Operations on Sets (Union, Intersection, Difference)

For all sets, A ∩ B ⊆ A ⊆ A ∪ B. Therefore the equality A ∩ B = A ∪ B ∪ C implies that A ∪ B is also contained in A ∩ B. Since A ∩ B is always contained in A ∪ B, we obtain A ∩ B = A ∪ B, which is possible exactly when A = B. The equality also forces every element of C to lie in A ∩ B, so C ⊆ A ∩ B.

First form the union A ∪ B by listing every element appearing in either set: A ∪ B = {1,2,3,4}. Set difference removes from this result every element that belongs to C. Since 2 and 3 are in C, remove them; 1 and 4 are not in C and remain. Thus (A ∪ B) \ C = {1,4}, so option A is correct.

The union A ∪ B is divided into three pairwise disjoint regions: elements only in A, counted by |A \ B|; elements only in B, counted by |B \ A|; and common elements, counted by |A ∩ B|. Hence 70 = 25 + 30 + |A ∩ B|. Solving gives |A ∩ B| = 70 − 55 = 15, so option A is correct.

Every element of A is either outside B or inside B. Therefore A can be partitioned as A = (A \ B) ∪ (A ∩ B). Both parts are given to be subsets of C, and the union of subsets of C is also a subset of C. Consequently, A ⊆ C. No condition controls elements in B \ A, so B ⊆ C and A ∪ B ⊆ C do not necessarily follow.

The condition A \ B = ∅ means that A has no element outside B; equivalently, A ⊆ B. When A is a subset of B, their union is simply B, so A ∪ B = B. The problem also states A ∪ B = U. Combining these equalities gives B = U. The other statements may occur in special cases but are not forced by the conditions.

An element belongs to the left side exactly when it is in A but not in B, and also in C. Thus it is simultaneously in A and C and outside B. The right side describes precisely the same condition: it is in A and in C \ B, meaning it belongs to C but not to B. Since both sides contain exactly the same elements, the identity is true for all sets A, B, and C.

The difference A \ B contains all elements that belong to A but do not belong to B. Set A includes every real number from -2 through 5, including both endpoints. Set B contains numbers greater than 1 and up to 7, but it does not include 1. Therefore, the part of A removed is (1, 5], while -2 through 1 remains. Hence A \ B = [-2, 1]. The closed endpoint at 1 is important because 1 is not in B.

The elements of A are the multiples of 6 from 1 to 90, so |A| = floor(90/6) = 15. The elements of B are the multiples of 15, so |B| = floor(90/15) = 6. Elements in both sets are multiples of lcm(6, 15) = 30; there are floor(90/30) = 3 such elements. By inclusion-exclusion, |A ∪ B| = |A| + |B| − |A ∩ B| = 15 + 6 − 3 = 18.

An element belongs to (A ∪ B) \ C exactly when it is in A or B and is not in C. This is equivalent to saying that it is either in A \ C or in B \ C. Hence (A ∪ B) \ C = (A \ C) ∪ (B \ C), which is the distributive law of set difference over union. The other statements confuse union and intersection laws.

To find the number of students in at least one of the two sets, use the inclusion-exclusion formula: |M ∪ P| = |M| + |P| − |M ∩ P|. Thus |M ∪ P| = 72 + 64 − 28 = 108. The remaining students belong to neither set, so subtract the union from the total: 120 − 108 = 12. Therefore, 12 students are in neither mathematics nor physics.

There are floor(84/7) = 12 multiples of 7 in U, so |A| = 12. There are floor(84/12) = 7 multiples of 12, so |B| = 7. A number belongs to both sets when it is a multiple of lcm(7, 12). Since 7 and 12 are relatively prime, their least common multiple is 84, giving one common element. Therefore, |A ∪ B| = 12 + 7 − 1 = 18.

To find A \ B, retain elements of A that are not in B. Set B begins just greater than −1, so −1 itself is not in B and remains in the difference. Every number greater than −1 and below 6 belongs to B and must be removed. The left endpoint −4 is included in A, giving A \ B = [-4, −1].

The elements of A are divided into two disjoint parts: those also belonging to B, namely A ∩ B = C, and those not belonging to B, namely A \ B. Removing B from A therefore removes exactly the common part C. Consequently, A \ B = A \ C. The condition A ∪ B = U is not needed for this equality.

The union A ∪ B is divided into three mutually disjoint regions: elements that are only in A, elements that are only in B, and elements common to both sets. These regions have sizes |A \ B| = 17, |B \ A| = 23, and |A ∩ B| = 14. Since they do not overlap with one another, add them directly: |A ∪ B| = 17 + 23 + 14 = 54. Thus option A is correct.

For any sets X and Y, the equality X ∩ Y = X holds exactly when every element of X is also an element of Y; in other words, X ⊆ Y. Here X is A and Y is B ∪ C. Thus every element of A must belong to B or C, so A ⊆ B ∪ C. No reverse inclusion is required.

First form the union A ∪ B by listing every element that occurs in either A or B, without repeating any element. This gives {2, 3, 4, 6, 8, 9, 10, 12, 15}. Now intersect this set with C = {6, 12, 18}; intersection keeps only elements common to both sets. The common elements are 6 and 12, while 18 is absent from A ∪ B. Therefore, (A ∪ B) ∩ C = {6, 12}.

The set A \ B contains elements of A that are not in B. Since |A| = 41 and |A \ B| = 19, the elements common to A and B number |A ∩ B| = 41 − 19 = 22. Apply the inclusion–exclusion formula: |A ∪ B| = |A| + |B| − |A ∩ B|. Hence |A ∪ B| = 41 + 36 − 22 = 55. Subtracting the intersection prevents the common elements from being counted twice.

The symmetric difference A △ B consists of the elements belonging to exactly one of the two sets. The union contains those elements together with the common elements in A ∩ B. Therefore |A ∪ B| = |A △ B| + |A ∩ B|. Substituting the values gives 51 = 34 + |A ∩ B|, so |A ∩ B| = 17.

To prove the first inclusion, take any element x in A ∪ C. Then x belongs either to A or to C. If x ∈ A, the condition A ⊆ B gives x ∈ B, so x ∈ B ∪ D. If x ∈ C, the condition C ⊆ D gives x ∈ D, so again x ∈ B ∪ D. Thus every element of A ∪ C is in B ∪ D, proving A ∪ C ⊆ B ∪ D. The reverse and difference/intersection statements are not guaranteed.

The equality A ∪ B = A means that every element of B is already in A, so B ⊆ A. The condition A \ B = ∅ means that no element of A lies outside B; therefore A ⊆ B. Since each set is a subset of the other, the two sets are equal. Hence A = B, and no claim that either set is empty is justified.