Venn Diagrams

The set A contains its exclusive part A − B and its common part A ∩ B. Hence, n(A ∩ B) = n(A) − n(A − B) = 58 − 21 = 37. Using the inclusion–exclusion formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 58 + 49 − 37 = 70. Therefore, the complement has 110 − 70 = 40 elements. Thus, option A is correct.

The expression A − (B ∪ C) contains elements that belong to A but do not belong to the union B ∪ C. Not belonging to the union means that an element is outside both B and C. Therefore, the required Venn region is the portion inside A alone, excluding every overlap with B or C. Hence option A is correct.

For two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B), because the common elements are counted twice when n(A) and n(B) are added. Substituting the given expressions gives 30 = (3x + 2) + (2x + 5) − (x + 1) = 4x + 6. Therefore, 4x = 24 and x = 6. Hence, option A is correct.

The union A ∪ B is partitioned into three disjoint regions: A − B, B − A and A ∩ B. Therefore, 45 = (2x+3) + (x+7) + (x−1) = 4x+9. Subtracting 9 gives 4x = 36, and dividing by 4 gives x = 9. Option A is correct. The other values do not satisfy the stated union equation.

The universal set U contains every element under consideration. If n(A ∪ B) = n(U), then A ∪ B contains all elements of U. Therefore, no element lies outside A ∪ B, so its complement (A ∪ B)' is the empty set. The intersection and the two difference regions may still contain elements; the given condition only guarantees that the outside region is empty.

The statements A ∩ B = ∅ and B ∩ C = ∅ only say that B has no common element with A and no common element with C. They do not compare A directly with C. For example, let A = {1}, C = {1}, and B = {2}; then both given intersections are empty, but A ∩ C = {1}, which is not empty.

De Morgan’s law states that the complement of a union equals the intersection of the complements: (A ∪ B)' = A' ∩ B'. The condition that A and B are disjoint is not needed for this identity; it is true for any two subsets of the same universal set. The complement consists of elements outside both A and B.

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). Thus, the union is 82+76+69−34−29−27+12=149. The people in none of the sets are outside the union, so the required number is n(U)−n(A∪B∪C)=180−149=31. Therefore, option A is correct.

The governing Venn-diagram relation is n(A ∪ B) = n(A) + n(B) − n(A ∩ B), since the common region is counted twice in the first two terms. Substitution gives 72 = (4x+5) + (3x+8) − (2x+1). Simplifying, 72 = 5x + 12, so 5x = 60 and x = 12. The values then give valid nonnegative region counts, confirming the result. Hence option A is uniquely correct.

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.

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.

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.

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 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.

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.

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.

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). Substituting the given values gives 160 = 70 + 75 + 80 − S + 18, where S is the required pairwise-intersection sum. Thus 160 = 243 − S, so S = 83.

Both the assertion and the reason are true. When n(A) and n(B) are added, every element common to A and B appears once in n(A) and once in n(B), so the intersection is counted twice. The union should count each element only once; therefore n(A ∩ B) must be subtracted once. This gives the inclusion–exclusion formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B), so the reason correctly explains the assertion.

Both statements are true, and the reason correctly explains the assertion. For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). If A and B are disjoint, their intersection is the empty set, so n(A ∩ B) = 0. Consequently, no element is counted in both sets, and the union contains the sum of the elements in A and B. Hence n(A ∪ B) = n(A) + n(B).

“Exactly one set” means elements that belong to A only, B only, or C only. It excludes every pairwise-overlap-only region and the central region common to all three sets. Therefore add only the three single-set regions: 26 + 31 + 24 = 81. The values 14, 16, 12, and 9 describe elements in at least two sets and must not be included.