Venn Diagrams

Use the inclusion-exclusion identity for two sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, n(A ∩ B) = 35 + 30 − 50 = 15. The intersection is the part counted in both sets, so it is the amount that must be removed from the sum of the two separate set sizes to obtain the union.

First calculate the number in at least one group: n(A ∪ B) = 45 + 37 − 18 = 64. Students in neither group are outside the union but still inside the universal set U. Hence, neither = n(U) − n(A ∪ B) = 80 − 64 = 16. Therefore, option A is correct; 64 is the number in at least one group, not the number outside both.

The total of A consists of two parts: the elements only in A and the elements in the intersection A ∩ B. Therefore, only A = n(A) − n(A ∩ B) = 32 − 14 = 18. Option C gives the common region, while option D gives the total of A, including the common region. Thus, option A correctly represents the exclusive A region.

The total number of elements in B includes the elements only in B together with the common elements in A ∩ B. Therefore, the exclusive B region is n(B) − n(A ∩ B) = 41 − 17 = 24. Option C is only the intersection, and option D is the complete set B. Since the shared elements must be removed from B, option A is correct.

The union of two sets is partitioned into three non-overlapping regions: A − B, A ∩ B, and B − A. Therefore, n(A ∪ B) = n(A − B) + n(A ∩ B) + n(B − A) = 12 + 6 + 9 = 27. Since these regions do not overlap, their sizes can be added directly. Hence, option A is correct.

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.

“At least one drink” means the union of the tea and coffee groups. People who like both drinks are included in both totals, so they have been counted twice. Subtract the overlap once: n(Tea ∪ Coffee) = 40 + 32 − 18 = 54. Thus, 54 people like tea, coffee, or both. The word “at least” includes those who like both.

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.

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 30+28+24−10−8−6+4=62. The triple intersection is added once because it was subtracted twice while removing the pairwise overlaps.

To obtain the region belonging only to A, begin with all elements of A. Remove the elements shared by A and B and those shared by A and C. The elements in A ∩ B ∩ C were removed twice in those two subtractions, so add them back once. Therefore, Only A = n(A)−n(A ∩ B)−n(A ∩ C)+n(A ∩ B ∩ C).

The only-A region is calculated by n(A)−n(A ∩ B)−n(A ∩ C)+n(A ∩ B ∩ C). Hence, Only A = 42−15−18+7 = 16. The triple intersection must be added once because its elements were included in both pairwise intersections and would otherwise be subtracted twice. Thus option A is correct.

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 consists of elements that belong to exactly one of the two sets. Algebraically, A △ B=(A−B)∪(B−A). Thus it includes the non-overlapping part of A together with the non-overlapping part of B, but excludes A ∩ B because common elements belong to both sets rather than exactly one.

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.