Venn Diagrams

The universal set is divided into four mutually exclusive Venn-diagram regions: A ∩ Bᶜ, Aᶜ ∩ B, A ∩ B, and Aᶜ ∩ Bᶜ. The first three regions contain 48 + 44 + 28 = 120 elements. Since the universal set contains 160 elements, the remaining region is n(Aᶜ ∩ Bᶜ) = 160 − 120 = 40. Thus, option B is correct.

For two finite sets, the inclusion-exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 96 = 58 + 49 − n(A ∩ B). Therefore, n(A ∩ B) = 107 − 96 = 11. The subtraction removes the double counting of common elements, so option A is correct.

First find the number in at least one of the two sets using inclusion-exclusion: n(A ∪ B) = 52 + 47 − 24 = 75. Students in neither set are outside this union, so their number is n(U) − n(A ∪ B) = 90 − 75 = 15. Equivalently, this is n((A ∪ B)ᶜ). Therefore, option A is correct.

For three sets, inclusion-exclusion gives 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 40 + 38 + 35 − 14 − 12 − 10 + 5 = 82. The triple intersection is added once because it was subtracted too many times.

The set A contains four disjoint regions: the region belonging only to A, the region belonging to A and B but not C, the region belonging to A and C but not B, and the central region common to all three sets. Therefore, n(A)=18+9+5+7=39. The B∩C-only region is not included because it lies outside A.

The union A ∪ B ∪ C represents everyone who belongs to at least one of the three sets. The people in none of the sets are outside this union, so their number is found by subtracting the union from the universal set: n(U) − n(A ∪ B ∪ C) = 200 − 164 = 36. Therefore, 36 people belong to none of the three sets.

“At least one set” means the union A∪B∪C. The given categories are mutually exclusive Venn-diagram regions: only A, only B, only C, exactly two sets, and all three sets. Hence every element in the union is counted exactly once. Therefore, n(A∪B∪C)=16+19+14+27+8=84, so option A is correct.

The 39 students who chose Physics include both the students who chose only Physics and the 22 students who chose both subjects. Therefore, the number choosing only Physics is 39−22=17. The value 24 is only Mathematics, while 63 is the number choosing at least one subject, not only Physics. Hence option A is correct.

Because A ⊆ B, every element of A lies inside B. The difference B − A therefore contains the elements that belong to B but not to A. For finite sets, n(B − A) = n(B) − n(A) = 64 − 28 = 36. Thus option A is correct. The values 28 and 64 refer to A and B themselves, while 92 incorrectly adds them.

A∩B=∅ means that A and B are disjoint, so they have no common elements. For any two sets, n(A∪B)=n(A)+n(B)−n(A∩B). Substituting the values gives 31+44−0=75. Since no element is counted in both sets, simple addition is valid. Therefore, option A is the correct union cardinality.

The union A∪B is partitioned into three non-overlapping regions: elements in A but not B, elements in B but not A, and elements common to both A and B. Therefore, n(A∪B)=n(A−B)+n(B−A)+n(A∩B)=23+17+12=52. Each element is counted exactly once in this sum, so option A is correct.

In a three-set Venn diagram, A−B means all elements that are in A and not in B. This is not necessarily the same as the region only in A, because some elements may belong to both A and C while still not belonging to B. Thus A−B can contain the only-A region and the A∩C-only region. The same careful interpretation applies to B−C and C−A, so option A is correct.

For two sets, the union consists of three non-overlapping regions: the elements only in A, represented by A − B; the elements only in B, represented by B − A; and the common elements, represented by A ∩ B. Hence n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the values gives 70 = 24 + 31 + n(A ∩ B), so n(A ∩ B) = 15. Option A is correct.

Each pairwise intersection includes the elements common to all three sets. If the three pairwise intersections are added, every element in the triple intersection is counted three times, although it belongs to none of the exactly-two regions. Subtracting 3n(A ∩ B ∩ C) removes these triple counts and leaves only elements in exactly two sets. Hence option A is correct.

The sum of the three pairwise intersections is 20+18+16=54. However, every element in the triple intersection has been counted in all three pairwise intersections. Since the triple intersection contains 6 elements, it contributes three times, so subtract 3×6. The number in exactly two sets is 54−18=36. Therefore option A is correct.

The pairwise intersections include the seven people who belong to all three sets. Thus the A∩B-only region is 18−7=11, and the C∩A-only region is 16−7=9. Set A contains its A-only region, these two pairwise-only regions, and the triple region. Hence A-only=45−11−9−7=18. Therefore option A 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 70 + 62 + 58 − 30 − 24 − 26 + 12 = 122. The elements in none of the sets are outside the union, so their number is n(U) − n(A ∪ B ∪ C) = 130 − 122 = 8. Option A is correct.

When the circle representing A lies completely inside the circle representing B, every element of A is also an element of B. This is precisely the definition of A being a subset of B, written A ⊆ B. The reverse inclusion is not necessarily true because B may contain additional elements outside A. The sets are not disjoint, so option A is correct.

The statements A ∩ B = ∅ and B ∩ C = ∅ only say that A has no common element with B and C has no common element with B. They do not compare A directly with C. For example, let A = {1}, B = {2}, and C = {1, 3}. Then both given intersections are empty, but A ∩ C = {1}, which is not empty. Therefore, A and C may overlap, so option A is correct.

The difference A − B consists of all elements that belong to A but do not belong to B. In a Venn diagram, this is the portion of circle A excluding the common overlap with circle B. A ∩ B represents the shared region, B − A represents the part only in B, and A ∪ B represents everything in either set. Hence option A is correct.