Operations on Sets (Union, Intersection, Difference)

The union B ∪ C contains every Venn-diagram region that lies in B or in C. These regions are only B, only C, only A ∩ B, only B ∩ C, only C ∩ A, and the triple intersection. Therefore, n(B ∪ C) = 23 + 17 + 8 + 10 + 6 + 5 = 69. The only region excluded is the part belonging exclusively to A, which is 19. Hence option C is correct.

The union A ∪ B is divided into three disjoint regions: the elements only in A, counted by n(A − B) = 47; the elements only in B, counted by n(B − A) = 39; and the common elements, counted by n(A ∩ B). Therefore, 126 = 47 + 39 + n(A ∩ B), so n(A ∩ B) = 126 − 86 = 40. Hence, option A is correct.

The complement Bᶜ consists of all elements in the universal set U that are not in B. Therefore, its cardinality is found by subtracting the number of elements in B from the number of elements in U: n(Bᶜ) = n(U) − n(B) = 130 − 91 = 39. The information A ⊆ B and n(A) = 38 is not needed for this calculation. Thus, option A is correct.

The intersection \(A\cap B\) contains numbers that are multiples of both 2 and 3. Such numbers are multiples of \(\operatorname{LCM}(2,3)=6\). Within the universal set from 1 to 30, they are 6, 12, 18, 24, and 30. There are therefore five elements, so \(n(A\cap B)=5\). Option A is correct; the other choices do not count the common multiples accurately.

There are \(\lfloor40/4\rfloor=10\) multiples of 4 and \(\lfloor40/5\rfloor=8\) multiples of 5. Numbers counted in both sets are multiples of \(\operatorname{LCM}(4,5)=20\), namely 20 and 40, so there are 2 common elements. By inclusion-exclusion, \(n(A\cup B)=10+8-2=16\). Therefore, option B is correct.

A prime number has exactly two positive divisors: 1 and itself. Every even number is divisible by 2. The number 2 is the only even number that is prime, because every other even number has at least the divisors 1, 2, and itself. Therefore, A ∩ B = {2}. Note that 1 is not prime.

A number in all three sets must be divisible by 2, 3, and 5. Since these factors are pairwise coprime, their least common multiple is \(\operatorname{LCM}(2,3,5)=30\). The multiples of 30 in \(\{1,\ldots,60\}\) are 30 and 60, so the intersection contains two elements. Hence, \(n(A\cap B\cap C)=2\), making option B correct.

The quantities n(A − B), n(B − C), and n(C − A) describe only three difference regions. They do not reveal the sizes of all pairwise intersections, the triple intersection, or portions such as A ∩ B but not C. Consequently, the total size of A ∪ B ∪ C cannot always be determined from these three numbers alone. Options B, C, and D are valid definitions.

The set A is divided into two disjoint parts: A − B and A ∩ B. Therefore n(A) = n(A − B) + n(A ∩ B). Substituting the given values gives 48 = 17 + n(A ∩ B), so n(A ∩ B) = 31. This is confirmed independently from B: 52 − 21 = 31. Thus option A is correct; the other values result from subtracting the wrong region or adding unrelated parts.

Partition the union into three non-overlapping groups: exactly one set, exactly two sets, and all three sets. Let x be the number in exactly two sets. Then 128 = 63 + x + 11, because each element is counted once in these exclusive categories. Solving gives x = 128 − 63 − 11 = 54. The triple-intersection is subtracted only once; subtracting it twice would incorrectly produce 43.

Use the distributive law for sets: \((A\cap B)\cup(A\cap B^c)=A\cap(B\cup B^c)\). A set and its complement together form the universal set, so \(B\cup B^c=U\). Hence the expression becomes \(A\cap U=A\). In a Venn diagram, the two terms divide A into its parts inside and outside B, so option A is the only correct answer.

The intersection A ∩ B is the region shared by both sets. Removing this common region from A leaves only those elements that belong to A but do not belong to B. By the definition of set difference, this remaining region is A \ B. In a two-circle Venn diagram, it is the part of circle A outside the overlap. Therefore, option A is the only correct answer.

Because A is a subset of C, every element of A lies inside C. The difference C − A therefore contains all elements of C except the 37 elements belonging to A. Hence n(C − A) = n(C) − n(A) = 132 − 37 = 95. The value 48 would come from subtracting B from C, but the question specifically asks for C − A. The nesting information confirms that the subtraction is valid.

The elements in A that also lie in B or C are counted by n(A ∩ (B ∪ C)). Using the two-set union rule inside A, n(A ∩ (B ∪ C)) = n(A ∩ B) + n(A ∩ C) − n(A ∩ B ∩ C) = 40 + 37 − 16 = 61. The triple intersection is subtracted once because it appears in both pairwise intersections. Hence, elements only in A = n(A) − 61 = 94 − 61 = 33. Therefore, option C is correct.

A number in A ∩ B must be divisible by both 2 and 3. Such numbers are multiples of lcm(2, 3) = 6. Within U, the multiples are 6, 12, 18, 24, 30, and 36. There are 36 ÷ 6 = 6 such numbers, so n(A ∩ B) = 6. The other choices count a different set or do not use the common condition.

The intersection A ∩ B contains numbers that satisfy both conditions: they must be prime and even. Every even number greater than 2 has at least the divisors 1, 2, and itself, so it is composite. The number 2 is the only even prime. Therefore A ∩ B = {2}, making option A correct; option C contains many composite even numbers.

The meaning of set difference is fixed: A − B contains elements in A but not B, and similarly for the other differences, so B, C, and D are always true. However, the three given counts do not reveal the pairwise overlaps, the triple overlap, or several exclusive regions. These unknown parts can change the union, so n(A ∪ B ∪ C) cannot always be determined. Therefore A is correct.

By De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. First find the intersection using n(A ∪ B) = n(A) + n(B) − n(A ∩ B): 139 = 82 + 76 − n(A ∩ B), so n(A ∩ B) = 19. The complement of this intersection in the universal set therefore has 190 − 19 = 171 elements. Hence option D is correct.

The set A ∩ B contains precisely those elements common to A and B. Subtracting this common part from A leaves the elements that are in A but not in B. By definition, that remainder is A − B. Algebraically, A − (A ∩ B) = A ∩ (A ∩ B)ᶜ = A ∩ Bᶜ = A − B. Thus option A is the only correct answer; the intersection is the part removed, not the part retained.

Each pairwise intersection includes the 16 elements that lie in all three sets. Therefore, the pair-only regions are n(A ∩ B only) = 45 − 16 = 29, n(A ∩ C only) = 38 − 16 = 22, and n(B ∩ C only) = 35 − 16 = 19. These regions are disjoint, so the number in exactly two sets is 29 + 22 + 19 = 70. Option C is correct; adding 45 + 38 + 35 would count the central region repeatedly.