Operations on Sets (Union, Intersection, Difference)

Correct answer: C. ∅

Explanation: The intersection A ∩ B contains only the elements that are present in both A and B. Set A contains 1, 2, and 3, while set B contains 4 and 5. There is no common element between the two sets. Therefore, A and B are disjoint sets, and their intersection is the empty set, written as ∅. Hence, option C is correct.

Correct answer: B. 20

Explanation: For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since A ∩ B = ∅, its cardinality is zero. Thus n(A ∪ B) = 9 + 11 − 0 = 20. Equivalently, because the sets are disjoint, no element is counted twice, so their cardinalities can be added directly. Therefore, option B is correct.

Correct answer: A. {a, e}

Explanation: The vowels in the English alphabet are A = {a, e, i, o, u}. Set B contains {a, b, c, d, e}. The intersection contains only letters present in both sets. The common letters are a and e; i, o, and u are vowels but do not belong to B. Therefore, A ∩ B = {a, e}, making option A correct.

Correct answer: B. 6

Explanation: The union contains every distinct element appearing in either set. Combining A and B gives A ∪ B = {2, 3, 5, 7, 11, 13}, which has six elements. The same result follows from n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 4 + 4 − 2 = 6, because 5 and 7 are common and must not be counted twice. Hence, option B is correct.

Correct answer: C. \(U\)

Explanation: The complement \(A'\) consists of all elements of the universal set \(U\) that are not in \(A\). Therefore, every element of \(U\) lies either in \(A\) or in \(A'\), and hence belongs to their union. Consequently, \(A\cup A'=U\). This is one of De Morgan’s basic complement laws; in contrast, \(A\cap A'=\varnothing\).

Correct answer: D. \(\varnothing\)

Explanation: By definition, \(A'\) contains exactly those elements of \(U\) that are not in \(A\). Thus no element can belong to both \(A\) and \(A'\) simultaneously. Their common part is therefore empty, giving \(A\cap A'=\varnothing\). This conclusion remains valid for every set \(A\) relative to the same universal set \(U\), including when \(A\) itself is empty or equal to \(U\).

Correct answer: B. 9

Explanation: Using the convention \(\mathbb N=\{1,2,3,\ldots\}\), the set \(A=\{1,2,\ldots,12\}\) has 12 elements. The perfect squares not exceeding 12 are \(B=\{1,4,9\}\), which has 3 elements. Removing these three elements from \(A\) leaves \(|A-B|=12-3=9\). Thus option B is correct; option C would incorrectly ignore the subtraction.

Correct answer: C. A − B = B − A

Explanation: Union and intersection are commutative operations, so changing the order of the sets does not change the result: A ∪ B = B ∪ A and A ∩ B = B ∩ A. Also, the union of any set with the empty set is the set itself. However, set difference depends on order. In general, A − B contains elements of A that are not in B, whereas B − A contains elements of B that are not in A. Therefore, A − B = B − A is generally false, although it may happen for special sets such as A = B.

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

Explanation: The union of two sets contains every element that belongs to at least one of them. The empty set ∅ contains no elements, so it contributes nothing when forming a union. Consequently, A ∪ ∅ = A. Since A = {1, 2, 3}, the result is {1, 2, 3}. Option A would be the result of confusing union with intersection in this situation, while options C and D introduce or omit elements without justification. This property is called the identity property of union.

Correct answer: C. ∅

Explanation: The intersection of two sets consists of elements common to both sets. Although A contains 1, 2, and 3, the empty set ∅ contains no elements at all. Therefore, there cannot be any element common to A and ∅, and A ∩ ∅ = ∅. Options A and B incorrectly treat intersection like union, while option D invents an element that is not present in either set. This is the empty-set property of intersection and is valid for every set A.

Correct answer: A. 17

Explanation: Let H be the set of students studying Hindi and E the set studying English. Students counted in H ∩ E study both languages, so they must be removed from the total Hindi group to find only Hindi: n(H − E) = n(H) − n(H ∩ E) = 24 − 7 = 17. Therefore, 17 students study only Hindi. The value 11 is the number studying only English, 7 is the number studying both, and 35 is an incorrect addition that double-counts the overlap.

Correct answer: A. 5; A ∩ B = {6, 12, 18, 24, 30}

Explanation: The intersection A ∩ B contains numbers that belong to both sets, so each number must be divisible by both 2 and 3. Such numbers are multiples of lcm(2, 3) = 6. The multiples of 6 from 1 through 30 are 6, 12, 18, 24, and 30. Therefore, A ∩ B has 5 elements, so option A is correct.

Correct answer: A. 34

Explanation: Disjoint sets have no common elements, so A ∩ B = ∅ and n(A ∩ B) = 0. The general formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives n(A ∪ B) = 15 + 19 − 0 = 34. Thus every element from both sets is counted exactly once in the union.

Correct answer: A. B ⊆ A

Explanation: The equality A ∪ B = A means that adding every element of B to A introduces no new element. Therefore, each element of B must already belong to A, which is exactly the definition of B ⊆ A. The sets need not be equal, disjoint, or complements; B may be a proper subset of A or may equal A.

Correct answer: A. \(B\subseteq A\)

Explanation: The intersection \(A\cap B\) contains elements common to both sets. If this intersection is exactly \(B\), then every element of \(B\) must also belong to \(A\). Therefore, \(B\) is a subset of \(A\), written as \(B\subseteq A\). The statement does not necessarily imply that every element of \(A\) belongs to \(B\), so \(A\subseteq B\) is not required.

Correct answer: A. \(\{3,5,9\}\)

Explanation: A three-set intersection contains only elements that occur in all three sets. The elements 3, 5, and 9 appear in \(A\), \(B\), and \(C\). Every other listed element is absent from at least one set. Hence \(A\cap B\cap C=\{3,5,9\}\), making option A correct.

Correct answer: A. \((A,\varnothing,A)\)

Explanation: The empty set has no elements. Therefore, taking the union of \(A\) with the empty set adds nothing, so \(A\cup\varnothing=A\). Their intersection has no common element, so \(A\cap\varnothing=\varnothing\). Subtracting the empty set removes nothing, giving \(A-\varnothing=A\). Hence option A is correct.

Correct answer: A. \(\{7,11,13\}\)

Explanation: Set \(B\) is formed by selecting from \(A\) only those elements strictly greater than 5. Testing the elements gives \(B=\{7,11,13\}\); the value 5 is excluded because 5 is not greater than 5. Since \(B\subseteq A\), the intersection \(A\cap B\) equals \(B\), namely \(\{7,11,13\}\).

Correct answer: A. \(\{1,2,3,4,6,12\}\)

Explanation: The positive divisors of 36 are \(\{1,2,3,4,6,9,12,18,36\}\), while those of 48 are \(\{1,2,3,4,6,8,12,16,24,48\}\). The common elements are therefore \(1,2,3,4,6,12\). Equivalently, common divisors are the divisors of \(\gcd(36,48)=12\), giving option A.

Correct answer: A. \(\{1,2,3,4,6,8,9,12,18,24\}\)

Explanation: The positive divisors of 18 are \(\{1,2,3,6,9,18\}\), and the positive divisors of 24 are \(\{1,2,3,4,6,8,12,24\}\). A union contains every element belonging to at least one set, with repeated elements written only once. Combining these lists gives \(\{1,2,3,4,6,8,9,12,18,24\}\), so option A is correct.

Correct answer: A. \(\{1,4,7,8\}\)

Explanation: First find the union of A and B by listing every element that occurs in either set: \(A\cup B=\{1,2,3,4,5,6,7,8\}\). Next, take the intersection with C, which means retain only the elements common to this union and C. Since 1, 4, 7, and 8 all belong to the union, the result is \(\{1,4,7,8\}\). Therefore, option A is correct; option B wrongly omits 8.

Correct answer: A. \(B-A\)

Explanation: The union \(A\cup B\) contains every element of A and every element of B. When A is removed from this union, all elements belonging to A disappear, including those that may also be in B. The elements left are precisely the elements that belong to B but do not belong to A. Therefore, \((A\cup B)-A=B-A\), so option A is correct. This is a standard identity involving union and set difference.

Correct answer: A. \(\varnothing\)

Explanation: To determine A, solve \(x^2-1=0\), which factors as \((x-1)(x+1)=0\). Thus, for integer x, \(x=1\) or \(x=-1\), so \(A=\{-1,1\}\). The condition defining B is already \(x^2=1\), giving the same set \(B=\{-1,1\}\). Since every element of A is also in B, no element remains after subtracting B from A. Hence \(A-B=\varnothing\), so option A is correct.

Correct answer: A. 44

Explanation: Every element of A belongs either to A − B or to A ∩ B. These two parts are disjoint: an element cannot be outside B and simultaneously inside B. Thus A is partitioned into the two given parts, so n(A) = n(A − B) + n(A ∩ B) = 28 + 16 = 44. Therefore, option A is correct.

Correct answer: A. \(\{4,5\}\)

Explanation: The condition defining \(A\) gives all integers from \(-3\) through \(5\), so \(A=\{-3,-2,-1,0,1,2,3,4,5\}\). For \(B\), the inequality \(x^2\le9\) is equivalent to \(|x|\le3\), or \(-3\le x\le3\). Therefore, \(B=\{-3,-2,-1,0,1,2,3\}\). The difference \(A\setminus B\) consists only of elements in \(A\) that are not in \(B\). Removing the seven elements of \(B\) from \(A\) leaves \(\{4,5\}\), so option A is correct. Option C is wrong because \(-3\in B\).