Question 1

Odd One Out — For 22:0, 24:12, 23:5, 24:18, three satisfy “right = (ones digit)^2 − (tens digit)^2”. Identify the exception.

Accepted Answer

Introduction / Context: Some pair-mappings depend on digit-wise operations. Here, the operative rule for three items is: if x has tens digit t and ones digit u, then right = u^2 − t^2. Given Data / Assumptions: Two-digit x with tens t and ones u. Compute u^2 − t^2 and compare with the given right-hand value. Concept / Approach: Apply the digit-square-difference rule to each pair. Step-by-Step Solution: 22 → t=2, u=2 → u^2 − t^2 = 4 − 4 = 0 → matches 22:0.24 → t=2, u=4 → u^2 − t^2 = 16 − 4 = 12 → matches 24:12.23 → t=2, u=3 → u^2 − t^2 = 9 − 4 = 5 → matches 23:5.24 → t=2, u=4 → u^2 − t^2 = 12 (not 18) → 24:18 is the exception. Verification / Alternative check: The same rule can be written as (u − t)*(u + t). For 24, that is (4 − 2)*(4 + 2) = 2*6 = 12, reconfirming the mismatch for 24:18. Why Other Options Are Wrong: 22:0 — correct by rule. 24:12 — correct by rule. 23:5 — correct by rule. Common Pitfalls: Confusing t^2 − u^2 with u^2 − t^2, which flips the sign and invalidates matches. Final Answer: 24 : 18

Question 2

Odd One Out — From the pairs (25, 36), (9, 64), (100, 121), (144, 169), pick the pair that is not a consecutive-squares pair.

Accepted Answer

Introduction / Context: Square recognition and adjacency (n^2 and (n+1)^2) provide a clean basis for classification. Three pairs are consecutive squares; one is not. Given Data / Assumptions: 25 = 5^2, 36 = 6^2 → consecutive squares. 100 = 10^2, 121 = 11^2 → consecutive squares. 144 = 12^2, 169 = 13^2 → consecutive squares. 9 = 3^2, 64 = 8^2 → not consecutive. Concept / Approach: Express each number as an integer square and check whether the exponents differ by 1. Step-by-Step Solution: (25, 36) → (5^2, 6^2) ✓(100, 121) → (10^2, 11^2) ✓(144, 169) → (12^2, 13^2) ✓(9, 64) → (3^2, 8^2) ✗ (gap of 5 in the bases) Verification / Alternative check: Compute differences: (n+1)^2 − n^2 = 2n + 1; for (5,6) that is 11; for (10,11) that is 21; for (12,13) that is 25. The non-consecutive pair will not fit this form consistently. Why Other Options Are Wrong: (25, 36): consecutive squares. (100, 121): consecutive squares. (144, 169): consecutive squares. Common Pitfalls: Assuming any two squares constitute a valid pair. Here, adjacency of bases is the key. Final Answer: 9, 64

Question 3

Odd One Out — In the twin-like pairs (919, 949), (646, 686), (828, 848), (434, 464), identify which pair does not obey “middle digit doubled” while first and last stay fixed.

Accepted Answer

Introduction / Context: Three pairs hold outer digits constant while doubling the middle digit on the right value; one pair violates this. Detecting a consistent digit operation is essential. Given Data / Assumptions: We compare left and right numbers digit-wise: left = x y z → right = x (2y) z. Doubling means numeric doubling (e.g., 3 → 6; 4 → 8). Concept / Approach: Check whether the right number keeps the same first and last digits and doubles the middle digit of the left number. Step-by-Step Solution: 646 → 686: outer digits 6 and 6 fixed; middle 4 → 8 (doubled) ✓828 → 848: outer digits 8 and 8 fixed; middle 2 → 4 (doubled) ✓434 → 464: outer digits 4 and 4 fixed; middle 3 → 6 (doubled) ✓919 → 949: outer digits 9 and 9 fixed; middle 1 → 4 (quadrupled, not doubled) ✗ Verification / Alternative check: There is no carry effect because all doubled middles (2→4, 3→6, 4→8) remain single-digit. Only 1→4 fails the doubling criterion (should be 2). Why Other Options Are Wrong: (646, 686): follows the doubling rule. (828, 848): follows the doubling rule. (434, 464): follows the doubling rule. Common Pitfalls: Reading speed errors (confusing 2→4 with 1→4). Always track digit-by-digit. Final Answer: 919, 949

Question 4

Odd One Out — In 83–75, 58–50, 49–42, 25–17, three differences are 8. Find the exception.

Accepted Answer

Introduction / Context: Constant-difference checks quickly reveal outliers in paired numbers. Here, the majority difference is 8. Given Data / Assumptions: Pairs: 83–75, 58–50, 49–42, 25–17. Compute absolute differences for consistency. Concept / Approach: Calculate |a − b| for each pair and compare values. Step-by-Step Solution: |83 − 75| = 8.|58 − 50| = 8.|49 − 42| = 7.|25 − 17| = 8.Only 49–42 deviates from the common difference of 8. Verification / Alternative check: Because all are descending pairs, direct subtraction b = a − d yields the same conclusion with d = 8 except for 49–42 (d = 7). Why Other Options Are Wrong: 83–75: follows difference 8. 58–50: follows difference 8. 25–17: follows difference 8. Common Pitfalls: Simple arithmetic slips. Recheck subtraction carefully for 49–42. Final Answer: 49-42

Question 5

Odd One Out — For the pairs (9, 49), (13, 121), (10, 61), (7, 25), choose the only pair that mixes an even number with an odd number.

Accepted Answer

Introduction / Context: Parity (even vs odd) is a fundamental classification tool. Three pairs are odd–odd; one is even–odd. Given Data / Assumptions: (9, 49): both odd. (13, 121): both odd (121 = 11^2). (7, 25): both odd. (10, 61): 10 is even, 61 is odd. Concept / Approach: Identify the pair whose members have different parity. Step-by-Step Solution: Check each pair’s parity: odd–odd, odd–odd, even–odd, odd–odd.The single even–odd mix is (10, 61). Verification / Alternative check: Other traits (e.g., perfect squares present) do not yield a unique outlier here because multiple pairs include squares. Parity isolates one pair cleanly. Why Other Options Are Wrong: (9, 49): odd–odd. (13, 121): odd–odd. (7, 25): odd–odd. Common Pitfalls: Overlooking that 121 is odd even though it is a square; all odd squares remain odd. Final Answer: 10, 61

Question 6

Odd One Out — Among the quadruples [77, 70, 49, 56], [56, 64, 96, 48], [66, 44, 23, 60], [18, 27, 63, 81], select the one that does not consist of multiples of a single common base.

Accepted Answer

Introduction / Context: Grouping by a shared factor is a common pattern in set-classification problems. Three sets here are built from clean multiples of a single number; one set is not. Given Data / Assumptions: [77, 70, 49, 56]: all multiples of 7. [56, 64, 96, 48]: all multiples of 8. [18, 27, 63, 81]: all multiples of 9. [66, 44, 23, 60]: 66 and 44 are multiples of 11; 60 is not; 23 is prime and not a multiple of 11. Concept / Approach: For each set, check divisibility by a candidate base. A valid set will see all members divisible by that base. Step-by-Step Solution: Divisible by 7? 77, 70, 49, 56 — all yes.Divisible by 8? 56, 64, 96, 48 — all yes.Divisible by 9? 18, 27, 63, 81 — all yes.Divisible by 11? 66 (yes), 44 (yes), 23 (no), 60 (no) — mixed. Verification / Alternative check: Compute gcd of each set: gcd(77,70,49,56)=7; gcd(56,64,96,48)=8; gcd(18,27,63,81)=9; gcd(66,44,23,60)=1 → confirms the outlier set. Why Other Options Are Wrong: [77, 70, 49, 56]: valid “all multiples of 7.” [56, 64, 96, 48]: valid “all multiples of 8.” [18, 27, 63, 81]: valid “all multiples of 9.” Common Pitfalls: Picking a base that only fits some elements (e.g., “multiples of 2” would match many numbers across sets but not isolate a single outlier). Final Answer: 66, 44, 23, 60

Question 7

Odd One Out — In the triplets (14, 17, 23), (19, 22, 28), (17, 20, 26), (21, 23, 30), three follow +3 then +6. Identify the exception.

Accepted Answer

Introduction / Context: Step-patterns in triplets are common: here, the dominant pattern is “add 3, then add 6.” We must locate the lone triplet that does not adhere. Given Data / Assumptions: Triplets: (14, 17, 23), (19, 22, 28), (17, 20, 26), (21, 23, 30). We compare consecutive differences. Concept / Approach: Compute d1 = second − first and d2 = third − second for each triplet; compare against the target (+3, +6). Step-by-Step Solution: (14, 17, 23): +3, +6 ✓(19, 22, 28): +3, +6 ✓(17, 20, 26): +3, +6 ✓(21, 23, 30): +2, +7 ✗ Verification / Alternative check: Any uniform alternative (e.g., +4, +6) fails at least two triplets. The +3, +6 profile fits three and isolates one. Why Other Options Are Wrong: (14, 17, 23): matches +3, +6. (19, 22, 28): matches +3, +6. (17, 20, 26): matches +3, +6. Common Pitfalls: Summing first and second to get the third; that does not hold here and obscures the simple difference pattern. Final Answer: 21, 23, 30

Question 8

Odd One Out — Among 2–4–8, 4–16–32, 3–9–27, 5–25–125, select the only triplet that is not (n, n^2, n^3).

Accepted Answer

Introduction / Context: Power progressions are a classic theme: many triplets take the form (n, n^2, n^3). We identify the single triplet that does not fit. Given Data / Assumptions: 2–4–8 → 2, 2^2, 2^3. 3–9–27 → 3, 3^2, 3^3. 5–25–125 → 5, 5^2, 5^3. 4–16–32 → 4, 4^2, but 4^3 = 64 (not 32). Concept / Approach: Compute squares and cubes of the first element and compare with the second and third elements of each triplet. Step-by-Step Solution: 2^2 = 4, 2^3 = 8 → matches.3^2 = 9, 3^3 = 27 → matches.5^2 = 25, 5^3 = 125 → matches.4^2 = 16, 4^3 = 64 → third is 32 ✗ Verification / Alternative check: Check ratios: in a true (n, n^2, n^3), ratios are n and n; only the 4-based triplet shows 16/4 = 4 and 32/16 = 2, not equal. Why Other Options Are Wrong: 2–4–8: valid power triplet. 3–9–27: valid power triplet. 5–25–125: valid power triplet. Common Pitfalls: Assuming geometric progression suffices. A GP with ratio 2 can also appear, but (n, n^2, n^3) imposes equal successive ratios (both equal to n). Final Answer: 4-16-32

Question 9

Odd One Out — For the quartets (11, 17, 23), (12, 14, 16), (8, 10, 12), (18, 36, 72), three are arithmetic progressions; one is a geometric progression. Identify it.

Accepted Answer

Introduction / Context: Sequence classification frequently contrasts arithmetic progression (AP) with geometric progression (GP). We locate the single GP among three APs. Given Data / Assumptions: AP: constant difference between consecutive terms. GP: constant ratio between consecutive terms. Concept / Approach: Compute differences and ratios for each sequence and classify accordingly. Step-by-Step Solution: (11, 17, 23): differences +6 and +6 → AP.(12, 14, 16): differences +2 and +2 → AP.(8, 10, 12): differences +2 and +2 → AP.(18, 36, 72): ratios ×2 and ×2 → GP. Verification / Alternative check: No uniform difference fits (18, 36, 72); no uniform ratio fits the other three. This cleanly separates one GP from the APs. Why Other Options Are Wrong: (11, 17, 23): AP. (12, 14, 16): AP. (8, 10, 12): AP. Common Pitfalls: Mistaking “even-only” as a defining feature; parity does not determine AP vs GP. Final Answer: 18, 36, 72

Question 10

Odd One Out — In the triples 3–4–14, 56–57–67, 102–103–113, 22–23–13, three end with a prime; one ends with a composite. Identify the exception.

Accepted Answer

Introduction / Context: Prime identification within short triples is a common classification trick. Here, the last term is decisive: three are primes; one is composite. Given Data / Assumptions: Candidates for the third term: 14, 67, 113, 13. Primality checks: 67 (prime), 113 (prime), 13 (prime), 14 (composite = 2 * 7). Concept / Approach: Check whether the final number of each triple is prime. If only one is composite, it is the odd one out. Step-by-Step Solution: 56–57–67 → 67 is prime.102–103–113 → 113 is prime.22–23–13 → 13 is prime.3–4–14 → 14 is composite (2 * 7). Verification / Alternative check: Quick divisibility rules: 14 is even → divisible by 2; the others are not divisible by small primes (3, 5, 7) within their bounds, confirming primality. Why Other Options Are Wrong: 56–57–67: ends with a prime. 102–103–113: ends with a prime. 22–23–13: ends with a prime. Common Pitfalls: Overanalyzing the first two terms. The discriminant here is solely the primality of the third term. Final Answer: 3-4-14

Question 11

Classification – Alphanumeric codes: identify the odd one out by checking whether the middle number equals the sum of the alphabetical positions of the first and last letters (A=1 … Z=26). Exactly three obey this rule; one does not. Which code is di…

Accepted Answer

Introduction / Context: Many verbal-reasoning “classification” items hide a deterministic rule. Here, each code has the pattern Letter–Number–Letter. A frequent device is to relate the number to the alphabetical positions of the two letters. We must find which code violates the common rule. Given Data / Assumptions: Alphabet index uses A=1 to Z=26. Codes given: C9F, H20L, N31Q, E29Y. Task: test if the middle number equals position(first letter) + position(last letter). Concept / Approach: Compute the position of each letter and compare the sum with the embedded number. The odd one will be the single code where the equality fails. Step-by-Step Solution: C9F: C=3, F=6 → 3+6=9 → matches.H20L: H=8, L=12 → 8+12=20 → matches.N31Q: N=14, Q=17 → 14+17=31 → matches.E29Y: E=5, Y=25 → 5+25=30 ≠ 29 → mismatch. Verification / Alternative check: Re-run the additions or cross-verify with a quick alphabet index list; only E29Y fails the equality, so it must be the outlier. Why Other Options Are Wrong: C9F, H20L, N31Q: In each case, the number equals the sum of the letter positions, so they belong to the majority pattern. Common Pitfalls: Confusing zero-based indexing with A=1, or attempting multiplication rather than addition. Also avoid off-by-one errors near the end of the alphabet. Final Answer: E29Y

Question 12

Classification – Letter pairs: choose the odd one out by comparing the forward shift from the first letter to the second (A=1 … Z=26). Three pairs advance by +4 positions; one pair advances by a different amount. Which pair is different?  Options: G…

Accepted Answer

Introduction / Context: Letter-pair classifications often rely on a constant positional shift from the first to the second letter. Detecting the uniform step highlights the majority; any deviation becomes the odd one out. Given Data / Assumptions: Alphabet index: A=1 … Z=26. Pairs: GK, MQ, PU, SW. Rule candidate: second letter position = first letter position + k (modulo 26), testing constant k. Concept / Approach: Compute the forward step for each pair and compare. Three pairs should share the same step; one will differ. Step-by-Step Solution: G(7) → K(11) = +4.M(13) → Q(17) = +4.S(19) → W(23) = +4.P(16) → U(21) = +5 (different). Verification / Alternative check: Reverse-check: subtract the second letter index from the first; only PU yields a +5 step, confirming the anomaly. Why Other Options Are Wrong: GK, MQ, SW: Each advances exactly +4 positions, so they fit the majority pattern. Common Pitfalls: Wrapping the alphabet unnecessarily (no wrap occurs here) or miscounting inclusive/exclusive steps. Final Answer: PU

Question 13

Classification – Letter triplets: pick the odd one out where the first to second jump is +2 and the second to third jump is +3 (A=1 … Z=26, cyclic if needed). Three triplets follow +2 then +3; one does not. Which triplet is different?  Options: CEH,…

Accepted Answer

Introduction / Context: Triplet sequences frequently embed two fixed jumps. Recognizing the increment sizes (+2 then +3 here) helps isolate the single violator. Given Data / Assumptions: Alphabet index A=1 … Z=26; treat Z→A with wrap for addition. Triplets: CEH, KMP, XZC, NPT. Target rule: first→second = +2; second→third = +3. Concept / Approach: Compute consecutive jumps per triplet and compare with the rule. Step-by-Step Solution: C(3)→E(5)=+2; E(5)→H(8)=+3 ✔K(11)→M(13)=+2; M(13)→P(16)=+3 ✔X(24)→Z(26)=+2; Z(26)→C(3)=(+3 with wrap) ✔N(14)→P(16)=+2; P(16)→T(20)=+4 ✖ Verification / Alternative check: Recompute with a quick index list; only NPT breaks the +3 second jump. Why Other Options Are Wrong: CEH, KMP, XZC: All satisfy +2 then +3 (with cyclic wrap in XZC). Common Pitfalls: Forgetting cyclic wrap at Z, or miscounting steps by including the starting letter twice. Final Answer: NPT

Question 14

Classification – Letter pairs: identify the odd one out using the rule “sum of alphabetical positions = 27.” Three pairs satisfy first+second = 27 (A=1 … Z=26); one pair does not. Which pair is different?  Options: DW, LO, JR, HS.

Accepted Answer

Introduction / Context: Complementary letter pairs often sum to a fixed constant (A+Z=27, B+Y=27, etc.). This is a common test in pair-classification problems. Given Data / Assumptions: A=1 … Z=26; target sum=27. Pairs: DW, LO, JR, HS. Concept / Approach: Compute the sum of positions for each pair; the outlier will not total 27. Step-by-Step Solution: D(4)+W(23)=27 ✔L(12)+O(15)=27 ✔H(8)+S(19)=27 ✔J(10)+R(18)=28 ✖ Verification / Alternative check: Check complements: D↔W, L↔O, H↔S are standard complement pairs; J↔Q would be 27, not J↔R. Why Other Options Are Wrong: DW, LO, HS: Each forms a valid 27-sum complementary pair. Common Pitfalls: Adding letter indices with off-by-one mistakes; ensure A=1 (not 0). Final Answer: JR

Question 15

Classification – Letter triplets with fixed jumps: identify the odd one out. In three triplets the pattern is “+4 then −2” over successive letters (A=1 … Z=26); one triplet violates this. Which triplet is different?  Options: PUS, HLJ, UYW, BFD.

Accepted Answer

Introduction / Context: Triplet puzzles commonly encode two-step movements. Our job is to find the majority pattern and the single exception. Given Data / Assumptions: Alphabet positions A=1 … Z=26. Triplets: PUS, HLJ, UYW, BFD. Candidate rule: first→second = +4; second→third = −2. Concept / Approach: Compute both jumps for each triplet, comparing with the proposed (+4, −2) scheme. Step-by-Step Solution: HLJ: H(8)→L(12)=+4; L(12)→J(10)=−2 ✔UYW: U(21)→Y(25)=+4; Y(25)→W(23)=−2 ✔BFD: B(2)→F(6)=+4; F(6)→D(4)=−2 ✔PUS: P(16)→U(21)=+5; U(21)→S(19)=−2 ✖ Verification / Alternative check: Re-evaluate PUS: the first jump is +5, not +4, confirming it as the violator. Why Other Options Are Wrong: HLJ, UYW, BFD: Each exactly follows +4 then −2, so they belong to the majority. Common Pitfalls: Forgetting to count strictly by positions or misreading U→S as −1 (it is −2). Final Answer: PUS

Question 16

Classification – Descending letter triplets: select the odd one out. In three triplets the pattern is “−1 then −3” across letters (A=1 … Z=26); one triplet breaks this. Which triplet is different?  Options: MLI, FEB, UTQ, SRN.

Accepted Answer

Introduction / Context: Another standard pattern uses fixed negative jumps. We check whether each triplet descends by −1 and then by −3; the deviant triplet is the answer. Given Data / Assumptions: Alphabet positions A=1 … Z=26. Triplets: MLI, FEB, UTQ, SRN. Target rule: first→second = −1; second→third = −3. Concept / Approach: Compute both jumps and compare with the target pattern. Step-by-Step Solution: MLI: M(13)→L(12)=−1; L(12)→I(9)=−3 ✔FEB: F(6)→E(5)=−1; E(5)→B(2)=−3 ✔UTQ: U(21)→T(20)=−1; T(20)→Q(17)=−3 ✔SRN: S(19)→R(18)=−1; R(18)→N(14)=−4 ✖ Verification / Alternative check: Re-check R→N: 18 to 14 is indeed −4, confirming the violation. Why Other Options Are Wrong: MLI, FEB, UTQ: Each satisfies −1 then −3, hence they belong together. Common Pitfalls: Counting letters inclusively (which would distort the step sizes) or mixing up N and O around 14–15. Final Answer: SRN

Question 17

Classification – Alphanumeric codes: choose the odd one out by inspecting the numeric part. Three codes contain a composite even number (non-prime); one code contains the only prime number. Which code is different?  Options: ZS12, PM4, RJ16, FD2.

Accepted Answer

Introduction / Context: When multiple rules could exist among the letters, the number itself sometimes provides the cleanest separator. Here we test primality/compositeness of the numeric component to find the outlier. Given Data / Assumptions: Codes: ZS12, PM4, RJ16, FD2. Numbers: 12, 4, 16, 2. Definition: a prime has exactly two positive divisors; composites have more than two. Concept / Approach: Classify each number as prime or composite and select the only prime as the odd one out. Step-by-Step Solution: 12 = 2 * 2 * 3 → composite.4 = 2 * 2 → composite.16 = 2^4 → composite.2 → prime (the only even prime). Verification / Alternative check: Quick divisor check confirms only 2 has exactly two divisors (1 and 2). Why Other Options Are Wrong: ZS12, PM4, RJ16: Their numeric parts are composite, so they form the majority group. Common Pitfalls: Misclassifying 1 (neither prime nor composite) or overlooking that 2 is the only even prime. Final Answer: FD2

Question 18

Classification – Perfect squares: pick the odd one out. Three numbers are perfect squares; one is not a perfect square. Which number is different?  Options: 144, 169, 256, 288.

Accepted Answer

Introduction / Context: Square numbers are a staple of numerical classification. Identifying which figures are exact squares separates the majority from the exception. Given Data / Assumptions: 144 = 12^2, 169 = 13^2, 256 = 16^2. 288 is between 16^2=256 and 17^2=289. Concept / Approach: Test each value against nearby integer squares or factorization to confirm square status. Step-by-Step Solution: Check 144 → 12 * 12 → perfect square.Check 169 → 13 * 13 → perfect square.Check 256 → 16 * 16 → perfect square.Check 288 → 16^2=256 and 17^2=289; 288 is not an exact square. Verification / Alternative check: Prime factors: 288 = 2^5 * 3^2 includes an odd exponent (5), hence not a perfect square. Why Other Options Are Wrong: 144, 169, 256: Each is an exact square and belongs to the majority. Common Pitfalls: Confusing 289 (17^2) with 288; a quick boundary check prevents this error. Final Answer: 288

Question 19

Classification – Domestic/working animals: identify the odd one out based on typical use. Three are common draught or riding animals; one is primarily a pet and not used for draught work. Which option is different?  Options: Camel, Horse, Bullock, C…

Accepted Answer

Introduction / Context: Animal-use classification distinguishes working/draught species from companion pets. The majority here are traditionally used for transport, riding, or pulling loads. Given Data / Assumptions: Camel: transport/draught in arid regions. Horse: riding/transport. Bullock (ox): draught animal for carts/ploughing. Cat: primarily a companion/pet, not a draught animal. Concept / Approach: Group by primary historical/typical human use: draught/riding vs pet/companion. Step-by-Step Solution: Tag Camel/Horse/Bullock as working animals.Tag Cat as a pet with no draught role.Hence, Cat is the odd one out. Verification / Alternative check: Cross-cultural usage records consistently place camels, horses, and oxen in transport/agriculture roles. Why Other Options Are Wrong: Camel/Horse/Bullock: All are typically used for riding or pulling loads. Common Pitfalls: Avoid edge cases like therapy animals or circus roles; the standard, widespread function is intended. Final Answer: Cat

Question 20

Classification – Spices vs oilseed: choose the odd one out. Three are culinary spices; one is primarily an oilseed/legume. Which option does not belong?  Options: Cumin, Groundnut, Clove, Peppier.

Accepted Answer

Introduction / Context: Food classifications often distinguish spices (aromatic flavoring agents) from seeds grown mainly for oil or protein. Identifying the category mismatch reveals the outlier. Given Data / Assumptions: Cumin: spice (seeds used for flavor). Clove: spice (flower buds). Pepper (spelled “Peppier” here): spice (peppercorns), but treated as spice. Groundnut: peanut; grown chiefly as an oilseed/legume, not a spice. Concept / Approach: Group by primary culinary role: spice vs oilseed/legume. Step-by-Step Solution: Mark Cumin/Clove/Pepper as spices.Mark Groundnut as oilseed/legume.Thus Groundnut is the odd one out. Verification / Alternative check: Market categories and cookery texts list cumin/clove/pepper under spices; groundnut under nuts/oilseeds. Why Other Options Are Wrong: Cumin/Clove/Pepper: All are classic spices used primarily for flavoring. Common Pitfalls: Do not confuse “seeds used in cooking” with the formal spice category; groundnut is not generally a spice. Final Answer: Groundnut

Classification Questions