Venn Diagrams

The intersection A ∩ B can contain only elements that are already in A. If n(A ∩ B) equals n(A), then the intersection contains as many elements as the whole set A. Therefore every element of A must also lie in B, which gives A ⊆ B. The sets do not have to be equal; B may contain additional elements outside A. Hence option A is necessary and correct.

The three categories are mutually exclusive and together cover every element in the union: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Therefore no inclusion–exclusion correction is needed. Add the category counts directly: n(A ∪ B ∪ C) = 71 + 46 + 15 = 132. Option B would omit the elements belonging to all three sets, so option C is the unique correct answer.

The union is partitioned into the disjoint regions \(A-B\), \(A\cap B\), and \(B-A\). Hence \(n(B-A)=104-37-26=41\). Set \(B\) consists of the region \(B-A\) together with the intersection, so \(n(B)=41+26=67\). Thus option C is correct. Option A is only the size of \(B-A\), while the other values do not follow from the partition.

The complement \(A^c\) contains all elements outside \(A\). In the Venn diagram, these are exactly the region \(B-A\) and the region outside \(A\cup B\). The intersection lies inside \(A\), so it must not be counted. Therefore \(n(A^c)=n(B-A)+n((A\cup B)^c)=30+15=45\). Option A is correct.

First use the inclusion–exclusion formula: n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 73 + 67 − 115 = 25. By De Morgan’s law, Aᶜ ∪ Bᶜ = (A ∩ B)ᶜ. Therefore, n(Aᶜ ∪ Bᶜ) = n(U) − n(A ∩ B) = 150 − 25 = 125. Thus, option D is correct; the original options did not include the mathematically correct value.

For three sets, the union can be divided into disjoint regions: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Hence, n(A ∪ B ∪ C) = exactly-one + exactly-two + exactly-three. Substitution gives 180 = exactly-one + 70 + 20, so exactly-one = 180 − 90 = 90. Therefore, option B is correct.

The union A ∪ B consists of three mutually disjoint regions: the elements only in A, the elements only in B, and the elements common to both sets. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 75 = x + 2x + 15, so 3x = 60 and x = 20. Hence, option B is correct.

The complement (A ∪ B)ᶜ represents the elements in the universal set that are outside both A and B. Thus, n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 210 − 58 = 152. For two sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, 152 = 96 + 88 − n(A ∩ B), which gives n(A ∩ B) = 32. Hence, option C is correct.

Let n(A ∩ B) = x. Students in exactly one set are counted in A only and B only, so their number is n(A) + n(B) − 2n(A ∩ B). Thus, 98 = 86 + 78 − 2x = 164 − 2x. Hence, 2x = 66 and x = 33. Equivalently, the common students are counted twice in n(A) + n(B), so twice the intersection must be removed. Option C is correct.

For three sets, the inclusion–exclusion formula is n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substituting the given values gives 162 = 82 + 76 + 70 − 34 − 30 − 27 + x = 147 + x. Therefore x = 15, so n(A ∩ B ∩ C) is 15. The triple intersection is added because it was subtracted three times in the pairwise terms.

The seven listed regions are mutually disjoint and together form A ∪ B ∪ C. Their total is 31 + 26 + 24 + 15 + 13 + 11 + 8 = 128. The complement contains all universal-set elements outside the union, so n((A ∪ B ∪ C)ᶜ) = n(U) − n(A ∪ B ∪ C) = 160 − 128 = 32. Therefore option A is correct.

A consists of every Venn-diagram region lying inside circle A. These are the part only in A, the part only in A ∩ B, the part only in A ∩ C, and the central part in all three sets. Therefore n(A) = 25 + 14 + 12 + 9 = 60. The pairwise values are explicitly labelled “only,” so they must not be counted again in any other way. Option B is correct.

For finite sets, the symmetric-difference formula is n(A △ B) = n(A) + n(B) − 2n(A ∩ B). Let x = n(A ∩ B). Then 104 = 88 + 82 − 2x = 170 − 2x. Thus 2x = 66 and x = 33. The common elements are subtracted twice because they are included in both n(A) and n(B), so option B is correct.

By De Morgan’s law, Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ. Therefore, the sets A ∪ B and Aᶜ ∩ Bᶜ are complementary regions of the universal set U. Their cardinalities add to n(U): 157 + 83 = 240. Hence option A is the only true relation; the other options confuse intersection, union, or equality of unrelated regions.

Use the inclusion–exclusion formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). There are 72 ÷ 6 = 12 multiples of 6 and 72 ÷ 8 = 9 multiples of 8. The common multiples are multiples of lcm(6, 8) = 24, giving 72 ÷ 24 = 3. Thus n(A ∪ B) = 12 + 9 − 3 = 18, so B is correct.

A number in A ∩ B ∩ C must be divisible by 2, 3, and 5 simultaneously. Since these numbers are pairwise coprime, their least common multiple is 2 × 3 × 5 = 30. The multiples of 30 in U are 30, 60, and 90. Therefore the triple intersection has 3 elements, so option B is correct.

The set A consists of the elements only in A, represented by A − B, together with the elements common to both sets, represented by A ∩ B. Therefore, n(A) = n(A − B) + n(A ∩ B), so n(A ∩ B) = 56 − 19 = 37. We can verify the result using set B: n(B) = n(B − A) + n(A ∩ B), giving 61 − 24 = 37. Since both calculations agree, the correct answer is option B, 37.

A three-set Venn diagram has seven regions inside the union: three regions belonging to exactly one set, three regions belonging to exactly two sets, and the central region belonging to all three sets. The six given regions total 38 + 34 + 29 + 21 + 18 + 16 = 156. Since the union contains all seven regions, the central region is 172 − 156 = 16. Therefore, n(A∩B∩C)=16.

The phrases “exactly one,” “exactly two,” and “all three” refer to mutually exclusive regions of the Venn diagram. Their union is the complete set of elements belonging to at least one of A, B, or C. Therefore, n(A∪B∪C)=84+63+18=165. No further inclusion–exclusion correction is needed because the three given counts are already disjoint categories.

The union of three sets can be partitioned into three mutually exclusive categories: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Let x be the number in exactly two sets. Then 156 = 74 + x + 17. Hence x = 156 − 74 − 17 = 65. Thus, 65 elements lie in exactly two of the three sets.

The union A ∪ B consists of three disjoint regions: elements only in A, elements common to A and B, and elements only in B. Since n(A − B) = 45 and n(A ∩ B) = 32, the number of elements only in B is 128 − 45 − 32 = 51. Therefore, n(B) = n(B − A) + n(A ∩ B) = 51 + 32 = 83. Hence, option C is correct.

De Morgan’s law states that Aᶜ∩Bᶜ=(A∪B)ᶜ. Thus, Aᶜ∩Bᶜ represents the elements in the universal set that lie outside A∪B. Since the universal set has 190 elements and the union contains 143 elements, the required number is 190−143=47. The individual values n(A) and n(B) are not needed after the union is given.

A two-set Venn diagram divides the universal set into four disjoint regions: A∩Bᶜ, Aᶜ∩B, A∩B, and Aᶜ∩Bᶜ. The first three regions contain 41+34+26=101 elements. Since all four regions together contain 130 elements, the remaining region is n(Aᶜ∩Bᶜ)=130−101=29. Therefore, option A is correct.

The union can be partitioned into three disjoint categories: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Let the first category have x elements. Then 205 = x + 83 + 24. Therefore x = 205 − 83 − 24 = 98. The phrase “exactly two sets” means the total of the three pair-only regions, while the 24 elements in all three sets form a separate category. Hence option B is correct.

The union of two sets is divided into three disjoint Venn-diagram regions: A − B, B − A, and A ∩ B. Therefore n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given expressions gives 120 = x + 3x + 24. Thus 4x = 96 and x = 24. Consequently, n(A − B) is 24 and n(B − A) is 72, whose total with the intersection is 120.