Venn Diagrams

The region outside A is A′, and the region outside B is B′. To be outside both sets simultaneously, an element must belong to A′ ∩ B′. By De Morgan’s law, this is also equal to (A ∪ B)′. In contrast, A ∪ B includes elements in at least one set, A ∩ B is the overlap, and A − B lies inside A. Therefore option A is correct.

The three regions A − B, B − A, and A ∩ B are mutually disjoint, and together they form A ∪ B. Therefore, (x + 5) + (2x − 3) + (x + 1) = 43. Simplifying gives 4x + 3 = 43, so 4x = 40 and x = 10. The three region sizes are then 15, 17, and 11, whose sum is 43. Thus option A is the only correct answer.

For two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B), because the common elements are counted twice in n(A) + n(B). Substitution gives (3x + 8) + (2x + 17) − (x + 5) = 60. Thus 4x + 20 = 60, so 4x = 40 and x = 10. Checking gives n(A) = 38, n(B) = 37, intersection = 15, and union = 38 + 37 − 15 = 60.

Use the two-set formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given expressions gives (4x + 6) + (3x + 9) − (2x + 5) = 80. Combining terms, 5x + 10 = 80, so 5x = 70 and x = 14. Verification gives n(A) = 62, n(B) = 51, and n(A ∩ B) = 33; therefore 62 + 51 − 33 = 80. Option A is correct.

Students choosing at least one option are counted by the union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus the number is 68 + 57 − 31 = 94. The required percentage is (94 ÷ 120) × 100 = 78.333..., which rounds to 78.33%. The subtraction prevents students who chose both options from being counted twice. Therefore option A is correct.

The percentage in at least one set is the percentage of the union. Use the inclusion-exclusion rule: percentage(A ∪ B) = percentage(A) + percentage(B) − percentage(A ∩ B). Therefore, 35% + 48% − 20% = 63%. The common 20% is subtracted once because it was included in both 35% and 48%. Hence option A is correct.

For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The given equality has no subtraction term, so n(A ∩ B) must be zero. Therefore, A and B have no common element; their circles do not overlap in the Venn diagram, and they are called disjoint sets.

When A ⊆ B, every element of A is already contained in B. Taking the union adds no new element to B, so A ∪ B = B. The elements common to A and B are exactly all elements of A, so A ∩ B = A. Therefore option A gives the two results in the required order; option B reverses them.

When two sets partially overlap, the diagram is divided into three relevant parts: the portion belonging only to A, the common overlapping portion, and the portion belonging only to B. These are A − B, A ∩ B, and B − A respectively. No element can belong to two of these regions at the same time, so they are pairwise disjoint.

The union A ∪ B is partitioned into three mutually exclusive regions: A ∩ B′, A′ ∩ B, and A ∩ B. The outside region A′ ∩ B′ is not part of the union. Therefore n(A ∪ B) = 29 + 34 + 21 = 84. Option A is correct; subtracting or omitting one region produces the distractor values.

A universal set divided by A and B has four disjoint regions: A ∩ B′, A′ ∩ B, A ∩ B, and A′ ∩ B′. The first three contain 18, 22, and 16 elements, so their total is 56. Since the universal set has 75 elements, the outside region is 75 − 56 = 19. Therefore, n(A′ ∩ B′) = 19.

Each pairwise intersection includes its exclusive pair region as well as the central region common to all three sets. Therefore, n(A ∩ B)=11+4=15, n(B ∩ C)=13+4=17, and n(C ∩ A)=9+4=13. Their sum is 15+17+13=45. The central region is counted three times because it belongs to every pair.

Use 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). Substituting gives 120 = 55+60+65−78+x = 102+x, so x = 18. Thus option A is correct; the triple intersection is added back once.

The quantities n(A − B), n(B − C), and n(C − A) describe only selected portions of a three-set Venn diagram. They do not tell us the sizes of several other regions, such as the pairwise-overlap parts, the central triple intersection, or possibly other exclusive regions. Since the union is the sum of every region inside at least one circle, different diagrams can satisfy the given values but have different union sizes. Therefore, the union cannot be determined from these data alone.

For each of the three sets, an element may be inside or outside that set. Thus there are 2 choices for each set and 2 × 2 × 2 = 2^3 = 8 membership patterns. One pattern means being outside A, outside B, and outside C simultaneously; it is the region outside all three circles. The remaining 8 − 1 = 7 patterns correspond to membership in at least one set, so they form the seven inner regions.

“Not in at least one set” means not belonging to the union, or the complement of A ∪ B ∪ C. Apply inclusion–exclusion: n(A ∪ B ∪ C) = 120 + 110 + 95 − 52 − 41 − 37 + 19 = 214. The pairwise intersections are subtracted because they were counted twice, and the triple intersection is added once to correct the over-subtraction. Hence the number outside the union is n(U) − 214 = 250 − 214 = 36.

The union A ∪ B consists of three disjoint parts: students who chose only A, students who chose only B, and students who chose both. Therefore, 72 = 28 + 34 + n(A ∩ B). Solving gives n(A ∩ B) = 72 − 28 − 34 = 10. The 100-student total is not needed for this calculation because the question already gives the number choosing at least one option, that is, the union size.

For two finite sets, the addition n(A) + n(B) counts the common elements twice, so the union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Rearranging gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 62 + 55 − 91 = 26. The universal-set size is not required because the union size is already supplied.

The set A contains both the students who belong only to A and the students common to A and B. Thus n(A) = n(only A) + n(A ∩ B). Therefore, n(only A) = 37 − 19 = 18. The universal-set size and the total size of B are unnecessary for this particular question. Subtracting the common region once prevents those students from being counted as exclusive members of A.

The complement of A ∪ B contains elements of the universal set that belong to neither A nor B. First calculate the union using n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 70 + 65 − 28 = 107. Then subtract this union from the universal set: n((A ∪ B)') = n(U) − n(A ∪ B) = 150 − 107 = 43. Hence option A is correct.