Operations on Sets (Union, Intersection, Difference)

The people counted in T include both those who drink only tea and those who drink both tea and coffee. Therefore, the number who drink only tea is |T \ C| = |T| − |T ∩ C| = 88 − 39 = 49. The total number surveyed and the coffee-only count are not needed for this particular question. Hence option A is correct.

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 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 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 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.

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 \(150-120=30\).