Venn Diagrams

The complement of A ∪ B contains elements outside both sets. Thus, n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 180 − 38 = 142. For two sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the values gives 142 = 92 + 84 − n(A ∩ B), so n(A ∩ B) = 176 − 142 = 34. Therefore, option B is correct.

Let x = n(A ∩ B). The students in exactly one set are those in A only or B only. Their number is [n(A) − x] + [n(B) − x] = n(A) + n(B) − 2x. Therefore, 86 = 74 + 68 − 2x = 142 − 2x. Thus 2x = 56 and x = 28. Hence, the intersection contains 28 students, so option C is correct.

The total of the two set sizes is 77 + 64 = 141, which is 21 greater than the 120 elements available in U. Those extra 21 memberships must occur in the overlap. Using the lower-bound formula, n(A ∩ B) ≥ n(A) + n(B) − n(U), the minimum is 77 + 64 − 120 = 21. Thus option C is correct.

The intersection A ∩ B consists of elements that belong to both A and B. It cannot contain more elements than the smaller of the two sets, because every common element must be an element of B as well as A. Therefore, n(A ∩ B) ≤ min(n(A), n(B)) = min(66, 59) = 59. This maximum is attainable by placing all 59 elements of B inside A, so option B 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 64+58+52−24−21−19+8=118. The pairwise intersections are subtracted because they were counted twice, while the triple intersection is added once because it was then subtracted too many times.

Let x=n(A∩B∩C). The inclusion–exclusion formula gives 140=70+65+60−28−24−22+x. The sum of the single-set cardinalities is 195, and the sum of the pairwise intersections is 74, so 140=121+x. Therefore x=19. The triple intersection must be added because subtracting all pairwise intersections removes the common central region too many times.

Use the partition of each set into its exclusive part and common part. From A, n(A ∩ B) = n(A) − n(A − B) = 55 − 20 = 35. From B, the same intersection must be n(B) − n(B − A) = 50 − 18 = 32. A single intersection cannot simultaneously have two different cardinalities, so the supplied figures cannot describe valid sets together. Therefore the data are inconsistent.

Apply inclusion–exclusion 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). Thus the union is 96 + 88 + 74 − 40 − 31 − 29 + 12 = 170. The universal set has 200 people, so those in none of the sets are 200 − 170 = 30. Hence option C is correct.

A three-set Venn diagram has seven mutually exclusive 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. The six stated regions total 32 + 28 + 24 + 18 + 16 + 14 = 132. Since the union contains 150 elements, the central triple-intersection is 150 − 132 = 18. Therefore option B is correct.

First count total memberships across the three sets: n(A) + n(B) + n(C) = 82 + 76 + 70 = 228. People in exactly one set contribute one membership each, those in exactly two sets contribute two each, and those in all three contribute three each. If x is the exactly-one count, then 228 = x + 2(54) + 3(18). Hence x = 228 − 108 − 54 = 66. Therefore option A is correct.

Let x be the number of elements belonging to exactly two sets. The sum of the three set cardinalities is 72 + 66 + 60 = 198. Elements in exactly one set contribute once, elements in exactly two sets contribute twice, and elements in all three sets contribute three times. Thus 198 = 84 + 2x + 3(10) = 114 + 2x, so 2x = 84 and x = 42. Therefore, option B is correct.

Use 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 72 + 66 + 59 − 28 − 24 − 22 + 9 = 132. The triple intersection is added once because it was subtracted too many times during the pairwise correction.

Each pairwise intersection includes the elements that lie in all three sets. Thus, the elements in exactly A and B are 36 − 14 = 22; exactly B and C are 33 − 14 = 19; and exactly C and A are 31 − 14 = 17. Adding these disjoint regions gives 22 + 19 + 17 = 58. Therefore option B is correct.

First find the elements in exactly two sets by removing the triple intersection from each pair: (48 − 17) + (41 − 17) + (39 − 17) = 31 + 24 + 22 = 77. The elements in all three sets, 17, must then be included because “at least two” means exactly two or exactly three. Thus the required number is 77 + 17 = 94, option B.

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.

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 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.

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 three sets, inclusion-exclusion gives 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 40 + 38 + 35 − 14 − 12 − 10 + 5 = 82. The triple intersection is added once because it was subtracted too many times.

In a three-set Venn diagram, A−B means all elements that are in A and not in B. This is not necessarily the same as the region only in A, because some elements may belong to both A and C while still not belonging to B. Thus A−B can contain the only-A region and the A∩C-only region. The same careful interpretation applies to B−C and C−A, so option A is correct.

Each pairwise intersection includes the elements common to all three sets. If the three pairwise intersections are added, every element in the triple intersection is counted three times, although it belongs to none of the exactly-two regions. Subtracting 3n(A ∩ B ∩ C) removes these triple counts and leaves only elements in exactly two sets. Hence option A is correct.

The sum of the three pairwise intersections is 20+18+16=54. However, every element in the triple intersection has been counted in all three pairwise intersections. Since the triple intersection contains 6 elements, it contributes three times, so subtract 3×6. The number in exactly two sets is 54−18=36. Therefore option A is correct.

The pairwise intersections include the seven people who belong to all three sets. Thus the A∩B-only region is 18−7=11, and the C∩A-only region is 16−7=9. Set A contains its A-only region, these two pairwise-only regions, and the triple region. Hence A-only=45−11−9−7=18. Therefore option A is correct.