Operations on Sets (Union, Intersection, Difference)

Use the complement rule first: n(A) = n(U) − n(A′) = 95 − 47 = 48, and n(B) = 95 − n(B′) = 95 − 52 = 43. Then apply inclusion–exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 48 + 43 − 18 = 73. Thus option A is correct. The other choices result from omitting or mishandling the complement or intersection.

The universal set has 95 elements. Therefore, n(A) = 95 − 47 = 48 and n(B) = 95 − 52 = 43. Applying the two-set union formula gives n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 48 + 43 − 18 = 73. The intersection is subtracted once because it was counted in both set totals. Hence option A is correct.

The union A ∪ B is divided into three mutually disjoint Venn regions: A only, represented by A ∩ B' and equal to 24; B only, represented by A' ∩ B and equal to 31; and the common region A ∩ B, equal to 16. Adding these non-overlapping parts gives 24 + 31 + 16 = 71. Therefore, option A is correct.

People who like at least one beverage are counted by the union formula: n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 52 + 47 − 25 = 74. The remaining people like neither tea nor coffee, so subtract the union from the total: 90 − 74 = 16. Thus option A is correct.

For three sets, inclusion–exclusion gives n(A ∪ B ∪ C) = n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). The triple intersection is empty, so its cardinality is 0. Substitution gives 30+34+36−8−10−6+0 = 76. Therefore, option A is correct; simply adding all set sizes would double-count pairwise overlaps.

Use the inclusion–exclusion principle, because multiples common to A and B would otherwise be counted twice. There are floor(30/2) = 15 multiples of 2 and floor(30/5) = 6 multiples of 5. The overlap consists of multiples of lcm(2,5) = 10, namely 10, 20, and 30, so it has 3 elements. Therefore n(A ∪ B) = 15 + 6 − 3 = 18. Option A is correct; adding without subtraction gives the distractor 21.

A union B is partitioned into three disjoint regions: elements only in A, elements only in B, and elements in both sets. Thus n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 84 = n(A − B) + n(B − A) + 27. Rearranging, the requested sum is 84 − 27 = 57. This is also the size of the symmetric difference, so option A is correct.

The union A ∪ B must contain every element of A and every element of B, so its size cannot be less than either set. Therefore n(A ∪ B) ≥ max(64, 58) = 64. This minimum is attainable because B, which has 58 elements, can be completely contained in A, which has 64 elements. Thus n(A ∪ B) = 64.

Use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Hence, 52 = 30 + 28 − n(A ∩ B), so n(A ∩ B) = 6. The set A − B contains elements that are in A but not in B. Therefore, n(A − B) = n(A) − n(A ∩ B) = 30 − 6 = 24. Thus option A is correct.

The quantity n(A ∩ B) = 14 includes every element common to A and B, including the elements that may also belong to C. The triple intersection A ∩ B ∩ C contains 6 of these elements. “Only in A ∩ B” means the elements common to A and B but outside C, so subtract the triple-overlap: 14 − 6 = 8. Therefore, option A is the unique correct answer.

Students studying only Mathematics = 56 − 27 = 29, and students studying only Science = 49 − 27 = 22. Therefore, the number studying exactly one subject is 29 + 22 = 51. Equivalently, n(M △ S) = n(M) + n(S) − 2n(M ∩ S) = 56 + 49 − 54 = 51. The value 78 is the union, not the exactly-one count.

Use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, n(A ∩ B) = 64 + 59 − 91 = 32. 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) = 91 − 32 = 59. Hence, option A is correct.

Since A and B are disjoint, they have no common elements, so n(A ∪ B) = n(A) + n(B) = 44 + 52 = 96. The complement of A ∪ B contains all elements of the universal set U that are outside the union. Thus, n((A ∪ B)′) = n(U) − n(A ∪ B) = 130 − 96 = 34. Therefore, option A is correct.

Apply the inclusion–exclusion formula for two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given expressions gives 78 = (5x + 4) + (4x + 7) − (2x + 3). Simplifying, 78 = 7x + 8, so 7x = 70 and x = 10. Substitution confirms the result, so option A is correct.

The three regions forming A ∪ B are A − B, B − A, and A ∩ B. These regions are pairwise disjoint, so their cardinalities can be added: 71 = (3x + 2) + (2x + 5) + (x + 4) = 6x + 11. Hence 6x = 60 and x = 10. With x = 10, the three region sizes are 32, 25, and 14, whose sum is 71; therefore option A is correct.

There are floor(90/5) = 18 multiples of 5 and floor(90/9) = 10 multiples of 9. Numbers counted in both sets are multiples of lcm(5, 9) = 45; within 1 to 90 these are 45 and 90, so there are 2. By inclusion-exclusion, n(A ∪ B) = 18 + 10 − 2 = 26. Thus option A is correct.

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 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.

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.

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.