Complement of a Set and Its Properties

The complement \(A'\) consists of all elements in the universal set \(U\) that are not in \(A\). Since \(A\) and \(A'\) together partition \(U\), their cardinalities satisfy \(n(A')=n(U)-n(A)\). Substitution gives \(72-29=43\). Therefore, option B is correct. The value 29 is the size of A itself, not its complement.

The complement is taken relative to the stated universal set U. First, A ∪ B = {a,b,c,e,g}, containing every element in at least one of the two sets. Removing these union elements from U={a,b,c,d,e,g} leaves only d. Therefore (A ∪ B)′ = {d}, so option A is correct. Option B is the union itself, while {c} is the intersection.

First calculate the size of the union using n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Hence n(A ∪ B) = 58 + 47 − 20 = 85. The complement of the union contains all elements of the universal set that are outside both A and B. Therefore n((A ∪ B)′) = n(U) − n(A ∪ B) = 110 − 85 = 25. Thus option A is correct.

The complement of A ∩ B contains every element of the universal set U that is not in the intersection. For any finite set X contained in U, n(X′) = n(U) − n(X). Taking X = A ∩ B gives n((A ∩ B)′) = 96 − 34 = 62. Therefore option B is correct. The value 34 is the intersection itself, while 130 cannot be a complement size because it exceeds the universal-set size.

The region containing elements that are neither in A nor in B is the complement of their union, written as (A ∪ B)′. The universal set contains 92 elements, while the union contains 67 elements. Therefore the outside region has n((A ∪ B)′) = n(U) − n(A ∪ B) = 92 − 67 = 25 elements. Thus option A is correct. The value 67 represents the union itself, not the region outside both sets.

First form the union: A ∪ B = {a, c, d, e, f}. The complement of this union is found relative to the universal set U, so we select the elements of U that are not in the union. Those elements are b and h. Therefore, (A ∪ B)′ = {b, h}. The complement must always be taken with respect to the stated universal set.

A set and its complement are disjoint and together contain every element of the universal set. Therefore, A ∪ Aᶜ = U and n(A) + n(Aᶜ) = n(U). Substituting the given values gives 38 + 32 = 70, which confirms option A. The other statements either subtract the two regions or incorrectly claim that their cardinalities are equal.

The complement \(A^c\) contains all elements of the universal set U that are not members of A. Starting with U = {1,2,3,4,5,6,7} and removing 1, 3, 5, and 7 leaves {2,4,6}. Hence \(A^c=\{2,4,6\}\). The complement depends on the chosen universal set, so it cannot be identified without considering U.

The complement of A ∩ B contains every element of the universal set U that is not in the intersection. For a finite universal set, the complement rule is n(Xᶜ) = n(U) − n(X). Taking X = A ∩ B gives n((A ∩ B)ᶜ) = 105 − 37 = 68. Hence option A is correct.

The governing concept is the complement of a set relative to a stated universal set. Aᶜ contains every element of U that is not in A. From U = {p, q, r, s, t, u, v}, remove p, r, t, and v, because those four elements belong to A. The elements left are q, s, and u, so Aᶜ = {q, s, u}. Option A is A itself, while C is U and D is not justified; therefore B is correct.

The elements belonging to none of A, B, or C are precisely the elements in the complement of their union within the universal set. Hence n((A∪B∪C)ᶜ)=n(U)−n(A∪B∪C). Substituting the given values gives 110−92=18. Therefore, 18 elements lie outside all three sets, so option B is correct.

The complement Aᶜ contains all elements of the universal set U that are not in A. The sets A and Aᶜ are disjoint and together make up U, so n(A) + n(Aᶜ) = n(U). Substituting the given values gives n(A) + 48 = 95, hence n(A) = 95 − 48 = 47. Therefore, option A is correct.

The complement Aᶜ consists of all elements in the universal set U that are not elements of A. Starting with U and removing m, o, q, and s leaves n, p, r, and t. Hence Aᶜ={n,p,r,t}, so option B is correct. The complement depends on the chosen universal set; it is not simply the set of elements absent from the written list without reference to U.

The universal set U contains 160 elements. The union A∪B∪C contains every element belonging to at least one of the three sets, namely 127 elements. The remaining elements belong to none of them, so they form the complement of the union. Therefore, n((A∪B∪C)′)=n(U)−n(A∪B∪C)=160−127=33.

A set and its complement divide the universal set into two non-overlapping parts. Hence n(A) + n(Aᶜ) = n(U). Substituting the given values, n(A) + 58 = 132, so n(A) = 132 − 58 = 74. Therefore, option B is correct. The complement is not subtracted from A; together, A and Aᶜ account for every element of U exactly once.

The complement Bᶜ contains every element of the universal set U that is not an element of B. Therefore, the complement formula gives n(Bᶜ) = n(U) − n(B) = 150 − 108 = 42. The information A ⊆ B is not needed for this calculation; it only describes the relationship between A and B. Thus, option A is the only correct answer.

The complement of A ∪ B contains the elements of the universal set U that do not belong to A ∪ B. A set and its complement partition the universal set, so n((A ∪ B)ᶜ) = n(U) − n(A ∪ B). Substitution gives 150 − 118 = 32. Therefore, option A is correct; 118 is the union’s size, not its complement.

The inclusions A⊆B and B⊆C show that every element of A and B is already an element of C. Therefore the union of the nested sets is the largest set: A∪B∪C=C. Taking the complement relative to the same universal set U gives (A∪B∪C)′=C′. The other complements are generally larger and need not equal the required set, while the empty set is not justified.

The complement of A is defined relative to the universal set U. It contains every element of U that is not an element of A. Starting with U = {1, 2, 3, 4, 5} and removing 1, 3, and 5 leaves {2, 4}. Therefore A′ = U \ A = {2, 4}, making option A correct. The universal set must always be identified before finding a complement.

The complement B′ consists of all elements in the universal set U that are not in B. From U = {a, b, c, d}, remove b and d, the elements of B. The remaining elements are a and c. Thus B′ = U \ B = {a, c}, so option A is the only correct answer. The complement depends on the specified universal set.

To find A′, take the elements of the universal set U that are absent from A. The universal set contains 2, 4, 6, 8, and 10; removing 4 and 8 leaves 2, 6, and 10. Therefore A′ = U \ A = {2, 6, 10}. Option B repeats A itself, and option D omits 2, so option A is correct.

Under the usual school convention in this question, N denotes the positive natural numbers. Thus U = {1, 2, 3, 4, 5, 6}. The complement A′ contains the members of U that are not in A = {1, 2, 3}. Removing those three elements leaves {4, 5, 6}; hence option A is correct. Zero is not included under the stated convention.

First list every integer in the universal set: U = {−2, −1, 0, 1, 2}. The complement A′ contains exactly those elements of U that are not present in A. Since A = {−1, 0, 1}, removing these three elements from U leaves −2 and 2. Therefore, A′ = {−2, 2}. The complement is always defined relative to the stated universal set.

The complement A′ of a set A contains all elements of the universal set U that do not belong to A. Here A is the empty set, so it contains no elements at all. Consequently, none of the elements p, q, or r is excluded from the complement. Every element of U therefore belongs to A′, giving A′ = U = {p, q, r}. This illustrates the standard identity ∅′ = U.

The complement of \(A\) is defined as \(A'=U\setminus A\), meaning that it contains all elements of the universal set that are not in \(A\). Taking the complement again gives \((A')'=U\setminus(U\setminus A)=A\). Thus, the double-complement law states that complementing a set twice returns the original set. Therefore, option A is correct.