Operations on Sets (Union, Intersection, Difference)

By definition, \(A-B\) contains elements that are in A but not in B. The element blue belongs to both sets, so it must be removed from A. Red and green are in A and absent from B, giving \(A-B=\{\text{red},\text{green}\}\). Option B is the intersection, option C is not a subset of A, and option D is the union.

For two finite sets, the cardinality of the union is found using \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Substituting the given values gives \(5+4-2=7\). We subtract 2 because the common elements would otherwise be counted once in A and again in B. Therefore, option A, 7, is correct.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, n(A ∩ B) = 8 + 6 − 10 = 4. Hence, four elements are common to A and B. Option 2 is merely the difference between the set sizes, 14 ignores the overlap, and 10 is the cardinality of the union.

The difference A − B contains elements that belong to A but do not belong to B. Removing 2, 4, and 6 from A leaves A − B = {1, 3, 5}. This set has three elements, so the answer is 3. The value 6 is the size of A, while 0 would apply only if every element of A were also in B.

The intersection A ∩ B consists only of elements common to both sets. Comparing the two lists, the common elements are 3 and 5, so A ∩ B = {3, 5}. Therefore, n(A ∩ B) = 2. The other choices do not represent the number of shared elements: 4 is the size of A, 6 is too large, and 0 would mean the sets had no common element.

The equation \(A\cup B=B\cup A\) shows that interchanging the order of the two sets does not change their union. This is called the commutative property of union, just as \(a+b=b+a\) is commutativity for addition. The associative property involves three sets and parentheses, the distributive property connects union with intersection, and the identity property involves the empty set or universal set.

The equation A ∩ B = B ∩ A says that the intersection remains unchanged when the order of A and B is reversed. Therefore, it expresses the commutative property of intersection. It does not describe difference, the empty set, or complements, because none of those operations is represented in the given equation.

The intersection contains elements present in both participating sets. When both sets are the same set A, every element of A is present in both copies, so A ∩ A = A. This is the idempotent law for intersection. The empty set, complement Aᶜ, and difference A − B are not generally equal to A without additional conditions.

Evaluate the expression in the indicated order. First, \(A\cup B=\{1,2,3,4\}\), because all distinct elements from A and B are included. Next, intersect this result with \(C=\{4,5\}\). The only element common to both sets is 4, so \((A\cup B)\cap C=\{4\}\). Option B is wrong because 3 is not in C; option C is a union rather than the requested intersection, and option D ignores the common element 4.

First evaluate the intersection because the expression contains parentheses: \(A\cap B\) consists of elements common to both A and B, so \(A\cap B=\{2,4\}\). Next take the union with C, which includes every distinct element in either set: \(\{2,4\}\cup\{1,2,6\}=\{1,2,4,6\}\). Thus option A is correct. Option B stops after the intersection, option C ignores the intersection, and option D incorrectly includes 3, which is not in the intermediate result or C.

The difference \(A-B\) keeps elements that are in A but not in B. Removing 4 and 8 from A gives \(A-B=\{2,6\}\). Now form the union with \(C=\{8,10\}\), taking every distinct element: \(\{2,6\}\cup\{8,10\}=\{2,6,8,10\}\). Therefore option A is correct. Option B omits C, option C gives elements removed from A, and option D incorrectly retains 4, although 4 belongs to B.

First calculate the difference. Since 2 and 5 are the elements of B that occur in A, they must be removed: \(A-B=\{1,3,4\}\). The next operation is intersection with \(C=\{5,6\}\). None of 1, 3, or 4 is present in C, so there is no common element. Hence \((A-B)\cap C=\varnothing\), making option A correct. Option B wrongly keeps 5 even though it was removed, while option D is not in \(A-B\).

The union contains every element appearing in either set, so \(A\cup B=\{1,2,3,4\}\). The intersection contains only elements common to both sets, so \(A\cap B=\{2,3\}\). Element 1 belongs to A and therefore to the union, but it does not belong to B and therefore is not in the intersection. Thus option A is correct. Elements 2 and 3 are in both sets, so they belong to the intersection and cannot satisfy the condition.

The difference \(A-B\) contains elements that are in A but not in B. If this difference is empty, then no element of A lies outside B. Therefore every element of A is also an element of B, which is exactly the definition of \(A\subseteq B\). The reverse inclusion is not guaranteed, and the sets need not be disjoint or empty. Thus option A is the only conclusion that must be true.

To calculate \(A-B\), inspect each element of A and retain it only if it is not in B. Elements 2 and 4 occur in B, so they are removed. Element 6 is in B but not in A, so it cannot appear in \(A-B\). The remaining elements are 1, 3, and 5; hence \(C=A-B=\{1,3,5\}\). Therefore option A is correct, while the other options represent retained, removed, or unchanged elements incorrectly.

Students who play both games must belong to both sets, so we find the intersection \(C\cap F\). Comparing the two sets, Ravi is the only name appearing in both; Amit and Sita appear only in C, while Neha appears only in F. Therefore \(C\cap F=\{Ravi\}\), making option A correct. Option D is the union, which lists students playing at least one game rather than both.

The phrase “at least one category” means that a book may be in H, in E, or in both; this is exactly the union. Combining all distinct elements gives \(H\cup E=\{h_1,h_2,h_3,e_1,e_2\}\). The repeated element \(h_3\) is written only once because sets do not contain duplicates. Option B gives only common books, while C and D give books exclusive to one category, so option A is correct.

The positive factors of 12 are A = {1, 2, 3, 4, 6, 12}, while the positive factors of 18 are B = {1, 2, 3, 6, 9, 18}. The intersection A ∩ B contains only elements present in both sets. Therefore, the common factors are {1, 2, 3, 6}, so option A is correct.

The positive factors of 10 are A = {1, 2, 5, 10}, and the positive factors of 15 are B = {1, 3, 5, 15}. The union A ∪ B contains every element that belongs to either set, writing repeated elements only once. Thus A ∪ B = {1, 2, 3, 5, 10, 15}, so option A is correct.

First evaluate the expression inside the parentheses: B ∪ C = {2, 4, 5, 6, 8}. Next, compare this set with A = {1, 2, 5, 7}. The elements common to both sets are 2 and 5. Hence A ∩ (B ∪ C) = {2, 5}. Option C is only the union and does not complete the outer intersection.

First find the union B ∪ C = {20, 30, 50, 60}. Set difference A − (B ∪ C) keeps the elements of A that do not occur in that union. From A = {10, 20, 30, 40}, remove 20 and 30; 50 and 60 are not in A and therefore have no effect. The result is {10, 40}.

The union of two sets contains every element that belongs to at least one of the sets. Combining A = {1, 2, 3} and B = {3, 4, 5} gives 1, 2, 3, 4, and 5. The element 3 occurs in both sets, but set notation lists an element only once. Therefore, A ∪ B = {1, 2, 3, 4, 5}.

The intersection operation selects elements common to both sets, not elements appearing in only one set or in either set. Set A contains 2, 4, 6, and 8; set B contains 1, 2, 3, and 4. The common elements are exactly 2 and 4. Thus A∩B={2,4}, so option B is correct; option D represents the union instead.

The difference A − B contains elements that belong to A but do not belong to B. Starting with A = {a, b, c, d}, remove b and d because both are also in B = {b, d, e}. The element e is not in A, so it cannot appear in A − B. The remaining elements are a and c, giving A − B = {a, c}.

Set difference B−A means elements that belong to B but do not belong to A; the order matters. Starting with B={10,20}, remove 10 because it is also in A={5,10,15}. The remaining element is 20, which is not in A. Hence B−A={20}, so option B is correct. A−B would be a different set, namely {5,15}.