Venn Diagrams

The union of the three pairwise intersections contains every element that belongs to at least two of the sets. The three exclusive pairwise regions contribute 14, 11, and 13 elements. The central region A ∩ B ∩ C belongs to all three pairwise intersections, but it must be counted only once in their union. Therefore, the required number is 14 + 11 + 13 + 9 = 47, so option A is correct.

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 of three sets contains every region lying inside at least one of the three circles. These seven disjoint regions contain 23, 17, 19, 8, 9, 10, and 5 elements. Adding each region once gives 23 + 17 + 19 + 8 + 9 + 10 + 5 = 91. Therefore, option A is correct.

Since A ∩ B = ∅, the sets A and B are disjoint, so they have no common elements. Thus, n(A ∪ B) = n(A) + n(B) = 41 + 39 = 80. The complement of A ∪ B contains all elements of the universal set that are outside both A and B. Therefore, n((A ∪ B)′) = n(U) − n(A ∪ B) = 100 − 80 = 20. Option A is correct.

The students studying exactly Mathematics are 72 − 38 = 34, because those studying both subjects must be removed. The students studying exactly Physics are 64 − 38 = 26. Adding these non-overlapping groups gives 34 + 26 = 60. Equivalently, n(A △ B) = n(A) + n(B) − 2n(A ∩ B) = 60. Thus, option A is correct.

The set A contains its exclusive part A − B and its common part A ∩ B. Hence, n(A ∩ B) = n(A) − n(A − B) = 58 − 21 = 37. Using the inclusion–exclusion formula, n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 58 + 49 − 37 = 70. Therefore, the complement has 110 − 70 = 40 elements. Thus, option A is correct.

The total n(M ∩ P) = 25 includes the students in all three subjects, and n(C ∩ M) = 20 also includes that same centre. To obtain only M, subtract both pairwise intersections from n(M), then add the centre once because it was subtracted twice: 60 − 25 − 20 + 10 = 25. Option A is correct.

The expression A − (B ∪ C) contains elements that belong to A but do not belong to the union B ∪ C. Not belonging to the union means that an element is outside both B and C. Therefore, the required Venn region is the portion inside A alone, excluding every overlap with B or C. 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.

For two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B), because the common elements are counted twice when n(A) and n(B) are added. Substituting the given expressions gives 30 = (3x + 2) + (2x + 5) − (x + 1) = 4x + 6. Therefore, 4x = 24 and x = 6. Hence, option A is correct.

The union A ∪ B is partitioned into three disjoint regions: A − B, B − A and A ∩ B. Therefore, 45 = (2x+3) + (x+7) + (x−1) = 4x+9. Subtracting 9 gives 4x = 36, and dividing by 4 gives x = 9. Option A is correct. The other values do not satisfy the stated union equation.

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 universal set U contains every element under consideration. If n(A ∪ B) = n(U), then A ∪ B contains all elements of U. Therefore, no element lies outside A ∪ B, so its complement (A ∪ B)' is the empty set. The intersection and the two difference regions may still contain elements; the given condition only guarantees that the outside region is empty.

The statements A ∩ B = ∅ and B ∩ C = ∅ only say that B has no common element with A and no common element with C. They do not compare A directly with C. For example, let A = {1}, C = {1}, and B = {2}; then both given intersections are empty, but A ∩ C = {1}, which is not empty.

De Morgan’s law states that the complement of a union equals the intersection of the complements: (A ∪ B)' = A' ∩ B'. The condition that A and B are disjoint is not needed for this identity; it is true for any two subsets of the same universal set. The complement consists of elements outside both A and B.

The union is partitioned into three mutually exclusive types of Venn regions: elements belonging to exactly one set, elements belonging to exactly two sets, and elements belonging to all three sets. Thus 140 = 65 + 51 + n(A ∩ B ∩ C). Solving gives n(A ∩ B ∩ C) = 140 − 116 = 24. Therefore 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). Thus, the union is 82+76+69−34−29−27+12=149. The people in none of the sets are outside the union, so the required number is n(U)−n(A∪B∪C)=180−149=31. Therefore, 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.

The governing Venn-diagram relation is n(A ∪ B) = n(A) + n(B) − n(A ∩ B), since the common region is counted twice in the first two terms. Substitution gives 72 = (4x+5) + (3x+8) − (2x+1). Simplifying, 72 = 5x + 12, so 5x = 60 and x = 12. The values then give valid nonnegative region counts, confirming the result. Hence option A is uniquely correct.