Venn Diagrams

Correct answer: B. 44

Explanation: 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.

Correct answer: D. 101

Explanation: 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.

Correct answer: B. 28

Explanation: 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.

Correct answer: C. The common central part of all three circles

Explanation: 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.

Correct answer: C. 89

Explanation: 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.

Correct answer: B. 57

Explanation: 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.

Correct answer: C. 57

Explanation: 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.

Correct answer: A. 44

Explanation: 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.

Correct answer: D. 11

Explanation: 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.

Correct answer: A. n(A ∩ B) = 13

Explanation: 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.

Correct answer: B. 14

Explanation: 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.

Correct answer: A. 38

Explanation: 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.

Correct answer: B. 5

Explanation: 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.

Correct answer: A. 34

Explanation: 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.

Correct answer: A. 86

Explanation: 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.

Correct answer: C. 89

Explanation: Because A ⊆ B ⊆ C, every element of A is also an element of C. The set C − A therefore contains all elements of C except those belonging to A. Hence, n(C − A) = n(C) − n(A) = 118 − 29 = 89. The value of n(B) is not needed for this particular difference, although it confirms the nested relationship.

Correct answer: A. 49

Explanation: Each pairwise intersection includes the elements that lie in all three sets. Therefore, the regions belonging to exactly two sets are A∩B only: 31−12=19, B∩C only: 29−12=17, and C∩A only: 25−12=13. Adding these disjoint regions gives 19+17+13=49. Thus exactly 49 elements belong to two sets and not to the third.

Correct answer: C. 27

Explanation: To obtain the region only in A, subtract from n(A) the elements shared with B and the elements shared with C. The triple intersection has been subtracted twice, so add it once: only A=80−35−32+14=27. Thus 27 elements are in A but not in B or C. Simply calculating 80−35−32=13 would incorrectly remove the triple-overlap twice.

Correct answer: B. 82

Explanation: The elements in exactly two sets are found by removing the triple intersection from each pairwise intersection: (42−15)+(36−15)+(34−15)=27+21+19=67. The elements in all three sets, numbering 15, must then be added once. Therefore, the number in at least two sets is 67+15=82. This wording includes both exactly two and all three sets.

Correct answer: A. 37

Explanation: The union A ∪ B is made up of two disjoint parts: the symmetric difference A △ B, containing elements in exactly one of the sets, and the intersection A ∩ B, containing elements common to both sets. Therefore, n(A ∪ B) = n(A △ B) + n(A ∩ B). Thus, n(A ∩ B) = 119 − 82 = 37, so option A is correct.

Correct answer: A. A Δ B

Explanation: The union A ∪ B contains every element that belongs to A, to B, or to both. Removing A ∩ B eliminates the elements common to both sets, leaving only elements that belong to exactly one of the two sets. This set is called the symmetric difference, written as A Δ B = (A − B) ∪ (B − A).

Correct answer: A. \(n(A\cup B)+n(A^c\cap B^c)=n(U)\)

Explanation: By De Morgan’s law, \((A\cup B)^c=A^c\cap B^c\). Thus, \(A\cup B\) and \(A^c\cap B^c\) are complementary regions of the universal set and their cardinalities add to \(n(U)\). Numerically, \(128+72=200\), so option A is true. The intersection count 36 is not complementary to the union.

Correct answer: A. C = A ∩ B

Explanation: A number is divisible by 10 exactly when it is divisible by both 2 and 5, because 10 = 2 × 5 and these factors are coprime. Thus the numbers common to A and B are precisely the numbers in C. Therefore, C = A ∩ B. For example, 10, 20, and 30 belong to all three sets, whereas 2 belongs to A but not to C.

Correct answer: A. A ∩ B = ∅

Explanation: The cardinality formula for two sets is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). In the question, the union equals the simple sum, so the subtracted intersection term must be zero. Therefore n(A ∩ B) = 0, which means that A and B have no common element. In a Venn diagram, their circles do not overlap; such sets are called disjoint sets.

Correct answer: A. B ⊆ A

Explanation: The union A ∪ B contains every element of A and every element of B. If its cardinality is exactly n(A), adding B has not introduced any new element beyond those already in A. Hence every element of B must already belong to A, so B ⊆ A. The sets need not be equal: B may be a proper subset of A. Therefore option A is the only necessary conclusion.