Operations on Sets (Union, Intersection, Difference)

The positive perfect squares not exceeding 40 are 1, 4, 9, 16, 25, and 36. Among these, the even numbers are 4, 16, and 36. The intersection requires both conditions simultaneously: being a perfect square and being even. Therefore A ∩ B = {4, 16, 36}. Option C contains odd squares as well and is therefore only A, not the intersection.

The three pairwise intersection conditions show that no element is shared by any two of the sets. Thus A, B, and C are pairwise disjoint. For disjoint finite sets, the cardinality of their union is the sum of their cardinalities, because no element is counted twice. Therefore |A ∪ B ∪ C| = 9 + 11 + 13 = 33, so option A is correct.

For any sets X and Y, X \ Y = ∅ exactly when every element of X belongs to Y, which is equivalent to X ⊆ Y. Applying this fact with X = A and Y = B ∩ C, the given condition implies A ⊆ B ∩ C. It does not require B ∩ C to be contained in A, nor does it imply disjointness or the stated union equality.

First calculate B ∩ C. The elements common to B = {2, 4, 6, 8} and C = {1, 4, 6, 9} are 4 and 6, so B ∩ C = {4, 6}. Set difference A \ (B ∩ C) keeps the elements of A that are not in {4, 6}. Removing 4 and 6 from A leaves {1, 2, 3, 5}. Therefore option A is correct.

The set A \ B contains elements of A that are outside B, whereas A ∩ B contains elements of A that are inside B. An element cannot simultaneously be outside B and inside B. Therefore (A \ B) ∩ (A ∩ B) = ∅, so the two parts are disjoint. Their union is A because every element of A is either in B or not in B.

Consider any element x. If x is not in B, the union equality gives x ∈ A exactly when x ∈ C. If x is in B, the intersection equality gives x ∈ A exactly when x ∈ C. Thus every element belongs to A precisely when it belongs to C, so A = C. The other options do not necessarily follow.

The union A ∪ B consists of the elements in A together with the elements of B that are outside A. Therefore, |A ∪ B| = |A| + |B \ A|. Substituting the given values gives 65 = 39 + |B \ A|, so |B \ A| = 26. The intersection value 18 is consistent but is not needed for this direct calculation.

Solve the first equation: x² − 9 = (x − 3)(x + 3) = 0, so A = {−3, 3}. The second equation is x² − 6x + 9 = (x − 3)² = 0, so B = {3}. A union contains every distinct element appearing in either set; because 3 is repeated, it is written only once. Hence A ∪ B = {−3, 3}.

Because A ⊆ B, every element of A is also an element of B. However, C \ B contains only those elements of C that are not in B. Therefore no element can belong to both A and C \ B. Their intersection is consequently the empty set, ∅, regardless of the particular sets C and B.

The sets A \ B and A ∩ B are always disjoint: an element outside B cannot at the same time be inside B. If two disjoint sets are equal, their common set must be empty. Thus both sides must be empty. In particular, A \ B = ∅ and A ∩ B = ∅ force A = ∅, so option A is the valid situation. The other choices leave a nonempty side or make the two sides different.

Every element of A ∩ B is, by the given inclusion, an element of A \ C. Membership in A \ C means being in A but not being in C. Therefore, no element of A ∩ B can belong to C. Equivalently, the intersection of A ∩ B with C is empty, so A ∩ B ∩ C = ∅. The other statements do not follow from the given inclusion.

Use the basic containment properties of union and intersection. Since A ∩ C is always a subset of A, the equality A ∪ B = A ∩ C gives A ∪ B ⊆ A. Because B ⊆ A ∪ B, it follows that B ⊆ A. Also A ⊆ A ∪ B = A ∩ C, so every element of A lies in C; hence A ⊆ C. Therefore option A is necessary. The other choices require relationships not forced by the equality.

Set A consists of the even numbers from 1 through 50, so it has 50 ÷ 2 = 25 elements. Elements in both A and B must be divisible by both 2 and 5, hence divisible by 10. There are 50 ÷ 10 = 5 such elements. Therefore, A \ B contains the even numbers that are not divisible by 5, and its size is 25 − 5 = 20.

Take any element x ∈ B \ C. Since x ∈ B, it belongs to A ∪ B. The given equality then places x in A ∪ C. But x ∉ C by the definition of B \ C. Therefore x must belong to A. Since every element of B \ C lies in A, we conclude B \ C ⊆ A. The disjointness condition A ∩ B = ∅ is not needed for this particular conclusion, though it is compatible with the data.

First calculate the intersection A ∩ C. The elements common to A and C are a and d, so A ∩ C = {a,d}. Next calculate B \ A, which contains elements present in B but absent from A. Since b and d are in A, they are removed, leaving {f,g}. Taking the union gives {a,d} ∪ {f,g} = {a,d,f,g}. Therefore, option A is correct.

Use the inclusion–exclusion formula for three sets: |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |B ∩ C| − |C ∩ A| + |A ∩ B ∩ C|. Substitution gives 30 + 27 + 25 − 12 − 10 − 8 + 4 = 56. The triple intersection is added back because it was subtracted three times but should be counted once. Hence option A is correct.

Every set can be partitioned into two disjoint parts relative to B: the elements outside B, represented by A \ B, and the elements inside B, represented by A ∩ B. The corresponding two parts of A and C are given to be equal. Therefore, A = (A \ B) ∪ (A ∩ B) and C = (C \ B) ∪ (C ∩ B) are unions of the same parts, so A = C. Thus option A is correct.

The equality A \ (B \ C) = A means that removing B \ C from A removes no element. Therefore, A has no element in common with B \ C, so A ∩ (B \ C) = ∅. If an element belongs to A ∩ B, it cannot be outside C; otherwise it would belong to A ∩ (B \ C), which is impossible. Hence every element of A ∩ B belongs to C, giving A ∩ B ⊆ C. Option A is correct.

The first equality says that within A, the elements belonging to B are exactly the same as those belonging to C. The second equality says that the elements of A outside B are exactly the same as the elements of A outside C. Hence no element of A can belong to exactly one of B and C. The symmetric difference B Δ C consists precisely of elements belonging to one set but not the other. Therefore A ∩ (B Δ C) = ∅, so option A is correct.

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.