Operations on Sets (Union, Intersection, Difference)

The difference \(A-B\) consists of elements that belong to A but do not belong to B. Since \(A=B\), every element of A is also in B, so there can be no element left in A after removing B. Symbolically, \(A-B=A\cap B^c=A\cap A^c=\varnothing\). Therefore, the correct answer is the empty set, option C. It is not A or B because the two sets are identical, and it is not the universal set U because no element outside the sets is included in the difference.

For three finite sets, the inclusion–exclusion principle is \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\). Substituting the given values gives \(12+14+10-5-3-4+2=26\). The pairwise intersections are subtracted because their elements were counted twice, and the triple intersection is added once because it was then subtracted too many times. Therefore, option B, 26, is correct.

The elements only in A are counted by n(A − B) = n(A) − n(A ∩ B) = 24 − 6 = 18. Similarly, the elements only in B are n(B − A) = n(B) − n(A ∩ B) = 20 − 6 = 14. These two regions do not overlap, so their combined number is 18 + 14 = 32. Thus option B is correct. The value 38 is the union, not the sum of the two exclusive regions.

Set A is divided into two disjoint parts: the elements only in A, represented by A − B, and the elements common to A and B, represented by A ∩ B. Therefore n(A) = n(A − B) + n(A ∩ B). Using the given values, 45 = 19 + n(A ∩ B), so n(A ∩ B) = 45 − 19 = 26. Hence option B is the only correct answer.

The set B consists of two disjoint regions: the elements only in B and the elements in the intersection A ∩ B. Thus n(B) = n(B − A) + n(A ∩ B). Since n(B) = 50 and n(B − A) = 28, the common part is n(A ∩ B) = 50 − 28 = 22. Therefore option B is correct. The value 28 describes only B, while 50 describes all of B.

Use the two-set union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the known values gives 69 = 42 + n(B) − 18. Since 42 − 18 = 24, we have 69 = 24 + n(B), so n(B) = 69 − 24 = 45. Equivalently, n(B) = 69 − 42 + 18 = 45. Hence option B is correct.

The difference A − B contains elements that are present in A but absent from B. The elements 5 and 7 occur in both sets, so they must be removed from A. The remaining elements are 1, 3, and 9; therefore A − B = {1, 3, 9}. Option B is the intersection, option C belongs only to B, and option D is the union of the two sets.

The total number of elements in A includes both the elements only in A and the elements common to A and B. Therefore, the number only in A is n(A) − n(A ∩ B) = 44 − 18 = 26. Hence option B is correct. The value 18 represents the intersection, while 44 represents all of A, not just its exclusive region.

Five students are in neither A nor B, so the union contains 60 − 5 = 55 students. Apply the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus 55 = 25 + 30 − n(A ∩ B), giving n(A ∩ B) = 0. Therefore, the two sets are disjoint in this situation.

Set A can be divided into two non-overlapping regions: the elements in A but not in B, represented by A − B, and the elements common to both sets, represented by A ∩ B. These two regions together make all of A. Therefore, n(A) = n(A − B) + n(A ∩ B) = 16 + 11 = 27. The overlap is added because it belongs to A.

When \(A\subseteq B\), every element of A is already included in B. Therefore, combining all elements of A and B adds nothing beyond B, so \(A\cup B=B\). Option B would be correct only when A and B are equal. Option C is the empty set, while option D represents elements of A outside B; that set is empty under the given condition, not the union.

The intersection of two sets consists only of elements present in both sets. Comparing A = {2,4,6,8,10} with B = {1,2,3,5,7}, the only common element is 2. Therefore, \(A\cap B=\{2\}\). The other even elements belong only to A, while 1, 3, 5, and 7 belong only to B, so they cannot be included in the intersection.

The union contains every distinct element that occurs in A or in B, including elements common to both. Combining A = {a,c,e} and B = {b,c,d} gives {a,b,c,d,e}; the shared element c is written only once because sets do not repeat elements. The element f is in the universal set but in neither A nor B, so it is not part of the union.

For \(B-A\), inspect the elements of B and keep only those that are not in A. B contains r, s, and t. Since r and s also occur in A, they are removed, leaving only t. Thus \(B-A=\{t\}\). The order matters: \(B-A\) is not the same operation as \(A-B\), and it is also not the union of the two sets.

The total number of elements in B consists of the elements only in B together with the elements in the intersection A∩B. Hence, n(only B)=n(B)−n(A∩B)=57−23=34. The number 23 is only the common region, and 57 is the entire set B, not just its exclusive part. Therefore, option A is correct.

Because A is a subset of B, every element of A is already included in B. The set B − A therefore contains the elements that belong to B but do not belong to A. No overlap is counted in this difference, so n(B − A) = n(B) − n(A) = 73 − 28 = 45. Thus, option B is correct.

The governing concept is the cardinality formula for the union of two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since A ∩ B = ∅, the sets are disjoint and their intersection has zero elements. Therefore, n(A ∪ B) = 33 + 41 − 0 = 74. Options C and D give the size of only one set, while option B is the difference, so option A is correct.

The governing concept is symmetric difference. The set A △ B contains elements belonging to exactly one of A or B, so A △ B = (A − B) ∪ (B − A). These two difference sets cannot overlap; hence their cardinalities are added. Thus n(A △ B) = 29 + 17 = 46. Options C and D represent only one part, and option A subtracts the parts incorrectly. Therefore, option B is correct.

The intersection of two sets contains exactly the elements that occur in both sets. Comparing the elements one by one, b appears in A and B, d appears in A and B, and h appears in A and B. The elements a and f occur only in A, while c and g occur only in B. Therefore, A ∩ B = {b, d, h}, so option C is correct.

The difference A − B contains elements that are in A but not in B. From A, the elements 6, 8, and 12 also occur in B, so they must be removed. The remaining elements are 2, 4, and 10. The element 14 is not in A, so it cannot belong to A − B. Therefore, option A is correct.