Venn Diagrams

Apply 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). Substituting the given values gives 160 = 70 + 75 + 80 − S + 18, where S is the required pairwise-intersection sum. Thus 160 = 243 − S, so S = 83.

Both the assertion and the reason are true. When n(A) and n(B) are added, every element common to A and B appears once in n(A) and once in n(B), so the intersection is counted twice. The union should count each element only once; therefore n(A ∩ B) must be subtracted once. This gives the inclusion–exclusion formula n(A ∪ B) = n(A) + n(B) − n(A ∩ B), so the reason correctly explains the assertion.

Both statements are true, and the reason correctly explains the assertion. For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). If A and B are disjoint, their intersection is the empty set, so n(A ∩ B) = 0. Consequently, no element is counted in both sets, and the union contains the sum of the elements in A and B. Hence n(A ∪ B) = n(A) + n(B).

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.

Let the seven disjoint Venn regions be the only-A, only-B, only-C, AB-only, BC-only, AC-only, and ABC regions. Since the total union is 155 and A ∪ B is 118, only-C = 155 − 118 = 37. Similarly, only-A = 155 − 124 = 31 and only-B = 155 − 121 = 34. The set B contains only-B plus the three regions involving B, so the sum of B-only, AB-only, BC-only, and ABC is 64. Combining these region equations with the total gives AC-only = n((A ∩ C) − B) = 23.

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 112 + 104 + 93 − 46 − 38 − 35 + 17 = 207. The elements in none of the sets are outside the union, so their number is n(U) − n(A ∪ B ∪ C) = 240 − 207 = 33.

The elements only in C are those in C but not in A or B. Start with n(C) = 71. Subtract n(B ∩ C) = 29 and n(C ∩ A) = 27, because these are the portions of C shared with the other sets. The central region A ∩ B ∩ C, containing 12 elements, was subtracted twice, so add it back once. Thus only-C = 71 − 29 − 27 + 12 = 27.

“Exactly one set” means elements that belong to A only, B only, or C only. It excludes every pairwise-overlap-only region and the central region common to all three sets. Therefore add only the three single-set regions: 26 + 31 + 24 = 81. The values 14, 16, 12, and 9 describe elements in at least two sets and must not be included.

The union A ∪ B consists of three disjoint Venn regions: A − B, B − A, and A ∩ B. Therefore n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values, 126 = 48 + 41 + n(A ∩ B). Thus n(A ∩ B) = 126 − 48 − 41 = 37. Each region is counted exactly once in this decomposition.

Use the complement relationship to find the sizes of A and B. Since n(A′) = n(U) − n(A), n(A) = 160 − 63 = 97. Similarly, n(B) = 160 − 72 = 88. Now apply the two-set union formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 97 + 88 − 39 = 146. Thus the union contains 146 elements.

The universal set is divided into four mutually exclusive Venn-diagram regions: A only, B only, A ∩ B, and outside A ∪ B. Their given cardinalities are 36, 44, 25, and 18 respectively. Therefore, n(U) = 36 + 44 + 25 + 18 = 123. Each element of U belongs to exactly one of these four regions, so no region is counted twice.

Let B denote the Biology group and C denote the Chemistry group. By the inclusion–exclusion principle, n(B ∪ C) = n(B) + n(C) − n(B ∩ C) = 68 + 57 − 32 = 93. Students studying neither subject lie outside this union, so their number is 110 − 93 = 17. Therefore, option A is correct.

Because A is a subset of B, the set B − A contains the elements of B that are not in A, so its size is 91 − 52 = 39. The complement B' contains elements outside B, so n(B') = 140 − 91 = 49. These two regions are disjoint, because one is inside B and the other is outside B. Therefore the required size is 39 + 49 = 88.

The number n(A ∩ C) = 31 includes every student who belongs to both A and C, including those who also belong to B. The triple intersection has 13 students, so these must be removed to obtain the pair-only region. Hence the number in A ∩ C but not B is 31 − 13 = 18. The value 31 would incorrectly include the triple intersection.

Between 1 and 96, the number of multiples of 6 is floor(96/6) = 16, and the number of multiples of 10 is floor(96/10) = 9. Elements divisible by both 6 and 10 are multiples of lcm(6,10) = 30; there are floor(96/30) = 3. By inclusion–exclusion, n(A ∪ B) = 16 + 9 − 3 = 22.

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.

By De Morgan’s law, (A ∪ B)' means the region outside both A and B. Intersecting this region with C restricts the answer to points that are inside C but outside A and also outside B. Therefore the represented region is C − (A ∪ B), or C ∩ A' ∩ B'. Hence option A is the only correct description.

The difference A − (B ∪ C) contains elements that belong to A but do not belong to the union B ∪ C. Not belonging to B ∪ C means being outside both B and C, which is represented by B' ∩ C'. Thus A − (B ∪ C) = A ∩ (B ∪ C)' = A ∩ B' ∩ C' by De Morgan’s law. Option A is correct.

If A is a subset of B, every element of A is also contained in B. Therefore, any element that is outside B cannot be in A; otherwise it would simultaneously be in A and outside B, contradicting A ⊆ B. Hence every element of B' belongs to A', so B' ⊆ A'. This is the contrapositive form of subset inclusion, and option A is correct.

The identity A ∩ B = A means that every element of A lies in B, so A ⊆ B. Similarly, B ∩ C = B means B ⊆ C. By transitivity of subset inclusion, A ⊆ C. When A is a subset of C, intersecting A with C leaves A unchanged; therefore A ∩ C = A. No equality between A and B or between A and C is required, so option A is correct.