Operations on Sets (Union, Intersection, Difference)

The multiples of 4 up to 25 are \(\{4,8,12,16,20,24\}\), so \(|A|=6\). The multiples of 6 are \(\{6,12,18,24\}\), so \(|B|=4\). Their common elements are \(\{12,24\}\), giving \(|A\cap B|=2\). By inclusion–exclusion, \(|A\cup B|=6+4-2=8\). The subtraction prevents 12 and 24 from being counted twice.

Using the convention \(\mathbb N=\{1,2,3,\ldots\}\), the set \(A=\{1,2,\ldots,12\}\) has 12 elements. The perfect squares not exceeding 12 are \(B=\{1,4,9\}\), which has 3 elements. Removing these three elements from \(A\) leaves \(|A-B|=12-3=9\). Thus option B is correct; option C would incorrectly ignore the subtraction.

The intersection \(A\cap B\) is defined as the set of elements that belong to both \(A\) and \(B\). Since every element in the intersection necessarily belongs to \(A\), it follows that \(A\cap B\subseteq A\) for all sets \(A\) and \(B\). The other statements are not universally true: the two differences may differ, equality of union and intersection is exceptional, and \(A\subseteq A-B\) generally fails when the sets overlap.

Union and intersection are commutative operations, so changing the order of the sets does not change the result: A ∪ B = B ∪ A and A ∩ B = B ∩ A. Also, the union of any set with the empty set is the set itself. However, set difference depends on order. In general, A − B contains elements of A that are not in B, whereas B − A contains elements of B that are not in A. Therefore, A − B = B − A is generally false, although it may happen for special sets such as A = B.

The union of two sets contains every element that belongs to at least one of them. The empty set ∅ contains no elements, so it contributes nothing when forming a union. Consequently, A ∪ ∅ = A. Since A = {1, 2, 3}, the result is {1, 2, 3}. Option A would be the result of confusing union with intersection in this situation, while options C and D introduce or omit elements without justification. This property is called the identity property of union.

The intersection of two sets consists of elements common to both sets. Although A contains 1, 2, and 3, the empty set ∅ contains no elements at all. Therefore, there cannot be any element common to A and ∅, and A ∩ ∅ = ∅. Options A and B incorrectly treat intersection like union, while option D invents an element that is not present in either set. This is the empty-set property of intersection and is valid for every set A.

The difference A − B contains elements that are in A but not in B. Since 3 and 4 are common to both sets, A − B = {1, 2}, so n(A − B) = 2. Similarly, B − A contains elements in B but not in A, giving B − A = {5, 6} and n(B − A) = 2. Hence the required sum is 2 + 2 = 4. The value 2 counts only one directional difference, while 6 and 8 result from overcounting or using the wrong set formula.

The union A ∪ B can be partitioned into three mutually disjoint parts: the elements only in A, represented by A − B; the elements only in B, represented by B − A; and the elements common to both, represented by A ∩ B. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 60 = 18 + 22 + n(A ∩ B), so n(A ∩ B) = 20. Thus option A is correct.

A contains all natural numbers up to 15, while B contains the odd natural numbers up to 15. Subtracting B from A removes 1, 3, 5, 7, 9, 11, 13, and 15. The elements left in A are precisely the even natural numbers up to 15: {2, 4, 6, 8, 10, 12, 14}. Therefore option B is correct. Option A is B itself, option C incorrectly includes 16, which is greater than 15, and option D would mean every element of A were odd.

Let H be the set of students studying Hindi and E the set studying English. Students counted in H ∩ E study both languages, so they must be removed from the total Hindi group to find only Hindi: n(H − E) = n(H) − n(H ∩ E) = 24 − 7 = 17. Therefore, 17 students study only Hindi. The value 11 is the number studying only English, 7 is the number studying both, and 35 is an incorrect addition that double-counts the overlap.

The intersection contains the real numbers that belong to both intervals. A includes every number from -1 through 4, including both endpoints. B contains numbers greater than 1 and less than 6; therefore, 1 is excluded, while 4 is included because A includes 4 and B also contains 4. Hence the common interval is (1, 4], which is option B.

Factor the first equation: x² − 7x + 12 = (x − 3)(x − 4) = 0, so A = {3, 4}. Factor the second equation: x² − 9x + 20 = (x − 4)(x − 5) = 0, so B = {4, 5}. The union contains every distinct element appearing in either set, so A ∪ B = {3, 4, 5}. Therefore, option A is correct.

A − B contains elements present in A but absent from B, so A − B = {1, 9, 25}. Similarly, B − A = {2, 8, 32}. Taking their union gives {1, 9, 25} ∪ {2, 8, 32} = {1, 2, 8, 9, 25, 32}. This operation is also called the symmetric difference because common elements 4 and 16 are excluded. Thus option B is correct.

The intersection A ∩ B contains numbers that belong to both sets, so each number must be divisible by both 2 and 3. Such numbers are multiples of lcm(2, 3) = 6. The multiples of 6 from 1 through 30 are 6, 12, 18, 24, and 30. Therefore, A ∩ B has 5 elements, so option A is correct.

Since x is an integer and −2 ≤ x < 5, A = {−2, −1, 0, 1, 2, 3, 4}. The inequality x² ≤ 9 is equivalent to −3 ≤ x ≤ 3, so B = {−3, −2, −1, 0, 1, 2, 3}. Set difference A − B retains members of A that are absent from B. Every member except 4 is removed, giving {4}; therefore option A is correct.

The union is partitioned into three disjoint parts: A − B, A ∩ B, and B − A. The given parts account for {1, 4} and {2, 5}. Removing these four elements from A ∪ B = {1, 2, 3, 4, 5, 6} leaves {3, 6}. These remaining elements must be in B but not in A, so B − A = {3, 6}. Option A is correct.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 63 = 45 + 38 − n(A ∩ B), or 63 = 83 − n(A ∩ B). Therefore n(A ∩ B) = 83 − 63 = 20. The overlap is subtracted because common elements are counted twice in n(A) + n(B), so option A is correct.

The relation A ⊆ B means every element of A is also an element of B. Consequently, A ∪ B = B and A ∩ B = A. Substituting these results into the expression gives (A ∪ B) − (A ∩ B) = B − A. Notice that A − B is empty because no element of A lies outside B. Therefore, option A is the only correct answer.

Evaluate the parentheses first. The union B ∪ C contains all distinct elements from B and C: {1, 2, 4, 6, 7, 8}. Now intersect this set with A = {1, 2, 3, 4, 5}. The common elements are 1, 2, and 4; elements 6, 7, and 8 are not in A. Therefore A ∩ (B ∪ C) = {1, 2, 4}, so option A is correct.

A − B = ∅ means that no element of A lies outside B; therefore, every element of A belongs to B, so A ⊆ B. The additional condition B − A ≠ ∅ says that at least one element belongs to B but not to A. Hence A is a proper subset of B, written A ⊂ B. Equality and the disjointness condition are therefore impossible.