Venn Diagrams

For two sets, the union consists of three non-overlapping regions: the elements only in A, represented by A − B; the elements only in B, represented by B − A; and the common elements, represented by A ∩ B. Hence n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the values gives 70 = 24 + 31 + n(A ∩ B), so n(A ∩ B) = 15. Option A is correct.

When the circle representing A lies completely inside the circle representing B, every element of A is also an element of B. This is precisely the definition of A being a subset of B, written A ⊆ B. The reverse inclusion is not necessarily true because B may contain additional elements outside A. The sets are not disjoint, so option A is correct.

The difference A − B consists of all elements that belong to A but do not belong to B. In a Venn diagram, this is the portion of circle A excluding the common overlap with circle B. A ∩ B represents the shared region, B − A represents the part only in B, and A ∪ B represents everything in either set. Hence option A is correct.

When A ⊆ B, every element of A is already contained in B. Taking the union adds no new element to B, so A ∪ B = B. The elements common to A and B are exactly all elements of A, so A ∩ B = A. Therefore option A gives the two results in the required order; option B reverses them.

People in none of the sets are outside the union A ∪ B ∪ C. In set notation, this group is (A ∪ B ∪ C)' within the universal set U. The universal set is divided into the union and its complement, so n(U) = n(A ∪ B ∪ C) + n((A ∪ B ∪ C)'). Therefore, n((A ∪ B ∪ C)') = 200 − 164 = 36. Hence option A is correct.

The union A ∪ B consists of every element that is in A, in B, or in both sets. In a Venn diagram, this means shading the complete interior of the A circle and the complete interior of the B circle, including their overlapping lens-shaped region. The outside region is excluded, so option A is correct.

The intersection A ∩ B consists of elements common to both sets. If its cardinality equals n(A), every element of A is included in the intersection and therefore also belongs to B. Hence A is a subset of B, written A ⊆ B. The condition n(A) > 0 confirms that A is non-empty. Option B reverses the inclusion, while C and D contradict the non-empty intersection condition.

The universal set U contains every region shown in the Venn diagram, including the region outside both A and B. Therefore, n(U) = 14 + 22 + 18 + 11 = 65. The outside region must not be omitted because it still contains elements of U that belong to neither set. Thus option A is correct.

The phrase “exactly two sets” refers only to the three pairwise-only regions: A∩B excluding C, B∩C excluding A, and C∩A excluding B. Therefore, add 13, 11, and 9: 13+11+9=33. The central region containing all three sets is not included because those elements belong to three sets, not exactly two.

“At least two sets” includes the three pairwise-only regions and the central region common to all three sets. It therefore excludes the single-set-only regions and includes 11 elements in only A∩B, 13 in only B∩C, 9 in only C∩A, and 6 in A∩B∩C. The total is 11+13+9+6=39. Hence option A is correct.

In a two-set Venn diagram, A ∪ B consists of three non-overlapping regions: A − B, B − A, and A ∩ B. Therefore n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting gives 112 = 43 + 37 + n(A ∩ B). Since 43 + 37 = 80, the intersection has 112 − 80 = 32 elements. Thus option A is the unique correct answer.

The positive multiples of 5 not exceeding 20 are 5, 10, 15, and 20. Of these, 5 and 15 are odd, whereas 10 and 20 are even. Therefore B ∩ C = {5, 15}, and its cardinality is 2. Set A gives the common upper bound but is not needed for this particular intersection.

The difference B−C means elements belonging to B but excluded from C. Intersecting this set with A keeps only elements that also belong to A. Therefore A ∩ (B−C) is precisely the part common to A and B while lying outside C. It excludes the central triple-overlap region.

The relation A⊆B⊆C means every element of A is in B and every element of B is in C. Consequently, all elements of A and B are already contained in C. Taking the union therefore adds nothing beyond C, while every element of C is included in the union. Hence A ∪ B ∪ C=C.

The universal set U consists of all elements that lie inside A ∪ B ∪ C and all elements outside that union. The complement (A ∪ B ∪ C)′ contains 25 elements, while the union contains 190 elements. These two disjoint regions together make the whole universal set. Therefore n(U) = n(A ∪ B ∪ C) + n((A ∪ B ∪ C)′) = 190 + 25 = 215. The information about exactly two sets is not needed.

The intersection A ∩ B contains numbers that are both perfect squares and even. The square numbers from 1 to 64 are 1, 4, 9, 16, 25, 36, 49, and 64. The even members are 4, 16, 36, and 64. Thus A ∩ B = {4, 16, 36, 64}, which has four elements, so n(A ∩ B) = 4. The other options do not count this set correctly.

When A and B are disjoint, they have no common elements. The union A ∪ B contains all elements of A together with all elements of B. Removing A from this union removes only the elements belonging to A, leaving every element of B. Thus (A ∪ B) − A = B. The disjointness condition ensures that no element of B is also removed as an element of A, so option A is correct.

A−B is empty exactly when every element of A belongs to B, so n(A−B)=0 implies A⊆B. Similarly, n(B−C)=0 implies B⊆C. Subset relations are transitive: if every element of A is in B and every element of B is in C, then every element of A is in C. Hence A⊆C must hold. The condition n(A)>0 does not change this conclusion.

By definition, the complement \(A^c\) consists of every element in the universal set \(U\) that does not belong to \(A\). In a Venn diagram, \(U\) is normally shown as a rectangle and \(A\) as a circle inside it. Therefore, \(A^c\) is the portion of the rectangle outside the circle, not the area outside the rectangle.