Venn Diagrams

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.

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.

The universal set is divided into four mutually exclusive Venn-diagram regions: A ∩ Bᶜ, Aᶜ ∩ B, A ∩ B, and Aᶜ ∩ Bᶜ. The first three regions contain 48 + 44 + 28 = 120 elements. Since the universal set contains 160 elements, the remaining region is n(Aᶜ ∩ Bᶜ) = 160 − 120 = 40. Thus, option B is correct.

First find the number in at least one of the two sets using inclusion-exclusion: n(A ∪ B) = 52 + 47 − 24 = 75. Students in neither set are outside this union, so their number is n(U) − n(A ∪ B) = 90 − 75 = 15. Equivalently, this is n((A ∪ B)ᶜ). Therefore, option A is correct.

The set A contains four disjoint regions: the region belonging only to A, the region belonging to A and B but not C, the region belonging to A and C but not B, and the central region common to all three sets. Therefore, n(A)=18+9+5+7=39. The B∩C-only region is not included because it lies outside A.

“At least one set” means the union A∪B∪C. The given categories are mutually exclusive Venn-diagram regions: only A, only B, only C, exactly two sets, and all three sets. Hence every element in the union is counted exactly once. Therefore, n(A∪B∪C)=16+19+14+27+8=84, so option A is correct.

The union A∪B is partitioned into three non-overlapping regions: elements in A but not B, elements in B but not A, and elements common to both A and B. Therefore, n(A∪B)=n(A−B)+n(B−A)+n(A∩B)=23+17+12=52. Each element is counted exactly once in this sum, so option A is correct.

The statements A ∩ B = ∅ and B ∩ C = ∅ only say that A has no common element with B and C has no common element with B. They do not compare A directly with C. For example, let A = {1}, B = {2}, and C = {1, 3}. Then both given intersections are empty, but A ∩ C = {1}, which is not empty. Therefore, A and C may overlap, so option A is correct.

The region outside A is A′, and the region outside B is B′. To be outside both sets simultaneously, an element must belong to A′ ∩ B′. By De Morgan’s law, this is also equal to (A ∪ B)′. In contrast, A ∪ B includes elements in at least one set, A ∩ B is the overlap, and A − B lies inside A. Therefore option A is correct.

Students choosing at least one option are counted by the union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus the number is 68 + 57 − 31 = 94. The required percentage is (94 ÷ 120) × 100 = 78.333..., which rounds to 78.33%. The subtraction prevents students who chose both options from being counted twice. Therefore option A is correct.

The percentage in at least one set is the percentage of the union. Use the inclusion-exclusion rule: percentage(A ∪ B) = percentage(A) + percentage(B) − percentage(A ∩ B). Therefore, 35% + 48% − 20% = 63%. The common 20% is subtracted once because it was included in both 35% and 48%. Hence option A is correct.

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The given equality has no subtraction term, so n(A ∩ B) must be zero. Therefore, A and B have no common element; their circles do not overlap in the Venn diagram, and they are called disjoint sets.

When two sets partially overlap, the diagram is divided into three relevant parts: the portion belonging only to A, the common overlapping portion, and the portion belonging only to B. These are A − B, A ∩ B, and B − A respectively. No element can belong to two of these regions at the same time, so they are pairwise disjoint.

The union A ∪ B is partitioned into three mutually exclusive regions: A ∩ B′, A′ ∩ B, and A ∩ B. The outside region A′ ∩ B′ is not part of the union. Therefore n(A ∪ B) = 29 + 34 + 21 = 84. Option A is correct; subtracting or omitting one region produces the distractor values.

A universal set divided by A and B has four disjoint regions: A ∩ B′, A′ ∩ B, A ∩ B, and A′ ∩ B′. The first three contain 18, 22, and 16 elements, so their total is 56. Since the universal set has 75 elements, the outside region is 75 − 56 = 19. Therefore, n(A′ ∩ B′) = 19.

For each of the three sets, an element may be inside or outside that set. Thus there are 2 choices for each set and 2 × 2 × 2 = 2^3 = 8 membership patterns. One pattern means being outside A, outside B, and outside C simultaneously; it is the region outside all three circles. The remaining 8 − 1 = 7 patterns correspond to membership in at least one set, so they form the seven inner regions.

The union A ∪ B consists of three disjoint parts: students who chose only A, students who chose only B, and students who chose both. Therefore, 72 = 28 + 34 + n(A ∩ B). Solving gives n(A ∩ B) = 72 − 28 − 34 = 10. The 100-student total is not needed for this calculation because the question already gives the number choosing at least one option, that is, the union size.

For two finite sets, the addition n(A) + n(B) counts the common elements twice, so the union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 62 + 55 − 91 = 26. The universal-set size is not required because the union size is already supplied.

The set A contains both the students who belong only to A and the students common to A and B. Thus n(A) = n(only A) + n(A ∩ B). Therefore, n(only A) = 37 − 19 = 18. The universal-set size and the total size of B are unnecessary for this particular question. Subtracting the common region once prevents those students from being counted as exclusive members of A.

The complement of A ∪ B contains elements of the universal set that belong to neither A nor B. First calculate the union using n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 70 + 65 − 28 = 107. Then subtract this union from the universal set: n((A ∪ B)') = n(U) − n(A ∪ B) = 150 − 107 = 43. Hence option A is correct.

Since A ∩ B = ∅, the sets A and B are disjoint, so they have no common elements. Thus, n(A ∪ B) = n(A) + n(B) = 41 + 39 = 80. The complement of A ∪ B contains all elements of the universal set that are outside both A and B. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 100 − 80 = 20. Option A is correct.

The students studying exactly Mathematics are 72 − 38 = 34, because those studying both subjects must be removed. The students studying exactly Physics are 64 − 38 = 26. Adding these non-overlapping groups gives 34 + 26 = 60. Equivalently, n(A △ B) = n(A) + n(B) − 2n(A ∩ B) = 60. Thus, option A is correct.