Operations on Sets (Union, Intersection, Difference)

The governing principle is the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Because A ∪ B is contained in U, its size cannot exceed 75. Thus 46 + 41 − n(A ∩ B) ≤ 75, so n(A ∩ B) ≥ 12. This bound is attainable when the union has 75 elements, making 12 the minimum. Zero ignores the limited size of U.

The intersection A ∩ B consists only of elements common to both sets. Therefore, it cannot contain more elements than either set, so n(A ∩ B) ≤ min(73, 58) = 58. This value is attainable if every element of B is also in A, meaning B is a subset of A. Hence the maximum is 58.

A union contains every element of each participating set, so it must contain the larger set. Therefore, n(A ∪ B) ≥ max[n(A), n(B)] = max(91, 86) = 91. This lower bound is possible if all 86 elements of B lie inside A, meaning B ⊆ A. Then A ∪ B = A and has 91 elements. Option B is too small, while 177 is the sum without accounting for overlap.

The set A' ∩ B contains elements that are in B but not in A. This is the same as B − A. Set B has 38 elements, and 17 of them are also in A, because n(A ∩ B) = 17. Removing these common elements leaves 38 − 17 = 21 elements. Therefore n(A' ∩ B) = 21.

The difference A − B represents the elements in A but not in B. Hence the number common to A and B is n(A ∩ B) = n(A) − n(A − B) = 55 − 23 = 32. Applying inclusion–exclusion, n(A ∪ B) = 55 + 62 − 32 = 85. The sum 117 would count the 32 common elements twice, so it is not the union size.

The union is partitioned into three disjoint regions: A − B, A ∩ B, and B − A. Therefore, n(A ∪ B) = n(A − B) + n(A ∩ B) + n(B − A). Substituting the given values gives n(A − B) + n(B − A) = 98 − 35 = 63. This is also the cardinality of the symmetric difference, so option A is correct.

Since A ⊆ B, every element of A is already in B, so A ∪ B = B. Substituting this into the expression gives (A ∪ B) − A = B − A. This is the portion of B outside A. It is not necessarily empty: it becomes empty only when A = B. Therefore, option A correctly applies the subset property and the definition of set difference.

The condition A ∩ B = ∅ says that A and B are disjoint, so they have no common elements. The difference A − B removes from A only elements that also belong to B. Because no element of A belongs to B, nothing is removed, and A − B remains A. It is not empty unless A itself is empty; neither B nor the union is generally equal to the difference.

For finite sets, A∪B contains all elements of A and any additional elements contributed by B. If n(A∪B) equals n(A), B contributes no element outside A. Thus every element of B is already in A, which means B⊆A. The equality does not necessarily mean A=B, because B may be a proper subset of A.

The symmetric difference A △ B contains elements that belong to exactly one of the two sets. Its cardinality is n(A △ B) = n(A) + n(B) − 2n(A ∩ B). Since the intersection has size zero, the sets are disjoint and n(A △ B) = 28 + 34 − 0 = 62. Options 28 and 34 count only one set, while 6 is their difference.

First find the number reading at least one publication: 120 − 24 = 96. Let N denote newspaper readers and M magazine readers. By inclusion–exclusion, n(N ∪ M) = n(N) + n(M) − n(N ∩ M). Therefore, 96 = 66 + 58 − n(N ∩ M), so n(N ∩ M) = 124 − 96 = 28. Hence 28 people read both; 96 is the union, not the intersection.

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 pairwise intersection A ∩ B includes every element common to A and B, including those also belonging to C. Therefore, the 10 elements in A ∩ B ∩ C must be removed to obtain the region only in A ∩ B. Calculation: 26 − 10 = 16. The value 26 includes the triple-overlap, while 10 counts only that central part, so neither is the requested exclusive pairwise region.

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.

First find the intersection using n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus n(A ∩ B) = 72 + 67 − 101 = 38. The symmetric difference A △ B contains elements belonging to exactly one of the two sets, so n(A △ B) = n(A ∪ B) − n(A ∩ B) = 101 − 38 = 63. Hence option A is correct.

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.

The multiples of 6 from 1 to 96 are counted by ⌊96/6⌋ = 16. The multiples of 10 are counted by ⌊96/10⌋ = 9. Common elements are multiples of lcm(6,10) = 30, and there are ⌊96/30⌋ = 3 of them. Therefore, n(A ∪ B) = 16 + 9 − 3 = 22, by inclusion–exclusion. Thus option A is correct.

An element in A ∩ B must be divisible by both 7 and 15. Because 7 and 15 are coprime, their least common multiple is 7 × 15 = 105. The positive multiples of 105 not exceeding 105 consist only of 105 itself. Therefore A ∩ B = {105}, and its cardinality is n(A ∩ B) = 1. Hence option A is correct.

To belong to A ∩ B, a number must satisfy both conditions: it must be prime and also a multiple of 5. The only prime multiple of 5 is 5 itself, because every other positive multiple of 5 is divisible by 5 and has at least one additional factor. Therefore A ∩ B = {5}. The universal set ending at 50 does not change this conclusion.