ICMA Foundation- FBMS- Categorised PYQ- Class 11-Maths-Chapter-1-Sets -

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ICMA Foundation- FBMS- Categorised PYQ- Class 11-Maths-Chapter-1-Sets

Fundamentals of Business Management and Statistics

❓ 9 Questions

📅 Dec 2023, June 2024, Dec 2024, June 2025, Dec 2025, June 2026

📌 Answers are locked once submitted — results and explanations appear at the end.

ICMA Foundation- FBMS- Categorised PYQ- Class 11-Maths-Chapter-1-Sets

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QUESTION 1 OF 9

If P and Q are two sets then P Δ Q is equal to: [PYQ Dec 2023]

Answer Review

1 If P and Q are two sets then P Δ Q is equal to: [PYQ Dec 2023]

A P ∩ Q

B P − Q

C (P − Q) ∪ (Q − P)

D (P − Q) ∩ (Q − P)

💡 Short Explanation

� P Δ Q represents the symmetric difference between two sets. • It includes all elements that are in P or Q, but NOT in both. • Mathematically, it combines the elements exclusive to P and exclusive to Q.

✅ Correct Answer Explanation

�� The symmetric difference of two sets P and Q, denoted by P Δ Q, is defined as the set of elements which belong to either P or Q but not to their intersection. This is obtained by taking the union of the differences: (P − Q) which are elements only in P, and (Q − P) which are elements only in Q. Therefore, P Δ Q = (P − Q) ∪ (Q − P).

❓ Why Other Options Are Incorrect

� Option A → P ∩ Q represents the intersection (elements common to both), which is exactly what symmetric difference excludes. • Option B → P − Q only represents elements strictly in P, completely ignoring elements strictly in Q. • Option D → The intersection of (P − Q) and (Q − P) is always an empty set (null set) since they are mutually disjoint.

🎯 Strategy

Used • Substitution Application: → Substitute the mathematical definition of symmetric difference to find the exact equivalent expression among the options. Final Logic: → Symmetric difference combines exclusive parts of both sets via a union.

🧠 Memory Trick

�� Δ means "Different". Elements in P different from Q, UNION elements in Q different from P.

📚 Sets – Topic: Operations on Sets-Page 16

2 If the sets A = {2, 4, 6}, B = {2, 3, 5} and C = {1, 2, 5, 6}, then A ∪ (B ∩ C) is: [PYQ Dec 2023]

A {2, 4, 5, 6}

B {2, 5}

C {2}

D {1, 2, 3, 4, 5, 6}

💡 Short Explanation

� Find the intersection of B and C first (B ∩ C). • Then find the union of A with the result. • Do not duplicate elements when writing the final set.

✅ Correct Answer Explanation

�� Step 1: Compute (B ∩ C), which is the set of elements common to both B and C. B = {2, 3, 5} and C = {1, 2, 5, 6}, so (B ∩ C) = {2, 5}. Step 2: Compute A ∪ (B ∩ C), which combines all elements from A and the result of Step 1 without repetition. A = {2, 4, 6} and (B ∩ C) = {2, 5}. Therefore, {2, 4, 6} ∪ {2, 5} = {2, 4, 5, 6}.

❓ Why Other Options Are Incorrect

� Option B → Represents only (B ∩ C), ignoring the union with A. • Option C → Represents A ∩ B ∩ C, the elements common to all three sets. • Option D → Represents A ∪ B ∪ C, the union of all three sets.

🎯 Strategy

Used • Substitution Application: → Solve the brackets first (BODMAS for sets), substitute the resulting set back into the main expression, and compute. Final Logic: → B ∩ C = {2, 5}, and adding {2, 4, 6} via union yields {2, 4, 5, 6}.

🧠 Memory Trick

�� Brackets first: Intersect B & C, then Unite with A.

📚 Sets – Topic: Operations on Sets-Page 15

3 If A and B be two sets such that n(A) = 70, n(B) = 60 and n(A ∪ B) = 110, then n(A ∩ B) is [PYQ Dec 2024]

A 240

B 50

C 40

D 20

💡 Short Explanation

� Uses the principle of inclusion-exclusion. • Formula: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). • Substitute the given values to find the intersection.

✅ Correct Answer Explanation

�� According to the cardinal number formula for sets: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting the given values: 110 = 70 + 60 − n(A ∩ B). This simplifies to 110 = 130 − n(A ∩ B). Therefore, n(A ∩ B) = 130 − 110 = 20.

❓ Why Other Options Are Incorrect

� Option A → 240 is obtained by incorrectly adding all given numbers (70+60+110). • Option B → 50 is an arbitrary miscalculation. • Option C → 40 is an arbitrary miscalculation.

🎯 Strategy

Used • Substitution Application: → Plug the values into the standard set theory formula and solve for the unknown. Final Logic: → 130 - 110 = 20.

🧠 Memory Trick

�� Union = Sum of Sets - Intersection.

📚 Sets – Topic: Operations on Sets-Page 23

4 If A and B be any two sets, then (A ∩ B) ∪ (A ∩ Bᶜ) is [PYQ Dec 2024]

A A

B B

C Universal set

D Null set

💡 Short Explanation

� The expression can be simplified using the Distributive Law. • B ∪ Bᶜ equals the Universal set. • The intersection of any set with the Universal set is the set itself.

✅ Correct Answer Explanation

�� Using the Distributive Law of sets in reverse: (A ∩ B) ∪ (A ∩ Bᶜ) = A ∩ (B ∪ Bᶜ). According to complement laws, B ∪ Bᶜ equals the Universal set (U). Finally, the intersection of set A with the Universal set U is simply A (A ∩ U = A). Therefore, the entire expression simplifies to set A.

❓ Why Other Options Are Incorrect

� Option B → The expression distributes A, not B. • Option C → The Universal set is the result of B ∪ Bᶜ, not the final intersection. • Option D → The Null set would be the result if it were A ∩ Aᶜ.

🎯 Strategy

Used • Option Grouping Application: → Group the common set A using the distributive property to simplify the algebraic set expression. Final Logic: → A ∩ (B ∪ Bᶜ) = A ∩ U = A.

🧠 Memory Trick

�� A intersected with B, plus A intersected with "not B" just covers all parts of A.

📚 Sets – Topic: Properties of Operations on Sets-Page 16

5 The number of proper subsets of the set {a, e, i, o, u} is [PYQ Dec 2025]

A 32

B 31

C 30

D 28

💡 Short Explanation

� Count the number of elements (n) in the set. • The total number of subsets is 2ⁿ. • A proper subset excludes the set itself, so subtract 1.

✅ Correct Answer Explanation

�� The set {a, e, i, o, u} contains 5 elements (n = 5). The total number of subsets for a set with n elements is 2ⁿ = 2⁵ = 32. A proper subset is defined as any subset of a set that is not equal to the set itself. Therefore, the number of proper subsets is 2ⁿ − 1 = 32 − 1 = 31.

❓ Why Other Options Are Incorrect

� Option A → 32 is the total number of subsets, including the improper subset (the set itself). • Option C → 30 incorrectly subtracts 2, perhaps confusing it with non-empty proper subsets. • Option D → 28 is an arbitrary miscalculation.

🎯 Strategy

Used • Substitution Application: → Substitute n=5 into the proper subset formula 2ⁿ - 1. Final Logic: → 2⁵ - 1 = 32 - 1 = 31.

🧠 Memory Trick

�� Proper subsets = Total subsets - 1 (the set itself).

📚 Sets – Topic: Subsets-Page 10

6 A and B are subsets of a universal set U such that n(U) = 800, n(A) = 300, n(B) = 400 and n(A ∩ B) = 100. Then the number of elements in the set (Aᶜ ∩ Bᶜ) is [PYQ June 2024]

A 150

B 200

C 350

D 400

💡 Short Explanation

� Use De Morgan's Law: Aᶜ ∩ Bᶜ = (A ∪ B)ᶜ. • Find n(A ∪ B) using the inclusion-exclusion principle. • Subtract n(A ∪ B) from the Universal set n(U) to get the complement.

✅ Correct Answer Explanation

�� Step 1: By De Morgan's Law, the set (Aᶜ ∩ Bᶜ) is equal to (A ∪ B)ᶜ. Step 2: Calculate n(A ∪ B) using the formula n(A) + n(B) − n(A ∩ B). So, n(A ∪ B) = 300 + 400 − 100 = 600. Step 3: The number of elements in the complement (A ∪ B)ᶜ is the Universal set minus n(A ∪ B). n(U) − n(A ∪ B) = 800 − 600 = 200.

❓ Why Other Options Are Incorrect

� Option A → 150 is a miscalculation. • Option C → 350 is a miscalculation. • Option D → 400 is the value of n(B), not the requested set.

🎯 Strategy

Used • Substitution Application: → Apply De Morgan's Laws to transform the unknown into a calculable format. Final Logic: → 800 - (300 + 400 - 100) = 200.

🧠 Memory Trick

�� Outside A and outside B means completely outside their Union.

📚 Sets – Topic: Complement of a Set / De Morgan's Laws-Page 19

7 If A = {1, 2, 3, 4} and B = {3, 5, 7}, then (A − B) ∪ (B − A) is [PYQ June 2024]

A {1, 2, 3}

B {3}

C {1, 2, 3, 4, 5, 7}

D {1, 2, 4, 5, 7}

💡 Short Explanation

� First, find (A − B), which are elements exclusively in A. • Second, find (B − A), which are elements exclusively in B. • Take the union of both resulting sets.

✅ Correct Answer Explanation

�� (A − B) is the set of elements in A but not in B. Since 3 is in both, A − B = {1, 2, 4}. (B − A) is the set of elements in B but not in A. Again, skipping 3, B − A = {5, 7}. The union of these two sets, (A − B) ∪ (B − A), combines these exclusive elements into a single set: {1, 2, 4} ∪ {5, 7} = {1, 2, 4, 5, 7}.

❓ Why Other Options Are Incorrect

� Option A → Represents the set A minus the number 4, which has no direct relevance. • Option B → Represents A ∩ B, the intersection, which is the exact opposite of what is being asked (symmetric difference). • Option C → Represents A ∪ B, the union of all elements, failing to subtract the intersection {3}.

🎯 Strategy

Used • Elimination Application: → Identify the common element (3) and eliminate any option containing it, as difference operations remove common elements. Only D is left. Final Logic: → {1,2,3,4} - {3,5,7} = {1,2,4}. {3,5,7} - {1,2,3,4} = {5,7}. Union = {1,2,4,5,7}.

🧠 Memory Trick

�� Symmetric difference strictly punishes "sharing". Delete the shared items and combine the rest!

📚 Sets – Topic: Difference of Sets-Page 16

8 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} be the universal set, P = {1, 2, 5} and Q = {6, 7} be the two subsets. The set (P ∩ Q′) is [PYQ June 2025]

A P

B Q

C P′

D Q′

💡 Short Explanation

� Determine Q′ (elements in U but not in Q). • Intersect P with Q′. • Observe that P and Q are entirely disjoint.

✅ Correct Answer Explanation

�� The set Q′ is the complement of Q, containing all elements of U except 6 and 7. Thus, Q′ = {1, 2, 3, 4, 5, 8, 9, 10}. The intersection of P and Q′ (P ∩ Q′) looks for common elements between P = {1, 2, 5} and Q′. Since all elements of P ({1, 2, 5}) are present in Q′, P ∩ Q′ = {1, 2, 5}, which is exactly set P. Conceptually, since P and Q are mutually exclusive (disjoint), P is entirely contained within the complement of Q (Q′).

❓ Why Other Options Are Incorrect

� Option B → Represents set Q, which is disjoint from P. • Option C → Represents P′, the complement of P, which would be {3,4,6,7,8,9,10}. • Option D → Represents Q′, which is much larger than the intersection.

🎯 Strategy

Used • Contextual / Tonal Matching Application: → Since P and Q share no elements (disjoint sets), P must be completely inside the "not Q" set. Therefore, their intersection is just P. Final Logic: → Disjoint sets mean P ⊂ Q′, so P ∩ Q′ = P.

🧠 Memory Trick

�� If you have nothing in common with Q, you are already completely inside "Not Q".

📚 Sets – Topic: Complement of a Set-Page 18

9 If P be the set of all prime numbers and N = {x : 0 ≤ x ≤ 9}, then N – (P ∩ N) is [PYQ June 2026]

A {0, 4, 6, 8}

B {1, 2, 4, 6, 8}

C {1, 4, 6, 8, 9}

D {0, 1, 4, 6, 8, 9}

💡 Short Explanation

� List the elements of set N based on the given condition. • Identify the prime numbers within that range (P ∩ N). • Subtract the prime numbers from set N.

✅ Correct Answer Explanation

�� Set N contains all integers from 0 to 9 inclusive: N = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Set P is the set of all prime numbers. The intersection P ∩ N filters out only the prime numbers between 0 and 9: {2, 3, 5, 7}. The operation N − (P ∩ N) means we remove {2, 3, 5, 7} from set N. Removing these leaves {0, 1, 4, 6, 8, 9}.

❓ Why Other Options Are Incorrect

� Option A → Excludes 1 and 9, which are not prime and should remain in the set. • Option B → Excludes 0 and 9, and incorrectly leaves 2, which is a prime number. • Option C → Excludes 0, which is not a prime number and must be retained in the final set.

🎯 Strategy

Used • Elimination Application: → 0 and 1 are neither prime nor composite. They are in N. Therefore, any answer missing 0 or 1 is wrong. Option D is the only one with both 0 and 1. Final Logic: → N = {0..9}, Primes in N = {2,3,5,7}. Difference = {0,1,4,6,8,9}.

🧠 Memory Trick

�� 0 and 1 are the "forgotten numbers" in prime tests. Always keep them when removing primes!

📚 Sets – Topic: Difference of Sets-Page 16

ICMA Foundation- FBMS- Categorised PYQ- Class 11-Maths-Chapter-1-Sets

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