Operations on Sets (Union, Intersection, Difference)

The equality A ∪ B = A means that every element of B is already in A; otherwise the union would contain an additional element. Thus B ⊆ A. Similarly, B ∪ C = C means that every element of B is already in C, so B ⊆ C. Being a subset of both A and C means being a subset of their intersection. Therefore B ⊆ A ∩ C, making option A correct.

A ∩ B contains natural numbers not exceeding 25 that are both odd and divisible by 3. The multiples of 3 up to 25 are 3, 6, 9, 12, 15, 18, 21, and 24. Among these, the odd multiples are 3, 9, 15, and 21. There are four such numbers, so |A ∩ B| = 4. Therefore, option A is correct.

The equality A \ C = B \ C says that A and B contain exactly the same elements outside C. Thus any element of A that is not in B cannot be outside C; otherwise it would appear in A \ C but not in B \ C. Therefore every element of A lies in B ∪ C, which is equivalent to A \ (B ∪ C) = ∅. By the same reasoning, every element of B lies in A ∪ C, so B \ (A ∪ C) = ∅. Hence option A is always true.

The hypothesis says every element of \(A\setminus B\) belongs to the intersection \(A\cap C\). An element of an intersection belongs to both component sets, so every such element must belong to \(C\). Hence, by transitivity of inclusion, \(A\setminus B\subseteq C\). No relation between all of \(A\), \(B\), and \(C\) is forced.

The difference \(A\setminus B\) consists of elements in \(A\) that are not in \(B\). Removing 3 and 9 from \(A\) gives \(A\setminus B=\{1,5,7,11\}\). Now intersect this result with \(C=\{5,9,13\}\). The only common element is 5, so \((A\setminus B)\cap C=\{5\}\). Although 9 belongs to both \(A\) and \(C\), it is removed because it belongs to \(B\).

For any sets \(X\) and \(A\), the equality \(A\cup X=A\) holds exactly when every element of \(X\) is already in \(A\), that is, \(X\subseteq A\). Here \(X=B\cap C\), so the required conclusion is \(B\cap C\subseteq A\). The equality does not require either \(B\) or \(C\) individually to be contained in \(A\).

The union \(B\cup C\) begins at 3 because 3 is included in \(B\). The intervals overlap from 5 onward, and \(C\) continues to values just below 12, so \(B\cup C=[3,12)\). Taking the elements of \(A=[1,10]\) that are not in this union leaves the interval from 1 through values less than 3: \([1,3)\). The point 1 is included, while 3 is excluded because it belongs to \(B\).

The union is partitioned into three disjoint regions: \(A\setminus B\), \(A\cap B\), and \(B\setminus A\). Thus, \(|B\setminus A|=|A\cup B|-|A\setminus B|-|A\cap B|=92-31-27=34\). Set \(B\) consists of \(B\setminus A\) together with \(A\cap B\), so \(|B|=34+27=61\). Therefore, option A is correct.

Because \(B\subseteq C\), any element of \(A\) that lies in \(C\) but not in \(B\) would belong to \(A\setminus B\) but not to \(A\setminus C\), contradicting their equality. Thus every element of \(A\cap C\) is either in \(B\) or outside \(A\), which is written \(A\cap C\subseteq B\cup A^c\).

The inclusion–exclusion formula for three finite sets is \(|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|\). Since the triple intersection is empty, its cardinality is zero, so the final term contributes nothing. Therefore, the correct formula is option A. Pairwise intersections must be subtracted because their elements were counted twice in the initial sum.

The condition \(B\setminus A=\varnothing\) means that no element of \(B\) lies outside \(A\); hence \(B\subseteq A\). The condition \(A\setminus B\ne\varnothing\) means that at least one element of \(A\) is not in \(B\), so \(A\ne B\). Combining these facts gives the proper inclusion \(B\subset A\), making option A correct.

The set B \ C contains elements of B that are not in C. Therefore, A ∩ (B \ C) cannot contain any element of C. Since this set is equal to A ∩ B, no element common to A and B can belong to C. Hence (A ∩ B) ∩ C is empty, or A ∩ B ∩ C = ∅. Thus option A is the necessary conclusion.

The elements of A are the multiples of 4 from 4 through 60, so |A| = 60/4 = 15. An element belongs to both A and B precisely when it is divisible by both 4 and 6, hence by lcm(4,6) = 12. The multiples of 12 up to 60 are 12, 24, 36, 48, and 60, so |A ∩ B| = 5. Therefore |A \ B| = 15 − 5 = 10.

The union A ∪ B contains elements in A, in B, and also elements common to both sets. The symmetric difference A △ B contains only elements belonging to exactly one of the two sets; it excludes A ∩ B. Therefore the two sets can be equal only when there are no common elements. Hence A ∩ B = ∅, so option B is correct.

First find B \ C. The interval B is (−1, 6), while C contains every number from 2 through 8, including 2. Removing C from B leaves (−1, 2), with both endpoints excluded: −1 was already excluded from B and 2 belongs to C. This remaining interval lies completely inside A = [−3, 4], so its intersection with A is (−1, 2). Therefore option C is correct.

The difference \(U-A\) consists of elements that belong to \(U\) but do not belong to \(A\). Starting with \(U=\{2,4,6,8,10,12\}\), remove the elements \(4,8,12\) listed in \(A\). The elements left are \(2,6,10\). Thus, \(U-A=\{2,6,10\}\), making option A correct.

First convert the complement cardinalities: \(|A|=150-92=58\) and \(|B|=150-76=74\). By inclusion-exclusion, \(|A\cup B|=|A|+|B|-|A\cap B|=58+74-31=101\). Therefore, the complement of the union contains \(|U|-|A\cup B|=150-101=49\) elements. Subtracting the intersection avoids counting common elements twice.

Let \(M\) be the set of students studying mathematics and \(B\) the set studying biology. By inclusion–exclusion, \(n(M\cup B)=n(M)+n(B)-n(M\cap B)=82+74-36=120\). The students studying neither subject are outside this union, so their number is \(150-120=30\).