Venn Diagrams

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.

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.

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.

The governing concept is the inclusion–exclusion principle for three finite sets. It states n(A∪B∪C)=n(A)+n(B)+n(C)−[n(A∩B)+n(B∩C)+n(C∩A)]+n(A∩B∩C). Let x=n(A∩B∩C). Substitution gives 120=55+60+65−78+x. The sum of the individual cardinalities is 180, and 180−78=102, so 120=102+x. Therefore x=18. Option A is correct. The triple intersection is added back because elements common to all three sets are subtracted too many times during the pairwise subtraction. The other numerical choices do not satisfy the inclusion–exclusion equation.

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.

“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 of the three pairwise intersections contains every element that belongs to at least two of the sets. The three exclusive pairwise regions contribute 14, 11, and 13 elements. The central region A ∩ B ∩ C belongs to all three pairwise intersections, but it must be counted only once in their union. Therefore, the required number is 14 + 11 + 13 + 9 = 47, so option A is correct.

The union of three sets contains every region lying inside at least one of the three circles. These seven disjoint regions contain 23, 17, 19, 8, 9, 10, and 5 elements. Adding each region once gives 23 + 17 + 19 + 8 + 9 + 10 + 5 = 91. Therefore, option A is correct.

The total n(M ∩ P) = 25 includes the students in all three subjects, and n(C ∩ M) = 20 also includes that same centre. To obtain only M, subtract both pairwise intersections from n(M), then add the centre once because it was subtracted twice: 60 − 25 − 20 + 10 = 25. Option A is correct.

The union is partitioned into three mutually exclusive types of Venn regions: elements belonging to exactly one set, elements belonging to exactly two sets, and elements belonging to all three sets. Thus 140 = 65 + 51 + n(A ∩ B ∩ C). Solving gives n(A ∩ B ∩ C) = 140 − 116 = 24. Therefore option A is correct.

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 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.

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.

Let the seven disjoint Venn regions be the only-A, only-B, only-C, AB-only, BC-only, AC-only, and ABC regions. Since the total union is 155 and A ∪ B is 118, only-C = 155 − 118 = 37. Similarly, only-A = 155 − 124 = 31 and only-B = 155 − 121 = 34. The set B contains only-B plus the three regions involving B, so the sum of B-only, AB-only, BC-only, and ABC is 64. Combining these region equations with the total gives AC-only = n((A ∩ C) − B) = 23.

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 112 + 104 + 93 − 46 − 38 − 35 + 17 = 207. The elements in none of the sets are outside the union, so their number is n(U) − n(A ∪ B ∪ C) = 240 − 207 = 33.

The elements only in C are those in C but not in A or B. Start with n(C) = 71. Subtract n(B ∩ C) = 29 and n(C ∩ A) = 27, because these are the portions of C shared with the other sets. The central region A ∩ B ∩ C, containing 12 elements, was subtracted twice, so add it back once. Thus only-C = 71 − 29 − 27 + 12 = 27.

Because A is a subset of B, the set B − A contains the elements of B that are not in A, so its size is 91 − 52 = 39. The complement B' contains elements outside B, so n(B') = 140 − 91 = 49. These two regions are disjoint, because one is inside B and the other is outside B. Therefore the required size is 39 + 49 = 88.

The number n(A ∩ C) = 31 includes every student who belongs to both A and C, including those who also belong to B. The triple intersection has 13 students, so these must be removed to obtain the pair-only region. Hence the number in A ∩ C but not B is 31 − 13 = 18. The value 31 would incorrectly include the triple intersection.

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 97 + 88 + 82 − 41 − 36 − 34 + 18 = 174. The triple intersection is added once at the end because it was subtracted too many times. Thus option A is correct.

The governing concept is inclusion–exclusion for three sets. For sets A, B, and C, n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C). Let x represent the triple intersection. Substituting the data gives 151=82+77+69−32−28−25+x. The calculation on the right before x is 228−85=143, so 151=143+x and x=8. Therefore option A is correct. The triple intersection must be added once because its elements were included in all three single-set counts and then removed through the three pairwise intersections. The other choices fail the stated formula.

Let S=n(A∩B)+n(B∩C)+n(C∩A). Inclusion–exclusion for three sets states n(A∪B∪C)=n(A)+n(B)+n(C)−S+n(A∩B∩C). Substituting gives 184=86+91+88−S+23=288−S. Hence S=288−184=104. Therefore option A is correct; 81 would result from mishandling the triple-intersection term.

Each difference includes more than one possible Venn region; for example, A−B may contain the part only in A and the part in A∩C but not B. The three given counts therefore do not specify all single, pairwise, and triple regions. Different common-region counts can produce different unions, so the union cannot be determined uniquely.

Use inclusion–exclusion. There are floor(180/12) = 15 multiples of 12, floor(180/15) = 12 multiples of 15, and floor(180/20) = 9 multiples of 20. The pairwise common counts are 3, 3, and 3, while the common triple count is 3. Thus the union has 15 + 12 + 9 − 3 − 3 − 3 + 3 = 30 elements. Its complement has 180 − 30 = 150 elements.