Operations on Sets (Union, Intersection, Difference)

Let H be the set of students studying Hindi and E be the set studying English. The phrase “Hindi or English” means the union H ∪ E, including students who study both subjects. Therefore, n(H ∪ E) = n(H) + n(E) − n(H ∩ E) = 28 + 22 − 10 = 40. The subtraction prevents the ten students studying both subjects from being counted twice.

Use the two-set cardinality formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substituting the values gives 35 + 27 − 50 = 12. Thus, twelve elements belong to both A and B, so option A is correct. The result is reasonable because the union is smaller than 35 + 27 due to overlap.

The set A can be divided into two non-overlapping parts: the elements in A \ B and the elements in A ∩ B. Therefore, n(A) = n(A \ B) + n(A ∩ B). Using the given values, 20 = 13 + n(A ∩ B), so n(A ∩ B) = 20 − 13 = 7. Option 13 represents only the difference, while 33 is impossible because it exceeds the total size of A.

The union A ∪ B contains every element that belongs to A, to B, or to both, with repeated elements written only once. Combining A = {1,3,5,7} and B = {2,3,5,8} gives {1,2,3,5,7,8}. The elements 3 and 5 are common, but they are listed once. The universal set U provides the surrounding set of possible elements but does not change the union calculation.

The difference A \ B consists of elements that are in A but not in B. Begin with A = {1,2,4,8}; the elements 2, 4, and 8 also occur in B, so they must be removed. The only remaining element is 1. Therefore, A \ B = {1}. The set {2,4,8} is actually A ∩ B, whereas {6} belongs to B but not to A.

Two sets are called disjoint sets when they have no common element. The statement A ∩ B = ∅ precisely says that the intersection of A and B is empty, so no element belongs to both sets. Therefore, A and B are disjoint. They need not have the same number of elements, and neither set must contain the other. Hence, option A is the only correct description.

Because x must be an integer, list the values in each set. A contains −2, −1, 0, 1, 2, and 3, while B contains 0, 1, 2, 3, 4, and 5. The elements common to both lists are 0, 1, 2, and 3. Therefore, A ∩ B = {0,1,2,3}, which is option B. The endpoints are included because the inequalities use ≤.

The prime numbers not exceeding 10 are A = {2,3,5,7}. The odd natural numbers not exceeding 10 are B = {1,3,5,7,9}. To find A \ B, retain elements of A that do not belong to B. The numbers 3, 5, and 7 are odd and occur in B, while 2 is not odd and does not occur in B. Hence, A \ B = {2}.

The multiples of 3 not exceeding 15 are A = {3, 6, 9, 12, 15}, while the multiples of 5 not exceeding 15 are B = {5, 10, 15}. The union contains every element that belongs to A or B, but repeated elements are written only once. Therefore, A ∪ B = {3, 5, 6, 9, 10, 12, 15}. The common element 15 is included only once.

First evaluate the expression inside the parentheses. B ∪ C = {r, s, t, u}, because all distinct elements from B and C are included. Now intersect this result with A = {p, q, r, s}. The elements common to A and {r, s, t, u} are r and s only. Hence, A ∩ (B ∪ C) = {r, s}. The parentheses must be handled before taking the intersection.

First find the union B ∪ C. Combining B = {2, 4, 6} and C = {1, 4, 7} gives B ∪ C = {1, 2, 4, 6, 7}. The difference A \ (B ∪ C) consists of elements that are in A but not in this union. Removing 1, 2, and 4 from A = {1, 2, 3, 4, 5} leaves {3, 5}. Therefore, option A is correct.

The displayed identity is the distributive law of intersection over union. It has the same pattern as algebraic distribution: A is distributed across the union inside the parentheses, producing (A ∩ B) ∪ (A ∩ C). The commutative law changes the order of sets, the associative law changes grouping, and the complement law involves complements such as A′. Therefore, option A is the only correct answer.

Evaluate the parenthesized union first. A ∪ B = {1, 2, 3}, since the repeated element 2 is written only once. Next, intersect this result with C = {3, 4}. The only element common to {1, 2, 3} and {3, 4} is 3. Thus, (A ∪ B) ∩ C = {3}. The answer is not {2}, because 2 is not an element of C.

First calculate A ∩ B, the set of elements common to A and B. The common elements are 3 and 5, so A ∩ B = {3, 5}. Now take the union of this result with C = {3, 4, 5}. Including every distinct element gives {3, 5} ∪ {3, 4, 5} = {3, 4, 5}. Therefore, option B is correct. Elements such as 2, 7, 1, and 9 are not included because they are not in the intermediate intersection or in C.

The difference A \ B contains elements present in A but absent from B, so A \ B = {1, 2}. Similarly, B \ A contains elements present in B but absent from A, giving B \ A = {5, 6}. Their union is {1, 2} ∪ {5, 6} = {1, 2, 5, 6}. Thus the expression collects the elements belonging to exactly one of the two sets and excludes the common elements 3 and 4.

Taking N as the positive natural numbers, A = {1, 2, 3, 4, 5, 6, 7} because x < 8. The conditions 4 < x < 10 give B = {5, 6, 7, 8, 9}. Set difference A \ B keeps only the elements of A that do not occur in B. Removing 5, 6, and 7 from A leaves {1, 2, 3, 4}. Hence, option A is correct. The strict inequalities exclude 4 from B and 10 from consideration.

The union A ∪ B contains every element of A together with every element of B. If this union is exactly A, adding B has introduced no new element. Therefore every element of B must already belong to A, which means B ⊆ A. The other statements do not necessarily follow from the given equality.

The intersection A ∩ B contains exactly the elements common to A and B. If the intersection equals all of A, then every element of A must also be present in B. This is precisely the definition of A being a subset of B, so A ⊆ B. The equality does not require B to be empty or equal to A.

A \ B consists of elements that are in A but not in B. Removing 2, 4, and 6 from A leaves {1, 3, 5}. For B \ A, we look for elements in B that are absent from A. Since B = {2, 4, 6} is completely contained in A, no such elements exist, so B \ A = ∅. Hence the ordered pair is ({1, 3, 5}, ∅), making option A correct.

Because x is an integer and x² ≤ 4, we have |x| ≤ 2, so -2 ≤ x ≤ 2. Therefore A = {-2, -1, 0, 1, 2}. The intersection keeps only elements common to A and B. Comparing B = {-1, 0, 1, 3} with A, the common elements are -1, 0, and 1; 3 is not in A. Hence A ∩ B = {-1, 0, 1}, so option B is correct.