Operations on Sets (Union, Intersection, Difference)

The total number of elements in B includes both the elements only in B and the common elements in A ∩ B. Hence n(B only) = n(B) − n(A ∩ B) = 66 − 31 = 35. Option A is the intersection itself, not the exclusive B region. Option C is the complete size of B, and option D is an incorrect addition. Therefore, option B is correct.

The symmetric difference A△B contains elements that belong to A or B but not to both. It can be written as (A−B)∪(B−A); these two parts are disjoint. Therefore n(A△B)=n(A−B)+n(B−A)=38+27=65. The common intersection is excluded, so option B is the correct answer.

The elements belonging to exactly one set are n(A−B)+n(B−A). In terms of the given totals, this is n(A)+n(B)−2n(A∩B), because the common elements are counted in both n(A) and n(B). Thus 73=54+61−2n(A∩B)=115−2n(A∩B). Therefore 2n(A∩B)=42 and n(A∩B)=21.

The set A consists of two non-overlapping parts: the elements that are in A but not in B, represented by A − B, and the elements common to both A and B, represented by A ∩ B. Hence n(A) = n(A − B) + n(A ∩ B) = 41 + 22 = 63. Therefore, option B is correct. The value 41 counts only the exclusive part of A, while 22 counts only the common part.

The union is partitioned into three disjoint regions: elements only in A, elements only in B, and elements in both sets. Thus, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substitution gives 104 = 46 + n(B − A) + 25, so n(B − A) = 33. Option A is correct.

The complement of A ∪ B contains the elements of the universal set that are outside both A and B. Thus n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 125 − 24 = 101. Using the inclusion–exclusion formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, 101 = 62 + 55 − n(A ∩ B), giving n(A ∩ B) = 117 − 101 = 16. Hence option A is correct.

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.

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.

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 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.

Elements in exactly one set are those in A but not B together with those in B but not A. Thus only A has 60−25=35 elements and only B has 55−25=30 elements. Their total is 35+30=65. Equivalently, the formula is n(A)+n(B)−2n(A∩B)=60+55−50=65. The union, 90, would include the intersection as well.

The region A′ ∩ B′ contains elements outside both A and B. By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′, so n(A ∪ B) = n(U) − n(A′ ∩ B′) = 100 − 18 = 82. Using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we get 82 = 48 + 52 − n(A ∩ B). Therefore, n(A ∩ B) = 18. Thus, option A is correct.

The symmetric difference A Δ B contains the elements belonging to exactly one of the two sets. It is the union of the two exclusive regions, (A − B) ∪ (B − A), and does not include the common region A ∩ B. The union A ∪ B contains those exclusive elements as well as the intersection. Therefore, n(A ∪ B) = n(A Δ B) + n(A ∩ B) = 54 + 19 = 73. Option A is correct.

The union is divided into three disjoint parts: A − B, B − A, and A ∩ B. Hence n(A ∪ B) = (x + 4) + (2x − 1) + (x + 3). Substituting 46 gives 4x + 6 = 46, so 4x = 40 and x = 10. The corresponding regions are 14, 19, and 13, and 14 + 19 + 13 = 46, confirming the result. Therefore option A is correct.

Exactly one means belonging to A only or B only, but not to both. The A-only percentage is 60% − 25% = 35%, and the B-only percentage is 50% − 25% = 25%. Adding these disjoint parts gives 35% + 25% = 60%. Equivalently, exactly one = A + B − 2(A ∩ B) = 60 + 50 − 2(25) = 60%. The 85% option is the union and incorrectly includes the common group once.

First split set A into the disjoint regions A − B and A ∩ B. Thus n(A) = 35 + 26 = 61. Now use n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the known values gives 88 = 61 + n(B) − 26, so n(B) = 53. Option B is n(A), not n(B), making option A the correct answer.

Because A is a subset of B, every element of A is already included in B. Removing A from B therefore leaves the elements that belong only to B. For finite sets, n(B − A) = n(B) − n(A) = 57 − 32 = 25. The universal-set size is not needed for this calculation, so option A is correct.