Operations on Sets (Union, Intersection, Difference)

Let C be the set of cricket players and F the set of football players. Students who play at least one game belong to C ∪ F. By the inclusion–exclusion formula, n(C ∪ F) = n(C) + n(F) − n(C ∩ F) = 18 + 12 − 5 = 25. The five students who play both were counted twice, so they must be subtracted once.

Let M represent students who like Mathematics and S represent students who like Science. The 20 students in M include the 8 students who like both subjects. To count only Mathematics, remove the overlap: n(M − S) = n(M) − n(M ∩ S) = 20 − 8 = 12. Thus, 12 students like Mathematics but do not like Science.

The difference A − B keeps elements that are in A but not in B. Starting with A = {a, b, d}, remove b because b also belongs to B = {b, c}. The element c is not in A and therefore has no effect, while a and d remain. Thus A − B = {a, d}. The universal set U is only the surrounding reference set.

The difference B − C contains the elements that are in B but not in C. Since B = {3, 4, 5, 6} and C = {5, 6, 7}, removing 5 and 6 from B gives B − C = {3, 4}. Both 3 and 4 are also elements of A, so A ∩ (B − C) = A ∩ {3, 4} = {3, 4}. Therefore, option A is correct.

First find the union: A ∪ B = {1, 2, 5, 8, 9, 11}. The common elements are 5 and 11, so A ∩ B = {5, 11}. Removing these common elements from the union leaves {1, 2, 8, 9}. Thus the expression represents the elements belonging to exactly one of the two sets, and option B is correct.

For two finite sets, the addition rule is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 21 = 16 + 13 − n(A ∩ B). Therefore n(A ∩ B) = 29 − 21 = 8. The subtraction is necessary because elements common to A and B are counted twice when n(A) and n(B) are added.

The 18 students counted in the English group include the 10 students who chose both Hindi and English. To obtain the number who chose English only, remove the overlap: only English = n(English) − n(Hindi ∩ English) = 18 − 10 = 8. Therefore, 8 students chose only English. The Hindi total is not needed for this particular calculation.

The difference A − B contains elements of the first set A that are not present in the second set B. Starting with A = {3, 6, 9, 12, 15}, remove 6 and 12 because they belong to B. The elements 3, 9, and 15 remain; 18 is not in A and therefore cannot be included. Thus A − B = {3, 9, 15}.

The intersection contains only elements common to both sets. Set A contains the even numbers 2, 4, and 6, while set B contains the odd numbers 1, 3, and 5. No number appears in both sets, so the sets are disjoint. Consequently, their intersection has no elements and is the empty set: A ∩ B = ∅.

The union of two sets contains every element that belongs to at least one of the sets, without repeating any element. Here A = {r, s} and B = {r, s, t, u}. Since every element of A is already in B, combining the sets gives A ∪ B = {r, s, t, u}. Thus option C is correct. Option A omits t and u, option B omits r and s, and option D incorrectly represents the empty set.

The intersection of two sets consists only of the elements that are common to both sets. Comparing A = {5, 10, 15, 20} with B = {10, 20}, the common elements are 10 and 20. Therefore A ∩ B = {10, 20}, so option B is correct. The elements 5 and 15 occur only in A, while the empty set would be correct only if the sets had no common element.

The empty set contains no elements, so taking the union of A with the empty set does not add anything to A. The identity property of union is A ∪ ∅ = A. Since A = {11, 22, 33}, the result is {11, 22, 33}. Therefore option B is correct. Notice that ∅ is not the same as {0}; the former has no elements, whereas the latter has one element, namely 0.

An intersection contains elements common to both sets. The empty set has no elements at all, so no element of A can be common to A and ∅. Hence the identity property is A ∩ ∅ = ∅. The correct answer is option C. Option A is A itself, option B is the singleton set containing zero, and option D contains an element that is not present in the empty set.

The difference A − B contains the elements of A that are not in B. In this problem, B is the empty set, which contains no elements to remove from A. Therefore no element is deleted, and A − ∅ = A = {9, 18, 27}. Thus option A is correct. The result would be empty only if every element of A were removed by the second set.

For a difference X − Y, we retain elements that are in X but not in Y. Here the first set is ∅, which has no elements from the beginning. Since there is nothing in the first set to retain or remove, ∅ − A remains ∅, regardless of the elements in A. Therefore option B is correct. It is important not to reverse the order and confuse ∅ − A with A − ∅.

First calculate each difference separately. A − B contains elements in A but not B, so A − B = {1, 2}. Similarly, B − A = {5, 6}, because 5 and 6 are not in A. Their union is therefore {1, 2} ∪ {5, 6} = {1, 2, 5, 6}. The common elements 3 and 4 are excluded, so option A is correct.

Use the inclusion–exclusion formula for two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 14 + 9 − 4 = 19. The intersection is subtracted because its four elements were counted once in n(A) and again in n(B). Therefore option A, 19, is correct.

For two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearrange the formula to find the intersection: n(A ∩ B) = n(A) + n(B) − n(A ∪ B). Substitution gives 12 + 11 − 20 = 3. Thus the two sets have three common elements, and option C is the unique correct answer.

The set A is divided into two disjoint parts: the elements in A ∩ B and the elements in A − B. Therefore n(A) = n(A ∩ B) + n(A − B). Rearranging gives n(A − B) = 18 − 7 = 11. Hence option C is correct. The value 7 counts the common elements, while 18 counts all elements of A, so neither is the required difference.

The difference B − A contains those elements that belong to B but do not belong to A. The intersection A ∩ B contains the 6 elements common to both sets. Since B has 15 elements in total, remove its 6 common elements: n(B − A) = n(B) − n(A ∩ B) = 15 − 6 = 9. Therefore, option A is correct.