Operations on Sets (Union, Intersection, Difference)

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}.

First expand the set-builder description: the positive even numbers less than 10 are A={2,4,6,8}. Set B={2,3,5,7}. Intersection keeps only values occurring in both lists, and the sole common value is 2. Therefore A∩B={2}. Option B is A itself, option C is the odd part of B, and the empty set incorrectly assumes no common element.

The intersection contains elements common to both sets. Set A contains only odd numbers, while set B contains only even numbers. No number can be both one of the listed odd elements and one of the listed even elements. Therefore, A and B are disjoint sets, and their intersection is the empty set: A ∩ B = ∅.

The union A ∪ B contains every element that belongs to A, to B, or to both sets. Here, A = {1, 2} and B already contains 1, 2, 3, and 4. Therefore, combining the elements without repeating any element gives A ∪ B = {1, 2, 3, 4}. Since A is a subset of B, their union is simply B.

The intersection A ∩ B consists only of elements common to both A and B. The elements p and q occur in A and also occur in B, while r occurs only in B. Hence A ∩ B = {p, q}. Because A is a subset of B, the intersection of the two sets is the smaller set A.

The difference A − B contains elements that are in A but are not in B. Starting with A = {1, 2, 3, 4, 5}, remove the elements 2 and 4 because they belong to B. The elements left are 1, 3, and 5, so A − B = {1, 3, 5}. The order of elements in a set does not matter.

A − B contains elements that belong to A but do not belong to B. Here A and B are exactly equal, so every element of A is also present in B. After removing all elements of B from A, no element remains. Therefore A − B = ∅. This is the standard identity X − X = ∅ for every set X.

The union of two sets contains every element found in either set. The empty set B = ∅ contributes no elements at all, so combining A = {m, n, o} with B does not add or remove anything. Consequently, A ∪ ∅ = A = {m, n, o}. This is called the identity property of the empty set under union.

The intersection A ∩ ∅ contains elements that must belong to both A and the empty set. However, the empty set has no elements, so there cannot be any element common to A and ∅. Therefore A ∩ ∅ = ∅, regardless of which elements are present in A. This is the zero property of the empty set for intersection.

The difference A − A asks for elements that are in the first copy of A but not in the second copy. Since both copies are the same set, every element 1, 2, and 3 is removed by the subtraction. No element remains, so A − A = ∅. This is the general difference identity X − X = ∅.

Use the union formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 12 = 7 + 8 − n(A ∩ B), or 12 = 15 − n(A ∩ B). Rearranging, n(A ∩ B) = 15 − 12 = 3. Therefore, option C is correct and the two sets have three common elements.

The set A is divided into two disjoint parts: the elements common to A and B, represented by A ∩ B, and the elements of A outside B, represented by A − B. Thus n(A) = n(A ∩ B) + n(A − B). Substituting gives 10 = 4 + n(A − B), so n(A − B) = 6. Option C is correct.

The difference B − A contains the elements that are present in B but absent from A. Here, B = {2, 4, 6}, and every one of these elements is also present in A. Therefore, no element remains after removing from B the elements common with A, so B − A = ∅. Option B is only B itself and ignores the subtraction operation.