Operations on Sets (Union, Intersection, Difference)

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Here A ∩ B is empty, so its cardinality is zero. Therefore n(A ∪ B) = 15 + 18 − 0 = 33. Because the sets are disjoint, no element is counted twice. Options B and C give the size of only one set, while option D gives their difference.

Set A contains only even numbers, whereas set B contains only odd numbers. No number appears in both sets, so the sets are disjoint and their intersection is empty: A ∩ B = ∅. The union is not empty, and each set has elements that are absent from the other, so neither A \ B nor B \ A is empty. Hence option A is uniquely correct.

The displayed identity shows union distributed over intersection: the set A is combined separately with B and with C, and the resulting expressions are intersected. This is the distributive law of set algebra. The commutative law changes order, the associative law changes grouping, and the idempotent law has the form A ∪ A = A or A ∩ A = A. Therefore option A is correct.

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.

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.

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.

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.

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.

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.

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.

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.

An intersection contains only elements common to both sets. Comparing A = {a,c,e,g} and B = {b,c,e,f}, the shared elements are c and e. Therefore A ∩ B = {c,e}. The elements a and g occur only in A, b and f occur only in B, and d belongs to U but to neither A nor B.

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.

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.

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.

The union is A ∪ B = {2,3,4,5,6,8}, while the intersection is A ∩ B = {4,6}. The element 5 belongs to A and therefore to the union, but it is not common to both sets, so it is absent from the intersection. Thus option A is correct. Elements 4 and 6 are in the intersection, and 10 is in neither set.

The difference \(A\setminus B\) contains elements that belong to \(A\) but do not belong to \(B\). Starting with \(A=\{1,3,6,9\}\), remove 3 and 6 because both are also in \(B\). This gives \(A\setminus B=\{1,9\}\). Therefore, 9 is the only listed element in the difference. Options 3 and 6 are excluded because they are common to both sets, while 12 is not an element of \(A\).

The intersection contains elements common to both sets. Comparing \(A\) and \(B\), we obtain \(A\cap B=\{5,10\}\). A set is a subset of another set when every element of the first set is contained in the second set. Since the only element of \(\{5\}\) is 5, and 5 belongs to \(\{5,10\}\), option A is correct. The other options contain 2 or 15, which are not in the intersection.

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.

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.