Venn Diagrams

Among the integers from 1 to 60, there are floor(60/3)=20 multiples of 3 and floor(60/4)=15 multiples of 4. Multiples of both 3 and 4 are multiples of 12, so there are floor(60/12)=5 common elements. Therefore n(A∪B)=20+15−5=30. The complement contains the remaining 60−30=30 elements, so option A is correct.

Apply inclusion–exclusion to find the number in at least one set: n(A∪B∪C)=96+88+74−39−31−28+14=174. Pairwise intersections are subtracted because they were counted twice, and the triple intersection is added once because it was over-subtracted. The number outside all three sets is 210−174=36. Hence option A is correct.

To obtain the region belonging only to A, subtract from n(A) the elements shared with B and the elements shared with C, then add the triple intersection once because it was subtracted twice. Thus, only A = n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C) = 78−30−21+9 = 36. The B∩C value is not needed for this particular region. Therefore, option A is correct.

“At least two sets” includes the three pairwise-only regions and the central region common to all three sets. It therefore excludes the single-set-only regions and includes 11 elements in only A∩B, 13 in only B∩C, 9 in only C∩A, and 6 in A∩B∩C. The total is 11+13+9+6=39. Hence option A is correct.

In a two-set Venn diagram, A ∪ B consists of three non-overlapping regions: A − B, B − A, and A ∩ B. Therefore n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting gives 112 = 43 + 37 + n(A ∩ B). Since 43 + 37 = 80, the intersection has 112 − 80 = 32 elements. Thus option A is the unique correct answer.

Use the complement relation n(A)=n(U)−n(A′) and n(B)=n(U)−n(B′). Thus n(A)=140−58=82 and n(B)=140−71=69. Now apply the two-set union formula: n(A∪B)=n(A)+n(B)−n(A∩B)=82+69−34=117. The common elements are subtracted once to avoid double counting, so option A is correct.

The universal set is partitioned into four mutually disjoint regions: A∩B′, A′∩B, A∩B, and the outside region (A∪B)′. Every element of U belongs to exactly one of these regions. Therefore, n(U)=29+33+17+21=100. This is a partition count, so no inclusion–exclusion subtraction is needed.

To find the region belonging only to B, begin with all 68 members of B. Subtract the 29 members common to A and B and the 24 members common to B and C. The 11 members in all three clubs were subtracted twice, so add them back once: only B = 68 − 29 − 24 + 11 = 26. Thus option A is correct.

The union of three sets is partitioned into three mutually exclusive types of regions: elements in exactly one set, elements in exactly two sets, and elements in all three sets. Therefore, 135 = 58 + 49 + n(A ∩ B ∩ C). Solving gives n(A ∩ B ∩ C) = 135 − 58 − 49 = 28. Hence option A is correct.

The complement of A ∪ B contains elements of U that are neither multiples of 7 nor multiples of 4. There are 84/7 = 12 multiples of 7 and 84/4 = 21 multiples of 4. Their overlap consists of multiples of lcm(7,4) = 28, giving 84/28 = 3 elements. Thus n(A ∪ B) = 12 + 21 − 3 = 30, and n((A ∪ B)') = 84 − 30 = 54. Therefore option A is correct.

The positive multiples of 5 not exceeding 20 are 5, 10, 15, and 20. Of these, 5 and 15 are odd, whereas 10 and 20 are even. Therefore B ∩ C = {5, 15}, and its cardinality is 2. Set A gives the common upper bound but is not needed for this particular intersection.

Set difference X − Y means the elements that belong to X but do not belong to Y. Here X is A and Y is B ∩ C, the region common to B and C. Thus shade all of A, then exclude the portion of A lying in both B and C. The remaining region is the part of A outside B ∩ C. The triple intersection is removed, not selected.

The intersection A∩B contains exactly the elements common to A and B. If this intersection equals A, then every element of A must also lie in B. Therefore A is contained in B, so A⊆B. The condition A≠∅ is not needed for the implication, but it confirms that A is nonempty.

The union of the three pairwise intersections contains every element that belongs to at least two sets. Its disjoint Venn regions are: only A∩B with 14, only B∩C with 12, only C∩A with 15, and the central A∩B∩C region with 8. Add each once: 14+12+15+8=49.

The union contains every disjoint Venn-diagram region exactly once. Therefore, add the three only-set regions, the total of the exactly-two-set regions, and the three-set intersection: 31 + 27 + 25 + 42 + 11 = 136. Since these regions do not overlap, no subtraction or repeated counting is required.

Let x=n(A ∩ B ∩ C). Inclusion–exclusion gives 128=70+65+60−27−24−22+x, so x=52. However, the triple intersection must be contained in every pairwise intersection, so it cannot exceed n(C ∩ A)=22, n(B ∩ C)=24, or n(A ∩ B)=27. Since 52 is impossible, the supplied data are inconsistent and no such sets exist.

The region A ∩ (B ∪ C) consists of three mutually exclusive parts: the elements in only A ∩ B, the elements in only A ∩ C, and the central region A ∩ B ∩ C. Therefore, 44 = 18 + 16 + n(A ∩ B ∩ C). Hence, n(A ∩ B ∩ C) = 44 − 34 = 10. The central region is counted once in this partition.

The difference B−C means elements belonging to B but excluded from C. Intersecting this set with A keeps only elements that also belong to A. Therefore A ∩ (B−C) is precisely the part common to A and B while lying outside C. It excludes the central triple-overlap region.

Because A, B and C are mutually disjoint, no element belongs to more than one of these sets. Thus, the size of their union is the sum of their individual sizes: n(A ∪ B ∪ C) = 18 + 24 + 29 = 71. The complement contains all elements of the universal set that are outside this union. Therefore, n((A ∪ B ∪ C)') = 90 − 71 = 19.

The relation A⊆B⊆C means every element of A is in B and every element of B is in C. Consequently, all elements of A and B are already contained in C. Taking the union therefore adds nothing beyond C, while every element of C is included in the union. Hence A ∪ B ∪ C=C.