Operations on Sets (Union, Intersection, Difference)

The complement Aᶜ contains all elements that are not in A. Therefore, B ∩ Aᶜ contains those elements that belong to B and do not belong to A. This is exactly the definition of the set difference B − A, also written as B \ A, so option B is correct. In a Venn diagram, shade set B and remove its overlapping part with A. Option A represents elements in A but not B, while option C represents the common elements of A and B.

At least one of the two arts means the union of the music and painting groups. By the two-set inclusion–exclusion formula, n(M∪P)=n(M)+n(P)−n(M∩P). Therefore, n(M∪P)=58+49−22=85. The 22 students who learn both are subtracted once because they were counted in both 58 and 49.

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.

Two sets are called disjoint when they have no common element. The set containing elements common to both sets is their intersection, so the definition is written as A intersection B = the empty set. Disjoint sets need not themselves be empty; for example, {1, 2} and {3, 4} are disjoint. Therefore, A union B need not be empty, the sets need not be equal, and neither set must be a subset of the other. Option A is correct.

The set A can be divided into two non-overlapping regions: A − B, containing elements that are in A but not in B, and A intersection B, containing elements common to both sets. Their union is A, so their cardinalities add. Thus n(A) = n(A − B) + n(A intersection B) = 32 + 18 = 50. Subtracting the values would be incorrect because the two given regions are separate parts of A, not quantities to be removed from one another. Therefore, option B is correct.

The union of two sets consists of three disjoint Venn-diagram regions: A − B, A intersection B, and B − A. Thus n(A union B) = n(A − B) + n(A intersection B) + n(B − A). Substitution gives 92 = 35 + 24 + n(B − A). Solving, n(B − A) = 92 − 35 − 24 = 33. The value 59 is only 35 + 24 and does not represent the missing B-only region. Hence option A is correct.

Use the cardinality formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 76 = 31 + n(B) − 0. Therefore, n(B) = 76 − 31 = 45. The zero intersection means that A and B are disjoint, so no common elements are counted twice. Option A is n(A), option C is the union, and option D incorrectly adds 31 and 76.

Apply n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substitution yields n(A ∩ B) = 68 + 53 − 91 = 121 − 91 = 30. The intersection is the part counted in both sets, so it is obtained by removing the union count from the sum of the two individual counts.

The total number of elements in B includes both the elements only in B and the common elements in A ∩ B. Hence n(B only) = n(B) − n(A ∩ B) = 66 − 31 = 35. Option A is the intersection itself, not the exclusive B region. Option C is the complete size of B, and option D is an incorrect addition. Therefore, option B is correct.

The symmetric difference A△B contains elements that belong to A or B but not to both. It can be written as (A−B)∪(B−A); these two parts are disjoint. Therefore n(A△B)=n(A−B)+n(B−A)=38+27=65. The common intersection is excluded, so option B is the correct answer.

The intersection A∩B is the set of elements that occur in both A and B. Comparing the two listed sets, 4, 6, and 12 appear in each set, while 2, 8, and 10 occur only in A and 3 and 9 occur only in B. Thus A∩B={4,6,12}, so option A is correct. Option D is the union, not the intersection.

The union A∪B contains every distinct element that belongs to A or B. Combining the two sets gives a, b, c, d, e, g, i, and j. The repeated elements c and g are written only once because sets do not repeat members. Hence A∪B={a,b,c,d,e,g,i,j}, option B.

The difference A−B contains elements that are in A but not in B. From A, the elements 10, 20, and 30 also occur in B, so they must be removed. The remaining elements are 5, 15, and 25. Therefore A−B={5,15,25}, making option C correct. Option A is the common part, and option D is the union.

The elements belonging to exactly one set are n(A−B)+n(B−A). In terms of the given totals, this is n(A)+n(B)−2n(A∩B), because the common elements are counted in both n(A) and n(B). Thus 73=54+61−2n(A∩B)=115−2n(A∩B). Therefore 2n(A∩B)=42 and n(A∩B)=21.

“At least one art” means the union of the dance and music groups. For two sets, n(D∪M)=n(D)+n(M)−n(D∩M), because students learning both are counted twice in the simple sum. Therefore n(D∪M)=63+58−26=95. Thus, 95 students learn at least one of the two arts, so option A is correct.

The 52 people who like apples include both people who like only apples and people who like both apples and bananas. Since 19 people like both fruits, remove that overlap from the apple total: only apples=n(A)−n(A∩B)=52−19=33. Therefore, 33 people like only apples, so option A is correct.

The set A consists of two non-overlapping parts: the elements that are in A but not in B, represented by A − B, and the elements common to both A and B, represented by A ∩ B. Hence n(A) = n(A − B) + n(A ∩ B) = 41 + 22 = 63. Therefore, option B is correct. The value 41 counts only the exclusive part of A, while 22 counts only the common part.

The union 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). Substitution gives 104 = 46 + n(B − A) + 25, so n(B − A) = 33. Option A is correct.

The complement of A ∪ B contains the elements of the universal set that are outside both A and B. Thus n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 125 − 24 = 101. Using the inclusion–exclusion formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, 101 = 62 + 55 − n(A ∩ B), giving n(A ∩ B) = 117 − 101 = 16. Hence option A is correct.

The general formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since n(A ∩ B) = 0, the sets are disjoint and the formula becomes 92 = 40 + n(B). Thus n(B) = 92 − 40 = 52. Option A merely repeats n(A), while option C is the union total, not the size of B. Therefore, B is the only correct answer.