Operations on Sets (Union, Intersection, Difference)

To find \(B\setminus C\), retain the elements of B that do not occur in C. The elements of B are 2, 4, 6, and 8. Since 2 and 4 are also in \(C=\{1,2,3,4\}\), they must be removed. The elements 6 and 8 are not in C, so \(B\setminus C=\{6,8\}\). Therefore, option A is correct; the set A is extra information and is not needed for this calculation.

First form the union: \(A\cup B=\{1,2,4,7,8,10\}\), containing every distinct element in A or B. Next intersect this result with \(C=\{4,10,12\}\), retaining only elements common to both sets. The common elements are 4 and 10, while 12 is absent from the union. Thus \((A\cup B)\cap C=\{4,10\}\), so option A is correct.

Evaluate the parentheses first. The elements common to B and C are only s and u, so \(B\cap C=\{s,u\}\). Now remove from A every element that belongs to this intersection. The element s is in A and must be removed, but u is not in A and therefore has no effect. The remaining set is \(\{p,q,r,t\}\), which makes option A correct.

By definition, \(A\setminus B\) is formed by selecting some elements of A and removing those that also occur in B. Every element that remains therefore already belongs to A. Consequently, \(A\setminus B\subseteq A\) is always true for any sets A and B, including when the difference is empty. For these particular sets, \(A\setminus B=\{1,3\}\), which visibly confirms the statement.

Since A contains the natural numbers from 1 through 10, write \(A=\{1,2,3,4,5,6,7,8,9,10\}\). First calculate the union: \(B\cup C=\{1,2,4,6,7,8,10\}\). Remove these union elements from A. The numbers left are 3, 5, and 9, so \(A\setminus(B\cup C)=\{3,5,9\}\). Option B is the removed union, not the required difference.

List the positive divisors: \(A=\{1,2,3,5,6,10,15,30\}\) and \(B=\{1,3,5,9,15,45\}\). Their intersection is \(\{1,3,5,15\}\), and their union is \(\{1,2,3,5,6,9,10,15,30,45\}\). Removing the intersection from the union leaves the elements that belong to exactly one set: \(\{2,6,9,10,30,45\}\). This is the symmetric difference, so option A is correct.

First calculate the intersection of A and B. The elements common to both sets are A ∩ B = {2,4,6,8}. Set difference means retaining elements of the first set that do not occur in the second set. Since 4 and 8 are also in C = {1,4,8,12}, remove them from the intersection. The remaining elements are {2,6}, so option A is correct. Option D stops before subtraction, while option B contains the removed elements.

Students taking only online classes are 58 − 21 = 37, because the 21 students doing both activities must be excluded. Students visiting only the library are 46 − 21 = 25. Therefore, the number doing exactly one activity is 37 + 25 = 62. Equivalently, use 58 + 46 − 2(21) = 62.

A \ B = ∅ means that no element of A lies outside B, so every element of A belongs to B; hence A ⊆ B. Similarly, B \ A = ∅ gives B ⊆ A. Since each set is contained in the other, the two sets have exactly the same elements, and therefore A = B. The other conclusions do not follow.

The integers strictly between -5 and 4 are A = {-4,-3,-2,-1,0,1,2,3}. The even members in this interval are B = {-4,-2,0,2}. The difference A \ B keeps the elements of A that are not in B, so the odd integers remain: {-3,-1,1,3}. Therefore option A is correct. Option B lists the removed even elements, option C is the entire set A, and option D would incorrectly remove every element.

The intersection P ∩ Q contains only elements that occur in both sets. Comparing P = {a,b,c,d} with Q = {b,d,e}, the common elements are b and d. Therefore, P ∩ Q = {b,d}. Option A contains elements not common to both sets, option C is the union, and option D would apply only if there were no common elements.

The notation B \ A means that we begin with B and remove every element that is also present in A. Starting with B = {1,2,3,4}, the elements 2 and 4 are common to A and B, so they are removed. The elements 1 and 3 are not in A and therefore remain. Thus B \ A = {1,3}, making option A correct. Option C is A ∩ B, option B is A \ B, and option D is A ∪ B.

First form the union A ∪ B by listing every element appearing in either set without repetition: A ∪ B = {1,2,3,4,5}. Now intersect this result with C = {4,5,6}. The common elements are 4 and 5, so (A ∪ B) ∩ C = {4,5}. Therefore option A is correct. Option B includes 3, which is not in C; option D omits 4; and option C is the union of all listed elements rather than the requested intersection.

Evaluate the parentheses first: B ∩ C = {3,5}, because 3 and 5 are the only elements common to B and C. Then remove these elements from A: A − {3,5} = {1,7,9}. Thus option A is correct. Option B is only the intermediate intersection, while option C fails to remove anything.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The intersection is counted in both n(A) and n(B), so it must be subtracted once to correct the double count. Substituting the given values gives 18 + 14 − 6 = 26. Hence n(A ∪ B) = 26, so option A is correct. Option B forgets the overlap, while the other values result from incorrect arithmetic or sign usage.

Use the two-set inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substituting the data, n(A ∩ B) = 28 + 25 − 45 = 53 − 45 = 8. Thus option A is correct. The value 12 does not follow from the formula, 17 is an incorrect subtraction, and 53 is merely n(A) + n(B) before correcting for overlap.

“At least one subject” means the union of the Mathematics and Physics sets. By the inclusion–exclusion principle, n(M ∪ P) = n(M) + n(P) − n(M ∩ P). Therefore, n(M ∪ P) = 30 + 22 − 12 = 40. We subtract the 12 students studying both subjects because they were counted twice in 30 + 22. Thus, option A is correct.

Set A contains all integers from −2 through 4: {−2, −1, 0, 1, 2, 3, 4}. Set B contains all odd integers. The odd members of A are −1, 1, and 3, so A ∩ B = {−1, 1, 3}. Option B lists the even members of A, option C lists all of A, and option D incorrectly includes 5, which is outside A.

The intersection contains numbers that belong to both closed intervals. Its left endpoint is the larger of the two left endpoints, max(1, 3) = 3, and its right endpoint is the smaller of the two right endpoints, min(5, 8) = 5. Since both original intervals include their endpoints, the answer is the closed interval [3, 5]. Therefore, option B is correct.

The intervals overlap from 5 to 7, so together they form one continuous interval. Set A begins just greater than 2, so 2 is excluded. Set B ends at 9 and includes 9, so 9 is included. Every number between these endpoints belongs to at least one set. Hence A ∪ B = (2, 9], making option A correct.