Operations on Sets (Union, Intersection, Difference)

Use the inclusion–exclusion principle, because multiples common to A and B would otherwise be counted twice. There are floor(30/2) = 15 multiples of 2 and floor(30/5) = 6 multiples of 5. The overlap consists of multiples of lcm(2,5) = 10, namely 10, 20, and 30, so it has 3 elements. Therefore n(A ∪ B) = 15 + 6 − 3 = 18. Option A is correct; adding without subtraction gives the distractor 21.

The governing concept is intersection together with the least common multiple. A number in A ∩ B must be divisible by both 4 and 6, so it must be a multiple of lcm(4,6) = 12. The positive multiples of 12 not exceeding 50 are 12, 24, 36, and 48. There are exactly four such numbers, hence n(A ∩ B) = 4. Therefore option A is correct; the larger choices incorrectly count separate multiples rather than common ones.

A union B is partitioned into three disjoint regions: elements only in A, elements only in B, and elements in both sets. Thus n(A ∪ B) = n(A − B) + n(B − A) + n(A ∩ B). Substituting the given values gives 84 = n(A − B) + n(B − A) + 27. Rearranging, the requested sum is 84 − 27 = 57. This is also the size of the symmetric difference, so option A is correct.

“At least one language” means the union of the Hindi and English groups. By the inclusion–exclusion principle, n(H ∪ E) = n(H) + n(E) − n(H ∩ E), because students studying both languages would otherwise be counted twice. Therefore, n(H ∪ E) = 45 + 40 − 30 = 55. Hence, option A is correct.

For two subsets of a universal set, n(A ∪ B) cannot exceed n(U). Using n(A ∪ B) = n(A) + n(B) − n(A ∩ B), we require 60 ≥ 35 + 32 − n(A ∩ B). Hence n(A ∩ B) ≥ 7. This bound is attainable when the union contains all 60 elements, so the minimum possible intersection is 7.

The union A ∪ B must contain every element of A and every element of B, so its size cannot be less than either set. Therefore n(A ∪ B) ≥ max(64, 58) = 64. This minimum is attainable because B, which has 58 elements, can be completely contained in A, which has 64 elements. Thus n(A ∪ B) = 64.

The intersection A ∩ B contains elements common to both sets, so its cardinality cannot be greater than the cardinality of either set. Therefore, n(A ∩ B) ≤ min[n(A), n(B)] = min(46, 39) = 39. This maximum is attainable when every element of the smaller set B is also an element of A, so B is completely contained in A.

Use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Hence, 52 = 30 + 28 − n(A ∩ B), so n(A ∩ B) = 6. The set A − B contains elements that are in A but not in B. Therefore, n(A − B) = n(A) − n(A ∩ B) = 30 − 6 = 24. Thus option A is correct.

The values n(A − B), n(B − C), and n(C − A) describe only selected portions of the three sets. They do not reveal the sizes of regions shared by two sets or by all three sets, and those regions contribute to the union. Different Venn diagrams can therefore have the same three given differences but different union sizes. Additional common-region information is necessary.

The quantity n(A ∩ B) = 14 includes every element common to A and B, including the elements that may also belong to C. The triple intersection A ∩ B ∩ C contains 6 of these elements. “Only in A ∩ B” means the elements common to A and B but outside C, so subtract the triple-overlap: 14 − 6 = 8. Therefore, option A is the unique correct answer.

Students studying only Mathematics = 56 − 27 = 29, and students studying only Science = 49 − 27 = 22. Therefore, the number studying exactly one subject is 29 + 22 = 51. Equivalently, n(M △ S) = n(M) + n(S) − 2n(M ∩ S) = 56 + 49 − 54 = 51. The value 78 is the union, not the exactly-one count.

Use the inclusion–exclusion formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Therefore, n(A ∩ B) = 64 + 59 − 91 = 32. The symmetric difference A △ B contains elements belonging to exactly one of the two sets, so n(A △ B) = n(A ∪ B) − n(A ∩ B) = 91 − 32 = 59. Hence, option A is correct.

Because A is a subset of B, every element of A lies inside B. The set B − A therefore consists of the elements belonging to B but not to A. Its cardinality is n(B − A) = n(B) − n(A) = 83 − 47 = 36. The universal-set size is not needed for this difference, so option A is the only correct answer.

Since A and B are disjoint, they have no common elements, so n(A ∪ B) = n(A) + n(B) = 44 + 52 = 96. The complement of A ∪ B contains all elements of the universal set U that are outside the union. Thus, n((A ∪ B)′) = n(U) − n(A ∪ B) = 130 − 96 = 34. Therefore, option A is correct.

Apply the inclusion–exclusion formula for two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given expressions gives 78 = (5x + 4) + (4x + 7) − (2x + 3). Simplifying, 78 = 7x + 8, so 7x = 70 and x = 10. Substitution confirms the result, so option A is correct.

The three regions forming A ∪ B are A − B, B − A, and A ∩ B. These regions are pairwise disjoint, so their cardinalities can be added: 71 = (3x + 2) + (2x + 5) + (x + 4) = 6x + 11. Hence 6x = 60 and x = 10. With x = 10, the three region sizes are 32, 25, and 14, whose sum is 71; therefore option A is correct.

An element of A ∩ B must be divisible by both 6 and 8. Therefore all common elements are multiples of lcm(6,8) = 24. Within U = {1,2,...,72}, the relevant multiples are 24, 48, and 72. Equivalently, their number is floor(72/24) = 3. Hence n(A ∩ B) = 3 and option A is correct; counting all multiples of either number would incorrectly produce a larger value.

There are floor(90/5) = 18 multiples of 5 and floor(90/9) = 10 multiples of 9. Numbers counted in both sets are multiples of lcm(5, 9) = 45; within 1 to 90 these are 45 and 90, so there are 2. By inclusion-exclusion, n(A ∪ B) = 18 + 10 − 2 = 26. Thus option A is correct.

The intersection A ∩ B contains numbers that satisfy both conditions: they must be prime and even. Every even number greater than 2 is divisible by 2 and therefore has at least two factors, so it is not prime. The number 2 itself has exactly two positive factors, 1 and 2, and is the only even prime. Hence A ∩ B = {2}.

The intersection requires numbers that are both perfect squares and multiples of 3. The square numbers from 1 through 36 are 1, 4, 9, 16, 25, and 36. Checking divisibility by 3 leaves 9 and 36 only; the other squares are not multiples of 3. Hence A ∩ B = {9,36}, which has cardinality 2. Therefore option A is correct; counting every square or every multiple of 3 would answer a different question.