Operations on Sets (Union, Intersection, Difference)

The intersection A ∩ B consists of elements present in both sets: {3, 9}. The union A ∪ B contains every element appearing in either set: {1, 3, 5, 6, 7, 9}. Among the choices, 5 is in A and therefore in the union, but it is not in B, so it is not in the intersection. Thus 5 is correct.

Every element of A ∩ B belongs to both A and B. Since the union A ∪ B contains every element that belongs to at least one of the two sets, every element of the intersection must also be in the union. Therefore A ∩ B is always a subset of A ∪ B, regardless of whether the sets are equal, disjoint, finite, or empty.

An element belongs to A ∩ B ∩ C only if it appears in all three sets. First, the common elements of A and B are {2, 4}. Checking these against C = {4, 8, 12}, only 4 is present; 2 is absent. Although 8 is in B and C, it is not in A. Thus the three-set intersection is {4}, so option B is correct.

First calculate the union: A ∪ B = {m, n, o, p, q}. The difference (A ∪ B) \ A removes every element that belongs to A, namely m, n, and o. The remaining elements are p and q. Equivalently, since A is already included in the union, (A ∪ B) \ A = B \ A. Therefore the answer is {p, q}, which is option A; the other choices retain elements that should be removed.

The union contains all distinct elements: A ∪ B = {1, 2, 3, 4, 5, 6, 7}. Subtracting B removes 4, 5, 6, and 7 from this union, leaving {1, 2, 3}. This is also the identity (A ∪ B) \ B = A \ B, because all elements of B are removed. Therefore option A is correct; option C gives B \ A instead, while D is the un-subtracted union.

The union is divided into three mutually disjoint regions: elements only in A, elements in both A and B, and elements only in B. Their cardinalities are 9, 6, and 11 respectively. Since these regions together contain every element of A ∪ B, add them: n(A ∪ B) = 9 + 6 + 11 = 26. Therefore option C is correct. Adding only two regions would omit one part of the union.

The 45 people who like tea include the 20 people who like both tea and coffee. To find those who like tea only, subtract the intersection from the tea group: 45 − 20 = 25. The number 18 represents coffee only, while 63 represents the total who like at least one beverage, not tea only. Therefore the answer is 25.

The 24 football players include the 12 students who play both football and cricket. Therefore, the number who play football only is found by subtracting the intersection: 24 − 12 = 12. The value 20 is the number who play cricket only, calculated as 32 − 12. Thus only football corresponds to option A.

Set B is partitioned into two disjoint parts: the elements of B that are outside A, namely B \ A, and the elements common to both sets, namely A ∩ B. Thus n(B) = n(B \ A) + n(A ∩ B). Substituting the given values gives 35 = 24 + n(A ∩ B), so n(A ∩ B) = 35 − 24 = 11. Hence option A is correct; the union and n(A) values are unnecessary for this calculation.

The set B is the disjoint union of B \ A and A ∩ B: every element of B is either outside A or common to both sets. Therefore n(B) = n(B \ A) + n(A ∩ B). Using n(B) = 35 and n(B \ A) = 24, we get 35 = 24 + n(A ∩ B), so n(A ∩ B) = 11. Option A is correct. The value 24 describes only B \ A, not the intersection.

The intersection contains only elements common to both sets. Comparing A and B gives A ∩ B = {2, 3}. A set is a subset when every one of its elements belongs to the given set. Since 2 belongs to {2, 3}, {2} is a subset. The other options contain 4 or elements not common to both sets.

The governing concepts are set difference and intersection. In B \ C, remove from B every element that also occurs in C. Since 3 and 5 occur in C, B \ C = {2, 7}. Now intersect this result with A, retaining only elements common to both sets. The element 2 belongs to A, but 7 does not. Therefore A ∩ (B \ C) = {2}, so option A is correct; option D incorrectly keeps 7.

Use the definitions of union and intersection systematically. First, A ∪ B = {a, b, c, d, e}, because all distinct elements from both sets are included. Next, B ∪ C = {b, d, e, f}. The common elements in these two unions are b, d, and e. Hence (A ∪ B) ∩ (B ∪ C) = {b, d, e}. Option A is correct; option D is the full union, not the intersection.

The operation inside parentheses is performed first. The elements common to A and B are 2, 4, and 8, so A ∩ B = {2, 4, 8}. Set difference then removes from this result every element found in C. Because 4 and 8 are in C, they are deleted, while 2 remains because it is not in C. Thus (A ∩ B) \ C = {2}, making option A correct.

A is the set of natural numbers from 1 through 12. First form the union B ∪ C, which contains every element appearing in either set: {2, 3, 4, 6, 8, 9, 10, 12}. Set difference A \ (B ∪ C) keeps elements of A that are absent from this union. Checking 1 through 12 leaves 1, 5, 7, and 11. Therefore option A is correct.

The condition 2 divides x gives A = {2, 4, 6, 8, 10, 12, 14, 16}. The condition 4 divides x gives B = {4, 8, 12, 16}. In A \ B, remove the multiples of 4 from all even numbers. The remaining even numbers are 2, 6, 10, and 14. Thus option A is correct; option B lists the removed set, while option C is all of A.

The equality A ∪ B = A means that adding B to A introduces no new elements. Consequently, every element of B must already belong to A, which is precisely B ⊆ A. The second condition, A ∩ B = B, gives the same conclusion because every element of B is common to A and B. The sets need not be equal or empty, so option A is the only valid relation.

The union is partitioned into three disjoint regions: elements only in A, counted by n(A \ B); elements only in B, counted by n(B \ A); and common elements, counted by n(A ∩ B). Therefore n(A ∪ B) = 18 + 14 + 9 = 41. The common part is added once only, so option C is correct. Options A and B omit a region, while D overcounts.

First find A \ B by removing every element of B from A. This gives A \ B = {1, 3, 5, 7}, because 2, 4, and 6 are removed while 8 was not in A. Next intersect this result with C = {1, 4, 7, 10}. The elements common to both sets are 1 and 7, so (A \ B) ∩ C = {1, 7}. Option C stops after the difference and does not perform the intersection.

Because x is an integer and −3 ≤ x ≤ 2, A = {−3, −2, −1, 0, 1, 2}. The condition x² ≤ 4 means −2 ≤ x ≤ 2, so B = {−2, −1, 0, 1, 2}. Every element of B is already in A. Therefore their union contains all distinct elements of A and B, which is A ∪ B = {−3, −2, −1, 0, 1, 2}. Option B gives only B, not the union.