Venn Diagrams

A two-set Venn diagram is divided into four mutually exclusive regions: A−B, A∩B, B−A, and the region outside A∪B. The universal set contains every one of these regions. Therefore, n(U)=18+14+21+7=60. Since no region overlaps another in this partition, each value is added exactly once. Thus, option C is correct.

The universal set is divided into four regions: A only, A ∩ B, B only, and outside both sets. The three regions inside the union contain 24 + 13 + 26 = 63 elements. Therefore, the outside region contains n(U) − n(A ∪ B) = 88 − 63 = 25 elements. Hence, option A is correct.

The union contains every element that is in A, in B, or in both. The elements belonging to exactly one set are the two non-overlapping parts, A − B and B − A. Removing the common part from the union gives their total: 86 − 31 = 55. Therefore, option C is correct; adding the values would count the intersection twice.

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 35+32+29−12−10−9+4=69. Pairwise overlaps are subtracted because they were counted twice, while the triple overlap is added once because it was subtracted too many times.

For each pairwise intersection, remove the elements that are also in the third set. Thus, elements only in A and B are 18−6=12, only in B and C are 16−6=10, and only in C and A are 14−6=8. These three disjoint regions contain exactly two-set members, so the total is 12+10+8=30. The triple intersection must be removed from every pair.

To count elements only in A, begin with all 50 elements of A. Subtract the 21 elements shared by A and B and the 19 shared by A and C. The 8 elements in all three sets were subtracted twice, so add them back once: 50−21−19+8=18. Therefore, 18 elements are in A but in neither B nor C.

The value n(A∩B)=24 includes two kinds of elements: those in A and B but not C, and those in all three sets. The latter group has 9 elements. Therefore, the number only in A and B is n(A∩B)−n(A∩B∩C)=24−9=15. Thus option C is correct; simply using 24 would incorrectly include the triple intersection.

The symmetric difference A△B contains elements that belong to exactly one of the two sets. Its standard expression is (A−B)∪(B−A). Thus the two non-overlapping outer regions of the circles are included, while the common intersection A∩B is excluded. Therefore, option C correctly describes the shaded part of the Venn diagram.

The symmetric difference A △ B consists of the elements that are in A but not in B together with the elements that are in B but not in A. These two parts are disjoint, so n(A △ B) = n(A − B) + n(B − A). Substituting the given values gives 64 = 27 + n(B − A). Hence n(B − A) = 64 − 27 = 37, so option A is correct. The common elements of A and B are excluded from the symmetric difference.

Let x = n(A ∩ B). The elements that lie in exactly one of the two sets are the elements in A but not B plus the elements in B but not A. Their number is (39 − x) + (44 − x) = 83 − 2x. Since this number is 51, we obtain 83 − 2x = 51, so 2x = 32 and x = 16. Therefore, n(A ∩ B) = 16 and option B is correct. This is also consistent with the formula for symmetric difference.

Let M be the set of students who like Mathematics and P the set who like Physics. By the inclusion–exclusion principle, n(M union P) = n(M) + n(P) − n(M intersection P) = 34 + 31 − 13 = 52. Therefore, the number who like neither subject is the total number minus the union: 72 − 52 = 20. Hence option B is correct.

The people who like tea consist of two disjoint parts: those who like only tea and those who like both tea and coffee. Therefore, n(only tea) = n(tea) − n(both) = 45 − 15 = 30. The coffee total is not needed for this calculation. The value 66 is the number who like at least one beverage, since 45 + 36 − 15 = 66, not the number who like only tea. Thus option B is correct.

A subset of B means every element of A is already contained in B. In a Venn diagram, the circle representing A lies completely inside the circle representing B. Taking the union collects all elements belonging to A or B, but A contributes no elements outside B. Therefore, A union B = B. The other choices do not generally follow from the subset relation, so option B is the only correct answer.

Because A and B are disjoint, no element is counted in both sets. Consequently, every element of A belongs to exactly one set, and every element of B also belongs to exactly one set. The number of elements in exactly one of the two sets is therefore n(A) + n(B) = 36 + 42 = 78. Equivalently, the symmetric difference has size 78 because the intersection has size zero. Hence option D is correct.

A union B is partitioned into exactly three mutually disjoint regions: the A-only region A − B, the common region A intersection B, and the B-only region B − A. Therefore, n(A union B) = n(A − B) + n(A intersection B) + n(B − A). Substituting the values gives 70 = 26 + n(A intersection B) + 19. Hence n(A intersection B) = 70 − 26 − 19 = 25. Option A is correct.

The complement of A union B contains the elements outside both A and B. Therefore, n(A union B) = n(U) − n((A union B) complement) = 100 − 18 = 82. For two finite sets, n(A union B) = n(A) + n(B) − n(A intersection B). Substituting gives 82 = 48 + 46 − n(A intersection B), so n(A intersection B) = 94 − 82 = 12. Thus option B is correct.

The symbol A ∩ B ∩ C denotes the intersection of three sets. Therefore, it contains elements that belong simultaneously to A, B, and C. In a Venn diagram with three overlapping circles, this intersection is shown in the central region common to all three circles. The number 11 means that this common central region contains 11 elements; it does not refer to a region belonging to only one set or to the outside of the circles.

The union A ∪ B ∪ C consists of every region lying inside at least one of the three sets. Since the question gives the seven mutually exclusive interior regions, we add them once each: 14 + 18 + 16 + 7 + 6 + 5 + 4 = 70. The triple-intersection value is already one separate region, so it must not be added again through another formula.

To find n(A), include every Venn-diagram region that lies inside set A. These regions are the part only in A, the part only in A ∩ B, the part only in C ∩ A, and the central part A ∩ B ∩ C. Hence n(A) = 20 + 8 + 6 + 5 = 39. Regions that lie outside A are not included in the cardinality of A.

The cardinality n(B) is obtained by adding all mutually exclusive regions contained in B. These are the region only in B, the region only in A ∩ B, the region only in B ∩ C, and the central triple-intersection region. Therefore, n(B) = 17 + 9 + 11 + 6 = 43. No region outside B should be counted.