Venn Diagrams

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.

Apply the distributive identity \((X\cup Y)\cap(X\cup Z)=X\cup(Y\cap Z)\). Taking \(X=A\), \(Y=B\), and \(Z=B^c\), the expression becomes \(A\cup(B\cap B^c)\). A set and its complement are disjoint, so \(B\cap B^c=\varnothing\). Therefore the result is \(A\cup\varnothing=A\), making option A correct.

Let x be the number of elements belonging to exactly two sets. The sum of the three set cardinalities is 72 + 66 + 60 = 198. Elements in exactly one set contribute once, elements in exactly two sets contribute twice, and elements in all three sets contribute three times. Thus 198 = 84 + 2x + 3(10) = 114 + 2x, so 2x = 84 and x = 42. Therefore, option B 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.

Use the inclusion–exclusion formula for three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substitution gives 72 + 66 + 59 − 28 − 24 − 22 + 9 = 132. The triple intersection is added once because it was subtracted too many times during the pairwise correction.

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.

Each pairwise intersection includes the elements that lie in all three sets. Thus, the elements in exactly A and B are 36 − 14 = 22; exactly B and C are 33 − 14 = 19; and exactly C and A are 31 − 14 = 17. Adding these disjoint regions gives 22 + 19 + 17 = 58. Therefore option B is correct.

First find the elements in exactly two sets by removing the triple intersection from each pair: (48 − 17) + (41 − 17) + (39 − 17) = 31 + 24 + 22 = 77. The elements in all three sets, 17, must then be included because “at least two” means exactly two or exactly three. Thus the required number is 77 + 17 = 94, option B.

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.

The symmetric difference A Δ B consists of elements that are in A or B but not in both. Thus, A ∪ B is partitioned into two disjoint parts: A Δ B and A ∩ B. Therefore, n(A ∪ B) = n(A Δ B) + n(A ∩ B). Substituting the values gives 137 = 96 + n(A ∩ B), so n(A ∩ B) = 137 − 96 = 41. Hence, 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.