Venn Diagrams

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.

The union A ∪ B contains B and possibly some additional elements from A. If its cardinality is exactly equal to n(B), there can be no elements of A outside B. Equivalently, using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), the given equality gives n(A) = n(A ∩ B), so every element of A lies in B. Therefore A ⊆ B, making option A correct.

The symmetric difference A △ B consists of elements that belong to A or B but not to both. Since n(A ∩ B) = 0, the sets are disjoint, so every element of A and every element of B belongs to the symmetric difference. Therefore n(A △ B) = n(A) + n(B) = 39 + 48 = 87. Option A is correct. In general, n(A △ B) = n(A) + n(B) − 2n(A ∩ B).

The number who read at least one of the two types is 150 − 28 = 122, because 28 students read neither. Let N be the novel readers and P the poetry readers. By inclusion-exclusion, n(N ∪ P) = n(N) + n(P) − n(N ∩ P). Therefore 122 = 83 + 69 − n(N ∩ P), so n(N ∩ P) = 30. Hence option A is correct.

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 97 + 88 + 82 − 41 − 36 − 34 + 18 = 174. The triple intersection is added once at the end because it was subtracted too many times. Thus option A is correct.

The phrase “at least two sets” includes every element lying in exactly two sets as well as every element lying in all three sets. The relevant Venn regions are therefore the three pairwise-only regions and the central triple-intersection region. Adding them gives 19 + 17 + 21 + 11 = 68. The central region is included only once because each element must be counted once.

For three finite sets, inclusion–exclusion gives 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 151=82+77+69−32−28−25+x=143+x. Hence x=151−143=8. The triple intersection is therefore 8, while the other choices do not satisfy the formula.

By distributivity, A∩(B∪C)=(A∩B)∪(A∩C). The stated region contains three disjoint parts: the A∩B-only region with 23 elements, the A∩C-only region with 18 elements, and the central triple intersection with x elements. Therefore 52=23+18+x, so x=11. Hence option A is correct; 41 is only the sum of the two exclusive parts.

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.

Let S=n(A∩B)+n(B∩C)+n(C∩A). Inclusion–exclusion for three sets states n(A∪B∪C)=n(A)+n(B)+n(C)−S+n(A∩B∩C). Substituting gives 184=86+91+88−S+23=288−S. Hence S=288−184=104. Therefore option A is correct; 81 would result from mishandling the triple-intersection term.

Each difference includes more than one possible Venn region; for example, A−B may contain the part only in A and the part in A∩C but not B. The three given counts therefore do not specify all single, pairwise, and triple regions. Different common-region counts can produce different unions, so the union cannot be determined uniquely.

Both the assertion and the reason are true. Since A ⊆ B, every element of A already belongs to B. The intersection A ∩ B contains elements common to both sets. Therefore, every element of A is retained and no element is added from B, so A ∩ B = A. The reason directly explains why the assertion holds.

The universal set U is divided into two disjoint regions: the union A ∪ B ∪ C and its complement (A ∪ B ∪ C)′. Hence n(U) = n(A ∪ B ∪ C) + n((A ∪ B ∪ C)′) = 215 + 35 = 250. The information that exactly two sets have 62 elements is irrelevant to this particular question because the union size is already given.

The quantity n(B ∩ C) = 34 includes every element common to B and C, including those that also belong to A. The central three-set intersection contains 16 elements. Therefore, the region belonging only to B and C is n(B ∩ C) − n(A ∩ B ∩ C) = 34 − 16 = 18. Thus the correct answer is 18.

Use inclusion–exclusion. There are floor(180/12) = 15 multiples of 12, floor(180/15) = 12 multiples of 15, and floor(180/20) = 9 multiples of 20. The pairwise common counts are 3, 3, and 3, while the common triple count is 3. Thus the union has 15 + 12 + 9 − 3 − 3 − 3 + 3 = 30 elements. Its complement has 180 − 30 = 150 elements.

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.