Operations on Sets (Union, Intersection, Difference)

Because A is a subset of B, every element of A is already included in B. The set B − A therefore contains the elements that belong to B but do not belong to A. No overlap is counted in this difference, so n(B − A) = n(B) − n(A) = 73 − 28 = 45. Thus, option B is correct.

The governing concept is the cardinality formula for the union of two finite sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since A ∩ B = ∅, the sets are disjoint and their intersection has zero elements. Therefore, n(A ∪ B) = 33 + 41 − 0 = 74. Options C and D give the size of only one set, while option B is the difference, so option A is correct.

The intersection of two sets contains exactly the elements that occur in both sets. Comparing the elements one by one, b appears in A and B, d appears in A and B, and h appears in A and B. The elements a and f occur only in A, while c and g occur only in B. Therefore, A ∩ B = {b, d, h}, so option C is correct.

The difference A − B contains elements that are in A but not in B. From A, the elements 6, 8, and 12 also occur in B, so they must be removed. The remaining elements are 2, 4, and 10. The element 14 is not in A, so it cannot belong to A − B. Therefore, option A is correct.

At least one of the two arts means the union of the music and painting groups. By the two-set inclusion–exclusion formula, n(M∪P)=n(M)+n(P)−n(M∩P). Therefore, n(M∪P)=58+49−22=85. The 22 students who learn both are subtracted once because they were counted in both 58 and 49.

Two sets are called disjoint when they have no common element. The set containing elements common to both sets is their intersection, so the definition is written as A intersection B = the empty set. Disjoint sets need not themselves be empty; for example, {1, 2} and {3, 4} are disjoint. Therefore, A union B need not be empty, the sets need not be equal, and neither set must be a subset of the other. Option A is correct.

The intersection A∩B is the set of elements that occur in both A and B. Comparing the two listed sets, 4, 6, and 12 appear in each set, while 2, 8, and 10 occur only in A and 3 and 9 occur only in B. Thus A∩B={4,6,12}, so option A is correct. Option D is the union, not the intersection.

The union A∪B contains every distinct element that belongs to A or B. Combining the two sets gives a, b, c, d, e, g, i, and j. The repeated elements c and g are written only once because sets do not repeat members. Hence A∪B={a,b,c,d,e,g,i,j}, option B.

The difference A−B contains elements that are in A but not in B. From A, the elements 10, 20, and 30 also occur in B, so they must be removed. The remaining elements are 5, 15, and 25. Therefore A−B={5,15,25}, making option C correct. Option A is the common part, and option D is the union.

“At least one art” means the union of the dance and music groups. For two sets, n(D∪M)=n(D)+n(M)−n(D∩M), because students learning both are counted twice in the simple sum. Therefore n(D∪M)=63+58−26=95. Thus, 95 students learn at least one of the two arts, so option A is correct.

The 52 people who like apples include both people who like only apples and people who like both apples and bananas. Since 19 people like both fruits, remove that overlap from the apple total: only apples=n(A)−n(A∩B)=52−19=33. Therefore, 33 people like only apples, so option A is correct.

The general formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Since n(A ∩ B) = 0, the sets are disjoint and the formula becomes 92 = 40 + n(B). Thus n(B) = 92 − 40 = 52. Option A merely repeats n(A), while option C is the union total, not the size of B. Therefore, B is the only correct answer.

Use the distributive law for sets: \((A\cap B)\cup(A\cap B^c)=A\cap(B\cup B^c)\). A set and its complement together form the universal set, so \(B\cup B^c=U\). Hence the expression becomes \(A\cap U=A\). In a Venn diagram, the two terms divide A into its parts inside and outside B, so option A is the only correct answer.

The intersection A ∩ B is the region shared by both sets. Removing this common region from A leaves only those elements that belong to A but do not belong to B. By the definition of set difference, this remaining region is A \ B. In a two-circle Venn diagram, it is the part of circle A outside the overlap. Therefore, option A is the only correct answer.

The intersection A ∩ B contains numbers that satisfy both conditions: they must be prime and even. Every even number greater than 2 has at least the divisors 1, 2, and itself, so it is composite. The number 2 is the only even prime. Therefore A ∩ B = {2}, making option A correct; option C contains many composite even numbers.

“Only in A” means the elements belonging to A but not to B, represented by A − B or A ∩ Bᶜ. The total number in A includes the 38 students who are also in B. Therefore, subtract the intersection from A: n(A − B) = n(A) − n(A ∩ B) = 72 − 38 = 34. Thus, option A is correct.

The phrase “at least one” means belonging to A or B or both, which is the union A ∪ B. The problem already gives three separate, non-overlapping regions: only A has 40 students, only B has 28, and both sets have 12. Therefore, n(A ∪ B) = 40 + 28 + 12 = 80. Option A is correct.

The governing concept is set intersection: B ∩ C contains only numbers satisfying both conditions. The even natural numbers up to 12 are 2, 4, 6, 8, 10, and 12, while the multiples of 3 are 3, 6, 9, and 12. Their common elements are therefore B ∩ C = {6, 12}. Since this set has two elements, n(B ∩ C) = 2, so option A is correct. Options B, C, and D do not match the actual intersection.

The governing concept is set difference: A − B consists of elements that belong to A but do not belong to B. The primes from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19. All primes except 2 are odd and therefore lie in B. Since 2 is even, it is the only prime excluded from B, giving A − B = {2}. Thus option A is correct; option C is A’s odd-prime part, not the difference.

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.

“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.

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.

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.

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.

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}.