Operations on Sets (Union, Intersection, Difference)

Use the standard decomposition \(B=(B\setminus A)\cup(A\cap B)\). From \(A\cup B=A\cup C\), the parts outside \(A\) are equal: \(B\setminus A=C\setminus A\). The additional condition \(A\cap B=A\cap C\) makes the parts inside \(A\) equal as well. Thus both disjoint components of \(B\) and \(C\) match, so \(B=C\). The other statements do not generally determine equality.

An element of \(A\setminus(B\setminus C)\) is in \(A\), but it is not in \(B\setminus C\). Being outside \(B\setminus C\) means either being outside \(B\), or being in \(C\). Therefore the result consists of elements in \(A\setminus B\) together with elements in \(A\cap C\). Hence \(A\setminus(B\setminus C)=(A\setminus B)\cup(A\cap C)\), making option A correct.

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.

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.

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.

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.

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.

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 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.

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.

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 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 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 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.

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.