Operations on Sets (Union, Intersection, Difference)

Correct answer: A. 2

Explanation: The intersection requires numbers that are both perfect squares and multiples of 3. The square numbers from 1 through 36 are 1, 4, 9, 16, 25, and 36. Checking divisibility by 3 leaves 9 and 36 only; the other squares are not multiples of 3. Hence A ∩ B = {9,36}, which has cardinality 2. Therefore option A is correct; counting every square or every multiple of 3 would answer a different question.

Correct answer: A. 58

Explanation: The intersection A ∩ B consists only of elements common to both sets. Therefore, it cannot contain more elements than either set, so n(A ∩ B) ≤ min(73, 58) = 58. This value is attainable if every element of B is also in A, meaning B is a subset of A. Hence the maximum is 58.

Correct answer: A. 91

Explanation: A union contains every element of each participating set, so it must contain the larger set. Therefore, n(A ∪ B) ≥ max[n(A), n(B)] = max(91, 86) = 91. This lower bound is possible if all 86 elements of B lie inside A, meaning B ⊆ A. Then A ∪ B = A and has 91 elements. Option B is too small, while 177 is the sum without accounting for overlap.

Correct answer: A. A

Explanation: The condition A ∩ B = ∅ says that A and B are disjoint, so they have no common elements. The difference A − B removes from A only elements that also belong to B. Because no element of A belongs to B, nothing is removed, and A − B remains A. It is not empty unless A itself is empty; neither B nor the union is generally equal to the difference.

Correct answer: A. {5}

Explanation: To belong to A ∩ B, a number must satisfy both conditions: it must be prime and also a multiple of 5. The only prime multiple of 5 is 5 itself, because every other positive multiple of 5 is divisible by 5 and has at least one additional factor. Therefore A ∩ B = {5}. The universal set ending at 50 does not change this conclusion.

Correct answer: A. 63

Explanation: The intersection A ∩ B contains only elements that belong to both sets, so it cannot contain more elements than either A or B. Consequently, n(A ∩ B) ≤ min(n(A), n(B)) = min(82, 63) = 63. This maximum is possible when every element of B is also an element of A, meaning B is a subset of A. Therefore option A is correct.

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

Explanation: The union A ∪ B is the set of all elements that belong to A or to B or to both. Combining A = {1, 2, 3} and B = {3, 4, 5} gives 1, 2, 3, 4, and 5. The common element 3 is written only once because a set does not repeat elements. Therefore, A ∪ B = {1, 2, 3, 4, 5}.

Correct answer: A. {2, 4}

Explanation: The intersection A ∩ B contains only those elements that are present in both A and B. Checking the elements of A, 2 occurs in B and 4 also occurs in B, whereas 6 and 8 do not. Therefore, the common-element set is A ∩ B = {2, 4}. The other options represent unrelated combinations or an empty intersection.

Correct answer: A. {a, c}

Explanation: The difference A − B consists of elements that belong to A but do not belong to B. In A = {a, b, c, d}, the elements b and d are also in B, so they are removed. The elements a and c are not in B and remain. Hence A − B = {a, c}. Notice that set difference is order-sensitive; B − A would give a different result.

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

Explanation: First convert P from set-builder form into roster form. Taking N as the positive natural numbers, x < 5 gives P = {1, 2, 3, 4}. The union includes every distinct element of P and Q. Since 2 and 4 are already present in P, they are not repeated; adding 6 gives P ∪ Q = {1, 2, 3, 4, 6}.

Correct answer: A. ∅

Explanation: The intersection contains elements common to both sets. The elements of A are 5, 10, and 15, while the elements of B are 20 and 25. There is no number appearing in both lists, so the sets are disjoint. Therefore, their intersection is the empty set: A ∩ B = ∅. The union would contain all five elements, but that is not being asked.

Correct answer: A. \(\{1,2,3,4\}\)

Explanation: The union of sets contains every distinct element that belongs to at least one of the sets. Starting with \(A\cup B\), we get \(\{1,2,3\}\). Taking the union of this result with \(C=\{3,4\}\) adds only 4, because 3 is already present. Therefore, \(A\cup B\cup C=\{1,2,3,4\}\). Repeated elements are written only once in a set, so option D is not acceptable.

Correct answer: A. \(\{3,5\}\)

Explanation: An intersection contains only elements common to every set involved. The elements 3 and 5 occur in \(A\), in \(B\), and in \(C\). The elements 2 and 7 occur only in \(A\), while 1 and 9 occur only in \(B\), and 11 occurs only in \(C\). Hence the common intersection is \(A\cap B\cap C=\{3,5\}\), making option A correct.

Correct answer: A. \(\varnothing\)

Explanation: The difference \(B\setminus A\) consists of elements that belong to \(B\) but do not belong to \(A\). Here, both elements of \(B\), namely 2 and 4, are already elements of \(A\). Therefore, no element remains after removing from \(B\) the elements common with \(A\), so \(B\setminus A=\varnothing\). Option A is correct.

Correct answer: A. \(\{1,2,3,4\}\)

Explanation: The union \(A\cup B\) contains every distinct element found in either \(A\) or \(B\). Set \(A\) contributes 1, 2, and 3; set \(B\) contributes 3 and 4. Since 3 is common, it is listed only once, giving \(A\cup B=\{1,2,3,4\}\). The universal set \(U\) provides the surrounding context but does not mean that all elements of \(U\) belong to the union.

Correct answer: A. \(\{2,6,10\}\)

Explanation: The set difference \(A\setminus B\) keeps elements that are in \(A\) but not in \(B\). From \(A\), the elements 4 and 8 are also present in \(B\), so they must be removed. The elements 2, 6, and 10 are not in \(B\), while 12 is not even in \(A\). Hence \(A\setminus B=\{2,6,10\}\), so option A is correct.

Correct answer: A. \(\{1,2,3\}\)

Explanation: The empty set \(\varnothing\) contains no elements. Therefore, taking the union of \(A\) with the empty set adds nothing to \(A\), and the result remains unchanged: \(A\cup\varnothing=A\). Since \(A=\{1,2,3\}\), the answer is \(\{1,2,3\}\). This is called the identity property of union; it should not be confused with intersection with the empty set, which is empty.

Correct answer: A. \(\varnothing\)

Explanation: An intersection contains elements common to both sets. The empty set \(\varnothing\) has no elements at all, so there cannot be any element common to \(A=\{4,5,6\}\) and \(\varnothing\). Therefore, \(A\cap\varnothing=\varnothing\). Notice that this differs from union with the empty set: \(A\cup\varnothing=A\), whereas intersection with it always produces the empty set.

Correct answer: A. \(\varnothing\)

Explanation: The difference \(A\setminus A\) consists of elements that belong to the first copy of \(A\) but do not belong to the second copy. Since both sets are identical, every element \(p\), \(q\), and \(r\) is removed. No element can remain in one copy while being absent from the other, so \(A\setminus A=\varnothing\). This is a general identity for every set.

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

Explanation: First, form the union: \(A\cup B=\{1,2,3,4,5,6\}\). Next, find the intersection: \(A\cap B=\{3,4\}\). Subtracting the intersection from the union removes the elements common to both sets, leaving \(\{1,2,5,6\}\). Thus option A is correct. This result is also called the symmetric difference because it contains elements belonging to exactly one of the two sets.

Correct answer: A. \(\{a,e,i,o,u\}\)

Explanation: The vowels in English are \(A=\{a,e,i,o,u\}\). The set \(B=\{a,e,i\}\) is a subset of \(A\), so every element of \(B\) is already present in \(A\). Taking the union adds no new element; therefore \(A\cup B=A=\{a,e,i,o,u\}\). Option B is only the smaller subset, not the union.

Correct answer: A. \(\{2,4,6\}\)

Explanation: Assuming the standard school convention \(\mathbb{N}=\{1,2,3,\ldots\}\), we have \(A=\{1,2,3,4,5,6\}\). The even natural numbers not exceeding 8 are \(B=\{2,4,6,8\}\). The common elements are 2, 4, and 6, so \(A\cap B=\{2,4,6\}\). The element 8 is excluded because it is not in A.

Correct answer: A. \(A-B=\{0,1\}\) और \(B-A=\{4,5\}\)

Explanation: For \(A-B\), retain elements of A that are not in B. Since 2 and 3 are common, they are removed from A, giving \(A-B=\{0,1\}\). For \(B-A\), remove 2 and 3 from B, giving \(B-A=\{4,5\}\). Hence option A is correct. This also shows that set difference is generally not commutative.

Correct answer: A. {1, 4, 16}

Explanation: The governing concept is set intersection. The intersection A∩B contains exactly those elements that occur in both A and B, not elements that occur in only one set. Compare the members of A={1,4,9,16} with B={1,2,4,8,16}. The number 1 appears in both, 4 appears in both, and 16 appears in both. The number 9 occurs only in A, while 2 and 8 occur only in B. Therefore A∩B={1,4,16}, so option A is correct. Option B incorrectly selects the element unique to A. Option C lists elements unique to B. Option D contains every element from either set, so it represents the union A∪B rather than the intersection. Repeated elements are written only once in a set.

Correct answer: A. \(\{\text{red},\text{blue},\text{green},\text{yellow}\}\)

Explanation: The union contains every distinct element appearing in either set. From A we take red, blue, and green; from B we add yellow. Blue appears in both sets, but a set lists an element only once. Therefore, \(A\cup B=\{\text{red},\text{blue},\text{green},\text{yellow}\}\). Option B is only the intersection, not the union.