Operations on Sets (Union, Intersection, Difference)

The elements of B consist of two non-overlapping groups: those only in B and those in the intersection \(A\cap B\). Therefore, \(n(B-A)=n(B)-n(A\cap B)=41-17=24\). Option B is correct. The number 17 counts only the shared elements, and 41 counts the whole of B, so neither represents the only-B region.

Every element of A is either exclusive to A, belonging to \(A-B\), or common to both sets, belonging to \(A\cap B\). These two parts are disjoint and together form A. Thus, \(n(A)=n(A-B)+n(A\cap B)=16+11=27\). Option C is correct; 16 counts only the exclusive portion and does not include the shared elements.

The set B is partitioned into two disjoint regions: elements only in B, represented by \(B-A\), and elements common to A and B, represented by \(A\cap B\). Therefore, \(n(B)=n(B-A)+n(A\cap B)=21+13=34\). Option C is correct. The value 21 omits the common elements, while 8 and 273 have no valid relation to the given partition.

The intersection A ∩ B is the set of elements appearing in both A and B. Comparing the members, 3 and 5 occur in each set. The elements 2, 7, and 11 occur only in A, while 1 and 9 occur only in B. Hence A ∩ B = {3,5}, making option A correct. Option D is the union, not the intersection.

“Only in A” means the set difference A−B: retain elements of A and remove every element that also appears in B. Since 6 and 10 are common to A and B, removing them from A={4,6,8,10} leaves {4,8}. Thus option A is correct; option B is the intersection, C belongs only to B, and D is the union.

“Only in B” means the set difference B − A: retain elements of B and remove every element also found in A. Since q and r are common to both sets, removing them from B={q,r,s,t} leaves {s,t}. Thus B − A = {s,t}, so option C is correct. Option B is the intersection, and option D is the union; neither represents elements exclusive to B.

The union A∪B contains every distinct element that belongs to A or to B, without repeating common elements. Combining A={1,2,3,4} with B={3,4,5,6} gives 1, 2, 3, 4, 5, and 6. Hence A∪B={1,2,3,4,5,6}, so option D is correct. Option A is only the intersection.

First find the number who listen to at least one medium: n(R ∪ P) = 46 + 38 − 16 = 68. The overlap is subtracted because people who listen to both were counted twice. Therefore, those who listen to neither are 90 − 68 = 22. Thus, option A is correct.

The painting group has 33 members in total, including the 12 members who participate in both singing and painting. Therefore, members only in painting = 33 − 12 = 21. The club size and singing total are not required for this direct difference calculation. Hence option B is correct.

Since \(A\cap B=\varnothing\), the sets are disjoint and have no common elements. Therefore, every element in A and every element in B is counted exactly once in their union. The cardinality rule is \(n(A\cup B)=n(A)+n(B)\). Hence, \(n(A\cup B)=24+37=61\). Options 24 and 37 represent the sizes of the individual sets, not their union, while 13 has no basis in the given information. Thus, option D is correct.

The statement \(A\subseteq B\) means that every element of A is already contained in B. When A is united with B, no new elements are added beyond those already in B; therefore, \(A\cup B=B\). Consequently, \(n(A\cup B)=n(B)=46\). The value 18 is only the cardinality of A, 28 is \(46-18\), the number of elements in \(B\setminus A\), and 64 incorrectly adds the two set sizes without removing the overlap. Hence, option C is correct.

Because \(A\subseteq B\), every element of A is also an element of B. The elements common to A and B are therefore exactly the elements of A, so \(A\cap B=A\). It follows that \(n(A\cap B)=n(A)=15\). The value 52 is the size of B, not the intersection; 37 is the difference \(52-15\), not a common part; and 67 is impossible because an intersection cannot contain more elements than either original set. Thus, option A is correct.

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.

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.

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.