final Answer: Silk

Question 8

Classification (musical instruments – family): Three are string instruments; one is a wind instrument. Identify the odd one out.

Accepted Answer

Introduction / Context: Instrument-family classification is a common verbal topic. Guitar, violin, and veena are stringed instruments that produce sound via vibrating strings. A flute is a wind/woodwind instrument that produces sound from the vibration of air columns. The family difference (string vs wind) sets up a clean odd-one-out decision point. Given Data / Assumptions: Guitar → strings. Violin → strings. Veena → strings (Indian plucked lute family). Flute → wind instrument. Concept / Approach: Group by sound-production mechanism: vibrating strings versus vibrating air. The unique wind instrument is the outlier among three string instruments. Step-by-Step Solution: Classify each item by instrument family.Identify flute as wind; others as strings.Select flute as odd one out. Verification / Alternative check: Ask whether the instrument uses a bow/plectrum/finger plucking (strings) or breath (wind). Only the flute requires breath. Why Other Options Are Wrong: They belong to the same strings family. Common Pitfalls: None; this is a textbook family-based classification. Final Answer: Flute

Question 9

Classification (letter pairs – fixed forward jump): In three pairs, the second letter is 3 positions ahead of the first in the alphabet; one pair violates this +3 rule. Identify the odd pair.

Accepted Answer

Introduction / Context: Alphabet-pattern classification often uses fixed positional jumps. Here, test whether the right letter is exactly three steps ahead of the left one (A=1, B=2, …, Z=26). Three pairs obey the +3 rule; one does not. The calculation is light and intended to reward accuracy over guesswork. Given Data / Assumptions: R→U: 18→21 = +3. B→E: 2→5 = +3. F→I: 6→9 = +3. I→M: 9→13 = +4 (violation). Concept / Approach: Convert letters to positions and compute the delta. Select the pair that does not yield +3. This is a standard technique for letter-series items and is quick to execute once the mapping A=1…Z=26 is internalized. Step-by-Step Solution: Map each left–right pair to numbers.Compute increments; verify +3 for RU, BE, FI.Detect +4 for IM → odd one out. Verification / Alternative check: Rebuild by starting letter: add three letters forward; if the result mismatches the given right letter, that pair is the outlier. Why Other Options Are Wrong: RU, BE, FI adhere exactly to the +3 rule. Common Pitfalls: Off-by-one counting (forgetting to exclude the starting letter when stepping). Always count moves, not letters listed. Final Answer: IM

Question 10

Classification (lexical category – predominant use): Three words are primarily common nouns in everyday use; one is frequently used as a verb as well as a noun. Identify the odd one out.

Accepted Answer

Introduction / Context: Word-class classification can hinge on predominant grammatical use. “Teeth,” “sheep,” and “greed” function primarily as common nouns in neutral contexts (teeth = plural noun; sheep = plural-invariant noun; greed = abstract noun). “Sleep,” however, freely functions both as a noun (get some sleep) and as a verb (I sleep early). Reasoning tests often treat a flexible part-of-speech item as the odd one against items with predominantly nominal usage. Given Data / Assumptions: TEETH → common plural noun; not typically a verb. SHEEP → common noun; plural-invariant; not a verb in standard usage. GREED → abstract noun; not a verb. SLEEP → can be both a common noun and a verb in everyday usage. Concept / Approach: Identify which word exhibits dual major category behavior (noun and verb) in general English usage. With three items strictly nominal and one flexibly nominal/verbal, the latter is the outlier. This retains objectivity and avoids subtle semantic debates about countability or irregular plurals. Step-by-Step Solution: Tag each option by typical part(s) of speech.Find the only word commonly used as a verb in base form without derivation (“to sleep”).Select SLEEP as the odd one out. Verification / Alternative check: Substitute in frames: “to ____” works naturally for sleep; “to teeth/sheep/greed” does not function as standard English without specialized contexts. Why Other Options Are Wrong: TEETH, SHEEP, GREED are used almost exclusively as nouns in ordinary contexts. Common Pitfalls: Overanalyzing morphological quirks (sheep’s invariant plural) instead of focusing on part-of-speech flexibility. Final Answer: SLEEP

Question 11

Choose the odd letter-triplet: In each three-letter group, check the movement in the English alphabet from the 1st to the 2nd letter and from the 2nd to the 3rd letter. Three groups follow the fixed pattern “+3 then -5” (forward by 3 positions, then…

Accepted Answer

Introduction / Context: Letter-classification questions commonly encode arithmetic jumps on alphabet positions (A=1, B=2, …, Z=26). Here, the governing pattern is “+3 then -5” across each three-letter group. We must find the set that fails the rule. Given Data / Assumptions: Alphabet indices: A=1 … Z=26. Groups: LOJ, FID, RUP, ILN. Target rule: first→second = +3; second→third = -5. Concept / Approach: Translate letters to indices, apply the difference check (+3, then -5), and flag any violation. Wrap-around is not required here because all letters remain within A–Z bounds. Step-by-Step Solution: LOJ: L(12)→O(15) = +3; O(15)→J(10) = -5 → fits.FID: F(6)→I(9) = +3; I(9)→D(4) = -5 → fits.RUP: R(18)→U(21) = +3; U(21)→P(16) = -5 → fits.ILN: I(9)→L(12) = +3; L(12)→N(14) = +2 → violates. Verification / Alternative check: Reverse-check by recomputing: only ILN shows +3 followed by +2, not -5, so it breaks the rule. Why Other Options Are Wrong: LOJ, FID, and RUP implement the exact +3 then -5 transitions. Common Pitfalls: Counting letter positions off by one or forgetting that J=10, N=14, etc. Always map letters to numbers before comparing. Final Answer: ILN is the odd triplet because it violates the “+3 then -5” pattern.

Question 12

Identify the odd word: Three words describe extent/measurement in the vertical or linear sense (height/length), while one relates to brightness/illumination. Choose the mismatching word.

Accepted Answer

Introduction / Context: Verbal classification often groups words by domain of meaning. Here, three adjectives belong to physical extent (height/length) while one belongs to light/visibility. Given Data / Assumptions: Words: Long, Tall, High, Dim. Category 1: measurement of extent; Category 2: brightness. Concept / Approach: Map each word to its primary semantic field and separate the outlier that does not describe linear or vertical extent. Step-by-Step Solution: Long → primarily length.Tall → primarily height.High → vertical position/elevation.Dim → degree of light/brightness (not extent). Verification / Alternative check: Try substituting context sentences: “a long rope,” “a tall building,” “a high wall” are natural; “a dim building” shifts to lighting, confirming different semantics. Why Other Options Are Wrong: Long, Tall, and High share the measurement domain. Common Pitfalls: Assuming High and Tall are always interchangeable. They are related, but both stay within the extent/height family, unlike Dim. Final Answer: Dim is the odd word (brightness term, not an extent term).

Question 13

Choose the odd number: Among 78, 48, 72, and 54, exactly one number has 13 as a prime factor. Identify that distinct number.

Accepted Answer

Introduction / Context: Several classification items rely on a specific hidden factor. Here, we test for the presence of the prime 13 in the factorization. Given Data / Assumptions: Numbers: 78, 48, 72, 54. Check prime factor 13. Concept / Approach: Factor each number or check divisibility by 13. Because 13 * 6 = 78, this quickly singles out the outlier. Step-by-Step Solution: 78 = 13 * 6 → contains factor 13.48 = 2^4 * 3 → no factor 13.72 = 2^3 * 3^2 → no factor 13.54 = 2 * 3^3 → no factor 13. Verification / Alternative check: Dividing each by 13: only 78/13 = 6 is integer; others are not. Why Other Options Are Wrong: 48, 72, and 54 share small prime factors 2 and 3 but lack 13. Common Pitfalls: Using only parity or sum-of-digits rules; those do not detect the presence of 13. Final Answer: 78 is the only number with prime factor 13 and is the odd number.

Question 14

Identify the odd word: Three choices (Cake, Pastry, Bread) are edible bakery items. One is an adjective describing taste rather than a food item. Choose the outlier.

Accepted Answer

Introduction / Context: Word-classification problems hinge on semantic categories. Here, three nouns name baked foods, while one describes flavor. Given Data / Assumptions: Options: Sweet, Cake, Pastry, Bread. We are grouping by “edible bakery item.” Concept / Approach: Distinguish parts of speech and semantic role: adjective vs concrete food nouns. Step-by-Step Solution: Cake → bakery food.Pastry → bakery food.Bread → bakery staple.Sweet → flavor descriptor (adjective/quality), not a specific food type. Verification / Alternative check: Substitute into “I ate …”: “I ate cake/pastry/bread” is normal; “I ate sweet” is ungrammatical unless used as a count noun (“a sweet”), which changes the word form. Why Other Options Are Wrong: They are concrete foods and belong to the same category. Common Pitfalls: Interpreting “sweet” as a noun (“a sweet”)—that is a different lexical form; here the given form functions adjectivally. Final Answer: Sweet is the odd word.

Question 15

Choose the odd number: In 39, 83, 78, and 52, three numbers are multiples of 13; one number is not and is also prime. Identify this distinct number.

Accepted Answer

Introduction / Context: Spotting a specific divisor often solves a classification item quickly. Here we focus on 13. Given Data / Assumptions: Numbers: 39, 83, 78, 52. Fact: 39 = 3 * 13, 78 = 6 * 13, 52 = 4 * 13. Concept / Approach: Identify multiples of 13 and isolate the remainder. Step-by-Step Solution: 39, 78, 52 → all divisible by 13.83 → not divisible by 13 and has no smaller prime divisors → prime. Verification / Alternative check: 13 * 6 = 78, 13 * 4 = 52, 13 * 3 = 39. Nearest to 83 are 78 and 91; hence 83 is not 13k, and trial division up to sqrt(83) confirms primality. Why Other Options Are Wrong: They are exact multiples of 13 and so share a property that 83 lacks. Common Pitfalls: Assuming “larger” means composite; size alone is irrelevant. Final Answer: 83 is the odd number.

Question 16

Identify the odd number: Among 31, 39, 47, and 41, three are prime; one is composite. Choose the composite number.

Accepted Answer

Introduction / Context: Prime recognition is central to numeric classification. Only one selection is composite. Given Data / Assumptions: Set: 31, 39, 47, 41. Definition: prime has exactly two positive divisors. Concept / Approach: Apply small-prime divisibility tests, especially 3 and 5 for quick screening. Step-by-Step Solution: 31 → not divisible by 2, 3, 5 → prime.39 → 3 + 9 = 12 → divisible by 3 → composite.47 → not divisible by 2, 3, 5 → prime.41 → not divisible by 2, 3, 5 → prime. Verification / Alternative check: Direct factorization: 39 = 3 * 13 confirms compositeness. Why Other Options Are Wrong: 31, 47, 41 are primes. Common Pitfalls: Confusing proximity to a multiple of 3 with actual divisibility. Final Answer: 39 is the composite number and the odd one out.

Question 17

Choose the odd number: In 25, 64, 189, and 441, three are perfect squares (n^2) and one is not. Identify the non-square.

Accepted Answer

Introduction / Context: Perfect-square detection is a standard classification skill. Given Data / Assumptions: Numbers: 25 = 5^2, 64 = 8^2, 441 = 21^2, 189 = ? We test whether 189 equals n^2 for an integer n. Concept / Approach: Use known squares or check between consecutive squares. Step-by-Step Solution: 25 → perfect square (5^2).64 → perfect square (8^2).441 → perfect square (21^2).189 → lies between 13^2 = 169 and 14^2 = 196 → not a square. Verification / Alternative check: Prime factorization 189 = 3^3 * 7 has exponents not all even; hence not a square. Why Other Options Are Wrong: 25, 64, and 441 are exact squares. Common Pitfalls: Assuming “near a square” implies squareness; exact equality is required. Final Answer: 189 is the non-square.

Question 18

Identify the odd number: Three are squares of prime numbers (13^2, 17^2, 19^2 after repair), while one is a square of a composite number (15^2). Choose the square of the composite.

Accepted Answer

Introduction / Context: The original options contained a duplicate “225”. Under the Recovery-First policy, we minimally repair by treating the fourth entry as the next consistent prime square, 361 (which is 19^2). Now the set contrasts prime squares with one composite-base square. Given Data / Assumptions: 169 = 13^2 (prime base). 289 = 17^2 (prime base). 361 = 19^2 (prime base). 225 = 15^2 (composite base; 15 = 3 * 5). Concept / Approach: Classify by the nature of the base n whose square forms n^2: prime vs composite. Step-by-Step Solution: Check bases: 13, 17, 19 are prime; 15 is composite.Therefore 225 (15^2) is distinct from the others. Verification / Alternative check: Factor 225 = 3^2 * 5^2; its base (15) is composite. Why Other Options Are Wrong: 169, 289, and 361 are squares of primes, sharing the majority property. Common Pitfalls: Overlooking the duplicate and ending with two “correct” entries; minimal repair ensures a single unambiguous answer. Final Answer: 225 is the square of a composite and is the odd number.

Question 19

Choose the odd number: 27, 64, and 125 are perfect cubes (3^3, 4^3, 5^3). One number is not a perfect cube. Identify the non-cube.

Accepted Answer

Introduction / Context: Recognizing perfect cubes is a common categorization task. Given Data / Assumptions: 27 = 3^3, 64 = 4^3, 125 = 5^3. 384 = ? Concept / Approach: Compare each entry to nearby cubes or factorize to check whether all prime exponents are multiples of 3. Step-by-Step Solution: 27 → perfect cube.64 → perfect cube.125 → perfect cube.384 = 2^7 * 3 (not all exponents multiples of 3) → not a cube. Verification / Alternative check: Locate 7^3 = 343 and 8^3 = 512; 384 lies between them and is neither. Why Other Options Are Wrong: They equal exact cubes. Common Pitfalls: Assuming “even and large” implies a cube; exact equality to n^3 is required. Final Answer: 384 is the non-cube.

Question 20

Identify the odd number: Among 84, 120, 72, and 98, three are divisible by 6 (by 2 and by 3) while one is not. Choose the non-multiple of 6.

Accepted Answer

Introduction / Context: A number is a multiple of 6 if it is even and divisible by 3. We separate the one that fails this combined criterion. Given Data / Assumptions: Set: 84, 120, 72, 98. Digit-sum rule for 3 and parity for 2. Concept / Approach: Apply the two checks: evenness and divisibility by 3. Step-by-Step Solution: 84 → even; 8+4=12 → divisible by 3 → multiple of 6.120 → even; 1+2+0=3 → multiple of 6.72 → even; 7+2=9 → multiple of 6.98 → even; 9+8=17 (not multiple of 3) → not a multiple of 6. Verification / Alternative check: Direct division by 6 yields integer results for 84, 120, 72, but not for 98. Why Other Options Are Wrong: They satisfy both conditions and are multiples of 6. Common Pitfalls: Stopping after checking evenness only; divisibility by 3 is also required. Final Answer: 98 is the odd number (not a multiple of 6).

Classification Questions