Equal sets and Subsets

Correct answer: B. 3

Explanation: The condition A ⊂ X ⊆ B means that X must contain p and q, may contain r and s, and must be different from A. The four possible supersets formed from r and s are {p,q}, {p,q,r}, {p,q,s}, and {p,q,r,s}. Since A itself is not allowed by the proper-subset sign, remove {p,q}. Therefore, 4 − 1 = 3 sets satisfy the condition.

Correct answer: A. \(A\cap B=\emptyset\)

Explanation: The difference \(A\setminus B\) contains the elements of \(A\) that are not in \(B\). If removing the elements of \(B\) from \(A\) leaves all of \(A\) unchanged, then no element of \(A\) can belong to \(B\). Hence the two sets have no common element, so \(A\cap B=\emptyset\). This does not imply that either set is empty or that they are equal; it only establishes that they are disjoint.

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

Explanation: The set \(A\) is itself an element of its power set \(\mathcal{P}(A)\), because every set is a subset of itself. Given \(\mathcal{P}(A)\subseteq\mathcal{P}(B)\), it follows that \(A\in\mathcal{P}(B)\). Membership in \(\mathcal{P}(B)\) means being a subset of \(B\), so \(A\subseteq B\). The other options do not follow from the given power-set inclusion and can fail for ordinary nested sets.

Correct answer: A. A = B

Explanation: The statement A ⊆ B says that every element of A belongs to B. The condition B − A = ∅ says that B has no element outside A; therefore every element of B also belongs to A, so B ⊆ A. Since both A ⊆ B and B ⊆ A hold, the two sets are equal. Thus A = B, and option A is correct. No information requires A to be empty.

Correct answer: A. {2} ∈ A

Explanation: The set A has exactly three elements: 1, the set {2}, and 3. Therefore {2} is an element of A, so option A is true. However, 2 itself is not an element of A; it is inside the nested set {2}. Options C and D are also false because both require 2 to be an element of A, which it is not. This example shows why membership (∈) and subset (⊆) must be distinguished carefully, especially when a set contains another set as one of its elements.

Correct answer: A. Both A ∈ B and A ⊆ B are true

Explanation: Because A = {1,2} appears explicitly as one complete element inside B, A ∈ B is true. Also, the two elements of A, namely 1 and 2, both occur in B. Therefore every element of A belongs to B, so A ⊆ B is also true. Notice the distinction: A is present as a single set-element in B, while 1 and 2 are present individually, which establishes the subset relation. Thus both statements in option A are correct.

Correct answer: A. Both the assertion and the reason are true, and the reason correctly explains the assertion.

Explanation: By definition, X ⊆ A means that every element of X is also an element of A. The empty set has no elements, so it is impossible to find an element of ∅ that lies outside A. Thus the universal condition is automatically true for every set A; this is commonly called vacuous truth. The reason states exactly why the empty set is a subset of every set, so both the assertion and reason are true and the reason correctly explains the assertion.

Correct answer: A. A = B

Explanation: The positive natural-number divisors of 18 are 1, 2, 3, 6, 9, and 18. Therefore, the set-builder description of A gives A = {1, 2, 3, 6, 9, 18}, which is exactly the roster description of B. Since two sets are equal when they contain precisely the same elements, A = B. Neither set is a proper subset of the other, and their intersection is certainly not empty.

Correct answer: B. 8

Explanation: The element 2 is required, so it is fixed as included in every acceptable subset. The element 4 is forbidden, so it is fixed as excluded. The remaining elements 1, 3, and 5 are unrestricted; each may be selected or not selected independently. Thus there are 2 choices for each of three free elements, giving 2³ = 8 valid subsets. Therefore, option B is correct.

Correct answer: B. 7

Explanation: A has 7 elements, and a subset containing exactly 6 elements can be formed by choosing 6 of them. Therefore, the number is C(7,6) = 7. An even simpler view is that each such subset is obtained by omitting exactly one element from A, and there are 7 possible elements to omit. Hence option B is correct. The value 21 is C(7,2), not C(7,6).

Correct answer: B. 6

Explanation: The elements 1 and 2 are required, so they occupy two of the four positions. We must choose the remaining two elements from {3,4,5,6}, which has four elements. Using combinations, the count is C(4,2) = 4!/(2!2!) = 6. Thus option B is correct. Choosing C(4,1), C(5,2), or C(6,2) would not enforce both required elements correctly.

Correct answer: A. 68

Explanation: People with only a mobile phone number 92 − 51 = 41, because those with both devices must be removed. People with only a laptop number 78 − 51 = 27. Exactly one device means only mobile or only laptop, so the required number is 41 + 27 = 68. Therefore, option A is correct.

Correct answer: B. 5

Explanation: Factor the equation as a² - b² = (a - b)(a + b) = 0. Since a and b are both positive elements of {1,2,3,4,5}, the equation a + b = 0 is impossible. Therefore, a - b = 0, so a = b. The valid ordered pairs are (1,1), (2,2), (3,3), (4,4), and (5,5). There are exactly five such pairs, so option B is correct.

Correct answer: B. 3

Explanation: Check the condition separately for each x in A. If x = 1, even the largest sum is 1 + 8 = 9, so no pair qualifies. If x = 3, only y = 8 gives a sum greater than 9, contributing one pair. If x = 5, y = 6 and y = 8 give sums 11 and 13, contributing two pairs. The total is 0 + 1 + 2 = 3, so option B is correct.

Correct answer: A. 6

Explanation: Count the allowable values of \(y\) for each value of \(x\). If \(x=1\), then \(y\ge3\), so \(y=3,4,5\): three pairs. If \(x=2\), then \(y\ge4\), so \(y=4,5\): two pairs. If \(x=3\), then \(y\ge5\), so only \(y=5\): one pair. Hence the total is \(3+2+1=6\), making option A correct.