Operations on Sets (Union, Intersection, Difference)

First calculate the difference A − B. The elements 2 and 4 are common to A and B, so they are removed from A, giving A − B = {1, 3}. This result is already a subset of A. Therefore, intersecting it with A does not change it: A ∩ (A − B) = {1, 3}. Thus option A is the only correct answer.

The union A ∪ B contains every element from either set, so A ∪ B = {5, 6, 7, 8, 9}. Taking the intersection of this union with A selects the elements that are also in A. Since every element of A is already in A ∪ B, the result is A itself: A ∩ (A ∪ B) = {5, 6, 7}. This illustrates the absorption law.

The common elements of A and B are q and r, so A ∩ B = {q, r}. These elements are already contained in A = {p, q, r}. Taking the union of A with a subset of A adds nothing new. Therefore, A ∪ (A ∩ B) = A = {p, q, r}. This is the absorption identity A ∪ (A ∩ B) = A.

Let D be the set of students who like drawing and M the set who like music. Students liking at least one activity are in D ∪ M. By the inclusion-exclusion formula, n(D ∪ M) = n(D) + n(M) − n(D ∩ M) = 22 + 16 − 6 = 32. We subtract the 6 students who like both because they were counted twice.

The number who like only tea is obtained by removing those who like both tea and coffee from the total tea group. Thus, only tea = n(T) − n(T ∩ C) = 30 − 12 = 18. The value 13 represents only coffee, 43 represents the union of the two groups, and 55 incorrectly adds the totals without correcting for the overlap.

The natural-number factors of 12 are A = {1, 2, 3, 4, 6, 12}. The intersection A ∩ B contains only elements common to this factor set and B = {1, 2, 3, 5}. The common elements are 1, 2, and 3; 5 is not a factor of 12, while 4, 6, and 12 are not in B. Hence, A ∩ B = {1, 2, 3}.

The natural-number factors of 18 are A = {1, 2, 3, 6, 9, 18}. The difference A − B keeps elements of A that are not in B. Since 2, 3, 6, and 9 are removed, the elements left are 1 and 18. Therefore, A − B = {1, 18}. The operation does not remove elements that are outside A.

Evaluate the expression inside the parentheses first. Removing B = {2, 3, 5} from A = {1, 2, 3, 4, 5, 6} gives A − B = {1, 4, 6}. Now intersect this result with C = {3, 4, 6}. The elements common to {1, 4, 6} and C are 4 and 6. Therefore, (A − B) ∩ C = {4, 6}.

First form the union of A and B by listing each distinct element once: A ∪ B = {1, 2, 3, 4, 6, 8}. The intersection with C = {2, 6, 9} keeps only elements found in both sets. The common elements are 2 and 6; 9 is not in the union. Therefore, (A ∪ B) ∩ C = {2, 6}, making option B correct. Option A is the union's non-common portion, not the requested intersection.

Evaluate the parentheses first. The elements common to B = {2, 4, 6, 8} and C = {1, 4, 6} are B ∩ C = {4, 6}. Set difference A − (B ∩ C) means remove 4 and 6 from A. Starting with A = {1, 2, 3, 4, 5, 6}, the remaining elements are {1, 2, 3, 5}. Thus option A is correct; option B is the removed intersection, not the difference.

Let S be the set of students reading storybooks and P the set reading poetry. “At least one type” means n(S ∪ P) = 45. Using n(S ∪ P) = n(S) + n(P) − n(S ∩ P), we get 45 = 35 + 20 − n(S ∩ P). Hence n(S ∩ P) = 10. Therefore, 10 students read both types.

Compute both directed differences. A − B contains elements in A but not B, so A − B = {3, 9}. Likewise, B − A contains elements in B but not A, giving B − A = {18, 24}. These two sets are disjoint, meaning they have no common element. Their intersection is therefore the empty set, ∅, so option C is correct. Option D is only the first difference and is not the final intersection.

The union A ∪ B contains every distinct element that occurs in A or in B. Combining the two lists gives 1, 2, 3, 4, 5, and 6. Elements 2 and 5 appear in both sets, but set notation records each element only once. Therefore, A ∪ B = {1, 2, 3, 4, 5, 6}, so option A is correct. Option B is only the intersection, not the union.

The intersection A ∩ B consists of elements present in both A and B. Checking the lists, 6 appears in A and B, and 12 also appears in both. The elements 3 and 9 occur only in A, while 18 occurs only in B, so they are excluded. Hence A ∩ B = {6, 12}, and option B is correct. Option C incorrectly combines all elements as if the operation were union.

The difference A \ B contains elements that belong to A but do not belong to B. In A = {1, 4, 7, 10}, the elements 4 and 10 are also in B, so they must be removed. The elements 1 and 7 are not in B and remain in the result. Therefore, A \ B = {1, 7}, making option A correct. Option B lists the common elements, while option D is closer to a union.

For B \ A, retain elements of B that are not present in A. B contains 1, 5, 8, and 14. Since 5 and 8 also occur in A, remove them; 1 and 14 do not occur in A, so they remain. Thus B \ A = {1, 14}, which is option C. Option B is the intersection, option A comes from A \ B, and option D includes elements that should not be retained.

The union A ∪ B contains every distinct element from both sets, so A ∪ B = {a, b, c, d, e, f}. The intersection A ∩ B contains only the common elements, giving {c, d}. Set difference removes the elements of the second set from the first; therefore removing c and d from the union leaves {a, b, e, f}. Thus option A is correct. The other choices represent the intersection, the full union, or an empty result.

The empty set ∅ has no elements, so taking its union with A does not add anything. This follows from the identity property of union: for every set A, A ∪ ∅ = A. Substituting A = {1, 2, 3} gives A ∪ ∅ = {1, 2, 3}. Therefore option B is correct. Option A confuses union with intersection, while options C and D incorrectly add or remove elements.

The intersection A ∩ B contains only the elements that belong to both A and B. The empty set ∅ contains no elements, so there is no element that can be common to A and ∅. Therefore, A ∩ ∅ = ∅. Notice that {0} is not the empty set because it contains one element, namely 0. This is a standard property of set intersection.

For two finite sets, the number of elements in their union is given by n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 18 + 12 − 5 = 25. We subtract the intersection because the five common elements were counted once in n(A) and again in n(B). Hence, the correct answer is 25, option A.