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

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