Equal sets and Subsets

Correct answer: A. A = B

Explanation: The prime factorization of 18 is 18 = 2 × 3 × 3 = 2 × 3². When prime factors are written as a set, a repeated factor is listed only once. Thus A = {2, 3}, which is exactly the set B. The sets are equal, not proper subsets of one another, so option A is correct.

Correct answer: A. {2, 4, 8, 16}

Explanation: The positive divisors of 16 are 1, 2, 4, 8, and 16. The condition requires x to be even, so the odd divisor 1 must be removed. The remaining elements are 2, 4, 8, and 16; therefore A = {2, 4, 8, 16}, which is option A. Option B includes 1, option C includes 6, which is not a divisor of 16, and option D omits 2.

Correct answer: A. C ⊂ A and C ≠ A

Explanation: The natural-number divisors of 15 are A = {1, 3, 5, 15}. Every element of C = {1, 3, 5} is present in A, so C is a subset of A. However, C does not contain 15, which belongs to A, so the two sets are not equal. Hence C is a proper subset of A, written C ⊂ A and C ≠ A. Therefore option A is correct.

Correct answer: D. 7

Explanation: The symbol A ⊂ B denotes a proper subset, so A is contained in B but is not equal to B. For finite sets, a proper subset must have strictly fewer elements than the original set. Since B has 7 elements, n(A) may be 0, 4, or 6, depending on the subset, but it cannot be 7. If n(A) were 7, A and B would have equal finite cardinality and would be equal, contradicting A ⊂ B. Thus option D is impossible.

Correct answer: A. B ⊂ A

Explanation: The elements of B are 4 and 8, and both of them occur in A. Therefore B is a subset of A. Since A also contains 2, 6, and 10, which are not in B, the two sets are not equal; B is specifically a proper subset of A. Statement A is therefore correct. Statement B reverses the inclusion, statement C ignores the extra elements of A, and statement D is false because 10 is not in B.

Correct answer: A. {3, 5, 6}

Explanation: The condition 3 ≤ x < 7 gives A = {3, 4, 5, 6}. A proper subset must contain only elements of A and must omit at least one element of A. Option A, {3, 5, 6}, satisfies both conditions: all its elements belong to A, but it omits 4, so it is smaller than A. Option B equals A, while options C and D contain elements outside A. Hence option A is correct.

Correct answer: A. \(B\subset A\) and \(B\neq A\)

Explanation: Factor the quadratic as \(x^2-4x+3=(x-1)(x-3)\). Thus its natural-number solutions are 1 and 3, so \(A=\{1,3\}\). Since \(B=\{1\}\), every element of B belongs to A, but A has the additional element 3. Therefore B is a proper subset of A: \(B\subset A\) and \(B\neq A\). Option B is false because the sets do not have exactly the same elements, while option D is false because 3 is not in B.

Correct answer: D. Neither \(A\subseteq B\) nor \(B\subseteq A\)

Explanation: For \(A\subseteq B\), every element of A must belong to B. This fails because 1 belongs to A but not to B. Similarly, \(B\subseteq A\) is false because 4 belongs to B but not to A. The sets are also not equal, although they share the elements 2 and 3. Therefore neither set is a subset of the other, so option D is correct. Common elements alone do not establish a subset relation.

Correct answer: A. \(A=B\)

Explanation: The positive multiples of 6 that are less than 30 are obtained by using 6, 12, 18, and 24. The next multiple is 30, but it is excluded because the condition says less than 30, not less than or equal to 30. Therefore \(A=\{6,12,18,24\}=B\), so option A is correct. Options B and C incorrectly claim a proper-subset relation, while option D is false because 30 is not included.

Correct answer: A. \(A=B\)

Explanation: The positive multiples of 5 below 20 are 5, 10, and 15. The number 20 is not included because the inequality is strict: less than 20. Hence the set described by B is \(\{5,10,15\}\), exactly the same as A. Therefore A and B are equal sets, and option A is correct. Neither proper-subset option applies, and option D is false because 20 is excluded from B.

Correct answer: A. {-2, 0, 2}

Explanation: The condition x ∈ Z means that x must be an integer, and |x| ≤ 2 means that x can range from -2 to 2 inclusive. Therefore, A = {-2, -1, 0, 1, 2}. A set is a subset of A only when every one of its elements belongs to A. Option A contains -2, 0, and 2, all of which are in A, so it is the only valid subset.

Correct answer: A. B ⊂ A and C ⊂ A

Explanation: Every element of B, namely 2 and 4, belongs to A, and B has fewer elements than A; therefore B is a proper subset of A. Similarly, every element of C, namely 1 and 3, belongs to A, and C is also smaller than A. Hence both B ⊂ A and C ⊂ A are true. B and C are not equal because they contain different elements.

Correct answer: B. A ⊂ B and A ≠ B

Explanation: The empty set ∅ is a subset of every set, so ∅ ⊆ {0}. Since B contains the element 0 while A contains no elements, A and B are not equal. Therefore, A is a proper subset of B, written A ⊂ B. Also, 0 is not an element of A because A is empty, so option D is false.

Correct answer: A. A = B

Explanation: The prime digits less than 10 are 2, 3, 5, and 7. Therefore B = {2, 3, 5, 7}, which has exactly the same elements as A. Hence A = B. The digit 0 is not prime, and 1 is also not prime, so neither can be added to B. Options B and C incorrectly claim a proper-subset relationship, which requires one set to contain an element absent from the other. Option D is false because the intersection is the whole set, not {1}.

Correct answer: A. It is true, and A is a proper subset of B.

Explanation: Every element of A, namely 1, 2, and 3, is also an element of B. In addition, B contains the element 4, which is not in A. Thus A is contained in B but is not equal to B; this is exactly the definition of a proper subset, written A ⊂ B. Option B contradicts the extra element 4. Options C and D misunderstand the definition: the absence of 4 from A supports the proper-subset relation rather than disproving it.

Correct answer: A. A = B

Explanation: A positive divisor of 20 is a positive integer that divides 20 without a remainder. The complete list is 1, 2, 4, 5, 10, and 20. Thus A = {1, 2, 4, 5, 10, 20}, which is exactly the displayed set B. Therefore A = B. The list must include both 1 and 20: 1 divides every positive integer, and every positive integer divides itself. Removing either would make the set incomplete, so option D is also incorrect.

Correct answer: A. A = B

Explanation: In ordinary set notation, an element is recorded only once; repeated listings do not create new elements. Therefore A = {a, b, c} after removing the repeated a and b. This is exactly the set B, so A = B. Option B incorrectly treats a set like an ordered list or multiset. Option C is false because B is not a proper subset of A when the two sets are equal. Option D is false because their intersection is {a, b, c}, not the empty set.

Correct answer: A. A = B (A और B समान समुच्चय हैं)

Explanation: The positive multiples of 3 are 3, 6, 9, 12, 15, and so on. Applying the strict condition x < 15 excludes 15, leaving A = {3, 6, 9, 12}. This is exactly the set B, so A = B. Neither set contains an extra element, and option D is wrong because 15 does not satisfy x < 15. The symbols < and ≤ must be distinguished carefully.

Correct answer: A. A ⊂ B ⊂ C (A, B का और B, C का उचित उपसमुच्चय है)

Explanation: Every element of A, namely 1 and 2, is also present in B, and B contains the additional element 3. Thus A is a proper subset of B. Similarly, every element of B is present in C, while C has the additional element 4, so B is a proper subset of C. Therefore the correct chain is A ⊂ B ⊂ C. The inclusions are proper because the consecutive sets are not equal.

Correct answer: B. \(\{2,4,6\}\)

Explanation: Every set is a subset of itself, so \(A\subseteq A\). However, a proper subset must be strictly smaller than the original set and is usually written with \(\subset\) or \(\subsetneq\). The set \(\{2,4,6\}\) is exactly equal to \(A\), so it is a subset but not a proper subset. The other three choices are proper subsets because they contain fewer elements than \(A\). Therefore option B is correct.

Correct answer: A. \(A=B\)

Explanation: The prime factorization of 30 is \(30=2\times3\times5\). Therefore its prime divisors are exactly 2, 3, and 5, so \(A=\{2,3,5\}\). This is the same set as \(B\), which means \(A=B\), not merely a proper subset. The numbers 1 and 30 are not prime numbers, so option D is also incorrect. Hence option A gives the correct relation.

Correct answer: A. \(B=C\) और \(B\subset A\)

Explanation: To form \(C\), select from \(A\) only those elements that are even. The even elements of \(\{1,2,3,4,5,6\}\) are 2, 4, and 6, so \(C=\{2,4,6\}\). This is exactly the set \(B\), hence \(B=C\). Since 2, 4, and 6 belong to \(A\), and A also contains 1, 3, and 5, B is a proper subset of A. Therefore option A is correct.

Correct answer: A. A = B

Explanation: The condition x ∈ N and x divides 18 means that A contains all positive divisors of 18. These divisors are 1, 2, 3, 6, 9, and 18, so A = {1, 2, 3, 6, 9, 18}. This list is exactly B; hence A = B. Option B is false because 1, 6, 9, and 18 are not prime. Option C is false because the two sets have all the same elements. Option D is false because 18 divides itself.

Correct answer: A. {-3, -2, -1, 0, 1}

Explanation: The condition x ∈ Z means that only integers are allowed. The inequality -3 ≤ x includes -3, while x < 2 excludes 2. Listing all integers between these limits gives -3, -2, -1, 0, and 1. Thus A = {-3, -2, -1, 0, 1}, making option A correct. Option C incorrectly includes 2, and option D omits -3.

Correct answer: A. {2}

Explanation: An element of A must satisfy both conditions: it must be even and it must be prime, while also being less than 10. The only even prime number is 2, and 2 is less than 10. The numbers 4, 6, and 8 are even but composite; 3, 5, and 7 are prime but odd. Therefore A contains only 2, so A = {2}. Option A is correct.