Venn Diagrams

The intersection A ∩ B consists of elements that belong to both A and B. It cannot contain more elements than the smaller of the two sets, because every common element must be an element of B as well as A. Therefore, n(A ∩ B) ≤ min(n(A), n(B)) = min(66, 59) = 59. This maximum is attainable by placing all 59 elements of B inside A, so option B is correct.

Because A ⊆ B ⊆ C, every element of A is also an element of C. The set C − A therefore contains all elements of C except those belonging to A. Hence, n(C − A) = n(C) − n(A) = 118 − 29 = 89. The value of n(B) is not needed for this particular difference, although it confirms the nested relationship.

Apply 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 64+58+52−24−21−19+8=118. The pairwise intersections are subtracted because they were counted twice, while the triple intersection is added once because it was then subtracted too many times.

Let x=n(A∩B∩C). The inclusion–exclusion formula gives 140=70+65+60−28−24−22+x. The sum of the single-set cardinalities is 195, and the sum of the pairwise intersections is 74, so 140=121+x. Therefore x=19. The triple intersection must be added because subtracting all pairwise intersections removes the common central region too many times.

Each pairwise intersection includes the elements that lie in all three sets. Therefore, the regions belonging to exactly two sets are A∩B only: 31−12=19, B∩C only: 29−12=17, and C∩A only: 25−12=13. Adding these disjoint regions gives 19+17+13=49. Thus exactly 49 elements belong to two sets and not to the third.

To obtain the region only in A, subtract from n(A) the elements shared with B and the elements shared with C. The triple intersection has been subtracted twice, so add it once: only A=80−35−32+14=27. Thus 27 elements are in A but not in B or C. Simply calculating 80−35−32=13 would incorrectly remove the triple-overlap twice.

The elements in exactly two sets are found by removing the triple intersection from each pairwise intersection: (42−15)+(36−15)+(34−15)=27+21+19=67. The elements in all three sets, numbering 15, must then be added once. Therefore, the number in at least two sets is 67+15=82. This wording includes both exactly two and all three sets.

A consists of every region inside the circle representing A. These regions are only A, only A∩B, only A∩C, and the central triple intersection. Since the regions are disjoint, add their cardinalities: n(A)=18+12+10+6=46. The regions only in B or only in C are not part of A and must not be included.

The union A ∪ B is made up of two disjoint parts: the symmetric difference A △ B, containing elements in exactly one of the sets, and the intersection A ∩ B, containing elements common to both sets. Therefore, n(A ∪ B) = n(A △ B) + n(A ∩ B). Thus, n(A ∩ B) = 119 − 82 = 37, so option A is correct.

The union A ∪ B contains every element that belongs to A, to B, or to both. Removing A ∩ B eliminates the elements common to both sets, leaving only elements that belong to exactly one of the two sets. This set is called the symmetric difference, written as A Δ B = (A − B) ∪ (B − A).

By De Morgan’s law, \((A\cup B)^c=A^c\cap B^c\). Thus, \(A\cup B\) and \(A^c\cap B^c\) are complementary regions of the universal set and their cardinalities add to \(n(U)\). Numerically, \(128+72=200\), so option A is true. The intersection count 36 is not complementary to the union.

A number is divisible by 10 exactly when it is divisible by both 2 and 5, because 10 = 2 × 5 and these factors are coprime. Thus the numbers common to A and B are precisely the numbers in C. Therefore, C = A ∩ B. For example, 10, 20, and 30 belong to all three sets, whereas 2 belongs to A but not to C.

The cardinality formula for two sets is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). In the question, the union equals the simple sum, so the subtracted intersection term must be zero. Therefore n(A ∩ B) = 0, which means that A and B have no common element. In a Venn diagram, their circles do not overlap; such sets are called disjoint sets.

The union A ∪ B contains every element of A and every element of B. If its cardinality is exactly n(A), adding B has not introduced any new element beyond those already in A. Hence every element of B must already belong to A, so B ⊆ A. The sets need not be equal: B may be a proper subset of A. Therefore option A is the only necessary conclusion.

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.

Use the partition of each set into its exclusive part and common part. From A, n(A ∩ B) = n(A) − n(A − B) = 55 − 20 = 35. From B, the same intersection must be n(B) − n(B − A) = 50 − 18 = 32. A single intersection cannot simultaneously have two different cardinalities, so the supplied figures cannot describe valid sets together. Therefore the data are inconsistent.

Apply inclusion–exclusion 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). Thus the union is 96 + 88 + 74 − 40 − 31 − 29 + 12 = 170. The universal set has 200 people, so those in none of the sets are 200 − 170 = 30. Hence option C is correct.

A three-set Venn diagram has seven mutually exclusive 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. The six stated regions total 32 + 28 + 24 + 18 + 16 + 14 = 132. Since the union contains 150 elements, the central triple-intersection is 150 − 132 = 18. Therefore option B is 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.

First count total memberships across the three sets: n(A) + n(B) + n(C) = 82 + 76 + 70 = 228. People in exactly one set contribute one membership each, those in exactly two sets contribute two each, and those in all three contribute three each. If x is the exactly-one count, then 228 = x + 2(54) + 3(18). Hence x = 228 − 108 − 54 = 66. Therefore option A is correct.