Venn Diagrams

The set A consists of the elements only in A, represented by A − B, together with the elements common to both sets, represented by A ∩ B. Therefore, n(A) = n(A − B) + n(A ∩ B), so n(A ∩ B) = 56 − 19 = 37. We can verify the result using set B: n(B) = n(B − A) + n(A ∩ B), giving 61 − 24 = 37. Since both calculations agree, the correct answer is option B, 37.

For set A, the intersection would have size n(A) − n(A − B) = 64 − 22 = 42. For set B, it would have size n(B) − n(B − A) = 59 − 20 = 39. The same intersection cannot simultaneously have two different cardinalities. Therefore the supplied data are inconsistent, making option A correct; the other conclusions do not resolve this contradiction.

Apply the three-set inclusion–exclusion formula: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substitution gives 118 + 104 + 92 − 48 − 41 − 37 + 16 = 204. Those in none of the sets are 260 − 204 = 56, so option B is correct.

A three-set Venn diagram has seven regions inside the union: three regions belonging to exactly one set, three regions belonging to exactly two sets, and the central region belonging to all three sets. The six given regions total 38 + 34 + 29 + 21 + 18 + 16 = 156. Since the union contains all seven regions, the central region is 172 − 156 = 16. Therefore, n(A∩B∩C)=16.

The phrases “exactly one,” “exactly two,” and “all three” refer to mutually exclusive regions of the Venn diagram. Their union is the complete set of elements belonging to at least one of A, B, or C. Therefore, n(A∪B∪C)=84+63+18=165. No further inclusion–exclusion correction is needed because the three given counts are already disjoint categories.

The union of three sets can be partitioned into three mutually exclusive categories: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Let x be the number in exactly two sets. Then 156 = 74 + x + 17. Hence x = 156 − 74 − 17 = 65. Thus, 65 elements lie in exactly two of the three sets.

The total of the three set sizes counts each person according to the number of sets to which that person belongs. If x people belong to exactly one set, then 96+88+82 = x + 2(69) + 3(21). Thus 266 = x + 138 + 63, so x = 266 − 201 = 65. The required number of people in exactly one set is therefore 65.

The union A ∪ B consists of three disjoint regions: elements only in A, elements common to A and B, and elements only in B. Since n(A − B) = 45 and n(A ∩ B) = 32, the number of elements only in B is 128 − 45 − 32 = 51. Therefore, n(B) = n(B − A) + n(A ∩ B) = 51 + 32 = 83. Hence, option C is correct.

De Morgan’s law states that Aᶜ∩Bᶜ=(A∪B)ᶜ. Thus, Aᶜ∩Bᶜ represents the elements in the universal set that lie outside A∪B. Since the universal set has 190 elements and the union contains 143 elements, the required number is 190−143=47. The individual values n(A) and n(B) are not needed after the union is given.

A two-set Venn diagram divides the universal set into four disjoint regions: A∩Bᶜ, Aᶜ∩B, A∩B, and Aᶜ∩Bᶜ. The first three regions contain 41+34+26=101 elements. Since all four regions together contain 130 elements, the remaining region is n(Aᶜ∩Bᶜ)=130−101=29. Therefore, option A is correct.

The symmetric difference is defined as A △ B = (A ∩ Bᶜ) ∪ (Aᶜ ∩ B). These two regions are disjoint, so their cardinalities must be added: n(A △ B) = 52 + 39 = 91. Since the question states that n(A △ B) is 100, the two pieces contradict the stated symmetric difference. Therefore, the data are inconsistent. The intersection A ∩ B is not included in the symmetric difference.

Add the cardinalities of the three sets to obtain the total membership count: 84 + 78 + 72 = 234. If x elements belong to exactly two sets, then elements in exactly one set contribute 96, elements in exactly two sets contribute 2x, and elements in all three sets contribute 3 × 12. Hence 234 = 96 + 2x + 36, so 2x = 102 and x = 51. Therefore, option B is correct.

The union can be partitioned into three disjoint categories: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Let the first category have x elements. Then 205 = x + 83 + 24. Therefore x = 205 − 83 − 24 = 98. The phrase “exactly two sets” means the total of the three pair-only regions, while the 24 elements in all three sets form a separate category. Hence option B is correct.

The union of two sets is divided into three disjoint Venn-diagram regions: A − B, B − A, and A ∩ B. Therefore n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given expressions gives 120 = x + 3x + 24. Thus 4x = 96 and x = 24. Consequently, n(A − B) is 24 and n(B − A) is 72, whose total with the intersection is 120.

First calculate the union using the addition rule for two sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 91 + 87 − 34 = 144. The complement of A ∪ B consists of elements in the universal set that belong to neither A nor B. Hence n((A ∪ B)ᶜ) = n(U) − n(A ∪ B) = 190 − 144 = 46. Therefore option A is correct.

The symmetric difference A △ B contains the elements belonging to exactly one of the two sets, while A ∩ B contains the common elements. These regions are disjoint and together make the entire union: A ∪ B = (A △ B) ∪ (A ∩ B). Therefore n(A ∪ B) = 112 + 39 = 151. The common region must be added because it is excluded from the symmetric difference. Hence option C is correct.

Apply the inclusion-exclusion formula for three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substitution gives 90 + 84 + 78 − 36 − 32 − 30 + 13 = 167. The triple intersection is added once at the end because it was subtracted too many times. Thus option B is correct.

For two subsets of a universal set, n(A ∪ B) cannot exceed n(U). Since n(A ∪ B) = n(A) + n(B) − n(A ∩ B), the minimum intersection occurs when the union is as large as possible, namely 180. Therefore, n(A ∩ B) = 105 + 96 − 180 = 21. Thus, at least 21 elements must be common to A and B.

The union is divided into three disjoint regions: A − B, A ∩ B, and B − A. First, n(B − A) = 136 − 49 − 31 = 56. Set B consists of B − A together with A ∩ B, so n(B) = 56 + 31 = 87. Therefore, option C is correct; 56 counts only the non-common part of B.

There are floor(96/4) = 24 multiples of 4 and floor(96/6) = 16 multiples of 6 in U. Numbers counted in both sets are multiples of lcm(4,6) = 12, so there are floor(96/12) = 8 of them. By inclusion-exclusion, n(A ∪ B) = 24 + 16 − 8 = 32. Therefore, option A is correct.