Operations on Sets (Union, Intersection, Difference)

The symmetric difference is defined by A △ B = (A − B) ∪ (B − A). Since A ⊆ B, no element of A lies outside B, so A − B = ∅. Therefore A △ B = B − A. Given that A △ B = {7}, it follows immediately that B − A = {7}. Hence option A is correct; the empty-set option would contradict the given nonempty symmetric difference.

The union A union B is partitioned into three disjoint regions: A − B, A intersection B, and B − A. The given values provide n(A) = n(A − B) + n(A intersection B) = 63, while n(A union B) = n(A) + n(B − A) = 63 + 34 = 97. These equations are consistent for many intersection values, provided the corresponding A − B value changes. For example, an intersection of 10 gives A − B = 53, while an intersection of 20 gives A − B = 43. Hence the intersection cannot be determined uniquely; option D is correct.

The quantities n(A − B), n(B − C), and n(C − A) describe only three difference regions. They do not reveal the sizes of all pairwise intersections, the triple intersection, or portions such as A ∩ B but not C. Consequently, the total size of A ∪ B ∪ C cannot always be determined from these three numbers alone. Options B, C, and D are valid definitions.

The set A is divided into two disjoint parts: A − B and A ∩ B. Therefore n(A) = n(A − B) + n(A ∩ B). Substituting the given values gives 48 = 17 + n(A ∩ B), so n(A ∩ B) = 31. This is confirmed independently from B: 52 − 21 = 31. Thus option A is correct; the other values result from subtracting the wrong region or adding unrelated parts.

The meaning of set difference is fixed: A − B contains elements in A but not B, and similarly for the other differences, so B, C, and D are always true. However, the three given counts do not reveal the pairwise overlaps, the triple overlap, or several exclusive regions. These unknown parts can change the union, so n(A ∪ B ∪ C) cannot always be determined. Therefore A is correct.

By De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. First find the intersection using n(A ∪ B) = n(A) + n(B) − n(A ∩ B): 139 = 82 + 76 − n(A ∩ B), so n(A ∩ B) = 19. The complement of this intersection in the universal set therefore has 190 − 19 = 171 elements. Hence option D is correct.

Each pairwise intersection includes the 16 elements that lie in all three sets. Therefore, the pair-only regions are n(A ∩ B only) = 45 − 16 = 29, n(A ∩ C only) = 38 − 16 = 22, and n(B ∩ C only) = 35 − 16 = 19. These regions are disjoint, so the number in exactly two sets is 29 + 22 + 19 = 70. Option C is correct; adding 45 + 38 + 35 would count the central region repeatedly.

Let x = n(A ∩ B ∩ C). The three-set inclusion-exclusion formula gives 100 = 50 + 45 + 40 − 18 − 15 − 12 + x. The known terms simplify to 90, so 100 = 90 + x and x = 10. The triple intersection must be added back because pairwise intersections overlap there, making option A correct.

First use the two-set union formula: n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 45 + 52 − 73 = 24. The symmetric difference contains the elements in exactly one set, so it excludes the common part. Hence n(A △ B) = n(A ∪ B) − n(A ∩ B) = 73 − 24 = 49. Option A is correct.

By the distributive law, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). In the Venn diagram this includes the A-B-only region with 9 elements, the A-C-only region with 6 elements, and the central triple-overlap region with 4 elements. These regions are disjoint, so the total is 9 + 6 + 4 = 19. Option A is correct.

Use the three-set inclusion–exclusion formula. Let x = n(A ∩ B ∩ C). Then 118 = 45 + 50 + 55 − 18 − 20 − 16 + x. The known terms give 150 − 54 = 96, so 118 = 96 + x and x = 22. Therefore, the triple intersection contains 22 elements, making option A correct.

For two subsets of a universal set, n(A ∪ B) cannot exceed n(U). Using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we require 60 ≥ 35 + 32 − n(A ∩ B). Hence n(A ∩ B) ≥ 7. This bound is attainable when the union contains all 60 elements, so the minimum possible intersection is 7.

The values n(A − B), n(B − C), and n(C − A) describe only selected portions of the three sets. They do not reveal the sizes of regions shared by two sets or by all three sets, and those regions contribute to the union. Different Venn diagrams can therefore have the same three given differences but different union sizes. Additional common-region information is necessary.

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). Substituting the values gives 82 + 76 + 69 − 37 − 32 − 29 + 15 = 144. Thus, 144 students solved at least one question.

The three given differences describe only selected portions of the sets: A outside B, B outside C, and C outside A. They do not determine pairwise-overlap regions, the triple intersection, or other exclusive regions that may contribute to the union. Different Venn diagrams can have the same three difference counts but different union sizes, so the union cannot be determined uniquely.

Apply the two-set union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substitution gives 106 = (6x + 5) + (5x + 9) − (3x + 4). Simplifying, 106 = 8x + 10, so 8x = 96 and x = 12. Substitution confirms that the three cardinalities are 77, 69, and 40, whose union is 106.

The union is partitioned into three mutually disjoint regions: A − B, B − A, and A ∩ B. Therefore their cardinalities add directly: (4x + 1) + (3x + 6) + (2x + 5) = 120. This simplifies to 9x + 12 = 120, so 9x = 108 and x = 12. At x = 12, the three regions are 49, 42, and 29, which sum to 120.

For two subsets of a universal set, n(A ∪ B) = n(A) + n(B) − n(A ∩ B), and n(A ∪ B) cannot exceed n(U). To make the intersection as small as possible, the union must be as large as possible, namely 90. Hence 90 ≥ 58 + 49 − n(A ∩ B), which gives n(A ∩ B) ≥ 17. This bound is attainable, so the minimum is 17.

The union is partitioned into three disjoint parts: A − B, A ∩ B, and B − A. The given parts account for {1, 4} and {2, 5}. Removing these four elements from A ∪ B = {1, 2, 3, 4, 5, 6} leaves {3, 6}. These remaining elements must be in B but not in A, so B − A = {3, 6}. Option A is correct.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 63 = 45 + 38 − n(A ∩ B), or 63 = 83 − n(A ∩ B). Therefore n(A ∩ B) = 83 − 63 = 20. The overlap is subtracted because common elements are counted twice in n(A) + n(B), so option A is correct.

The relation A ⊆ B means every element of A is also an element of B. Consequently, A ∪ B = B and A ∩ B = A. Substituting these results into the expression gives (A ∪ B) − (A ∩ B) = B − A. Notice that A − B is empty because no element of A lies outside B. Therefore, option A is the only correct answer.

Evaluate the parentheses first. The union B ∪ C contains all distinct elements from B and C: {1, 2, 4, 6, 7, 8}. Now intersect this set with A = {1, 2, 3, 4, 5}. The common elements are 1, 2, and 4; elements 6, 7, and 8 are not in A. Therefore A ∩ (B ∪ C) = {1, 2, 4}, so option A is correct.

Consider any element x. If x belongs to A, the equality A ∩ B = A ∩ C forces x to belong to B exactly when it belongs to C. If x does not belong to A, the equality A ∪ B = A ∪ C forces the same conclusion. Thus every element belongs to B and C together or to neither, so B = C. The stronger claim A = B = C does not necessarily follow.

For every element belonging to A, the first equality says that membership in B and membership in C as excluded elements are identical. The second equality confirms that the elements lying in both A and B are exactly those lying in both A and C. Therefore, B and C behave identically when restricted to A. They may still differ outside A, so B = C is not necessary.

First apply the inclusion–exclusion formula: n(A ∪ B ∪ C) = 40 + 36 + 28 − 14 − 10 − 12 + 5 = 73. Since C is contained in the union, removing C removes exactly 28 elements. Therefore n((A ∪ B ∪ C) − C) = 73 − 28 = 45. The triple intersection has already been handled correctly by inclusion–exclusion.