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Operations on Sets (Union, Intersection, Difference)
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.
Practice questions
01 Which of the following set identities is always true?
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Answer and explanation
Correct answer: A. A ∪ B = B ∪ A
Explanation: Union is commutative: A ∪ B = B ∪ A for every pair of sets because an element belongs to the union when it belongs to A or B, and changing the order does not change that condition. The other statements are not always true. In general A − B differs from B − A, A ∩ ∅ = ∅, and A ∪ A = A. Hence option A is correct.
02 If A = {2, 3, 4, 5}, B = {1, 3, 5, 7}, and C = {0, 3, 6, 9}, what is A ∩ B ∩ C?
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Answer and explanation
Correct answer: A. {3}
Explanation: The intersection of several sets contains only elements common to every set. Comparing A and B first gives A ∩ B = {3, 5}. The set C contains 3 but does not contain 5. Thus, after checking all three sets, only 3 remains, so A ∩ B ∩ C = {3}. Option B is not correct because 5 is absent from C, and option C includes that same incorrect element. Therefore, option A is correct.
03 If A = {1, 2, 3, 4, 5, 6}, B = {2, 3}, and C = {5, 6, 7}, what is A − (B ∪ C)?
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Answer and explanation
Correct answer: A. {1, 4}
Explanation: First calculate the union inside the parentheses: B ∪ C = {2, 3, 5, 6, 7}. Now A − (B ∪ C) means retain elements of A that are absent from this union. Removing 2, 3, 5, and 6 from A leaves {1, 4}. Although 7 belongs to the union, it is not in A and cannot appear in the difference. Thus option A is correct.
04 If A = {a, b, c, d}, B = {b, d, f}, and C = {d, e, f}, what is A ∩ (B ∪ C)?
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Answer and explanation
Correct answer: A. {b, d}
Explanation: Use the order of operations for sets: calculate the union inside the parentheses first. B ∪ C = {b, d, e, f}. Now intersect this set with A = {a, b, c, d}; only elements appearing in both sets are retained. Those common elements are b and d. Hence A ∩ (B ∪ C) = {b, d}, so option A is correct. The other options either omit common elements or include elements not in A.
05 If A = {1, 3, 5, 7}, B = {3, 5, 9}, and C = {5, 7, 11}, what is (A ∩ B) ∪ C?
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Answer and explanation
Correct answer: A. {3, 5, 7, 11}
Explanation: Follow the parentheses and calculate the intersection first. A ∩ B contains the elements common to A and B, so A ∩ B = {3, 5}. Now take the union with C = {5, 7, 11}; include every element appearing in either set, without repeating 5. The result is {3, 5, 7, 11}, so option A is correct.
06 If A = {2, 4, 6, 8} and B = {1, 2, 3, 4, 5}, how many elements are in A ∪ B?
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Answer and explanation
Correct answer: C. 7
Explanation: The union contains every distinct element from both sets, with common elements counted only once. Here A ∩ B = {2, 4}, so there are two repeated elements. Using the formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we get 4 + 5 − 2 = 7. Indeed, A ∪ B = {1, 2, 3, 4, 5, 6, 8}, which has seven elements. Option C is correct.
07 If A = {12, 24, 36, 48} and B = {24, 48, 60}, how many elements are in A ∩ B?
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Answer and explanation
Correct answer: B. 2
Explanation: The intersection A ∩ B consists only of elements that occur in both A and B. Comparing the sets, 24 appears in both and 48 also appears in both. The elements 12 and 36 occur only in A, while 60 occurs only in B. Therefore, A ∩ B = {24, 48}, which has two elements. Hence option B is correct.
08 If \(A=\{1,8,27,64\}\) and \(B=\{8,64,125\}\), what is \(A-B\)?
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Answer and explanation
Correct answer: A. \(\{1,27\}\)
Explanation: The set difference \(A-B\) contains every element that belongs to \(A\) but does not belong to \(B\). In \(A\), the elements 8 and 64 are also present in \(B\), so they are removed. The elements 1 and 27 are not in \(B\), so they remain. Therefore, \(A-B=\{1,27\}\). Option B is the intersection, option C contains an element only from \(B\), and option D is the union.
09 If \(A=\{1,2,3,4,5\}\) and \(B=\{6,7,8\}\), what is \(A-B\)?
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Answer and explanation
Correct answer: B. \(\{1,2,3,4,5\}\)
Explanation: The difference \(A-B\) keeps elements of \(A\) that are not elements of \(B\). The sets \(A\) and \(B\) are disjoint because they have no common element: 1 through 5 occur only in \(A\), while 6 through 8 occur only in \(B\). Thus no element is removed from \(A\), and \(A-B=A=\{1,2,3,4,5\}\).
10 If \(A=\{\text{pen},\text{book},\text{bag}\}\) and \(B=\{\text{book},\text{desk}\}\), what is \(A\cap B\)?
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Answer and explanation
Correct answer: B. \(\{\text{book}\}\)
Explanation: The intersection \(A\cap B\) consists only of elements that occur in both sets. The word “book” appears in \(A\) and also in \(B\), while “pen” and “bag” occur only in \(A\), and “desk” occurs only in \(B\). Therefore, \(A\cap B=\{\text{book}\}\). Option D represents the union, not the intersection.
11 If \(A=\{\text{rose},\text{lily}\}\) and \(B=\{\text{lily},\text{lotus},\text{jasmine}\}\), what is \(A\cup B\)?
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Answer and explanation
Correct answer: A. \(\{\text{rose},\text{lily},\text{lotus},\text{jasmine}\}\)
Explanation: The union \(A\cup B\) contains every distinct element that belongs to either \(A\), \(B\), or both. The elements are rose, lily, lotus, and jasmine. Because lily is common to both sets, it is written only once; repetition is not used in a set. Hence \(A\cup B=\{\text{rose},\text{lily},\text{lotus},\text{jasmine}\}\).
12 If \(A=\{\text{north},\text{south},\text{east},\text{west}\}\) and \(B=\{\text{east},\text{west}\}\), what is \(A-B\)?
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Answer and explanation
Correct answer: A. \(\{\text{north},\text{south}\}\)
Explanation: To find \(A-B\), remove from \(A\) every element that is also in \(B\). The elements east and west occur in both sets, so they are deleted from \(A\). The remaining elements are north and south. Therefore, \(A-B=\{\text{north},\text{south}\}\). This is not the intersection, empty set, or complete original set.
13 If \(A=\{2,4,6,8\}\) and \(B=\{4,8,12\}\), which element belongs to \(A\cap B\)?
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Answer and explanation
Correct answer: B. 4
Explanation: An element belongs to \(A\cap B\) only when it is present in both \(A\) and \(B\). The number 4 appears in both sets, and 8 also appears in both, although 8 is not offered as an option. The number 2 and 6 occur only in \(A\), while 12 occurs only in \(B\). Therefore, among the given choices, 4 is the correct answer.
14 If \(A=\{3,6,9\}\) and \(B=\{9,12,15\}\), which element must belong to \(A\cup B\)?
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Answer and explanation
Correct answer: A. 6
Explanation: The union \(A\cup B\) contains every element that belongs to at least one of the two sets. Since 6 is an element of \(A\), it must be included in the union. In fact, \(A\cup B=\{3,6,9,12,15\}\). The numbers 18, 21, and 0 are not present in either set, so they cannot belong to the union.
15 If \(A=\{1,2,5,10\}\) and \(B=\{2,4,6,10\}\), which element belongs to \(A-B\)?
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Answer and explanation
Correct answer: A. 1
Explanation: The set difference \(A-B\) includes elements of \(A\) that are absent from \(B\). Here 2 and 10 occur in both sets, so they are excluded. The number 1 occurs in \(A\) but not in \(B\), so it belongs to the difference; 5 also belongs to the difference, although it is not offered. The number 4 is only in \(B\).
16 If \(U=\{1,2,3,4,5,6,7,8\}\), \(A=\{1,2,7\}\), and \(B=\{2,4,6\}\), what is \(A\cup B\)?
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Answer and explanation
Correct answer: A. \(\{1,2,4,6,7\}\)
Explanation: The union combines all distinct elements from \(A\) and \(B\), without adding elements that belong only to the universal set \(U\). Combining \(\{1,2,7\}\) and \(\{2,4,6\}\) gives \(\{1,2,4,6,7\}\); the common element 2 is written once. Elements 3, 5, and 8 are in \(U\) but in neither \(A\) nor \(B\).
17 If \(U=\{a,b,c,d,e,f\}\), \(A=\{a,c,e\}\), and \(B=\{b,c,e,f\}\), what is \(A\cap B\)?
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Answer and explanation
Correct answer: A. \(\{c,e\}\)
Explanation: The intersection \(A\cap B\) contains only elements common to both sets. Comparing \(A=\{a,c,e\}\) with \(B=\{b,c,e,f\}\), the common elements are c and e. Therefore, \(A\cap B=\{c,e\}\). The universal set is not automatically the answer; a letter belongs to the intersection only when it is listed in both A and B.
18 If U = {10, 20, 30, 40, 50}, A = {10, 20, 40}, and B = {20, 30}, what is A − B?
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Answer and explanation
Correct answer: A. {10, 40}
Explanation: The difference A − B contains the elements that are present in A but absent from B. Set A contains 10, 20, and 40, while set B contains 20 and 30. Therefore, remove 20 from A; 10 and 40 remain. Hence, A − B = {10, 40}. The universal set U is not needed for this particular difference operation.
19 If A = {1, 2, 3, 4} and B = {2, 4, 6}, what is A ∩ (A − B)?
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Answer and explanation
Correct answer: A. {1, 3}
Explanation: First calculate the difference A − B. The elements 2 and 4 are common to A and B, so they are removed from A, giving A − B = {1, 3}. This result is already a subset of A. Therefore, intersecting it with A does not change it: A ∩ (A − B) = {1, 3}. Thus option A is the only correct answer.
20 If A = {5, 6, 7} and B = {6, 7, 8, 9}, what is A ∩ (A ∪ B)?
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Answer and explanation
Correct answer: A. {5, 6, 7}
Explanation: The union A ∪ B contains every element from either set, so A ∪ B = {5, 6, 7, 8, 9}. Taking the intersection of this union with A selects the elements that are also in A. Since every element of A is already in A ∪ B, the result is A itself: A ∩ (A ∪ B) = {5, 6, 7}. This illustrates the absorption law.
21 If A = {p, q, r} and B = {q, r, s}, what is A ∪ (A ∩ B)?
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Answer and explanation
Correct answer: A. {p, q, r}
Explanation: The common elements of A and B are q and r, so A ∩ B = {q, r}. These elements are already contained in A = {p, q, r}. Taking the union of A with a subset of A adds nothing new. Therefore, A ∪ (A ∩ B) = A = {p, q, r}. This is the absorption identity A ∪ (A ∩ B) = A.
22 In a class, 22 students like drawing, 16 like music, and 6 like both. How many students like at least one activity?
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Answer and explanation
Correct answer: A. 32
Explanation: Let D be the set of students who like drawing and M the set who like music. Students liking at least one activity are in D ∪ M. By the inclusion-exclusion formula, n(D ∪ M) = n(D) + n(M) − n(D ∩ M) = 22 + 16 − 6 = 32. We subtract the 6 students who like both because they were counted twice.
23 In a group, 30 students like tea, 25 like coffee, and 12 like both. How many students like only tea?
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Answer and explanation
Correct answer: A. 18
Explanation: The number who like only tea is obtained by removing those who like both tea and coffee from the total tea group. Thus, only tea = n(T) − n(T ∩ C) = 30 − 12 = 18. The value 13 represents only coffee, 43 represents the union of the two groups, and 55 incorrectly adds the totals without correcting for the overlap.
24 If A = {x : x ∈ N and x is a factor of 12} and B = {1, 2, 3, 5}, what is A ∩ B?
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Answer and explanation
Correct answer: A. {1, 2, 3}
Explanation: The natural-number factors of 12 are A = {1, 2, 3, 4, 6, 12}. The intersection A ∩ B contains only elements common to this factor set and B = {1, 2, 3, 5}. The common elements are 1, 2, and 3; 5 is not a factor of 12, while 4, 6, and 12 are not in B. Hence, A ∩ B = {1, 2, 3}.
25 If A = {x : x ∈ N and x is a factor of 18} and B = {2, 3, 6, 9}, what is A − B?
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Answer and explanation
Correct answer: A. {1, 18}
Explanation: The natural-number factors of 18 are A = {1, 2, 3, 6, 9, 18}. The difference A − B keeps elements of A that are not in B. Since 2, 3, 6, and 9 are removed, the elements left are 1 and 18. Therefore, A − B = {1, 18}. The operation does not remove elements that are outside A.
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