Operations on Sets (Union, Intersection, Difference)

“Only in A” means the elements belonging to A but not to B, represented by A − B or A ∩ Bᶜ. The total number in A includes the 38 students who are also in B. Therefore, subtract the intersection from A: n(A − B) = n(A) − n(A ∩ B) = 72 − 38 = 34. Thus, option A is correct.

Let x = n(A ∩ B ∩ C). The three-set inclusion-exclusion formula gives 100 = 50 + 45 + 40 − 18 − 15 − 12 + x. The known terms simplify to 90, so 100 = 90 + x and x = 10. The triple intersection must be added back because pairwise intersections overlap there, making option A correct.

Elements in exactly one set are those in A but not B together with those in B but not A. Thus only A has 60−25=35 elements and only B has 55−25=30 elements. Their total is 35+30=65. Equivalently, the formula is n(A)+n(B)−2n(A∩B)=60+55−50=65. The union, 90, would include the intersection as well.

The region A′ ∩ B′ contains elements outside both A and B. By De Morgan’s law, A′ ∩ B′ = (A ∪ B)′, so n(A ∪ B) = n(U) − n(A′ ∩ B′) = 100 − 18 = 82. Using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we get 82 = 48 + 52 − n(A ∩ B). Therefore, n(A ∩ B) = 18. Thus, option A is correct.

The symmetric difference A Δ B contains the elements belonging to exactly one of the two sets. It is the union of the two exclusive regions, (A − B) ∪ (B − A), and does not include the common region A ∩ B. The union A ∪ B contains those exclusive elements as well as the intersection. Therefore, n(A ∪ B) = n(A Δ B) + n(A ∩ B) = 54 + 19 = 73. Option A is correct.

The union is divided into three disjoint parts: A − B, B − A, and A ∩ B. Hence n(A ∪ B) = (x + 4) + (2x − 1) + (x + 3). Substituting 46 gives 4x + 6 = 46, so 4x = 40 and x = 10. The corresponding regions are 14, 19, and 13, and 14 + 19 + 13 = 46, confirming the result. Therefore option A is correct.

Exactly one means belonging to A only or B only, but not to both. The A-only percentage is 60% − 25% = 35%, and the B-only percentage is 50% − 25% = 25%. Adding these disjoint parts gives 35% + 25% = 60%. Equivalently, exactly one = A + B − 2(A ∩ B) = 60 + 50 − 2(25) = 60%. The 85% option is the union and incorrectly includes the common group once.

First split set A into the disjoint regions A − B and A ∩ B. Thus n(A) = 35 + 26 = 61. Now use n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the known values gives 88 = 61 + n(B) − 26, so n(B) = 53. Option B is n(A), not n(B), making option A the correct answer.

Because A is a subset of B, every element of A is already included in B. Removing A from B therefore leaves the elements that belong only to B. For finite sets, n(B − A) = n(B) − n(A) = 57 − 32 = 25. The universal-set size is not needed for this calculation, so option A is correct.

First use the two-set union formula: n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 45 + 52 − 73 = 24. The symmetric difference contains the elements in exactly one set, so it excludes the common part. Hence n(A △ B) = n(A ∪ B) − n(A ∩ B) = 73 − 24 = 49. Option A is correct.

By the distributive law, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). In the Venn diagram this includes the A-B-only region with 9 elements, the A-C-only region with 6 elements, and the central triple-overlap region with 4 elements. These regions are disjoint, so the total is 9 + 6 + 4 = 19. Option A is correct.

Use the complement rule first: n(A) = n(U) − n(A′) = 95 − 47 = 48, and n(B) = 95 − n(B′) = 95 − 52 = 43. Then apply inclusion–exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 48 + 43 − 18 = 73. Thus option A is correct. The other choices result from omitting or mishandling the complement or intersection.

The universal set has 95 elements. Therefore, n(A) = 95 − 47 = 48 and n(B) = 95 − 52 = 43. Applying the two-set union formula gives n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 48 + 43 − 18 = 73. The intersection is subtracted once because it was counted in both set totals. Hence option A is correct.

The union A ∪ B is divided into three mutually disjoint Venn regions: A only, represented by A ∩ B' and equal to 24; B only, represented by A' ∩ B and equal to 31; and the common region A ∩ B, equal to 16. Adding these non-overlapping parts gives 24 + 31 + 16 = 71. Therefore, option A is correct.

People who like at least one beverage are counted by the union formula: n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 52 + 47 − 25 = 74. The remaining people like neither tea nor coffee, so subtract the union from the total: 90 − 74 = 16. Thus option A is correct.

The phrase “at least one” means belonging to A or B or both, which is the union A ∪ B. The problem already gives three separate, non-overlapping regions: only A has 40 students, only B has 28, and both sets have 12. Therefore, n(A ∪ B) = 40 + 28 + 12 = 80. 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). The triple intersection is empty, so its cardinality is 0. Substitution gives 30+34+36−8−10−6+0 = 76. Therefore, option A is correct; simply adding all set sizes would double-count pairwise overlaps.

Use the three-set inclusion–exclusion formula. Let x = n(A ∩ B ∩ C). Then 118 = 45 + 50 + 55 − 18 − 20 − 16 + x. The known terms give 150 − 54 = 96, so 118 = 96 + x and x = 22. Therefore, the triple intersection contains 22 elements, making option A correct.

The governing concept is set intersection: B ∩ C contains only numbers satisfying both conditions. The even natural numbers up to 12 are 2, 4, 6, 8, 10, and 12, while the multiples of 3 are 3, 6, 9, and 12. Their common elements are therefore B ∩ C = {6, 12}. Since this set has two elements, n(B ∩ C) = 2, so option A is correct. Options B, C, and D do not match the actual intersection.

The governing concept is set difference: A − B consists of elements that belong to A but do not belong to B. The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19. All primes except 2 are odd and therefore lie in B. Since 2 is even, it is the only prime excluded from B, giving A − B = {2}. Thus option A is correct; option C is A’s odd-prime part, not the difference.