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SubjectsMathematics
समुच्चयों पर संक्रियाएँ (संघ, प्रतिच्छेद और अंतर)
In Class 11 Mathematics, the Sets chapter introduces Operations on Sets (Union, Intersection, Difference). Students learn to combine sets using union, identify common elements through intersection, and find elements belonging to one set but not another using difference. They also apply these operations to subset relations, Venn diagrams, and problems involving the number of elements in sets.
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.
