Complement of a Set and Its Properties

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 seven listed regions are disjoint and together form A∪B∪C. Their total is 23+27+21+13+11+9+7=111. The complement contains the universal-set elements outside this union. Hence n((A∪B∪C)ᶜ)=n(U)−n(A∪B∪C)=130−111=19. The triple intersection is counted once because it is given as its own region.

De Morgan’s law states that the complement of an intersection is the union of the individual complements. Therefore, \((A\cap B\cap C)^c=A^c\cup B^c\cup C^c\). An element is outside the intersection if it fails to belong to at least one of the three sets. Hence, option A is the only correct expression.

De Morgan’s law states that the complement of a union equals the intersection of the complements. Applying it successively gives (A∪B∪C)ᶜ = Aᶜ∩Bᶜ∩Cᶜ. In words, an element is outside the union exactly when it is outside A, outside B, and outside C simultaneously. Option A incorrectly uses a union of complements, while C and D omit the required complementation. Therefore, option B is correct.

De Morgan’s law states that \(A^c\cap B^c=(A\cup B)^c\). Thus the required set contains precisely the elements of \(U\) that are outside the union. Its cardinality is \(n(U)-n(A\cup B)=160-125=35\). Therefore option A is correct. The separate values of \(n(A)\) and \(n(B)\) are not needed once the union is known.

By De Morgan’s law, \(A^c\cup B^c=(A\cap B)^c\). Therefore its cardinality must be \(n(U)-n(A\cap B)=170-44=126\). The stated value is 146, which differs from 126 by 20. Since both values cannot describe the same complement in the same universal set, the supplied data are inconsistent. Hence option A is correct.

De Morgan’s law gives \(A^c\cup B^c=(A\cap B)^c\). Thus every element of the universal set is included except the 22 elements in \(A\cap B\). Consequently, \(n(A^c\cup B^c)=n(U)-n(A\cap B)=120-22=98\). Option D is correct. The other regional counts are consistent but unnecessary for this particular identity.

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.

By De Morgan’s law, Aᶜ∪Bᶜ=(A∩B)ᶜ. The complement of A∩B in the universal set must therefore contain 210−52=158 elements. However, the question states that n(Aᶜ∪Bᶜ)=170. Because 158 and 170 are different, the two supplied values cannot describe the same sets; hence the data are inconsistent.

De Morgan’s law gives Aᶜ∪Bᶜ=(A∩B)ᶜ. Thus, the requested set consists of every universal-set element except those in the intersection A∩B. Since U has 150 elements and A∩B has 29 elements, n(Aᶜ∪Bᶜ)=150−29=121. The difference-region values are consistent with the diagram but are not needed for this calculation.

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.

First find the sizes of A and B from their complements: n(A) = 80 − 35 = 45 and n(B) = 80 − 42 = 38. Then n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 45 + 38 − 25 = 58. Since A′ ∩ B′ = (A ∪ B)′ by De Morgan’s law, n(A′ ∩ B′) = 80 − 58 = 22. Therefore, option A is correct.

Since A ∩ B = ∅, the sets are disjoint and have no common elements. Therefore, n(A ∪ B) = n(A) + n(B) = 57 + 46 = 103. The complement of A ∪ B contains all universal-set elements outside that union. Hence n((A ∪ B)′) = n(U) − n(A ∪ B) = 125 − 103 = 22, so option A is correct.

There are floor(120/8) = 15 multiples of 8 and floor(120/12) = 10 multiples of 12. Numbers counted in both sets are multiples of lcm(8, 12) = 24, and there are floor(120/24) = 5 of them. Thus n(A ∪ B) = 15 + 10 − 5 = 20. The complement within U therefore has 120 − 20 = 100 elements. Hence option A is correct.

Use the inclusion-exclusion principle. The set A has 15 multiples of 8, while B has 10 multiples of 12. Their overlap consists of multiples of lcm(8, 12) = 24, giving 5 common elements. Therefore n(A ∪ B) = 15 + 10 − 5 = 20. Since the universal set has 120 elements, n((A ∪ B)') = 120 − 20 = 100. Thus option A is unambiguously correct.

Mutually disjoint sets have no common elements, so their union has size equal to the sum of their sizes: n(A∪B∪C)=27+34+41=102. The complement of this union contains the elements of U outside all three sets. Using n(X′)=n(U)−n(X), we get 125−102=23. Thus option A is correct; 102 is the union 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 statement is De Morgan’s law for the complement of a union. An element is outside A∪B exactly when it is not in A and also not in B. The conditions “not in A” and “not in B” describe A′ and B′, and their simultaneous occurrence is A′∩B′. Thus both the assertion and its reason are true, and the reason explains the assertion.