Venn Diagrams

Use the two-set inclusion–exclusion identity n(A∪B)=n(A)+n(B)−n(A∩B). Rearranging gives n(A∩B)=n(A)+n(B)−n(A∪B). Substitution yields 52+47−79=99−79=20. The common elements must be subtracted because adding the two set totals counts them twice. Therefore, option B is the unique correct answer.

The total n(A)=44 includes two parts: the region only in A and the common region A∩B. Therefore, the only-A region is found by subtracting the overlap from the total of A: n(A only)=n(A)−n(A∩B)=44−19=25. The value 19 is the intersection, while 44 includes both parts. Hence option B is correct.

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 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 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.

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.

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.

A union B consists of three mutually exclusive regions in a Venn diagram: the elements only in A, the elements common to both A and B, and the elements only in B. Therefore, n(A ∪ B) = n(A − B) + n(A ∩ B) + n(B − A). Substituting the values gives 74 = 28 + n(A ∩ B) + 19, so n(A ∩ B) = 74 − 28 − 19 = 27. Hence, option C is correct.

The condition A∩B=∅ means that A and B are disjoint; they have no common elements. The general formula is n(A∪B)=n(A)+n(B)−n(A∩B). Since the intersection has zero elements, n(A∪B)=45+37−0=82. Therefore option D is correct. Adding the two cardinalities is valid here precisely because the sets do not overlap.

The set A∩B includes every element common to A and B, including those that may also lie in C. Therefore it consists of the region only in A and B together with the triple intersection A∩B∩C. To isolate the first region, subtract the triple intersection: 29−12 = 17. Option B counts the entire pairwise intersection, and option D counts only the triple part. Hence option A 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.

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.

A consists of every region inside the circle representing A. These regions are only A, only A∩B, only A∩C, and the central triple intersection. Since the regions are disjoint, add their cardinalities: n(A)=18+12+10+6=46. The regions only in B or only in C are not part of A and must not be included.

Apply the distributive identity \((X\cup Y)\cap(X\cup Z)=X\cup(Y\cap Z)\). Taking \(X=A\), \(Y=B\), and \(Z=B^c\), the expression becomes \(A\cup(B\cap B^c)\). A set and its complement are disjoint, so \(B\cap B^c=\varnothing\). Therefore the result is \(A\cup\varnothing=A\), making option A correct.

The symmetric difference A Δ B consists of elements that are in A or B but not in both. Thus, A ∪ B is partitioned into two disjoint parts: A Δ B and A ∩ B. Therefore, n(A ∪ B) = n(A Δ B) + n(A ∩ B). Substituting the values gives 137 = 96 + n(A ∩ B), so n(A ∩ B) = 137 − 96 = 41. Hence, option B is correct.

For two finite sets, the inclusion-exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values gives 96 = 58 + 49 − n(A ∩ B). Therefore, n(A ∩ B) = 107 − 96 = 11. The subtraction removes the double counting of common elements, so option A is correct.

The union A ∪ B ∪ C represents everyone who belongs to at least one of the three sets. The people in none of the sets are outside this union, so their number is found by subtracting the union from the universal set: n(U) − n(A ∪ B ∪ C) = 200 − 164 = 36. Therefore, 36 people belong to none of the three sets.

The 39 students who chose Physics include both the students who chose only Physics and the 22 students who chose both subjects. Therefore, the number choosing only Physics is 39−22=17. The value 24 is only Mathematics, while 63 is the number choosing at least one subject, not only Physics. Hence option A is correct.

Because A ⊆ B, every element of A lies inside B. The difference B − A therefore contains the elements that belong to B but not to A. For finite sets, n(B − A) = n(B) − n(A) = 64 − 28 = 36. Thus option A is correct. The values 28 and 64 refer to A and B themselves, while 92 incorrectly adds them.

A∩B=∅ means that A and B are disjoint, so they have no common elements. For any two sets, n(A∪B)=n(A)+n(B)−n(A∩B). Substituting the values gives 31+44−0=75. Since no element is counted in both sets, simple addition is valid. Therefore, option A is the correct union cardinality.