Venn Diagrams

Students studying at least one subject are counted by the union formula: n(S ∪ C) = n(S) + n(C) − n(S ∩ C). Substituting the values gives 34 + 27 − 11 = 50. The 11 students studying both are subtracted once because they were initially counted twice. Hence option A is correct.

The 32 mango lovers include the 9 people who like both mango and banana. To find those who like only mango, subtract the overlap from the mango total: 32 − 9 = 23. The banana total and group size are not needed for this particular calculation. Therefore, option B is correct.

Let S represent the students who passed Science and let M ∩ S represent those who passed both subjects. The students counted in Science only are obtained by removing the students who passed both from the total Science group: n(S only) = n(S) − n(M ∩ S) = 49 − 35 = 14. Therefore, option A is correct.

Let R be the set of people who visited the rides and G the set who visited the game stalls. The union counts everyone who visited at least one place. Since the 30 people who visited both places are included in both totals, they must be subtracted once: n(R ∪ G) = 70 + 52 − 30 = 92. Thus option A is correct.

The union A ∪ B contains every element from both A and B. If this union is exactly A, then adding the elements of B contributes nothing new; every element of B must already be in A. Therefore, B ⊆ A. The other relations are not necessarily true, so option B is the unique correct answer.

The intersection A ∩ B ∩ C consists of elements that belong simultaneously to A, B, and C. In a three-set Venn diagram, this is the central region where all three circles overlap. It is not a part belonging to only one set and cannot lie outside the circles. Thus option C is correct.

For three finite sets, the inclusion–exclusion formula is n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substituting the values gives 16 + 18 + 9 − 6 − 2 − 3 + 1 = 33. The pairwise intersections are subtracted because they were counted twice, while the triple intersection is added once because it was subtracted too many times.

By definition, A ∩ B contains precisely those elements that belong to both A and B. Every element belonging to both sets must certainly belong to A, so A ∩ B ⊆ A is always true. Option B reverses the usual inclusion relationship and is generally false. Option C is not generally valid because the two differences usually contain different elements. Option D is also not generally true because a set and its complement are normally different.

For two finite sets, the union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The intersection is subtracted once because its elements are included in both n(A) and n(B), so they would otherwise be counted twice. Substituting the values gives n(A ∪ B) = 36 + 28 − 12 = 52. Therefore option A is the only correct answer.

Since A is a subset of B, every element of A is already included in B. The set B − A contains the elements that belong to B but not to A. Therefore, its cardinality is n(B − A) = n(B) − n(A) = 45 − 18 = 27. In a Venn diagram, A lies inside B, and the region of B outside A contains these 27 elements.

The condition A ∩ B = ∅ means that A and B are disjoint sets; they have no common elements. For any two sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Here the intersection has cardinality zero, so n(A ∪ B) = 26 + 19 − 0 = 45. Thus all elements from both sets are counted exactly once.

The symbol ∩ denotes intersection. Therefore, A ∩ B is the set of all elements that belong to A and B simultaneously. In a Venn diagram, this set is represented by the overlapping region of the two circles. The portions belonging only to A or only to B are not included in the intersection.

The 27 students studying Mathematics include the 12 students who study both Mathematics and Science. To count only Mathematics students, remove the overlap from the Mathematics total: n(Mathematics only) = n(Mathematics) − n(Both) = 27 − 12 = 15. The Science total is not needed for this calculation, although it helps describe the complete Venn diagram.

For two sets, the union consists of three mutually exclusive regions: the elements only in A, the elements only in B, and the elements common to both sets. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 48 = 20 + 13 + n(A ∩ B), so n(A ∩ B) = 48 − 33 = 15. Hence, option A is correct.

The intersection A ∩ B ∩ C contains only those elements that belong to A, B, and C simultaneously. In a three-circle Venn diagram, the three circles overlap in one central region. That central region is counted in every one of the three sets, so it represents the required intersection. Regions belonging to only one set or to just two sets are excluded.

The set A ∩ B includes all elements common to A and B, including those that may also belong to C. The requested part excludes C, so remove the triple intersection: n((A ∩ B) − C) = n(A ∩ B) − n(A ∩ B ∩ C) = 14 − 5 = 9. Therefore option A is correct. The value 5 counts the part inside all three sets, not the required exclusive pairwise region.

The union A ∪ B ∪ C contains every element belonging to at least one of the three sets. Therefore, elements in none of the sets are in its complement within U. Using the complement-count rule, the number is n(U) − n(A ∪ B ∪ C) = 90 − 76 = 14. Hence option A is correct. The value 76 counts elements in at least one set, while 90 is the total universal-set size.

Set difference A−B means the elements that belong to A but do not belong to B. In the diagram, this is the portion of circle A outside the overlap with circle B. The common lens-shaped region is excluded because those elements are also in B. Therefore, only the non-overlapping part of A represents A−B.

The symmetric difference A △ B is the union of the two disjoint regions A−B and B−A. These regions cannot contain the same element, because the first excludes B and the second excludes A. Therefore, their cardinalities are added: n(A △ B)=n(A−B)+n(B−A)=23+31=54. Hence option A is correct.

The union A ∪ B contains elements in A only, B only, and both sets. To retain the elements in exactly one set, remove the common intersection from the union. Thus, n(exactly one)=n(A ∪ B)−n(A ∩ B)=63−17=46. Equivalently, the exactly-one region is (A−B)∪(B−A), so option A is correct.

Two sets are called disjoint when they have no element in common. The intersection records exactly the common elements, so for disjoint sets the intersection must be the empty set: \(A\cap B=\varnothing\). Their union need not be empty because both sets may contain many elements. Disjointness also does not imply equality or that one set is a subset of the other.

For any finite set X contained in the universal set U, the number of elements in its complement is \(n(X^c)=n(U)-n(X)\). Here, take \(X=A\cap B\). Therefore, \(n((A\cap B)^c)=75-18=57\). The value 18 is the intersection itself, while 75 is the entire universal set. Hence option A is correct.

The universal set contains every disjoint region shown in the two-set Venn diagram: the A-only region has 18 elements, the intersection has 12, the B-only region has 22, and the outside region has 8. Adding all regions gives \(n(U)=18+12+22+8=60\). Therefore, option A is correct.

The expression A ∩ Bᶜ means the elements that belong to A and do not belong to B. The region common to A and the complement of B is therefore the part of A lying outside B. By the definition of set difference, this region is A − B. It is not B − A, because that would represent elements in B but outside A.

For two finite sets, the union formula is n(A∪B)=n(A)+n(B)−n(A∩B). The intersection is subtracted because elements common to A and B are counted once in n(A) and once again in n(B). Therefore, n(A∪B)=46+38−16=68. The universal-set size is not needed for this calculation. Thus, option B is correct.