Operations on Sets (Union, Intersection, Difference)

Correct answer: A. 22

Explanation: 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.

Correct answer: A. 35

Explanation: 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.

Correct answer: A. A

Explanation: 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.

Correct answer: A. B ⊆ A

Explanation: The equality A ∩ B = B means that taking the elements common to A and B leaves all of B unchanged. Therefore every element of B must also belong to A, which is exactly the statement B ⊆ A. A ⊆ B is the reverse implication and is not required. B ⊂ A is too strong because A and B may be equal, and A ∩ B = A would instead imply A ⊆ B.

Correct answer: A. {1, 5, 9}

Explanation: First find the union: A ∪ B = {1, 3, 5, 7, 9}. Next find the intersection: A ∩ B = {3, 7}. Removing the common elements from the union leaves {1, 5, 9}. Thus option A is correct. This expression is the symmetric difference of A and B, containing elements that belong to exactly one of the two sets.

Correct answer: A. {1, 2, 7, 9}

Explanation: The elements of A that are not in B are A \ B = {2, 7}. The elements of B that are not in A are B \ A = {1, 9}. Taking their union gives {1, 2, 7, 9}. The common elements 3, 5, and 11 are excluded from both differences. This operation is called the symmetric difference of the two sets, so option A is correct.

Correct answer: A. {2,4,8,14}

Explanation: First calculate the intersection A ∩ B. The elements common to A and B are 4 and 8, so A ∩ B = {4,8}. Next take the union with C: {4,8} ∪ {2,8,14} = {2,4,8,14}. Repeated elements are written only once. Therefore option A is correct. Option C omits 4, while option D incorrectly includes elements that are not in the intermediate result.

Correct answer: A. {6,8}

Explanation: First form the union A ∪ B = {1,2,3,4,6,8,9}. Now intersect this set with C = {6,8,10}. Only 6 and 8 occur in both sets, so (A ∪ B) ∩ C = {6,8}. Option B is wrong because 2 is not in C, and option D is the union rather than the requested intersection.

Correct answer: A. {1,2,3,4,6,12}

Explanation: List the natural-number factors of each number. The factors of 24 are {1,2,3,4,6,8,12,24}, and the factors of 36 are {1,2,3,4,6,9,12,18,36}. Their intersection contains only the factors appearing in both lists: {1,2,3,4,6,12}. Hence option A is correct.

Correct answer: A. {3,6,7,9,12,14,15,18,21}

Explanation: The multiples of 3 not exceeding 21 are A = {3,6,9,12,15,18,21}. The multiples of 7 not exceeding 21 are B = {7,14,21}. A union contains every distinct element from both sets, so A ∪ B = {3,6,7,9,12,14,15,18,21}. The common element 21 is listed only once.

Correct answer: A. {1,2,4,5,6,8,10}

Explanation: A union B contains every element belonging to A or B, without repetition. Combining A = {1,2,5,10} with B = {2,4,6,8,10} gives {1,2,4,5,6,8,10}. Option B is only the intersection, option C is the complement of the union in U, and option D is the entire universal set rather than the requested union.

Correct answer: A. ∅

Explanation: First calculate the difference B \ A, which contains elements of B that are not in A. Since 3 and 4 are already in A, B \ A = {5,6}. Now intersect {5,6} with A = {1,2,3,4}. There are no common elements, so A ∩ (B \ A) = ∅. Option C is only the difference, not the final intersection.

Correct answer: A. {m,n,p,q,r,s}

Explanation: The difference B \ A contains the elements in B that are absent from A. Since n and q are common, B \ A = {r,s}. Taking the union with A adds r and s to all elements already in A: {m,n,p,q} ∪ {r,s} = {m,n,p,q,r,s}. Thus option A is correct; option B stops before the final union.

Correct answer: A. {1,2,3,4,5}

Explanation: Because B = {2,4} is a subset of A, removing B from A leaves A \ B = {1,3,5}. Taking the union of this remainder with B restores the removed elements: {1,3,5} ∪ {2,4} = {1,2,3,4,5} = A. Therefore option A is correct. This illustrates that (A \ B) ∪ B = A when B is a subset of A.

Correct answer: D. \(A\cup B=\{2,4\}\)

Explanation: The intersection consists of common elements, so \(A\cap B=\{2,4\}\). Removing the common elements from \(A\) gives \(A\setminus B=\{0,6\}\), and removing them from \(B\) gives \(B\setminus A=\{1,3\}\). However, the union must contain every distinct element from either set. Thus \(A\cup B=\{0,1,2,3,4,6\}\), not \(\{2,4\}\). Therefore, statement D is false.

Correct answer: A. \(\varnothing\)

Explanation: The composite natural numbers not exceeding 15 include \(4,6,8,9,10,12,14,15\) (and the convention that 1 is neither prime nor composite is used). Every element of \(B=\{4,6,8,10,12,14\}\) is therefore an element of \(A\), so \(B\subseteq A\). The difference \(B\setminus A\) contains elements in B that are absent from A; there are none. Hence \(B\setminus A=\varnothing\), making option A correct.

Correct answer: A. \(\{4,10\}\)

Explanation: 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.

Correct answer: A. \(\{p,q,r,t\}\)

Explanation: 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.

Correct answer: A. \(\{3,5,9\}\)

Explanation: 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.

Correct answer: A. \(\{2,6,9,10,30,45\}\)

Explanation: 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.

Correct answer: A. 62

Explanation: 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.

Correct answer: A. A = B

Explanation: 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.

Correct answer: A. {1,7,9}

Explanation: 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.

Correct answer: A. \(\{-3,-2,2,3\}\)

Explanation: First solve the equation defining each set. From \(x^2=9\), we obtain \(x=3\) or \(x=-3\), so \(A=\{-3,3\}\). From \(x^2-4=0\), we get \(x^2=4\), hence \(x=2\) or \(x=-2\), so \(B=\{-2,2\}\). The union contains every element that belongs to either set, without repeating any element. Therefore, \(A\cup B=\{-3,-2,2,3\}\), which is option A. Option B lists only A, option C lists only B, and option D omits 2.

Correct answer: B. \(\{2\}\)

Explanation: Factor the first quadratic: \(x^2-5x+6=(x-2)(x-3)\), so its roots are 2 and 3 and \(A=\{2,3\}\). Factor the second quadratic: \(x^2-3x+2=(x-1)(x-2)\), so its roots are 1 and 2 and \(B=\{1,2\}\). The intersection consists only of elements common to both sets. The only common element is 2; therefore, \(A\cap B=\{2\}\), which is option B. Option A combines elements from both sets rather than finding common elements, option C includes 3 although it is not in B, and option D is incorrect because the intersection is not empty.