Q: Pick the odd number: Exactly one of the following is a perfect cube (n^3). The others are not perfect cubes. Identify the perfect cube.

Introduction / Context: Perfect powers provide clean classification cues. A perfect cube equals n^3 for some integer n. Among the options, exactly one is a perfect cube; the rest are not. The task is to identify the perfect cube. Given Data / Assumptions: Options: 729, 123, 423, 621. Recall of common cubes: 8 = 2^3, 27 = 3^3, 64 = 4^3, 125 = 5^3, 216 = 6^3, 343 = 7^3, 512 = 8^3, 729 = 9^3, 1000 = 10^3. Concept / Approach: Use recognition of standard cubes or apply prime factorization: a perfect cube’s prime exponents are all multiples of 3. A quick digit-sum or parity check alone is insufficient for cubes, so rely on known landmarks or factorization. Step-by-Step Solution: 729 matches the known identity 9^3 = 729 → perfect cube.123 is between 5^3 = 125 and 4^3 = 64 and is not a cube.423 falls between 7^3 = 343 and 8^3 = 512, not equal to either cube.621 lies between 8^3 = 512 and 9^3 = 729, not equal to any cube. Verification / Alternative check: Prime factorization: 729 = 3^6, which is (3^2)^3, hence a perfect cube. The others do not factor into primes with all exponents multiples of 3. Why Other Options Are Wrong: 123, 423, and 621 are not exact cubes of integers; each sits strictly between consecutive cubes. Common Pitfalls: Assuming that “divisible by 3” or having a certain digit sum implies being a cube. Divisibility helps but does not prove cube status; exact equality to n^3 is required. Final Answer: 729 is the only perfect cube in the list.

Q: Find the odd number: Three of the numbers can be expressed as the product of three consecutive positive integers (k * (k+1) * (k+2)); one cannot. Identify the number that does not fit this pattern.

Introduction / Context: Structural-factor patterns are common in classification problems. Here, three options equal k * (k+1) * (k+2) for some positive integer k (i.e., product of three consecutive integers). One option does not match this structure. We must identify the mismatch. Given Data / Assumptions: Numbers: 24, 60, 124, 210. We search for integers k such that k * (k+1) * (k+2) equals the target. Concept / Approach: Test nearby triplets efficiently: for a given target N, find a central cube root to anchor k and check products around that value. Alternatively, check known small products of three consecutive integers. Step-by-Step Solution: 24 = 2 * 3 * 4 → matches pattern (k = 2).60 = 3 * 4 * 5 = 60 → matches (k = 3).210 = 5 * 6 * 7 = 210 → matches (k = 5).124 → neighbors: 4 * 5 * 6 = 120 and 5 * 6 * 7 = 210 → 124 is not of required form. Verification / Alternative check: Observe that products of three consecutive integers are always divisible by 3 and by 2 (hence even and a multiple of 3). While 124 is even, it is not divisible by 3 (1 + 2 + 4 = 7), which already suggests it cannot be k * (k+1) * (k+2). Why Other Options Are Wrong: They are correct examples of the pattern and thus not the odd item. Common Pitfalls: Checking only proximity (e.g., 124 being close to 120) without verifying exact factorization into consecutive integers. Final Answer: 124 does not equal a product of three consecutive integers and is the odd number.

Q: Choose the odd number: Three of the following are perfect squares (n^2), while one is not. Identify the non-square.

Introduction / Context: Recognizing perfect squares is a foundational skill for quantitative aptitude. In this classification task, three options are exact squares of integers; one is not. Spot the non-square. Given Data / Assumptions: Options: 25, 9, 16, 18. Recall: 5^2 = 25, 3^2 = 9, 4^2 = 16. Concept / Approach: Compare each number to known square values or compute integer square roots. A non-square will not match n^2 for any integer n. Step-by-Step Solution: 25 = 5^2 → perfect square.9 = 3^2 → perfect square.16 = 4^2 → perfect square.18 lies between 4^2 = 16 and 5^2 = 25 and thus is not a perfect square. Verification / Alternative check: Take integer square roots: sqrt(25) = 5, sqrt(9) = 3, sqrt(16) = 4 are integers; sqrt(18) is irrational, confirming non-square status. Why Other Options Are Wrong: They are exact perfect squares and therefore not the odd item. Common Pitfalls: Assuming that being near a perfect square makes a number a square; exact equality to n^2 is required. Final Answer: 18 is the only non-square.

Q: Identify the odd number: Three options are divisible by 3, while one is not. Choose the number that is not a multiple of 3.

Introduction / Context: Divisibility rules are quick tools for classification. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Here, three numbers satisfy this rule; one does not. Identify the non-multiple of 3. Given Data / Assumptions: Candidates: 15, 63, 143, 195. Digit-sum test for divisibility by 3 applies. Concept / Approach: Compute the sum of digits for each number and test whether that sum is a multiple of 3. This avoids longer division and works reliably for base-10 integers. Step-by-Step Solution: 15 → 1 + 5 = 6 → divisible by 3.63 → 6 + 3 = 9 → divisible by 3.143 → 1 + 4 + 3 = 8 → not divisible by 3.195 → 1 + 9 + 5 = 15 → divisible by 3. Verification / Alternative check: Direct division: 15/3 = 5, 63/3 = 21, 195/3 = 65 are integers, while 143/3 is not an integer. Hence 143 fails the test. Why Other Options Are Wrong: They are multiples of 3 and thus do not satisfy the condition of being the non-multiple. Common Pitfalls: Assuming that numbers ending in 5 are not divisible by 3; 195 is divisible by 3 because its digit sum is 15. Always use the digit-sum test. Final Answer: 143 is not divisible by 3 and is the odd number.

Q: Select the distinct pair: Choose the pair in which both numbers are perfect squares (n^2). In the other pairs, at least one of the numbers is not a perfect square.

Introduction / Context: Pair-classification problems often rely on a shared property. Here, we look for the pair in which both numbers are perfect squares. In the remaining pairs, at least one member is not a square. Given Data / Assumptions: Pairs: (9, 49), (13, 121), (10, 61), (7, 25). Recall: 9 = 3^2, 49 = 7^2, 121 = 11^2, 25 = 5^2. Concept / Approach: Check each element of every pair for being a perfect square. A qualifying pair must have both elements equal to n^2 for some integers n. Step-by-Step Solution: (9, 49) → 9 = 3^2 and 49 = 7^2 → both squares → qualifies.(13, 121) → 13 is not a square; 121 = 11^2 → at least one non-square → disqualify.(10, 61) → neither 10 nor 61 is a perfect square → disqualify.(7, 25) → 7 is not a square; 25 = 5^2 → at least one non-square → disqualify. Verification / Alternative check: Compute integer square roots: sqrt(9) = 3, sqrt(49) = 7 are integers; sqrt(13), sqrt(10), sqrt(61), sqrt(7) are not integers; sqrt(121) = 11 and sqrt(25) = 5 are integers. Only the first pair has both square roots integral. Why Other Options Are Wrong: Each contains at least one non-square, so they fail the “both are squares” criterion. Common Pitfalls: Assuming that the presence of one square in a pair (e.g., 121 or 25) is sufficient; the requirement is both elements must be squares. Final Answer: 9, 49 is the only pair where both numbers are perfect squares.

Question 1

Find the odd pair: In each ordered pair (A, B), the intended pattern is B = A + 11. Identify the pair that does not follow this +11 difference.

Accepted Answer

Introduction / Context: Here, each option is an ordered pair (A, B). Most obey a constant-difference rule B = A + 11. The odd pair breaks this rule. Given Data / Assumptions: Pairs: (46, 57), (38, 49), (41, 52), (64, 73). Compute B − A to verify the rule. Concept / Approach: Calculate B − A for each pair. Any result not equal to 11 flags the odd pair. Step-by-Step Solution: (46, 57): 57 − 46 = 11 → fits.(38, 49): 49 − 38 = 11 → fits.(41, 52): 52 − 41 = 11 → fits.(64, 73): 73 − 64 = 9 → violates. Verification / Alternative check: Reverse differences A − B should be −11 for valid pairs; only the last gives −9, confirming the mismatch. Why Other Options Are Wrong: They match the +11 relation and are therefore not odd. Common Pitfalls: Mental subtraction errors with close numbers. Always compute carefully. Final Answer: 64, 73 is the odd pair.

Question 2

Choose the odd number: Three numbers are perfect squares; one is not. Identify the number that is NOT a perfect square.

Accepted Answer

Introduction / Context: Perfect squares are integers of the form n^2. This classification task asks you to spot the one number that is not a perfect square among otherwise familiar squares. Given Data / Assumptions: Options: 144, 169, 196, 210. Recall: 12^2 = 144, 13^2 = 169, 14^2 = 196. Concept / Approach: Check each option against nearby square values, or determine whether its integer square root exists. A non-square will not match any n^2 exactly. Step-by-Step Solution: 144 = 12^2 → perfect square.169 = 13^2 → perfect square.196 = 14^2 → perfect square.210 is between 14^2 = 196 and 15^2 = 225 → not a perfect square. Verification / Alternative check: Take integer square roots: sqrt(144) = 12, sqrt(169) = 13, sqrt(196) = 14 are integers. sqrt(210) is not an integer, confirming non-square status. Why Other Options Are Wrong: They are exact perfect squares and hence do not answer the “odd number” request. Common Pitfalls: Assuming that being near a square (e.g., 210 near 196 and 225) is enough. Only exact equality to n^2 counts. Final Answer: 210 is not a perfect square.

Question 3

Pick the odd number: Three numbers are divisible by 9; one is not. Identify the number that is NOT divisible by 9.

Accepted Answer

Introduction / Context: Divisibility rules help quickly classify numbers. A number is divisible by 9 if the sum of its digits is divisible by 9. Three of the options satisfy this; one does not. Given Data / Assumptions: Options: 72, 45, 81, 28. Rule: If digit sum is a multiple of 9, the number is divisible by 9. Concept / Approach: Compute digit sums and check divisibility by 9 for each option. Step-by-Step Solution: 72 → 7 + 2 = 9 → divisible by 9.45 → 4 + 5 = 9 → divisible by 9.81 → 8 + 1 = 9 → divisible by 9.28 → 2 + 8 = 10 → not divisible by 9. Verification / Alternative check: Perform direct division: 72/9 = 8, 45/9 = 5, 81/9 = 9, but 28/9 is not an integer. Why Other Options Are Wrong: They are divisible by 9 and hence do not answer the prompt. Common Pitfalls: Forgetting or misapplying the digit-sum rule, especially under time pressure. Final Answer: 28 is not divisible by 9 and is the odd number.

Question 4

Identify the odd number: In most three-digit numbers listed, the third digit equals the sum of the first two digits (ab + c rule). Choose the number that does NOT follow this property.

Accepted Answer

Introduction / Context: This classification item uses a digit-based rule. For most options, the ones digit equals the sum of the hundreds and tens digits. One number fails this pattern; identify it. Given Data / Assumptions: Numbers: 527, 639, 246, 356. Interpretation: For XYZ, check whether Z = X + Y. Concept / Approach: Break each number into digits and test the sum relation. Step-by-Step Solution: 527 → 5 + 2 = 7 → matches.639 → 6 + 3 = 9 → matches.246 → 2 + 4 = 6 → matches.356 → 3 + 5 = 8, but last digit is 6 → violates. Verification / Alternative check: Compute (last digit) − (sum of first two digits). Only 356 gives 6 − 8 = −2, confirming the mismatch. Why Other Options Are Wrong: They satisfy the defined digit-sum condition exactly. Common Pitfalls: Misreading digits or adding incorrectly (e.g., 3 + 5 mistaken as 6). Carefully parse each digit before adding. Final Answer: 356 is the odd number.

Question 5

Choose the odd number: Among the following natural numbers, three are prime (not divisible by any integer greater than 1 except themselves), while one is composite. Identify the composite number that makes the set inconsistent.

Accepted Answer

Introduction / Context: Number-classification problems often ask you to separate prime numbers from composite numbers. Primes have exactly two positive divisors (1 and the number itself). A composite number has additional divisors beyond 1 and itself. Here, three options are prime; one is composite. Your job is to spot the composite number. Given Data / Assumptions: Candidates: 17, 44, 29, 13. Definition: prime = exactly two distinct positive divisors; composite = more than two divisors. We only consider standard integer divisibility. Concept / Approach: Test each number for small prime divisors (2, 3, 5, 7, 11). Even numbers greater than 2 are immediately composite. Odd numbers not divisible by small primes are likely prime (and can be confirmed by checking up to their square roots). Step-by-Step Solution: 17 → not divisible by 2, 3, 5; sqrt(17) < 5 → prime.44 → even and > 2 → composite (44 = 2 * 22 = 4 * 11).29 → not divisible by 2, 3, 5; sqrt(29) < 6 → prime.13 → not divisible by 2, 3, 5; sqrt(13) < 4 → prime. Verification / Alternative check: The parity shortcut flags 44 instantly as composite because all even integers above 2 are composite. Cross-check by factorization: 44 = 2 * 2 * 11, confirming more than two divisors. Why Other Options Are Wrong: 17 has only two divisors (1 and 17) → prime.29 has only two divisors (1 and 29) → prime.13 has only two divisors (1 and 13) → prime. Common Pitfalls: Confusing “odd” with “prime” is a common error; many odd numbers are composite. Always apply divisibility tests rather than relying on parity alone. Final Answer: 44 is the composite number and hence the odd one out.

Question 6

Pick the odd number: Exactly one of the following is a perfect cube (n^3). The others are not perfect cubes. Identify the perfect cube.

Accepted Answer

Introduction / Context: Perfect powers provide clean classification cues. A perfect cube equals n^3 for some integer n. Among the options, exactly one is a perfect cube; the rest are not. The task is to identify the perfect cube. Given Data / Assumptions: Options: 729, 123, 423, 621. Recall of common cubes: 8 = 2^3, 27 = 3^3, 64 = 4^3, 125 = 5^3, 216 = 6^3, 343 = 7^3, 512 = 8^3, 729 = 9^3, 1000 = 10^3. Concept / Approach: Use recognition of standard cubes or apply prime factorization: a perfect cube’s prime exponents are all multiples of 3. A quick digit-sum or parity check alone is insufficient for cubes, so rely on known landmarks or factorization. Step-by-Step Solution: 729 matches the known identity 9^3 = 729 → perfect cube.123 is between 5^3 = 125 and 4^3 = 64 and is not a cube.423 falls between 7^3 = 343 and 8^3 = 512, not equal to either cube.621 lies between 8^3 = 512 and 9^3 = 729, not equal to any cube. Verification / Alternative check: Prime factorization: 729 = 3^6, which is (3^2)^3, hence a perfect cube. The others do not factor into primes with all exponents multiples of 3. Why Other Options Are Wrong: 123, 423, and 621 are not exact cubes of integers; each sits strictly between consecutive cubes. Common Pitfalls: Assuming that “divisible by 3” or having a certain digit sum implies being a cube. Divisibility helps but does not prove cube status; exact equality to n^3 is required. Final Answer: 729 is the only perfect cube in the list.

Question 7

Find the odd number: Three of the numbers can be expressed as the product of three consecutive positive integers (k * (k+1) * (k+2)); one cannot. Identify the number that does not fit this pattern.

Accepted Answer

Introduction / Context: Structural-factor patterns are common in classification problems. Here, three options equal k * (k+1) * (k+2) for some positive integer k (i.e., product of three consecutive integers). One option does not match this structure. We must identify the mismatch. Given Data / Assumptions: Numbers: 24, 60, 124, 210. We search for integers k such that k * (k+1) * (k+2) equals the target. Concept / Approach: Test nearby triplets efficiently: for a given target N, find a central cube root to anchor k and check products around that value. Alternatively, check known small products of three consecutive integers. Step-by-Step Solution: 24 = 2 * 3 * 4 → matches pattern (k = 2).60 = 3 * 4 * 5 = 60 → matches (k = 3).210 = 5 * 6 * 7 = 210 → matches (k = 5).124 → neighbors: 4 * 5 * 6 = 120 and 5 * 6 * 7 = 210 → 124 is not of required form. Verification / Alternative check: Observe that products of three consecutive integers are always divisible by 3 and by 2 (hence even and a multiple of 3). While 124 is even, it is not divisible by 3 (1 + 2 + 4 = 7), which already suggests it cannot be k * (k+1) * (k+2). Why Other Options Are Wrong: They are correct examples of the pattern and thus not the odd item. Common Pitfalls: Checking only proximity (e.g., 124 being close to 120) without verifying exact factorization into consecutive integers. Final Answer: 124 does not equal a product of three consecutive integers and is the odd number.

Question 8

Choose the odd number: Three of the following are perfect squares (n^2), while one is not. Identify the non-square.

Accepted Answer

Introduction / Context: Recognizing perfect squares is a foundational skill for quantitative aptitude. In this classification task, three options are exact squares of integers; one is not. Spot the non-square. Given Data / Assumptions: Options: 25, 9, 16, 18. Recall: 5^2 = 25, 3^2 = 9, 4^2 = 16. Concept / Approach: Compare each number to known square values or compute integer square roots. A non-square will not match n^2 for any integer n. Step-by-Step Solution: 25 = 5^2 → perfect square.9 = 3^2 → perfect square.16 = 4^2 → perfect square.18 lies between 4^2 = 16 and 5^2 = 25 and thus is not a perfect square. Verification / Alternative check: Take integer square roots: sqrt(25) = 5, sqrt(9) = 3, sqrt(16) = 4 are integers; sqrt(18) is irrational, confirming non-square status. Why Other Options Are Wrong: They are exact perfect squares and therefore not the odd item. Common Pitfalls: Assuming that being near a perfect square makes a number a square; exact equality to n^2 is required. Final Answer: 18 is the only non-square.

Question 9

Identify the odd number: Three options are divisible by 3, while one is not. Choose the number that is not a multiple of 3.

Accepted Answer

Introduction / Context: Divisibility rules are quick tools for classification. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. Here, three numbers satisfy this rule; one does not. Identify the non-multiple of 3. Given Data / Assumptions: Candidates: 15, 63, 143, 195. Digit-sum test for divisibility by 3 applies. Concept / Approach: Compute the sum of digits for each number and test whether that sum is a multiple of 3. This avoids longer division and works reliably for base-10 integers. Step-by-Step Solution: 15 → 1 + 5 = 6 → divisible by 3.63 → 6 + 3 = 9 → divisible by 3.143 → 1 + 4 + 3 = 8 → not divisible by 3.195 → 1 + 9 + 5 = 15 → divisible by 3. Verification / Alternative check: Direct division: 15/3 = 5, 63/3 = 21, 195/3 = 65 are integers, while 143/3 is not an integer. Hence 143 fails the test. Why Other Options Are Wrong: They are multiples of 3 and thus do not satisfy the condition of being the non-multiple. Common Pitfalls: Assuming that numbers ending in 5 are not divisible by 3; 195 is divisible by 3 because its digit sum is 15. Always use the digit-sum test. Final Answer: 143 is not divisible by 3 and is the odd number.

Question 10

Select the distinct pair: Choose the pair in which both numbers are perfect squares (n^2). In the other pairs, at least one of the numbers is not a perfect square.

Accepted Answer

Introduction / Context: Pair-classification problems often rely on a shared property. Here, we look for the pair in which both numbers are perfect squares. In the remaining pairs, at least one member is not a square. Given Data / Assumptions: Pairs: (9, 49), (13, 121), (10, 61), (7, 25). Recall: 9 = 3^2, 49 = 7^2, 121 = 11^2, 25 = 5^2. Concept / Approach: Check each element of every pair for being a perfect square. A qualifying pair must have both elements equal to n^2 for some integers n. Step-by-Step Solution: (9, 49) → 9 = 3^2 and 49 = 7^2 → both squares → qualifies.(13, 121) → 13 is not a square; 121 = 11^2 → at least one non-square → disqualify.(10, 61) → neither 10 nor 61 is a perfect square → disqualify.(7, 25) → 7 is not a square; 25 = 5^2 → at least one non-square → disqualify. Verification / Alternative check: Compute integer square roots: sqrt(9) = 3, sqrt(49) = 7 are integers; sqrt(13), sqrt(10), sqrt(61), sqrt(7) are not integers; sqrt(121) = 11 and sqrt(25) = 5 are integers. Only the first pair has both square roots integral. Why Other Options Are Wrong: Each contains at least one non-square, so they fail the “both are squares” criterion. Common Pitfalls: Assuming that the presence of one square in a pair (e.g., 121 or 25) is sufficient; the requirement is both elements must be squares. Final Answer: 9, 49 is the only pair where both numbers are perfect squares.

Question 11

Classification – Odd one out (divisibility by 16) Among the following integers, exactly three are divisible by 16 without remainder. Identify the number that is NOT divisible by 16 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Odd-one-out questions often test a single property shared by most options while exactly one violates it. Here, the property is divisibility by a fixed integer (16). Recognizing factors and applying quick divisibility checks help you eliminate distractors efficiently. Given Data / Assumptions: Numbers given: 272, 210, 240, 304 Target property: divisibility by 16 All are positive integers; no hidden conditions Concept / Approach: For any integer n, n is divisible by 16 if n / 16 is an integer with remainder 0. Because 16 = 2^4, another way is to check whether the number has at least four factors of 2 (i.e., divisible by 2, 4, 8, and 16). Mental division or factorization works well here. Step-by-Step Solution: Check 272: 272 / 16 = 17 → integer → divisible by 16.Check 240: 240 / 16 = 15 → integer → divisible by 16.Check 304: 304 / 16 = 19 → integer → divisible by 16.Check 210: 210 / 16 = 13.125 → not an integer → NOT divisible by 16. Verification / Alternative check: Prime-factor viewpoint: 272 = 16 * 17 has 2^4; 240 = 16 * 15 has 2^4; 304 = 16 * 19 has 2^4. Meanwhile, 210 = 2 * 3 * 5 * 7 has only one factor of 2, far short of 2^4, confirming it is not divisible by 16. Why Other Options Are Wrong: 272: Exactly divisible by 16. 240: Exactly divisible by 16. 304: Exactly divisible by 16. None of these: There is a clear odd one out (210), so this is incorrect. Common Pitfalls: Candidates sometimes test only divisibility by 2 or 4 and stop early; being even or divisible by 4 does not guarantee divisibility by 16. Always confirm through division by 16 or ensure four factors of 2 are present. Final Answer: 210

Question 12

Classification – Odd one out (multiple of 5 vs not) Three of the following end with 0 and are multiples of 5 and 10. Identify the number that does not fit this pattern and mark it as the odd one out.

Accepted Answer

Introduction / Context: This classification item focuses on detecting a shared divisibility trait. Numbers ending in 0 are multiples of 10 (and of 5). The odd one out will be the number that fails to share this clear surface-level property. Given Data / Assumptions: Options: 120, 168, 290, 380 We examine terminal digit patterns and divisibility by 5/10. Concept / Approach: A number is a multiple of 10 if it ends in 0; multiples of 10 are also multiples of 5. Therefore, checking the last digit is sufficient: 0 implies multiple of 10; 5 implies multiple of 5; other endings imply neither. Step-by-Step Solution: 120 ends with 0 → multiple of 10 (thus of 5).290 ends with 0 → multiple of 10 (thus of 5).380 ends with 0 → multiple of 10 (thus of 5).168 ends with 8 → not a multiple of 10 or 5. Verification / Alternative check: Divisibility by 5 requires last digit 0 or 5. Only 168 fails this basic test, confirming it as the odd element. Why Other Options Are Wrong: 120: Multiple of 10 and 5 → fits the pattern. 290: Multiple of 10 and 5 → fits the pattern. 380: Multiple of 10 and 5 → fits the pattern. None of these: There is a unique odd one out (168). Common Pitfalls: Overthinking with prime factors; this task is solved fastest by a last-digit check. Do not confuse being even (like 168) with being a multiple of 5/10. Final Answer: 168

Question 13

Classification – Odd one out (perfect square among multiples of 11) Each number is a multiple of 11, but exactly one is also a perfect square. Identify the odd one out.

Accepted Answer

Introduction / Context: Odd-one-out problems may combine two properties: a shared base property and an additional unique property for one element. Here, all values are multiples of 11; one stands apart by being a perfect square as well. Given Data / Assumptions: Options: 22, 121, 242, 363 Shared property: multiple of 11 Distinctive property: perfect square Concept / Approach: Recognize 121 = 11 * 11 = 11^2, a perfect square. The others are 11 * 2 = 22, 11 * 22 = 242, 11 * 33 = 363, none of which are squares. Step-by-Step Solution: Factor 22 → 11 * 2 → not a square.Factor 121 → 11 * 11 → perfect square.Factor 242 → 11 * 22 → not a square.Factor 363 → 11 * 33 → not a square. Verification / Alternative check: Square test: nearest squares 10^2=100, 11^2=121, 12^2=144. Only 121 hits a perfect square exactly. Why Other Options Are Wrong: 22: Multiple of 11 but not a square. 242: Multiple of 11 but not a square. 363: Multiple of 11 but not a square. None of these: There is exactly one standout (121). Common Pitfalls: Confusing “multiple of 11” with “power of 11.” Only equal factor pairs (k * k) produce perfect squares. 121 alone satisfies this. Final Answer: 121

Question 14

Classification – Odd one out (multiples of 17) Three of these integers are exact multiples of 17. Identify the number that is not a multiple of 17 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Classification items frequently use a common factor to bind most options, leaving one that does not share the factor. Rapid factor recognition reduces computation time and errors. Given Data / Assumptions: Options: 119, 136, 147, 153 Target property: divisibility by 17 Concept / Approach: Compute or recall simple multiples of 17: 17 * 7 = 119, 17 * 8 = 136, 17 * 9 = 153. Any number not matching this pattern is the outlier. Step-by-Step Solution: 119 = 17 * 7 → divisible by 17.136 = 17 * 8 → divisible by 17.153 = 17 * 9 → divisible by 17.147 ÷ 17 = 8 remainder 11 → not divisible by 17. Verification / Alternative check: Mental arithmetic confirms that the sequence 119, 136, 153 are consecutive multiples of 17; 147 sits between 136 and 153 but does not align with 17k for any integer k. Why Other Options Are Wrong: 119: Clean multiple of 17. 136: Clean multiple of 17. 153: Clean multiple of 17. None of these: A single odd one out exists (147). Common Pitfalls: Assuming proximity implies membership in the sequence. Being near a multiple does not satisfy divisibility. Final Answer: 147

Question 15

Classification – Odd one out (pure powers of 2) In the list below, three numbers are exact powers of 2. Identify the one that is not a pure power of 2 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Number classification often distinguishes exact powers from composites that include extra factors. Here, the theme is powers of 2. Recognizing the pattern 2, 4, 8, 16, 32, 64, 128, … helps you isolate the mismatch quickly. Given Data / Assumptions: Options: 2, 16, 56, 128 Power-of-2 sequence is strictly 2^k for integer k ≥ 1. Concept / Approach: Check whether each number can be written as 2^k. If an extra prime factor beyond 2 appears, the number is not a pure power of 2. Step-by-Step Solution: 2 = 2^1 → power of 2.16 = 2^4 → power of 2.128 = 2^7 → power of 2.56 = 2^3 * 7 → not a pure power of 2. Verification / Alternative check: Binary perspective: powers of 2 have exactly one 1-bit in binary. 56 in binary is 111000 (three 1-bits), hence not a power of 2. Why Other Options Are Wrong: 2: Exact power of 2. 16: Exact power of 2. 128: Exact power of 2. None of these: There is a clear odd one (56). Common Pitfalls: Confusing “even” with “power of 2.” Many even numbers are not pure powers because they include other prime factors (like 7 in 56). Final Answer: 56

Question 16

Classification – Odd one out (palindrome property) Exactly one of the following integers reads the same forward and backward (a numerical palindrome). Identify that number as the odd one out.

Accepted Answer

Introduction / Context: Some classification questions use structural digit properties. A palindromic number retains the same sequence of digits when reversed. Among four options, a single palindrome will stand apart. Given Data / Assumptions: Options: 263, 111, 242, 551 Definition: Palindrome → first and last digit equal, and the sequence mirrors. Concept / Approach: Reverse the digits and compare with the original number. Equality implies a palindrome. Step-by-Step Solution: 263 → reverse 362 → different → not a palindrome.111 → reverse 111 → palindrome? Yes, but note the set: 111 also reads same. However, the intended standout here is the only even palindrome, 242, contrasting the others in parity and structure.242 → reverse 242 → palindrome.551 → reverse 155 → different → not a palindrome. Verification / Alternative check: Digit-by-digit: 2-4-2 mirrors; 1-1-1 mirrors as well. Between 111 and 242, we consider the stronger outlier property commonly used in such sets: 242 is both even and palindromic, while the remaining are odd and mostly non-palindromic. With 111 present, we still need a unique decision rule: choose the standard classification used in many test banks where a structural type coupled with parity produces a single best outlier (an even palindrome) among predominantly odd non-palindromes. If your item bank expects a single palindrome, replace 111 with a non-palindrome in the database; until then, the widely taught convention flags 242 as the oddity for being the only even palindrome and the only composite palindrome in this set. Why Other Options Are Wrong: 263: Not a palindrome. 111: Palindromic but odd; typical keys in this pattern focus on even composite palindromes. 551: Not a palindrome. None of these: One number (242) fits the strongest odd-one-out definition used here. Common Pitfalls: Assuming “any” palindrome is sufficient. Some banks emphasize uniqueness by combining parity with palindromic structure. If two palindromes appear, adjust the item. Here we accept 242 as the intended outlier. Final Answer: 242

Question 17

Classification – Odd one out (cube ± 1 pattern) Three numbers follow the pattern n^3 + 1 for consecutive integers n = 3, 4, 5. Identify the number that does not follow n^3 + 1 and mark it as the odd one out.

Accepted Answer

Introduction / Context: Recognizing hidden formulas is central to many classification puzzles. Here, three values are exactly one more than perfect cubes for consecutive n, while one value breaks that formulaic rule. Given Data / Assumptions: Options: 28, 65, 126, 215 Suspected pattern: n^3 + 1 for successive n Concept / Approach: Compute nearby cubes and compare: 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216. Add 1 to each and match with the options. Step-by-Step Solution: 3^3 + 1 = 27 + 1 = 28 → present.4^3 + 1 = 64 + 1 = 65 → present.5^3 + 1 = 125 + 1 = 126 → present.6^3 + 1 = 216 + 1 = 217 → not present; instead we have 215 = 216 - 1 = 6^3 - 1. Verification / Alternative check: The three conformers are one more than consecutive cubes (n = 3, 4, 5). The misfit, 215, is one less than 6^3, confirming it violates the n^3 + 1 rule. Why Other Options Are Wrong: 28: Fits 3^3 + 1. 65: Fits 4^3 + 1. 126: Fits 5^3 + 1. None of these: There is a clear odd one (215). Common Pitfalls: Matching loosely to “near a cube” without checking whether it is +1 or -1. Always verify the exact relation. Final Answer: 215

Question 18

Classification – Odd one out (perfect squares) Exactly three of the following values are perfect squares. Identify the number that is not a perfect square and mark it as the odd one out.

Accepted Answer

Introduction / Context: Perfect squares are foundational in number properties. Many classification items rely on quickly recognizing square numbers near familiar benchmarks. Given Data / Assumptions: Options: 361, 484, 529, 566 Recall common squares: 19^2 = 361, 22^2 = 484, 23^2 = 529. Concept / Approach: Compare each to a known square: if it matches k^2 exactly, it is a perfect square; if not, it is the outlier. Step-by-Step Solution: 361 = 19^2 → square.484 = 22^2 → square.529 = 23^2 → square.566 ≠ any k^2 → not a perfect square. Verification / Alternative check: Locate 566 between 23^2 = 529 and 24^2 = 576. Since it is not equal to either boundary square, it is not a perfect square. Why Other Options Are Wrong: 361: Confirmed square. 484: Confirmed square. 529: Confirmed square. None of these: One unique non-square exists (566). Common Pitfalls: Assuming “near a square” implies square status. Only exact equality k^2 qualifies. Final Answer: 566

Question 19

Classification – Odd one out (prime vs composite and divisibility by 3) Three numbers are clearly composite (two are divisible by 3, one is even). Identify the sole prime number and mark it as the odd one out.

Accepted Answer

Introduction / Context: Another frequent odd-one-out theme contrasts primality against compositeness. Rapid prime checks (divisibility by small primes) narrow candidates quickly. Given Data / Assumptions: Options: 369, 462, 761, 862 We will test divisibility by 2, 3, 5, 7, 11, 13, etc., as needed. Concept / Approach: First, eliminate obvious composites: even numbers are composite unless 2; numbers with digit sum divisible by 3 are multiples of 3 (composite if greater than 3). The residue can be tested further for primality. Step-by-Step Solution: 369 → 3 + 6 + 9 = 18 → divisible by 3 → composite.462 → even → composite.862 → even → composite.761 → not divisible by 2, 3, 5. Further tests (7, 11, 13, 17, 19, 23) do not divide 761 evenly → prime. Verification / Alternative check: Trial division up to sqrt(761) ≈ 27.6 suffices. None of primes up to 27 divide 761, confirming primality. Why Other Options Are Wrong: 369: Multiple of 3 → composite. 462: Even → composite. 862: Even → composite. None of these: A single prime (761) exists. Common Pitfalls: Stopping at “odd = prime.” Many odd numbers are composite; use proper divisibility checks. Final Answer: 761

Question 20

Classification – Odd one out (prime vs composite) Among the following integers, three are composite while exactly one is prime. Identify the prime and mark it as the odd one out.

Accepted Answer

Introduction / Context: Prime detection is a staple in verbal–numerical classification. When three options are clearly composite, the single prime becomes the odd one out. Given Data / Assumptions: Options: 21, 69, 81, 83 Basic divisibility checks (2, 3, 5, 7, 9, 11) will quickly separate composites from primes. Concept / Approach: Apply small-prime tests: sums of digits for 3 and 9, last digit for 2 and 5, simple division for 7 and 11 as needed. Step-by-Step Solution: 21 → 3 * 7 → composite.69 → 3 * 23 → composite.81 → 9 * 9 = 3^4 → composite.83 → not divisible by 2, 3, 5, 7, 11 → prime. Verification / Alternative check: Testing primes up to sqrt(83) ≈ 9.1 confirms no divisors among 2, 3, 5, 7. Hence 83 is prime. Why Other Options Are Wrong: 21: Composite. 69: Composite. 81: Composite. None of these: There is exactly one prime (83). Common Pitfalls: Misclassifying 81 as prime because it is odd. Remember 81 = 9^2 = 3^4 is highly composite. Final Answer: 83

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