Operations on Sets (Union, Intersection, Difference)

Correct answer: A. 63

Explanation: First find the intersection using n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus n(A ∩ B) = 72 + 67 − 101 = 38. The symmetric difference A △ B contains elements belonging to exactly one of the two sets, so n(A △ B) = n(A ∪ B) − n(A ∩ B) = 101 − 38 = 63. Hence option A is correct.

Correct answer: A. 22

Explanation: The multiples of 6 from 1 to 96 are counted by ⌊96/6⌋ = 16. The multiples of 10 are counted by ⌊96/10⌋ = 9. Common elements are multiples of lcm(6,10) = 30, and there are ⌊96/30⌋ = 3 of them. Therefore, n(A ∪ B) = 16 + 9 − 3 = 22, by inclusion–exclusion. Thus option A is correct.

Correct answer: A. 1

Explanation: An element in A ∩ B must be divisible by both 7 and 15. Because 7 and 15 are coprime, their least common multiple is 7 × 15 = 105. The positive multiples of 105 not exceeding 105 consist only of 105 itself. Therefore A ∩ B = {105}, and its cardinality is n(A ∩ B) = 1. Hence option A is correct.

Correct answer: A. 102

Explanation: The set A is the disjoint union of A − B and A ∩ B. Thus n(A ∩ B) = n(A) − n(A − B) = 69 − 28 = 41. Now apply the union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 69 + 74 − 41 = 102. The value 143 incorrectly counts the common elements twice, while 41 is only the intersection. Therefore option A is correct.

Correct answer: A. 73

Explanation: The union consists of three disjoint parts: the elements only in A, the elements only in B, and the common elements in A ∩ B. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Hence the required value is 115 − 42 = 73. Thus option A is correct. This is also the cardinality of the symmetric difference A △ B.

Correct answer: A. {1, 2, 5, 6}

Explanation: First calculate each difference separately. A − B contains elements in A but not B, so A − B = {1, 2}. Similarly, B − A = {5, 6}, because 5 and 6 are not in A. Their union is therefore {1, 2} ∪ {5, 6} = {1, 2, 5, 6}. The common elements 3 and 4 are excluded, so option A is correct.

Correct answer: A. 19

Explanation: Use the inclusion–exclusion formula for two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 14 + 9 − 4 = 19. The intersection is subtracted because its four elements were counted once in n(A) and again in n(B). Therefore option A, 19, is correct.

Correct answer: C. 3

Explanation: For two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearrange the formula to find the intersection: n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substitution gives 12 + 11 − 20 = 3. Thus the two sets have three common elements, and option C is the unique correct answer.

Correct answer: C. 11

Explanation: The set A is divided into two disjoint parts: the elements in A ∩ B and the elements in A − B. Therefore n(A) = n(A ∩ B) + n(A − B). Rearranging gives n(A − B) = 18 − 7 = 11. Hence option C is correct. The value 7 counts the common elements, while 18 counts all elements of A, so neither is the required difference.

Correct answer: C. ∅

Explanation: Compute both directed differences. A − B contains elements in A but not B, so A − B = {3, 9}. Likewise, B − A contains elements in B but not A, giving B − A = {18, 24}. These two sets are disjoint, meaning they have no common element. Their intersection is therefore the empty set, ∅, so option C is correct. Option D is only the first difference and is not the final intersection.

Correct answer: A. {1, 2, 3, 4, 5, 6}

Explanation: The union A ∪ B contains every distinct element that occurs in A or in B. Combining the two lists gives 1, 2, 3, 4, 5, and 6. Elements 2 and 5 appear in both sets, but set notation records each element only once. Therefore, A ∪ B = {1, 2, 3, 4, 5, 6}, so option A is correct. Option B is only the intersection, not the union.

Correct answer: B. {6, 12}

Explanation: The intersection A ∩ B consists of elements present in both A and B. Checking the lists, 6 appears in A and B, and 12 also appears in both. The elements 3 and 9 occur only in A, while 18 occurs only in B, so they are excluded. Hence A ∩ B = {6, 12}, and option B is correct. Option C incorrectly combines all elements as if the operation were union.

Correct answer: A. {1, 7}

Explanation: The difference A \ B contains elements that belong to A but do not belong to B. In A = {1, 4, 7, 10}, the elements 4 and 10 are also in B, so they must be removed. The elements 1 and 7 are not in B and remain in the result. Therefore, A \ B = {1, 7}, making option A correct. Option B lists the common elements, while option D is closer to a union.

Correct answer: C. {1, 14}

Explanation: For B \ A, retain elements of B that are not present in A. B contains 1, 5, 8, and 14. Since 5 and 8 also occur in A, remove them; 1 and 14 do not occur in A, so they remain. Thus B \ A = {1, 14}, which is option C. Option B is the intersection, option A comes from A \ B, and option D includes elements that should not be retained.

Correct answer: A. {a, b, e, f}

Explanation: The union A ∪ B contains every distinct element from both sets, so A ∪ B = {a, b, c, d, e, f}. The intersection A ∩ B contains only the common elements, giving {c, d}. Set difference removes the elements of the second set from the first; therefore removing c and d from the union leaves {a, b, e, f}. Thus option A is correct. The other choices represent the intersection, the full union, or an empty result.

Correct answer: A. 25

Explanation: For two finite sets, the number of elements in their union is given by n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 18 + 12 − 5 = 25. We subtract the intersection because the five common elements were counted once in n(A) and again in n(B). Hence, the correct answer is 25, option A.

Correct answer: A. 40

Explanation: 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.

Correct answer: A. 12

Explanation: 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.

Correct answer: A. 7

Explanation: 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.

Correct answer: B. {0,1,2,3}

Explanation: 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 ≤.

Correct answer: A. {2}

Explanation: 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}.

Correct answer: A. {3, 5, 6, 9, 10, 12, 15}

Explanation: 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.

Correct answer: A. {r, s}

Explanation: 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.

Correct answer: A. {3, 5}

Explanation: 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.

Correct answer: A. Distributive law

Explanation: 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.