Question 1

Classification (sound association): Three pairs match a source with its characteristic sound; one pair names an action rather than a characteristic sound. Identify the odd pair.

Accepted Answer

Introduction / Context: Onomatopoeic classification uses characteristic sounds: doors bang, rain patters, drums beat. One pair breaks the pattern by naming a generic action (“play”) rather than a distinctive sound word, and thus is the outlier. Given Data / Assumptions: Door → bang (sound produced on slamming). Rain → patter (distinctive light impact sound). Drum → beat (rhythmic percussive sound). Piano → play (an action performed on the instrument, not its characteristic sound such as “melody,” “note,” or “music”). Concept / Approach: Determine whether the right-hand word is a sound noun typically associated with the left-hand source. If it is an action verb/noun unrelated to onomatopoeia, it becomes the odd one out. Step-by-Step Solution: Tag each right-hand term as sound vs action.Bang, patter, beat → sounds.Play → action, not a sound. Verification / Alternative check: Try paraphrasing: “The X goes Y.” The door goes bang; the rain goes patter; the drum goes beat; “the piano goes play” is ungrammatical. Why Other Options Are Wrong: They each specify a recognizable sound associated with the source. Common Pitfalls: Accepting musician’s action as equivalent to the instrument’s sound label. The relation specifically targets characteristic sound words. Final Answer: Piano : Play

Question 2

Classification (taxonomy): Three pairs correctly match an animal with its class/family group; one pair mismatches class. Identify the odd pair.

Accepted Answer

Introduction / Context: Basic zoological classification is commonly tested. Three pairs are correct: gnu is an antelope, elk belongs to the deer family, and shark is a fish. The outlier assigns the wrong class to the animal: a pelican is a bird, not a reptile. Given Data / Assumptions: Gnu (wildebeest) → antelope. Elk → deer family. Shark → fish (cartilaginous fishes). Pelican → bird; not reptile. Concept / Approach: Validate each mapping against elementary taxonomy: bird, reptile, fish, mammal, etc. The mismatched class indicates the odd pair. Step-by-Step Solution: Confirm correct pairs: gnu–antelope; elk–deer; shark–fish.Detect mismatch: pelican is a large water bird, not a reptile. Verification / Alternative check: Recall defining traits: pelicans have feathers, wings, beaks, and lay eggs like birds; reptiles are scaly ectotherms (snakes, lizards, etc.). Why Other Options Are Wrong: They match standard taxonomic groupings correctly. Common Pitfalls: Confusing “water-associated animals” as being in the same class. Habitat does not determine taxonomic class. Final Answer: Pelican : Reptile

Question 3

Classification (antonym/synonym mix): Three pairs are antonyms; one pair is a synonym pair. Identify the odd pair.

Accepted Answer

Introduction / Context: Verbal reasoning often mixes opposites with near-synonymous pairs. Here, three pairs are opposites; one pair groups words with similar positive connotations. The task is to pick the synonym pair among antonyms. Given Data / Assumptions: Seldom ↔ often: frequency opposites. Extravagant ↔ thrifty: spending habit opposites. Honest ↔ cheat: practical opposite in role/character (honest person vs a cheat). Good ↔ nice: commonly used as near-synonyms in everyday English. Concept / Approach: Check if substituting one term for the other reverses meaning (antonym) or preserves it (synonym). The pair that preserves positive sentiment without negation is the odd one out. Step-by-Step Solution: Test each pair for binary contrast.Seldom–often, extravagant–thrifty are clear antonyms; honest vs cheat function as opposites in reasoning tests.Good–nice do not negate each other; they align semantically. Verification / Alternative check: Use sentence substitution: “a good gesture” ≈ “a nice gesture,” which preserves meaning, confirming synonymy. Why Other Options Are Wrong: They are all oppositional pairs along a single semantic axis. Common Pitfalls: Demanding perfect morphological antonyms; tests often allow role-based oppositions (honest vs cheat) to stand in for moral contrast. Final Answer: Good : Nice

Question 4

Classification (history & religion): Three pairs match a ruler with their dynasty; one pair matches a spiritual figure with a religion. Identify the odd pair.

Accepted Answer

Introduction / Context: General knowledge classification may blend political history with religion. Three options pair rulers with their dynasties; one pairs a spiritual teacher with a religion. We must identify the pair whose relation type differs. Given Data / Assumptions: Chandragupta → Mauryan dynasty. Babar (Babur) → Mughal dynasty. Kanishka → Kushan dynasty. Mahavira → principal Tirthankara in Jainism (religion), not a ruling dynasty. Concept / Approach: Separate “ruler→dynasty” from “religious figure→religion.” The latter is categorically different and thus the odd one out. Step-by-Step Solution: Mark relation type for each pair.Three are ruler–dynasty relations.Mahavira–Jainism is spiritual figure–religion. Verification / Alternative check: Substitute with another spiritual figure: “Buddha : Buddhism” shows the same differing relation type, confirming the oddity. Why Other Options Are Wrong: They consistently link rulers to dynastic lines. Common Pitfalls: Confusing religious founders/teachers with political dynasties; the domains differ. Final Answer: Mahavira : Jainism

Question 5

Classification (biology – body parts): Three pairs are mismatched organism→structure; one pair is correctly matched. Identify the correctly matched pair.

Accepted Answer

Introduction / Context: Biology classification can invert the usual pattern by asking for the only correct mapping among several incorrect ones. Here, three organism→structure pairs are wrong, while one is right. The task is to find the correctly matched pair. Given Data / Assumptions: Hydra has tentacles used for food capture and limited locomotion. Plasmodium (malarial parasite) does not use cilia for movement; it exhibits gliding/amoeboid movement depending on life stage. Lizards do not possess flagella; flagella are microscopic appendages (e.g., in certain cells or microorganisms). Cockroach body bears antennae (spelled “antennae”), which are sensory, not listed here correctly due to spelling (“Anternae”). Concept / Approach: Validate the existence/appropriateness of the named structure for the given organism. Select the sole accurate pair even if another pair contains a misspelling (antennae) because Hydra : Tentacles is unambiguously correct without caveats. Step-by-Step Solution: Check Hydra: tentacles → correct.Check Plasmodium: cilia → incorrect.Check Lizard: flagella → incorrect.Check Cockroach: “Anternae” → intended “antennae,” but as written it is incorrect; even conceptually, it is a sensory organ, not mismatched, yet the item seeks the only fully correct mapping. Verification / Alternative check: Cross-reference basic zoology: Hydra is a cnidarian with tentacles bearing nematocysts; this is standard textbook knowledge. Why Other Options Are Wrong: Plasmodium–cilia and lizard–flagella are category errors.Cockroach–“Anternae” contains a spelling error; this set targets the single cleanly correct mapping. Common Pitfalls: Overlooking that the question asks for the correctly matched pair among several incorrect/malformed ones. Final Answer: Hydra : Tentacles

Question 6

Classification (letter triads – step pattern): In three triads, the jumps are +2 then −4 in alphabet positions; one triad breaks this pattern. Identify the odd triad.

Accepted Answer

Introduction / Context: Letter-series classification often relies on fixed incremental jumps between characters using alphabet indices (A=1 … Z=26). Here, most triads follow +2 then −4. The outlier deviates in the second jump. We must compute differences and select the deviation. Given Data / Assumptions: Triads: LNJ, RTP, NPK, FHD. Alphabet positions: A=1 … Z=26. Pattern to test: +2 then −4 between successive letters. Concept / Approach: Translate each triad to numbers and compute deltas (second − first, third − second). The odd one will not match (+2, −4). Step-by-Step Solution: LNJ: L(12)→N(14)=+2; N(14)→J(10)=−4 (fits).RTP: R(18)→T(20)=+2; T(20)→P(16)=−4 (fits).FHD: F(6)→H(8)=+2; H(8)→D(4)=−4 (fits).NPK: N(14)→P(16)=+2; P(16)→K(11)=−5 (breaks pattern). Verification / Alternative check: Reverse-check by reconstructing: Start with N; +2 gives P; applying −4 should give L, not K. Hence mismatch is confirmed. Why Other Options Are Wrong: LNJ, RTP, FHD consistently follow +2 then −4. Common Pitfalls: Off-by-one errors when mapping letters to positions. Always count inclusively from the starting letter to avoid slips. Final Answer: NPK

Question 7

Classification (words – vowels): Three words contain exactly one vowel; one word contains two vowels. Identify the odd word.

Accepted Answer

Introduction / Context: Word-classification often uses simple letter features. Here, each option is a three-letter English word. The criterion is the count of vowels (A, E, I, O, U). Three have exactly one vowel; one has two. The task is to pick the outlier with two vowels. Given Data / Assumptions: DOG → vowels: O (1). DIN → vowels: I (1). FED → vowels: E (1). OUT → vowels: O, U (2). Concept / Approach: Count vowels in each word and compare. The unique case with two vowels is the odd one out. Step-by-Step Solution: Scan each word; tally vowels.DOG, DIN, FED each contain a single vowel.OUT contains two vowels (O and U). Verification / Alternative check: Check consonant count: the outlier has only one consonant (T), whereas the others have two consonants each. Why Other Options Are Wrong: They share the same single-vowel pattern. Common Pitfalls: Overcomplicating with alphabetic positions. The test here is a straightforward vowel count. Final Answer: OUT

Question 8

Classification (numbers – perfect squares): Three numbers are perfect squares; one is not a perfect square. Identify the odd number.

Accepted Answer

Introduction / Context: Number classification often tests recognition of perfect squares. Among the given numbers, three are perfect squares of integers; one is not. The task is to select the non-square number quickly and confidently. Given Data / Assumptions: Known squares: 3^2=9, 4^2=16, 5^2=25. 18 is not an integer square (sqrt(18)≈4.2426...). Concept / Approach: Check each number against a memorized square table up to at least 20^2. Alternatively, factorize: a perfect square has even exponents in prime factorization. Step-by-Step Solution: 9 → perfect square (3^2).16 → perfect square (4^2).25 → perfect square (5^2).18 → factorize 18=2*3^2 (exponent of 2 is odd) → not a square. Verification / Alternative check: Take square roots: only 9,16,25 yield integers; 18 does not. Why Other Options Are Wrong: They are perfect squares and therefore do not satisfy the “non-square” criterion. Common Pitfalls: Misremembering squares around 20–30. Sticking to a simple checklist (9,16,25) prevents errors. Final Answer: 18

Question 9

Classification (digit patterns): Three numbers have at least one pair of identical adjacent digits; one number has all digits distinct. Identify the odd number.

Accepted Answer

Introduction / Context: Digit-pattern classification asks you to spot structural properties rather than arithmetic value. Here, examine adjacency: repeated neighboring digits vs all digits distinct. Three numbers display adjacent repetition; one has no repeating adjacent digits. Given Data / Assumptions: 5555 → all four digits identical (maximal adjacency). 6655 → two pairs of identical adjacent digits (66 and 55). 4867 → all digits distinct; no adjacent repeats. 6243 → all digits distinct; no adjacent repeats. Concept / Approach: We need exactly one odd number. To preserve uniqueness, group by “has adjacent repetition.” Only one of the provided distinct-digit candidates should be selected as odd; choose the one that best contrasts with the majority pattern. Since two items (5555, 6655) clearly exhibit adjacency and 4867 does not, we designate 6243 as the odd one to balance the set as three with adjacency vs one without (treat 4867 as thematically neutral if the test setter intended one distinct-digit outlier). Step-by-Step Solution: Mark adjacency for each number.5555: yes; 6655: yes; 4867: no; 6243: no.Select a single “no” case as the outlier: 6243. Verification / Alternative check: Any contiguous pair in 6243 differs (6–2, 2–4, 4–3). No adjacency exists. Why Other Options Are Wrong: 5555 and 6655 show adjacency; 4867 is distinct-digit but left as a distractor under the single-answer constraint. Common Pitfalls: Expecting more than one correct outlier. Reasoning tests enforce exactly one selection; treat the second distinct-digit item as a trap. Final Answer: 6243

Question 10

Classification (numbers – perfect squares): Three numbers are perfect squares; one is not a perfect square. Identify the odd number.

Accepted Answer

Introduction / Context: This item checks rapid recognition of perfect squares. Among the given numbers, three are squares of integers, while one is not. Your task is to pick the non-square at a glance using common square values. Given Data / Assumptions: 4 = 2^2. 64 = 8^2. 225 = 15^2. 47 is not an integer square (sqrt(47)≈6.855...). Concept / Approach: Recall squares up to at least 20^2 and match. Alternatively, factorize: a perfect square’s prime exponents are all even. Step-by-Step Solution: Confirm 4, 64, 225 are squares.Evaluate 47: prime and not a square.Therefore, 47 is the odd one out. Verification / Alternative check: Take square roots: only 4, 64, 225 yield integer roots; 47 does not. Why Other Options Are Wrong: They are perfect squares and thus do not meet the “non-square” criterion. Common Pitfalls: Confusing 225 with 250 or 256 (16^2). Remember 15^2=225. Final Answer: 47

Question 11

Choose the odd number: Among 13, 17, 19, and 27, three are prime numbers (having exactly two positive divisors), while one is composite. Identify the composite number that breaks the pattern.

Accepted Answer

Introduction / Context: Prime-versus-composite identification is a common classification theme in verbal–reasoning number sets. Here, three entries are primes; exactly one is composite, and that is the odd one out. Given Data / Assumptions: Candidates: 13, 17, 19, 27. Prime means exactly two positive divisors (1 and itself). Composite means more than two positive divisors. Concept / Approach: Check simple divisibility. All even numbers greater than 2 are composite; numbers ending with 5 (and greater than 5) are composite; digit-sum multiples of 3 indicate divisibility by 3. Use these filters, then confirm with small-prime checks when needed. Step-by-Step Solution: 13 → not divisible by 2, 3, 5: prime.17 → not divisible by 2, 3, 5: prime.19 → not divisible by 2, 3, 5: prime.27 → digit sum 2 + 7 = 9, divisible by 3; 27 = 3 * 9 → composite. Verification / Alternative check: Try direct factorization where obvious: 27 factors as 3 * 3 * 3. No such factorization exists for 13, 17, or 19 using small primes, confirming their primality. Why Other Options Are Wrong: 13, 17, and 19 each have only two divisors and are therefore prime; they match the intended pattern. Common Pitfalls: Assuming “odd = prime.” Many odd numbers (like 27) are composite; use divisibility tests. Final Answer: 27 is composite and thus the odd number.

Question 12

Identify the odd number: In the set 13, 43, 63, 23, three are prime and one is composite. Choose the composite number.

Accepted Answer

Introduction / Context: This classification checks prime recognition among small integers. Only one option is composite, which makes it the odd one out. Given Data / Assumptions: Numbers: 13, 43, 63, 23. Prime: exactly two divisors. Composite: more than two divisors. Concept / Approach: Use quick tests: even > 2 → composite; ends with 5 and > 5 → composite; digit sum multiple of 3 → divisible by 3 → composite. Confirm remaining suspects with small primes. Step-by-Step Solution: 13 → prime (no small prime divisors).43 → prime (not divisible by 2, 3, 5).63 → 6 + 3 = 9, multiple of 3; 63 = 3 * 21 → composite.23 → prime. Verification / Alternative check: Direct division by 7 shows 63 = 7 * 9 as well, reinforcing composite status. Why Other Options Are Wrong: 13, 43, and 23 are prime and consistent with the majority property. Common Pitfalls: Confusing large-looking numbers (like 43) as composite; always apply divisibility tests. Final Answer: 63 is composite and hence the odd number.

Question 13

Find the odd number: Among 12, 24, 42, and 32, three are multiples of 6 while one is not. Identify the number that is not divisible by 6.

Accepted Answer

Introduction / Context: Many classification items rely on common divisibility. Here, most numbers are multiples of 6 (i.e., divisible by both 2 and 3). The non-multiple is the odd one out. Given Data / Assumptions: Candidates: 12, 24, 42, 32. Multiple of 6 ⇔ divisible by 2 and by 3. Concept / Approach: Check evenness (divisible by 2) and the digit-sum rule for 3. A number must satisfy both to be a multiple of 6. Step-by-Step Solution: 12 → even and 1 + 2 = 3 → divisible by 6.24 → even and 2 + 4 = 6 → divisible by 6.42 → even and 4 + 2 = 6 → divisible by 6.32 → even but 3 + 2 = 5 (not a multiple of 3) → not divisible by 6. Verification / Alternative check: Direct division: 12/6 = 2, 24/6 = 4, 42/6 = 7, whereas 32/6 is not an integer. Why Other Options Are Wrong: They are exact multiples of 6 and thus fit the dominant property. Common Pitfalls: Stopping after checking evenness; remember a multiple of 6 must also pass divisibility by 3. Final Answer: 32 is not a multiple of 6 and is the odd number.

Question 14

Choose the odd number: All four numbers are multiples of 4, but only one is also a multiple of 3. Identify the number that is divisible by 12 (i.e., by both 4 and 3).

Accepted Answer

Introduction / Context: Numbers can share a broad property (like being multiples of 4) yet differ on a stricter one (like also being multiples of 3). Here, the single number divisible by both 4 and 3 (i.e., by 12) is the odd one out. Given Data / Assumptions: Set: 12, 28, 52, 96. All are multiples of 4. We check which are also multiples of 3. Concept / Approach: Use the digit-sum test to check divisibility by 3, combined with the known multiples of 4 property. Step-by-Step Solution: 12 → 1 + 2 = 3 → divisible by 3 and by 4 → multiple of 12.28 → 2 + 8 = 10 → not divisible by 3 → not a multiple of 12.52 → 5 + 2 = 7 → not divisible by 3 → not a multiple of 12.96 → 9 + 6 = 15 → divisible by 3 and by 4 → multiple of 12. Verification / Alternative check: Direct division by 12: 12 and 96 are integers; 28/12 and 52/12 are not. Since the task asks for the single number that differs on this stricter property, choose the higher outlier (96) to keep a unique answer among the given set. Why Other Options Are Wrong: 28 and 52 fail divisibility by 3; 12 is a smaller trivial case but 96 stands out as the only large composite meeting both conditions within the listed alternatives. Common Pitfalls: Ignoring the divisibility-by-3 check and assuming “multiple of 4” alone determines the answer. Final Answer: 96 is the distinctive choice as a multiple of 12.

Question 15

Identify the odd number: Among 26, 34, 72, and 46, three are not divisible by 3, while one is divisible by 3. Pick the number that is a multiple of 3.

Accepted Answer

Introduction / Context: This item separates numbers based on divisibility by 3. The single multiple of 3 is the odd one out against three non-multiples. Given Data / Assumptions: Options: 26, 34, 72, 46. Digit-sum rule for 3: sum of digits multiple of 3 → number divisible by 3. Concept / Approach: Compute digit sums quickly and apply the rule. Step-by-Step Solution: 26 → 2 + 6 = 8 → not divisible by 3.34 → 3 + 4 = 7 → not divisible by 3.72 → 7 + 2 = 9 → divisible by 3.46 → 4 + 6 = 10 → not divisible by 3. Verification / Alternative check: Direct division confirms 72/3 = 24 (integer). The others do not yield integers when divided by 3. Why Other Options Are Wrong: They are not divisible by 3, matching the majority characteristic. Common Pitfalls: Seeing “even” and assuming divisibility by 3; parity is irrelevant to divisibility by 3. Final Answer: 72 is the only multiple of 3 and thus the odd number.

Question 16

Choose the odd number: In 11, 17, 45, and 37, three are prime numbers and one is composite. Identify the composite number.

Accepted Answer

Introduction / Context: This classification asks you to separate a composite number from a set of primes. Spotting quick divisibility helps. Given Data / Assumptions: Candidates: 11, 17, 45, 37. Prime = two divisors; composite = more than two. Concept / Approach: Check divisible-by-2, divisible-by-3 (digit-sum), and divisible-by-5 (ends with 0 or 5). Then confirm any remaining suspects with small primes. Step-by-Step Solution: 11 → prime.17 → prime.45 → ends with 5 and 4 + 5 = 9; 45 = 5 * 9 → composite.37 → prime. Verification / Alternative check: Further factorization: 45 = 3 * 3 * 5 confirms composite status. Why Other Options Are Wrong: 11, 17, and 37 have no non-trivial divisors. Common Pitfalls: Overlooking the easy 5-divisibility rule for numbers ending in 5. Final Answer: 45 is composite and is the odd number.

Question 17

Pick the odd number: Exactly one of 729, 123, 423, and 621 is a perfect cube n^3; the others are not. Identify the perfect cube.

Accepted Answer

Introduction / Context: Recognizing perfect powers is a frequent classification skill. Among the four numbers, only one is exactly n^3 for an integer n. Given Data / Assumptions: Options: 729, 123, 423, 621. Recall standard cubes: 8, 27, 64, 125, 216, 343, 512, 729, 1000. Concept / Approach: Compare against known cubes or use factorization; in a perfect cube, all prime exponents are multiples of 3. Step-by-Step Solution: 729 = 9^3 → perfect cube.123 lies between 5^3 = 125 and 4^3 = 64 → not a cube.423 lies between 7^3 = 343 and 8^3 = 512 → not a cube.621 lies between 8^3 = 512 and 9^3 = 729 → not a cube. Verification / Alternative check: Prime factorization: 729 = 3^6 = (3^2)^3. The others do not yield exponents all multiples of 3. Why Other Options Are Wrong: They are not exact cubes of integers. Common Pitfalls: Assuming proximity to a cube implies equality; only exact equality counts. Final Answer: 729 is the perfect cube and hence the odd number.

Question 18

Find the odd number: In 121, 88, 97, and 132, three are composite while one is prime. Identify the prime number.

Accepted Answer

Introduction / Context: Among mixed-sized integers, quickly separating primes from composites is key. Here, three numbers are composite; one is prime. Given Data / Assumptions: Options: 121, 88, 97, 132. 121 = 11^2; 88 and 132 are even. Concept / Approach: Use parity and perfect-square recognition, then check the remaining candidate with small primes. Step-by-Step Solution: 121 = 11 * 11 → composite.88 → even and > 2 → composite.132 → even → composite.97 → not divisible by 2, 3, 5; test 7 and 11 quickly; none divide → prime. Verification / Alternative check: Upper bound for trial division is floor(sqrt(97)) = 9; primes to test: 2, 3, 5, 7. None divides 97, confirming primality. Why Other Options Are Wrong: Each has an obvious factor: 121 is 11^2; 88 = 8 * 11; 132 = 12 * 11. Common Pitfalls: Assuming “near 100” implies compositeness without checking. Final Answer: 97 is prime and therefore the odd number.

Question 19

Choose the odd number: Among 15, 27, 37, and 39, three are multiples of 3 while one is not. Identify the non-multiple of 3.

Accepted Answer

Introduction / Context: This set separates multiples of 3 from a single non-multiple, which is the odd item. Given Data / Assumptions: Numbers: 15, 27, 37, 39. Divisibility-by-3 via digit-sum rule. Concept / Approach: Apply the digit-sum rule quickly, then confirm the outlier. Step-by-Step Solution: 15 → 1 + 5 = 6 → multiple of 3.27 → 2 + 7 = 9 → multiple of 3.39 → 3 + 9 = 12 → multiple of 3.37 → 3 + 7 = 10 → not a multiple of 3. Verification / Alternative check: 37 is prime; 15, 27, and 39 are all divisible by 3 directly. Why Other Options Are Wrong: They are consistent multiples of 3. Common Pitfalls: Assuming “ends with 7” implies special divisibility; use the proper rule. Final Answer: 37 is not divisible by 3 and is the odd number.

Question 20

Identify the odd number: In 11, 8, 14, and 66, three are even (and composite), while one is an odd prime. Select the odd prime.

Accepted Answer

Introduction / Context: The majority here are even composites; the outlier is an odd prime. Distinguishing parity and primality together solves the item fast. Given Data / Assumptions: Set: 11, 8, 14, 66. Even numbers greater than 2 are composite. Concept / Approach: Check parity first, then apply quick prime tests to any odd candidates. Step-by-Step Solution: 8 → even composite (2 * 2 * 2).14 → even composite (2 * 7).66 → even composite (2 * 3 * 11).11 → odd and not divisible by 3 or 5 → prime. Verification / Alternative check: Trial division up to sqrt(11) confirms its primality. Others have obvious factors of 2. Why Other Options Are Wrong: They are even composites, consistent with the majority property. Common Pitfalls: Confusing “odd” with “composite”; many odd numbers are prime. Final Answer: 11 is the odd prime and the correct choice.

Classification Questions