Operations on Sets (Union, Intersection, Difference)

The set A consists of two disjoint parts: the elements that belong only to A, namely A \ B = {1, 7}, and the elements common to A and B, namely A ∩ B = {3, 5}. Hence A = (A \ B) ∪ (A ∩ B) = {1, 7} ∪ {3, 5} = {1, 3, 5, 7}. The elements 2 and 9 belong only to B, so they must not be included in A.

The common elements of A and B are 4 and 5, so A ∩ B = {4, 5}. Combining all distinct elements gives A ∪ B = {2, 3, 4, 5, 6, 7}. Removing B’s elements from A leaves A \ B = {2, 3}. However, B \ A means elements in B but not in A; these are 6 and 7, so B \ A = {6, 7}, not {2, 3}. Therefore statement D is false.

A union B contains every element that is in A, in B, or in both sets, with repeated elements written only once. Starting with A gives {2, 5, 8, 11}; adding the elements of B that are not already present, namely 1, 7, and 13, gives A ∪ B = {1, 2, 5, 7, 8, 11, 13}. Option B is the intersection, while C and D contain only exclusive portions of the sets.

The intersection A ∩ B contains precisely those elements that occur in both sets. Comparing the elements, 6 occurs in A and B, and 12 also occurs in A and B. The other elements occur in only one set. Hence A ∩ B = {6, 12}. Option B contains elements exclusive to A, option C contains elements exclusive to B, and option D is the union rather than the intersection.

A \ B contains elements that are in A but not in B. The elements c and g occur in both sets, so they must be removed from A. The remaining elements are a, e, and i; therefore A \ B = {a, e, i}. Option B is A ∩ B, the common part. Option C is B \ A, and option D is A ∪ B, so neither represents the required difference.

B \ A means that we keep the elements of B that are not present in A. The elements 4 and 16 occur in both A and B, so they are removed from B. The element 36 occurs in B but not in A, so it remains. Therefore B \ A = {36}. Option B is the intersection A ∩ B, while option D would be correct only if every element of B were also in A.

Assuming N contains the positive natural numbers, the odd numbers not exceeding 9 are A = {1, 3, 5, 7, 9}. The prime numbers less than 8 are B = {2, 3, 5, 7}; 1 is not prime. The elements common to both lists are 3, 5, and 7. Hence A ∩ B = {3, 5, 7}. Option B is A itself, and option C is B itself, so neither is the intersection.

Since \(x\) is a natural number and \(x^2\le49\), the possible values are \(1,2,3,4,5,6,7\). Thus, \(A=\{1,2,3,4,5,6,7\}\). The difference \(A\setminus B\) contains elements that belong to A but do not belong to B. Removing 2, 4, and 6 from A leaves \(\{1,3,5,7\}\). The element 8 in B is irrelevant because it is not in A. Therefore, option A is correct.

The intersection A ∩ B ∩ C contains only those elements that occur in all three sets simultaneously. The element 3 occurs in A, B, and C, and the element 6 also occurs in A, B, and C. The elements 1, 2, 5, and 9 fail to occur in at least one of the sets. Therefore, A ∩ B ∩ C = {3, 6}, so option A is correct.

First calculate the union inside the parentheses: B ∪ C = {s, u, v, w}. Next, retain only the elements that are also present in A = {r, s, t, u}. The common elements are s and u, while v and w are not in A and r and t are not in B ∪ C. Hence A ∩ (B ∪ C) = {s, u}; option A is correct.

First form the union B ∪ C = {1, 2, 5, 6, 8, 9}. The difference A \ (B ∪ C) consists of elements that belong to A but do not belong to this union. From A, the elements 1, 2, 5, and 6 must be removed; 3 and 4 remain. Therefore A \ (B ∪ C) = {3, 4}, making option A correct.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 26 + 19 − 8 = 37. The intersection is subtracted because the eight common elements were counted once in n(A) and once again in n(B). Thus the union contains 37 elements, so option A is correct.

Use the two-set formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the values gives 46 = 31 + 28 − n(A ∩ B), so n(A ∩ B) = 31 + 28 − 46 = 13. The common elements must be subtracted from the sum because they were counted twice. Therefore option A is correct.

The set B is divided into two disjoint parts: B \ A, which contains elements of B outside A, and A ∩ B, which contains elements common to A and B. Thus n(B) = n(B \ A) + n(A ∩ B). Consequently, 42 = 25 + n(A ∩ B), giving n(A ∩ B) = 17. Therefore option A is correct.

“At least one activity” means the union of the music and painting groups. Apply n(M ∪ P) = n(M) + n(P) − n(M ∩ P). Hence n(M ∪ P) = 36 + 29 − 14 = 51. The 14 students who learn both activities must be subtracted once because they were included in both original totals. Therefore option A is correct.

The number studying only mathematics is found by removing the students who study both subjects from the total mathematics group. Thus only mathematics = n(M) − n(M ∩ S) = 40 − 18 = 22. Option B, 16, is the number studying only science; option C is the union, and option D double-counts the overlap. Therefore option A is correct.

The number of people who travel only by metro is found by removing those who use both transport modes from the total metro users: 47 − 23 = 24. Therefore, option A is correct. Option B, 29, is obtained by subtracting the overlap from the bus total and represents only-bus users. Option D adds both totals without removing the overlap, so it double-counts 23 people.

The union A ∪ B contains three mutually disjoint regions: the elements only in A, the elements common to A and B, and the elements only in B. Hence n(A ∪ B) = 12 + 7 + 16 = 35. Option B omits the common region, while option C is only 12 + 7. Option D has no valid interpretation for the given partition.

Since A is a subset of B, every element of A is already contained in B. Therefore, A ∪ B = B. Intersecting this union with A gives B ∩ A = A, because all elements of A are in B. Thus (A ∪ B) ∩ A = A. Option B is the union before the final intersection, and options C and D are not generally equal to A.

The difference B \ A consists of elements that belong to B but do not belong to A. However, B ⊆ A means every element of B is already an element of A. Consequently, there is no element of B left outside A, so B \ A = ∅. The other options are not forced by the subset condition and may have completely different elements or sizes.