Operations on Sets (Union, Intersection, Difference)

First form the union: A ∪ B = {a, c, e} ∪ {b, e} = {a, b, c, e}. The complement contains the elements of U that are absent from this union. Since d is the only element of U not in A ∪ B, (A ∪ B)' = {d}. Therefore option A is correct. The complement must be taken relative to U.

Complements are taken relative to U. Thus A' = U − A = {2,4,6,8}, and B' = U − B = {1,4,6,7}. Their intersection contains the elements appearing in both lists, namely 4 and 6. Therefore A' ∩ B' = {4,6}, so option A is correct. Option B is A ∩ B, while the larger options confuse union with intersection.

In U, there are 15 multiples of 2 and 6 multiples of 5. Their overlap consists of multiples of lcm(2,5)=10, namely 10, 20, and 30, so there are 3 common elements. By inclusion-exclusion, n(A∪B)=15+6−3=18. Therefore the complement has n((A∪B)')=30−18=12 elements. Hence option A is correct.

Use the inclusion–exclusion identity n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substitution yields n(A ∩ B) = 20 + 15 − 28 = 7. The intersection therefore contains 7 elements. This subtraction prevents the elements common to both sets from being counted twice in the sum n(A) + n(B). Hence option B is correct.

The phrase “only in \(B\)” means that an element must belong to \(B\) and must be excluded from \(A\). Set difference expresses this condition as \(B-A\), or equivalently \(B\cap A'\). In contrast, \(A-B\) is only \(A\), \(A\cap B\) is the common region, and \(A'\cap B'\) is outside both sets. Hence option A is correct.

The set \(A\) is divided into two non-overlapping parts: the elements only in \(A\), represented by \(A-B\), and the common elements \(A\cap B\). Therefore, \(n(A)=n(A-B)+n(A\cap B)\). Rearranging gives \(n(A-B)=18-6=12\). Thus, option B is correct. The value 6 is only the overlap, while 18 includes both parts.

The elements of \(B\) consist of the elements only in \(B\) together with the common elements in \(A\cap B\). Hence, \(n(B)=n(B-A)+n(A\cap B)\). Substituting the given values gives \(n(B-A)=14-5=9\). Therefore, option B is correct. The value 5 counts the overlap, and 14 counts all of \(B\), not only its exclusive region.

The set A includes two parts: the elements only in A and the elements common to both A and B. Thus, n(A) = n(A only) + n(A ∩ B). Rearranging gives n(A only) = n(A) − n(A ∩ B) = 25 − 10 = 15. Therefore, option C is correct. The value 10 represents the common intersection, not the part belonging exclusively to A.

The total number of elements in B includes both the elements only in B and the elements shared with A. Hence, n(B only) = n(B) − n(A ∩ B). Substituting the given values gives 22 − 7 = 15. Therefore, option B is correct. The value 31 would be obtained from the union formula and does not represent the elements exclusively in B.

‘At least one game’ means the union of the cricket and football groups. By inclusion–exclusion, n(C ∪ F) = n(C) + n(F) − n(C ∩ F) = 18 + 12 − 5 = 25. The five students who play both are subtracted once because they were counted in both groups. Hence, option A is correct.

The 16 Hindi readers include the 6 people who read both Hindi and English. To find those who read only Hindi, subtract the intersection from the Hindi set: n(H only) = n(H) − n(H ∩ E) = 16 − 6 = 10. The total of 35 is not needed for this particular calculation. Therefore, option B is correct.

The intersection A ∩ B consists only of the elements that occur in both sets. Comparing A = {1, 2, 3, 4} with B = {3, 4, 5, 6}, the common elements are 3 and 4. Therefore, A ∩ B = {3, 4}. In a Venn diagram, these elements lie in the overlapping region of A and B.

The difference A − B contains every element of A that is not an element of B. In A = {2, 4, 6}, the elements 2 and 4 are also present in B, so they are removed. The element 6 is not present in B and remains. Hence, A − B = {6}. The order matters: B − A would give a different set.

Use the inclusion–exclusion formula for three finite sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substitution gives 10 + 8 + 6 − 3 − 2 − 1 + 0 = 18. Thus, the union contains 18 elements. Pairwise overlaps are subtracted to avoid counting common elements twice.

The governing concept is the union of a subset and its superset. Since A ⊆ B, every element of A is already contained in B. Therefore, forming A ∪ B adds no new element to B, so A ∪ B = B. Hence n(A ∪ B) = n(B) = 12. Option A gives n(A), option B is an irrelevant difference, and option D incorrectly adds the two cardinalities without considering overlap. Thus, option C is correct.

The governing concept is the intersection of a set with a superset. Because A ⊆ B, every element of A is common to A and B. Consequently, A ∩ B = A, and its cardinality is n(A ∩ B) = n(A) = 4. The value 10 describes the larger set B, not the intersection; 6 is a subtraction distractor, and 14 incorrectly adds the cardinalities. Therefore, option A is correct.

Set difference B − A means the elements that belong to B but do not belong to A. The complement A′ represents all elements outside A, so the required region is B ∩ A′. Since intersection is commutative, B ∩ A′ = A′ ∩ B. Option A represents A − B, option C is the union, and option D is the common region. Hence, option B is correct.

The universal set U contains every region shown in the rectangle: the A-only region, the intersection, the B-only region, and the region outside both circles. Hence, n(U) = 9 + 4 + 6 + 5 = 24. Therefore, option C correctly gives the total number of elements in U.

The universal set is partitioned into four disjoint Venn-diagram regions: outside both sets, only A, A ∩ B, and only B. Their cardinalities must add to n(U). Thus, 12 + 18 + n(A ∩ B) + 20 = 70. The known regions total 50, so n(A ∩ B) = 70 − 50 = 20. Therefore, option A is correct; the other values result from incomplete or incorrect addition and subtraction.

The union A ∪ B is made up of three mutually disjoint parts: the A-only region, the common region A ∩ B, and the B-only region. Therefore, its cardinality is the sum of these three parts: 36 = 14 + n(A ∩ B) + 11. Rearranging gives n(A ∩ B) = 36 − 14 − 11 = 11. Hence, option B is correct; 14 and 11 are already given region counts, while 25 is their sum.