Complement of a Set and Its Properties

Correct answer: B. {1, 3}

Explanation: The complement of A is found relative to the universal set U: A′ = U − A. Removing 2 and 4 from U gives A′ = {1, 3, 5}. A set is a subset when every one of its elements belongs to the larger set. Thus {1, 3} is a subset of {1, 3, 5}. The other choices contain 2 or 4, which are members of A and hence cannot belong to A′.

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

Explanation: Complements reverse the direction of inclusion. Since \(A\subseteq B\), every element outside \(B\) is certainly outside \(A\) as well. Thus, if \(x\in B^c\), then \(x\notin B\), which implies \(x\notin A\), so \(x\in A^c\). Therefore \(B^c\subseteq A^c\), making option A correct. Option B reverses this result, while option C requires \(A=B\). Option D is not generally true; it would impose an additional condition on the universal set.

Correct answer: A. {1, 4, 6, 8, 9, 10}

Explanation: Assuming N = {1,2,3,...}, the universal set is U = {1,2,3,4,5,6,7,8,9,10}. The primes in U are A = {2,3,5,7}. The complement A′ contains every member of U that is not prime, giving {1,4,6,8,9,10}. Number 1 is neither prime nor composite, so it correctly belongs to the complement. Option B is A itself, not A′.

Correct answer: B. B = {1, 3, 5}

Explanation: The complement of A relative to U is written A′ = U \ A. It contains every element of the universal set U that is not an element of A. Removing 0, 2, and 4 from U leaves 1, 3, and 5. Therefore A′ = {1, 3, 5}, which is exactly B. A itself contains the removed elements, U contains too many elements, and the empty set contains none.

Correct answer: A. A ⊆ U is false

Explanation: A complement relative to U is normally introduced for a set A that is contained in U. Here A contains d, but d is not an element of U; instead, U contains e, which is not in A. Thus A is not a subset of U, so the stated complement setup is invalid under the usual school-level definition. The other options are false: U is not a subset of A, A is not equal to U, and A is not empty.

Correct answer: B. 17

Explanation: To count students in A ∪ B, use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus, n(A ∪ B) = 18 + 22 − 7 = 33. The universal set has 50 students, so those outside the union number 50 − 33 = 17. The intersection is subtracted because students belonging to both sets were counted twice.

Correct answer: B. A = U

Explanation: The complement A′ consists of all elements of the universal set U that are not in A. If A′ = ∅, there is no element of U outside A. Since A is already a subset of U, it must contain every element of U. Therefore, A = U. The condition does not imply that either set is empty; it means A covers the whole universal set.

Correct answer: B. {1, 2, 4, 5, 7, 8, 10, 11}

Explanation: Taking N here as the positive natural numbers, U = {1,2,3,4,5,6,7,8,9,10,11,12}. The multiples of 3 in U are A = {3,6,9,12}. The complement A' contains every element of U that is not a multiple of 3, namely {1,2,4,5,7,8,10,11}. Therefore option B is correct; it has 12 − 4 = 8 elements.

Correct answer: A. {6}

Explanation: Complements must be taken relative to the universal set U. Thus A' = U − A = {4,5,6}, while B' = U − B = {1,2,6}. The elements common to both complements are found by intersection: {4,5,6} ∩ {1,2,6} = {6}. Therefore, A' ∩ B' = {6}. This also agrees with De Morgan's law: A' ∩ B' = (A ∪ B)'.

Correct answer: B. {a, c, d}

Explanation: First find the intersection of A and B. The only element common to A = {a,b} and B = {b,c} is b, so A ∩ B = {b}. The complement is taken relative to U = {a,b,c,d}; therefore, (A ∩ B)' consists of every element of U except b. Hence, (A ∩ B)' = {a,c,d}, which is option B.

Correct answer: C. 14

Explanation: Taking ℕ as the positive natural numbers, U = {1,2,...,20}, so n(U) = 20. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20, giving n(A) = 6. The complement A′ contains the members of U that are not divisors of 20. Hence n(A′) = 20 − 6 = 14, so option C is correct.

Correct answer: D. 12

Explanation: Assuming N denotes the positive natural numbers, U = {1,2,3,...,15}, so n(U) = 15. The multiples of 5 that lie in U are A = {5,10,15}, giving n(A) = 3. Since A′ contains the elements of U outside A, n(A′) = n(U) − n(A) = 15 − 3 = 12. Thus option D is correct.

Correct answer: B. \(A'\cup B'\)

Explanation: De Morgan’s law states that the complement of an intersection is the union of the complements: \((A\cap B)'=A'\cup B'\). An element is outside \(A\cap B\) whenever it fails to belong to at least one of the two sets. Therefore it belongs to \(A'\cup B'\). Option B is correct; option A incorrectly uses intersection.

Correct answer: B. \(A=U\)

Explanation: The complement is defined by \(A'=U\setminus A\). If \(A'=\varnothing\), then there is no element of \(U\) outside \(A\); hence every element of \(U\) belongs to \(A\), so \(U\subseteq A\). Since the question already gives \(A\subseteq U\), both inclusions imply \(A=U\). Therefore option B is correct.

Correct answer: C. 14

Explanation: The universal set U contains the integers 1 through 18, so it has 18 elements. The elements divisible by 4 are A = {4, 8, 12, 16}, giving |A| = 4. The complement A′ contains the elements of U that are not in A. Hence |A′| = |U| − |A| = 18 − 4 = 14, so option C is correct.

Correct answer: B. {p,q,s,t}

Explanation: First find the intersection. The only element common to A = {p,q,r} and B = {r,s} is r, so A ∩ B = {r}. The complement is taken in U, not in an unrestricted universe. Removing r from U = {p,q,r,s,t} leaves {p,q,s,t}. Hence (A ∩ B)′ = {p,q,s,t}, so option B is correct.

Correct answer: B. B' ⊆ A'

Explanation: Because A is a subset of B, every element of A is also an element of B. Now take any element x in B'. It is not in B. Since every element of A must lie in B, x cannot be in A either; therefore x belongs to A'. Hence B' ⊆ A'. This is the complement rule that reverses inclusion. Equality is not guaranteed unless A = B.

Correct answer: A. {4, 8, 9, 10}

Explanation: First find the union of A and B by listing every element appearing in either set: A ∪ B = {1, 2, 3, 5, 6, 7}. The complement is taken relative to U, so remove these union elements from U. The elements left are {4, 8, 9, 10}. Hence option A is correct. Option B is the union itself, while option C is the intersection, not the complement.

Correct answer: C. 22

Explanation: The universal set U contains the natural numbers 1 through 30, so n(U) = 30. The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30, giving n(A) = 8. The complement A' contains all elements of U that are not in A. Thus n(A') = n(U) − n(A) = 30 − 8 = 22. Hence option C is correct.

Correct answer: C. 5

Explanation: Use the complement-counting principle |A′| = |U| − |A|. The integers from −3 through 4 are U = {−3, −2, −1, 0, 1, 2, 3, 4}, so |U| = 8. The condition x² < 4 is equivalent to −2 < x < 2, giving A = {−1, 0, 1} and |A| = 3. Therefore |A′| = 8 − 3 = 5, so option C is correct.

Correct answer: A. ∅

Explanation: First find the union. A contains a, c, and e, while B contains b, c, and d. Together they contain every element of U, so A ∪ B = {a,b,c,d,e} = U. The complement is defined relative to U. Therefore (A ∪ B)′ = U \ (A ∪ B) = U \ U = ∅. Option B is not correct because c is already included in both A and B, so it cannot remain in the complement.

Correct answer: B. 4

Explanation: Use the inclusion–exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Therefore, |A ∪ B| = 12 + 9 − 5 = 16. The complement contains the elements of U that are not in A ∪ B, so |(A ∪ B)'| = |U| − |A ∪ B| = 20 − 16 = 4. Hence, option B is correct. Subtracting the intersection is essential because common elements would otherwise be counted twice.

Correct answer: A. {1,3,5,6,7}

Explanation: First find the intersection: A∩B contains the elements common to both sets, so A∩B={2,4}. The prime symbol denotes complement relative to the stated universal set U. Therefore (A∩B)'=U\(A∩B)={1,2,3,4,5,6,7}\{2,4}={1,3,5,6,7}. Option B is only the intersection, not its complement; the other options omit elements that must remain in the complement.

Correct answer: C. 4

Explanation: The union contains every element appearing in A or B: A∪B={2,3,4,6,8,9,10,12}. It has 8 elements. Since the universal set U has 12 elements, its complement has 12−8=4 elements, namely {1,5,7,11}. The same result follows from inclusion–exclusion: n(A∪B)=6+4−2=8 because 6 and 12 are counted in both sets.

Correct answer: B. Students studying neither subject

Explanation: A' consists of students who are not in A, so they do not study Mathematics. Similarly, B' consists of students who do not study Physics. Their intersection A'∩B' contains students satisfying both conditions simultaneously: they study neither Mathematics nor Physics. By De Morgan's law, A'∩B'=(A∪B)', which confirms that these are students outside the group studying at least one of the two subjects.