Venn Diagrams

The governing concept is the decomposition of symmetric difference: A△B = (A−B) ∪ (B−A). The two component sets are disjoint, so their cardinalities add. Hence 78 = 34 + n(B−A). Subtracting 34 from both sides gives n(B−A) = 78−34 = 44. Option A repeats the known first difference, option C repeats the total, and option D adds instead of subtracting. Therefore, option B is correct.

First find the number who like at least one subject: n(C∪B)=n(C)+n(B)−n(C∩B)=43+39−18=64. The students who like neither subject are outside this union. Hence neither = total students−students liking at least one subject=90−64=26. Therefore, option B is correct.

Since A ∩ B = ∅, the two sets are disjoint and have no common elements. Therefore, every element of A and every element of B belongs to exactly one of the two sets. The required number is n(A) + n(B) = 49 + 52 = 101. Thus, option D is correct; subtracting the two cardinalities or choosing only one set would not represent the total number of elements in exactly one set.

The union A ∪ B is divided into three mutually exclusive Venn-diagram regions: A − B, B − A, and A ∩ B. Therefore, n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 88 = 31 + 29 + n(A ∩ B), so n(A ∩ B) = 88 − 60 = 28. Hence option B is correct.

The intersection A ∩ B ∩ C contains elements that belong simultaneously to A, B, and C. In a Venn diagram with three overlapping circles, this is the central region common to all three circles. It does not mean an element is in only one set; the word ‘intersection’ requires membership in every listed set. Therefore, option C is correct.

The union of three sets includes every region inside at least one of the three circles. Because the values are given as ‘only’ regions, each region must be added exactly once: 18 + 22 + 19 + 9 + 8 + 7 + 6 = 89. Hence, n(A ∪ B ∪ C) = 89, so option C is correct.

To find n(A), add every disjoint Venn-diagram region that lies inside A. These regions are: only A, only A ∩ B, only A ∩ C, and the common region A ∩ B ∩ C. Therefore, n(A) = 27 + 10 + 12 + 8 = 57. The word “only” means the pairwise regions exclude the triple intersection, so the triple region must be added separately. Hence option B is correct.

The set B contains every Venn-diagram region located inside B. The stated regions are only B, only A ∩ B, only B ∩ C, and the triple intersection A ∩ B ∩ C. These regions are disjoint because the pairwise regions are specified as “only.” Therefore, n(B) = 24 + 11 + 13 + 9 = 57. Hence option C is correct.

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 76 + 69 + 62 − 30 − 27 − 24 + 10 = 136. Elements in none of the sets are 180 − 136 = 44. Hence, option A is correct.

For three sets, the inclusion–exclusion formula is 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 values gives 99 = 48 + 43 + 41 − 17 − 14 − 13 + x = 88 + x. Therefore, x = 11, so the central intersection contains 11 elements.

In a two-set Venn diagram, the union is the disjoint sum of the A-only region, the B-only region, and the common region. Thus n(A∪B) = n(A−B)+n(B−A)+n(A∩B). Substituting gives 73 = 28+32+n(A∩B), so n(A∩B) = 73−60 = 13. The value 60 is only the sum of the exclusive regions; therefore option A is correct.

The complement of A ∪ B contains the elements outside the union. Therefore, n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 150 − 32 = 118. Now apply n(A ∪ B) = n(A) + n(B) − n(A ∩ B): 118 = 70 + 62 − x. Hence x = 132 − 118 = 14. Therefore, n(A ∩ B) is 14, so option B is correct.

First separate set A into its exclusive and common parts: n(A) = n(A−B)+n(A∩B). Therefore n(A∩B) = 58−29 = 29. The union then contains A−B, A∩B, and B−A as disjoint regions. Hence 96 = 29+29+n(B−A), giving n(B−A) = 38. Equivalently, n(B−A)=n(A∪B)−n(A)=96−58=38. Thus option A is correct.

Use the three-set inclusion–exclusion formula: 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 91 = 46 + 40 + 37 − 14 − 12 − 11 + x = 86 + x. Hence x = 5. The central intersection is added back because it was subtracted too many times in the pairwise terms.

The difference A \ B means the set of elements that belong to A and do not belong to B. In a Venn diagram, it is the non-overlapping part of circle A, excluding the common region A ∩ B. By contrast, A ∩ B contains elements common to both sets, A ∪ B contains elements in either set, and B \ A is the part belonging only to B. Therefore, C is correct.

The notation C − B means the elements that belong to C but do not belong to B. Since B is a subset of C, all 61 elements of B lie inside C. Therefore, the elements remaining in C outside B are n(C) − n(B) = 95 − 61 = 34. The value of n(A) is not needed because A lies inside B and does not affect C − B.

Let C represent people who like cricket and F represent people who like football. The cricket-only group has 80 − 29 = 51 people, and the football-only group has 64 − 29 = 35 people. Exactly one sport means belonging to one set but not both, so the required total is 51 + 35 = 86. Equivalently, n(C) + n(F) − 2n(C ∩ F) = 80 + 64 − 58 = 86.

The complement of A ∪ B contains elements outside both sets. Thus, n(A ∪ B) = n(U) − n((A ∪ B)ᶜ) = 180 − 38 = 142. For two sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the values gives 142 = 92 + 84 − n(A ∩ B), so n(A ∩ B) = 176 − 142 = 34. Therefore, option B is correct.

Let x = n(A ∩ B). The students in exactly one set are those in A only or B only. Their number is [n(A) − x] + [n(B) − x] = n(A) + n(B) − 2x. Therefore, 86 = 74 + 68 − 2x = 142 − 2x. Thus 2x = 56 and x = 28. Hence, the intersection contains 28 students, so option C is correct.

The total of the two set sizes is 77 + 64 = 141, which is 21 greater than the 120 elements available in U. Those extra 21 memberships must occur in the overlap. Using the lower-bound formula, n(A ∩ B) ≥ n(A) + n(B) − n(U), the minimum is 77 + 64 − 120 = 21. Thus option C is correct.