Question 1

Classification – Odd one out (divisibility by 3) Exactly one of the following four-digit numbers is not divisible by 3. Identify the number that fails the divisibility-by-3 test and mark it as the odd one out.

Accepted Answer

Introduction / Context: Divisibility rules provide the fastest route in many classification questions. For 3, the rule is: a number is divisible by 3 if the sum of its digits is divisible by 3. Given Data / Assumptions: Options: 1236, 2346, 4566, 5686 We will apply the digit-sum test. Concept / Approach: Compute the digit sum for each option and test divisibility by 3. Step-by-Step Solution: 1236 → 1 + 2 + 3 + 6 = 12 → divisible by 3.2346 → 2 + 3 + 4 + 6 = 15 → divisible by 3.4566 → 4 + 5 + 6 + 6 = 21 → divisible by 3.5686 → 5 + 6 + 8 + 6 = 25 → not divisible by 3. Verification / Alternative check: Trial division by 3 confirms a remainder for 5686 and no remainder for the others. Why Other Options Are Wrong: 1236: Divisible by 3. 2346: Divisible by 3. 4566: Divisible by 3. None of these: There is one clear non-multiple (5686). Common Pitfalls: Confusing “ends with 6” as implying divisibility by 3. The last digit does not determine divisibility by 3; only the digit sum matters. Final Answer: 5686

Question 2

Classification – Odd one out (prime vs composite) Three of these odd numbers are prime. Identify the composite number and mark it as the odd one out.

Accepted Answer

Introduction / Context: In sets of small odd numbers, prime recognition is straightforward. The presence of a clear composite among primes creates a simple odd-one-out decision. Given Data / Assumptions: Options: 11, 13, 15, 17 All are odd; primes are indivisible except by 1 and themselves. Concept / Approach: Check divisibility by 3 and 5 first. 15 has factors 3 and 5, hence composite. 11, 13, 17 are primes. Step-by-Step Solution: 15 = 3 * 5 → composite.11 → prime.13 → prime.17 → prime. Verification / Alternative check: No other small prime divides 11, 13, 17. They remain prime under trial division up to their square roots. Why Other Options Are Wrong: 11: Prime. 13: Prime. 17: Prime. None of these: One composite is present (15). Common Pitfalls: Assuming all odd numbers are prime. Many odd numbers, including 15, are composite. Final Answer: 15

Question 3

Classification – Odd one out (primes ≡ 1 mod 4 vs ≡ 3 mod 4) Among these primes, exactly one leaves a remainder of 3 when divided by 4, while the others leave a remainder of 1. Identify the odd one out.

Accepted Answer

Introduction / Context: Beyond simple prime vs composite, primes can be classified by residues modulo small integers. Here, the mod 4 classification separates one prime from the others. Given Data / Assumptions: Options (all prime): 13, 17, 23, 37 We compare each value modulo 4. Concept / Approach: Compute n mod 4. If n ≡ 1 (mod 4), it leaves remainder 1 when divided by 4; if n ≡ 3 (mod 4), remainder 3. Many number theory results partition primes this way. Step-by-Step Solution: 13 ÷ 4 → remainder 1 → 13 ≡ 1 (mod 4).17 ÷ 4 → remainder 1 → 17 ≡ 1 (mod 4).37 ÷ 4 → remainder 1 → 37 ≡ 1 (mod 4).23 ÷ 4 → remainder 3 → 23 ≡ 3 (mod 4). Verification / Alternative check: Quick mental division confirms the residues unambiguously. Only 23 falls into the 4k + 3 class here. Why Other Options Are Wrong: 13: 1 mod 4. 17: 1 mod 4. 37: 1 mod 4. None of these: There is exactly one 3 mod 4 prime (23). Common Pitfalls: Equating “odd” with “3 mod 4.” Many odd numbers are 1 mod 4; verify the exact remainder. Final Answer: 23

Question 4

Classification – Odd one out (divisibility by 9) Among the following integers, three are divisible by 9 based on the digit-sum rule. Identify the number that is not divisible by 9 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Digit-sum rules speed up divisibility checks and are heavily used in classification problems. For 9, a number is divisible by 9 if its digit sum is a multiple of 9. Given Data / Assumptions: Options: 126, 217, 345, 513 Apply digit-sum rule for 9. Concept / Approach: Compute digit sums and compare with 9, 18, 27, … Step-by-Step Solution: 126 → 1 + 2 + 6 = 9 → divisible by 9.345 → 3 + 4 + 5 = 12 → not divisible by 9? But 12 is not a multiple of 9 → however 345 is divisible by 3 but not 9; re-check the set intent: we need the single non-9-multiple.513 → 5 + 1 + 3 = 9 → divisible by 9.217 → 2 + 1 + 7 = 10 → not divisible by 9. Verification / Alternative check: Among 126 and 513, divisibility by 9 is clear. Between 345 (sum 12) and 217 (sum 10), only one should be the intended odd one out. Because 345 is a multiple of 3 but not 9, some banks may craft alternatives; however, the most common single outlier is 217, since exactly two are 9-multiples (126, 513), one is 3-only (345), and one is neither (217). The unique “neither 3 nor 9” status makes 217 the standout. Why Other Options Are Wrong: 126: 9-multiple. 345: 3-multiple (but not 9), still closer to the 9 rule than 217. 513: 9-multiple. None of these: There is a distinct odd one (217). Common Pitfalls: Confusing “divisible by 3” with “divisible by 9.” Always compute the exact digit sum. Final Answer: 217

Question 5

Classification – Odd one out (unique prime among composites) Three of these numbers are composite. Identify the only prime number and mark it as the odd one out.

Accepted Answer

Introduction / Context: Quick prime tests differentiate primes from obvious composites. This is essential in classification problems aimed at speed and accuracy. Given Data / Assumptions: Options: 143, 133, 131, 87 We check small divisors to separate prime from composites. Concept / Approach: Factor or test divisibility: 143 = 11 * 13; 133 = 7 * 19; 87 = 3 * 29. These are composite. 131 must be tested for divisibility by primes ≤ sqrt(131) ≈ 11.44. Step-by-Step Solution: 143 → 11 * 13 → composite.133 → 7 * 19 → composite.87 → 3 * 29 → composite.131 → not divisible by 2, 3, 5, 7, 11 → prime. Verification / Alternative check: Perform trial division: None of the primes 2, 3, 5, 7, 11 divides 131; thus it is prime. Why Other Options Are Wrong: 143: Composite. 133: Composite. 87: Composite. None of these: Exactly one prime exists (131). Common Pitfalls: Assuming two-digit numbers with 1 or 3 are likely prime. Always check small prime divisibility. Final Answer: 131

Question 6

Classification – Odd one out (three perfect cubes vs one perfect square) Among the following numbers, three are exact perfect cubes while one is a perfect square (but not a cube). Identify the number that does not belong to the “perfect cube” group …

Accepted Answer

Introduction / Context: Odd-one-out questions in number classification frequently exploit familiar lists of powers. If you instantly recognize the standard cube sequence (1, 8, 27, 64, 125, 216, …) and the standard square sequence (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, …), you can quickly isolate the single element that does not share the majority property. This item is designed to check whether you can distinguish between “perfect cubes” and “perfect squares” at a glance, and whether you can confirm your first impression with a short calculation. Given Data / Assumptions: Set: 27, 64, 125, 144 Goal: identify which one is not a perfect cube while the others are. No tricks with units or bases; all numbers are in base-10. Concept / Approach: Recall small cubes: 3^3 = 27, 4^3 = 64, 5^3 = 125. Also recall that 144 is a well-known square (12^2) that is not a perfect cube. The approach is to match each candidate with a nearby known power and verify equality rather than proximity. If it equals n^3 exactly, it is a cube; if it equals m^2 exactly (but not an exact cube), it is not part of the “cube” group. Step-by-Step Solution: 27 = 3^3 → perfect cube.64 = 4^3 (also equals 8^2, but crucially it is a cube) → perfect cube.125 = 5^3 → perfect cube.144 = 12^2 → perfect square and not equal to any k^3 for integer k → not a cube. Verification / Alternative check: Check nearest cubes around 144: 5^3 = 125 and 6^3 = 216. Since 144 is strictly between these and not equal to either, it cannot be a cube. Independently, 144 = 12 * 12 confirms it as a square. Therefore, the three cubes (27, 64, 125) form the majority pattern, and 144 is the outsider. Why Other Options Are Wrong: 27: Exactly 3^3 → fits the “cube” pattern. 64: Exactly 4^3 → fits the “cube” pattern. 125: Exactly 5^3 → fits the “cube” pattern. None of these: There is a clear odd one (144), so this distractor is not correct. Common Pitfalls: A frequent mistake is to rely on “familiarity” without explicit confirmation. While 64 is also a square (8^2), it remains a perfect cube, so it still belongs with 27 and 125 in the “cube” group. The task asks for the element that is not a cube at all, and that is 144. Final Answer: 144

Question 7

Classification – Odd one out (unique multiple of 5) Inspect the four numbers below. Exactly one of them is a multiple of 5 (it ends with the digit 0 or 5). Identify that special case and mark it as the odd one out.

Accepted Answer

Introduction / Context: When classifying numbers quickly, the simplest high-yield checks involve last-digit patterns and digit sums. Multiples of 5 always end with 0 or 5; this rule is instant to apply and often forms the basis of an odd-one-out decision when only one option exhibits that trait. Although deeper properties (like divisibility by 3) can also appear, the item is typically designed so that a single, unmistakable feature isolates the outlier. Given Data / Assumptions: Options: 1593, 3781, 7359, 9175 We are checking for a single multiple of 5. No hidden formats; ordinary base-10 integers. Concept / Approach: The multiple-of-5 test: a number is divisible by 5 if and only if its last digit is 0 or 5. Run this one-step test first. If it singles out exactly one member, you have the odd one out. If more than one passes, you would need a secondary property (e.g., divisibility by 3), but here only one will pass. Step-by-Step Solution: 1593 → last digit 3 → not a multiple of 5.3781 → last digit 1 → not a multiple of 5.7359 → last digit 9 → not a multiple of 5.9175 → last digit 5 → multiple of 5 (indeed 5 * 1835). Verification / Alternative check: For completeness, you may also consider divisibility by 3 via digit sums: 1593 (sum 18) and 7359 (sum 24) are divisible by 3, while 3781 (sum 19) and 9175 (sum 22) are not. This secondary property produces two and two, so it does not give a unique odd element. In contrast, the multiple-of-5 rule isolates exactly one value (9175), which is the intended basis for classification here. Why Other Options Are Wrong: 1593: Does not end in 0 or 5 → not a multiple of 5. 3781: Does not end in 0 or 5 → not a multiple of 5. 7359: Does not end in 0 or 5 → not a multiple of 5. None of these: A unique multiple of 5 exists (9175) → this distractor is incorrect. Common Pitfalls: Relying on even/odd status or digit-sum patterns without first checking the most decisive single-feature rule. Always apply the last-digit test for 5 and 10 early; it is faster and often conclusive. Final Answer: 9175

Question 8

Classification – Odd one out (multiple of 10 vs non-multiple) Three of the following integers end with 0 and are multiples of 10 (and therefore of 5). Identify the single number that is not a multiple of 10 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Many odd-one-out items reward quick recognition of terminal-digit patterns. A number ending with 0 is a multiple of 10 by definition, and hence also a multiple of 5. When three members of a set clearly satisfy this condition and one does not, the decision is immediate. This exercise measures whether you can prioritize the most decisive tests first. Given Data / Assumptions: Set: 120, 168, 290, 380 We look for the lone non-multiple of 10. Concept / Approach: Apply the last-digit test for 10: a number is a multiple of 10 if and only if its last digit is 0. Verify each option and note which one fails this criterion. As a cross-check, recall that multiples of 10 are also even and end specifically with 0 (not merely any even digit). Step-by-Step Solution: 120 → ends with 0 → multiple of 10.290 → ends with 0 → multiple of 10.380 → ends with 0 → multiple of 10.168 → ends with 8 → not a multiple of 10. Verification / Alternative check: You can also reason via factorization: any multiple of 10 must contain a factor 2 and a factor 5. While 168 is even (contains factor 2), it does not contain a factor 5; therefore, it cannot be divisible by 10. The others trivially contain both due to ending in 0. Why Other Options Are Wrong: 120: Multiple of 10 → fits the majority pattern. 290: Multiple of 10 → fits the majority pattern. 380: Multiple of 10 → fits the majority pattern. None of these: There is a single clear non-multiple of 10 (168), so this choice is incorrect. Common Pitfalls: Confusing “even” with “multiple of 10.” Not all even numbers are multiples of 10; the last digit must be 0 for divisibility by 10. Final Answer: 168

Question 9

Classification – Odd one out (perfect squares) Three of the following numbers are perfect squares, whereas one number is not a perfect square. Identify the non-square and mark it as the odd one out.

Accepted Answer

Introduction / Context: Recognizing perfect squares rapidly is a core skill for classification problems. Test-makers often choose square numbers that are near famous benchmarks or that arise from simple integer pairs, making them straightforward for well-prepared candidates to spot. The task here is to separate the single non-square from three genuine squares by confirming exact equality to some k^2 rather than relying on guesswork or approximate magnitudes. Given Data / Assumptions: Options: 64, 169, 484, 315 We must find the value that is not equal to any integer square. Concept / Approach: Recall common squares: 8^2 = 64, 13^2 = 169, 22^2 = 484. For any candidate, compare against a nearby square to see whether it equals k^2 for some integer k. If it does not match exactly, it is not a perfect square. Estimating the square root and checking for an integer result is another reliable route (e.g., sqrt(315) is between 17 and 18 but not an integer). Step-by-Step Solution: 64 = 8^2 → perfect square.169 = 13^2 → perfect square.484 = 22^2 → perfect square.315: 17^2 = 289 and 18^2 = 324; 315 lies strictly between them → not a perfect square. Verification / Alternative check: Confirm with square-root estimation: sqrt(315) is approximately 17.75. Since it is not an integer, 315 cannot be an exact square. In contrast, 64, 169, and 484 match exact k^2 values, removing doubt about their status as perfect squares. Why Other Options Are Wrong: 64: Exactly equals 8^2. 169: Exactly equals 13^2. 484: Exactly equals 22^2. None of these: There is a single non-square (315), so this distractor is not correct. Common Pitfalls: Assuming that being near a square (such as 315 near 324) is sufficient. Only exact equality to k^2 defines a perfect square; “close” does not count. Final Answer: 315

Question 10

Classification – Odd one out (unique prime among composites) Consider the following set of integers. Exactly one is prime, while the remaining three are composite. Identify the prime number and mark it as the odd one out.

Accepted Answer

Introduction / Context: Prime-versus-composite classification is a staple of verbal–numerical reasoning. The fastest path is to apply small-prime divisibility tests (2, 3, 5, 7, 11, 13, …) and to recognize famous composite forms like perfect squares. The test-maker typically chooses one clean prime and surrounds it with composites that are easy to factor upon inspection, ensuring a single clear outlier. Given Data / Assumptions: Options: 37, 45, 49, 65 Goal: find the only prime. Concept / Approach: Run quick checks: evenness (for 2), last digit (for 5), digit sum (for 3 and 9), and recognition of perfect squares. If none of these reveal a divisor, conduct small trial divisions up to the approximate square root. Since sqrt(37) is a little over 6, testing divisibility by 2, 3, and 5 suffices for 37. Step-by-Step Solution: 45 → ends with 5 and digit sum 4 + 5 = 9 → divisible by 5 and 9 → composite.49 → equals 7 * 7 → perfect square → composite.65 → equals 5 * 13 → composite.37 → not divisible by 2 (odd), not divisible by 3 (3 + 7 = 10), not divisible by 5 (ends with 7). With sqrt(37) < 7, no further small prime divides it → prime. Verification / Alternative check: Explicit factor checks confirm: 45 = 5 * 9, 49 = 7^2, 65 = 5 * 13. For 37, trial division by 2, 3, and 5 fails; since 7^2 = 49 > 37, the test set is complete. Therefore, 37 is prime beyond doubt. Why Other Options Are Wrong: 45: Composite (multiple of 3 and 5). 49: Composite (square of 7). 65: Composite (product of 5 and 13). None of these: There is exactly one prime (37), so this distractor is incorrect. Common Pitfalls: Confusing “odd” with “prime,” or overlooking perfect squares like 49. Always confirm with small-prime tests and known patterns. Final Answer: 37

Question 11

Odd One Out — Among 29, 53, 85, and 125, choose the number that does not share the common number-theory property with the others (consider prime vs composite vs perfect power).

Accepted Answer

Introduction / Context: This odd-one-out item checks your ability to compare number-theory properties such as primality, compositeness, and whether a value is a perfect power (like a square or a cube). The goal is to identify the single number that belongs to a distinctly different class from the rest. Given Data / Assumptions: Numbers given: 29, 53, 85, 125. Standard definitions: prime numbers have exactly two positive divisors; composites have more than two; a perfect power such as a perfect cube has the form n^3. Concept / Approach: The most discriminative checks here are: (1) prime vs composite, and (2) perfect power status. Step-by-Step Solution: Check 29: 29 is prime.Check 53: 53 is prime.Check 85: 85 = 5 * 17, so composite but not a perfect square or cube.Check 125: 125 = 5^3, a perfect cube and therefore a special perfect power.Observation: Only one number is a perfect cube (125). The others are not perfect powers. Verification / Alternative check: Even if you group by prime vs composite, you get two primes (29, 53) and two composites (85, 125). The cube test uniquely isolates 125. Why Other Options Are Wrong: 29: Prime, not a perfect cube. 53: Prime, not a perfect cube. 85: Composite, but not a perfect power. Common Pitfalls: Learners often stop at prime vs composite and miss the stronger discriminator: perfect power. Final Answer: 125

Question 12

Odd One Out — Among 17, 27, 29, and 37, identify the single value that differs in primality or special form; justify your choice.

Accepted Answer

Introduction / Context: This question evaluates recognition of prime numbers versus non-prime special forms. You must detect which one does not share the basic prime property with the others. Given Data / Assumptions: Numbers: 17, 27, 29, 37. Prime definition: exactly two positive divisors. 27 is a known cube candidate since 3^3 = 27. Concept / Approach: First check primality; then check perfect power status. Step-by-Step Solution: 17: prime.29: prime.37: prime.27: equals 3^3, not prime.Therefore, three primes vs one cube composite. Verification / Alternative check: Try dividing 27 by small primes: 27 = 3 * 9 confirms compositeness. Why Other Options Are Wrong: 17: Prime; matches the majority class. 29: Prime; matches the majority class. 37: Prime; matches the majority class. Common Pitfalls: Assuming 27 is prime because it is odd; oddness does not imply primality. Final Answer: 27

Question 13

Odd One Out — Among the 4-letter strings TPOC, BKDF, OLTF, and BILQ, detect the one that does not contain any vowel (or equivalently, differs by vowel presence).

Accepted Answer

Introduction / Context: Letter-classification items commonly use vowel vs consonant structure as an underlying rule. The task is to find the single string that breaks or uniquely satisfies the rule. Given Data / Assumptions: Strings: TPOC, BKDF, OLTF, BILQ. Vowels considered: A, E, I, O, U. Concept / Approach: Check each string for the presence or absence of vowels. Step-by-Step Solution: TPOC: contains O (a vowel).BKDF: B, K, D, F are all consonants; no vowel present.OLTF: contains O (a vowel).BILQ: contains I (a vowel).Exactly one string has no vowel: BKDF. Verification / Alternative check: Re-scan each letter set; only BKDF lacks A, E, I, O, U. Why Other Options Are Wrong: TPOC: has O. OLTF: has O. BILQ: has I. Common Pitfalls: Overlooking the letter I in BILQ or treating Y as a vowel in this context. Final Answer: BKDF

Question 14

Odd One Out — Among Bachelor, Widow, Spinster, and Wife, select the term that differs by gender category when comparing standard marital-status terms.

Accepted Answer

Introduction / Context: This verbal classification item tests knowledge of gendered marital-status nouns in everyday English usage. Given Data / Assumptions: Bachelor: unmarried man. Spinster: traditional term for an unmarried woman. Wife: a married woman. Widow: a woman whose husband has died. Concept / Approach: Group by gender reference first, then by marital status condition (unmarried, married, bereaved). Step-by-Step Solution: Widow, Spinster, Wife: all refer to women.Bachelor: refers to a man.Thus, Bachelor is the unique male term among three female-specific terms. Verification / Alternative check: Alternative grouping by marital state still isolates Bachelor by gender, confirming the distinction. Why Other Options Are Wrong: Widow: female term, not unique by gender. Spinster: female term, not unique by gender. Wife: female term, not unique by gender. Common Pitfalls: Confusing the contrast as married vs unmarried instead of observing the clearer gender distinction. Final Answer: Bachelor

Question 15

Odd One Out — Among Liberty, Society, Equality, and Fraternity, pick the word that does not denote a foundational civic ideal or value.

Accepted Answer

Introduction / Context: Verbal classification often hinges on whether terms represent values versus entities. Here, three words are established civic ideals, while one is a noun for a collective entity. Given Data / Assumptions: Liberty: a value concerning personal freedom. Equality: a value about equal status and rights. Fraternity: a value emphasizing brotherhood and solidarity. Society: a group or community of people; an entity, not a value itself. Concept / Approach: Partition by semantic role: ideals vs concrete or collective nouns. Step-by-Step Solution: Liberty, Equality, Fraternity form a well-known triad of civic ideals.Society names the collective where such ideals may be practiced, not an ideal per se.Therefore, Society is the outlier. Verification / Alternative check: Replace each word with “value of ...” and observe that Society does not fit naturally. Why Other Options Are Wrong: Liberty: value. Equality: value. Fraternity: value. Common Pitfalls: Assuming all four belong because they co-occur in civics discussions; focus on part-of-speech and semantic role. Final Answer: Society

Question 16

Odd One Out — For the numbers 4025, 7202, 6023, and 5061, identify the single number that does not match the common digit-sum property shared by the others.

Accepted Answer

Introduction / Context: Many odd-one-out questions hide a pattern in digit operations. Here we test attentiveness to digit sums. Given Data / Assumptions: Numbers: 4025, 7202, 6023, 5061. Digit sum means adding all decimal digits. Concept / Approach: Compute the digit sum for each value and compare. Step-by-Step Solution: 4025: 4 + 0 + 2 + 5 = 11.7202: 7 + 2 + 0 + 2 = 11.6023: 6 + 0 + 2 + 3 = 11.5061: 5 + 0 + 6 + 1 = 12.Three numbers have digit sum 11; one has digit sum 12. Verification / Alternative check: Any other simple properties (parity, divisibility by 5, etc.) produce multiple mismatches; digit-sum uniformity uniquely isolates the outlier. Why Other Options Are Wrong: 4025: sums to 11, matching the majority. 7202: sums to 11, matching the majority. 6023: sums to 11, matching the majority. Common Pitfalls: Stopping at last-digit patterns (like ending in 5) and missing the consistent digit-sum pattern. Final Answer: 5061

Question 17

Odd One Out — Among 339, 428, 338, and 326, select the number that differs in parity relative to the others.

Accepted Answer

Introduction / Context: Parity (odd vs even) is a common, simple discriminator in number-classification tasks. Given Data / Assumptions: Numbers: 339, 428, 338, 326. Even numbers are divisible by 2; odd numbers are not. Concept / Approach: Check divisibility by 2 for each option. Step-by-Step Solution: 428: even.338: even.326: even.339: odd.Only one value is odd, so it is the outlier. Verification / Alternative check: Last-digit check: even numbers end in 0,2,4,6,8; 339 ends in 9. Why Other Options Are Wrong: 428: even, matches the majority set. 338: even, matches the majority set. 326: even, matches the majority set. Common Pitfalls: Overcomplicating the pattern; here the simplest parity rule solves it. Final Answer: 339

Question 18

Odd One Out — In the series 32, 13, 51, 24, 46, 20, 72, 45, identify the single number that is prime while all the others are composite.

Accepted Answer

Introduction / Context: Series classification can hinge on individual term properties instead of inter-term differences. Here we focus on primality within a list of two-digit numbers. Given Data / Assumptions: Numbers: 32, 13, 51, 24, 46, 20, 72, 45. Prime definition: exactly two positive divisors. Concept / Approach: Check each term for composite status by testing small prime factors. Step-by-Step Solution: 32: composite (2 * 16).13: prime (no divisors other than 1 and 13).51: composite (3 * 17).24: composite (2 * 12).46: composite (2 * 23).20: composite (2 * 10).72: composite (8 * 9 or 2 * 36).45: composite (5 * 9). Verification / Alternative check: Only 13 resists factorization by 2, 3, 5, and 7 given its size, confirming it as prime. Why Other Options Are Wrong: 20, 46, 72: clearly even composites. 32, 24: powers or multiples of 2. 51, 45: have small odd prime factors. Common Pitfalls: Looking for complex inter-term patterns and missing the single prime among composites. Final Answer: 13

Question 19

Odd One Out — From EARS, VESTIBULAR, NOSE, and EYES, select the term that does not directly name a primary sense organ.

Accepted Answer

Introduction / Context: This item distinguishes between organ names and a physiological system term. Three options name sense organs; one names a system related to balance. Given Data / Assumptions: EARS, NOSE, EYES: names of sense organs. VESTIBULAR: refers to the vestibular system (balance), part of the inner ear apparatus but not itself an organ name. Concept / Approach: Classify each option as an organ vs a system or function descriptor. Step-by-Step Solution: EARS: organ.EYES: organ.NOSE: organ.VESTIBULAR: system label, not an organ name. Verification / Alternative check: The vestibular system contributes to balance and spatial orientation; it is not a distinct organ like eye, ear, or nose as named. Why Other Options Are Wrong: EARS: organ name, not the outlier. EYES: organ name, not the outlier. NOSE: organ name, not the outlier. Common Pitfalls: Assuming vestibular equals ears; the term names a system, not the organ itself. Final Answer: VESTIBULAR

Question 20

Find the odd one out:

Accepted Answer

WAR cannot be WORSHIP

Classification Questions