Venn Diagrams

For two finite sets, the inclusion–exclusion principle gives n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The subtraction is necessary because the 8 elements common to A and B are counted once in n(A) and once in n(B). Thus n(A ∪ B) = 26 + 18 − 8 = 36. Therefore, option A is correct; 44 results from forgetting the overlap.

Elements that are neither in A nor in B lie outside the union A ∪ B, so they form the complement (A ∪ B)′ within the universal set U. The universal set is partitioned into the union and its complement. Therefore n((A ∪ B)′) = n(U) − n(A ∪ B) = 58 − 41 = 17. Option A is correct; 41 is the union itself and 58 is the total universe.

The complement A′ is defined relative to the universal set U. It consists of all elements that belong to U but do not belong to A. On a Venn diagram, this is the entire part of the rectangle outside circle A. Thus the correct representation is A′, option B.

The union A ∪ B includes every element belonging to A, to B, or to both. Thus the two circles, including their overlap, are shaded. The only region left unshaded inside U is outside both circles, which is the complement (A ∪ B)'. Hence option D is correct.

The middle overlapping region of two circles contains exactly those elements that belong to both A and B. This set is called their intersection and is written as A ∩ B. The union includes both circles, while A − B includes only the non-overlapping part of A. Therefore, option C is correct.

The union \(A\cup B\) contains every element that belongs to at least one of the two sets. The elements in neither set are outside this union but still inside the universal set. Hence their number is \(n(U)-n(A\cup B)=85-56=29\). Therefore, option A is correct. The value 56 counts the union itself, not the region outside both sets.

The union contains every element in at least one of the sets. Its three disjoint Venn-diagram regions are only A, the intersection, and only B. Therefore, \(n(A\cup B)=18+7+14=39\). Each element is counted exactly once because the regions are disjoint. Hence option C is correct; adding only the two exclusive regions would incorrectly omit the seven common elements.

The three listed regions inside \(A\cup B\) are disjoint, so their total is \(26+15+31=72\). The universal set has 94 elements, and the remaining elements are outside both sets. Hence, the number in neither set is \(94-72=22\). Option A is correct; 72 is the union count, not the outside count.

The universal set is divided into four disjoint regions: only A, the intersection, only B, and neither set. Their total is 76. Therefore, \(n(A\cap B)=76-(20+23+18)=76-61=15\). Option A is correct. The value 61 is only the sum of the three known regions, so it excludes the unknown intersection rather than representing it.

A two-set Venn diagram divides A ∪ B into three disjoint regions: only A, A ∩ B, and only B. Their counts must add to the union total. Therefore 63 = 24 + n(A ∩ B) + 27, so n(A ∩ B) = 63 − 24 − 27 = 12. Option A is correct. The numbers 24 and 27 are separate-only regions, not the common region.

Students studying at least one subject are counted by the union formula: n(S ∪ C) = n(S) + n(C) − n(S ∩ C). Substituting the values gives 34 + 27 − 11 = 50. The 11 students studying both are subtracted once because they were initially counted twice. Hence option A is correct.

The 32 mango lovers include the 9 people who like both mango and banana. To find those who like only mango, subtract the overlap from the mango total: 32 − 9 = 23. The banana total and group size are not needed for this particular calculation. Therefore, option B is correct.

Let S represent the students who passed Science and let M ∩ S represent those who passed both subjects. The students counted in Science only are obtained by removing the students who passed both from the total Science group: n(S only) = n(S) − n(M ∩ S) = 49 − 35 = 14. Therefore, option A is correct.

Let R be the set of people who visited the rides and G the set who visited the game stalls. The union counts everyone who visited at least one place. Since the 30 people who visited both places are included in both totals, they must be subtracted once: n(R ∪ G) = 70 + 52 − 30 = 92. Thus option A is correct.

The union A ∪ B contains every element from both A and B. If this union is exactly A, then adding the elements of B contributes nothing new; every element of B must already be in A. Therefore, B ⊆ A. The other relations are not necessarily true, so option B is the unique correct answer.

The intersection A ∩ B ∩ C consists of elements that belong simultaneously to A, B, and C. In a three-set Venn diagram, this is the central region where all three circles overlap. It is not a part belonging to only one set and cannot lie outside the circles. Thus option C is correct.

For three finite sets, the inclusion–exclusion formula is n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C). Substituting the values gives 16 + 18 + 9 − 6 − 2 − 3 + 1 = 33. The pairwise intersections are subtracted because they were counted twice, while the triple intersection is added once because it was subtracted too many times.

By definition, A ∩ B contains precisely those elements that belong to both A and B. Every element belonging to both sets must certainly belong to A, so A ∩ B ⊆ A is always true. Option B reverses the usual inclusion relationship and is generally false. Option C is not generally valid because the two differences usually contain different elements. Option D is also not generally true because a set and its complement are normally different.

For two finite sets, the union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The intersection is subtracted once because its elements are included in both n(A) and n(B), so they would otherwise be counted twice. Substituting the values gives n(A ∪ B) = 36 + 28 − 12 = 52. Therefore option A is the only correct answer.

For two finite sets, the union formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 28 + 22 − 10 = 40. The intersection is subtracted once because its elements were counted in both n(A) and n(B). Thus, the number of students in at least one of the two sets is 40.