NDA Sets previous year questions with solutions

Answer: (b) II only

Method. Membership is not transitive; subset is

I. x ∈ A and A ∈ B need not give x ∈ B. Take A = {1}, B = {{1}}, x = 1: 1 ∈ A, A ∈ B, but 1 ∉ B (B's only element is {1}). Incorrect.
II. A ⊂ B means every element of A is in B. If x ∉ B, then x cannot be in A. Correct (contrapositive).
II only.

Answer: (a) I only

Method. Finite or infinite sets

I. ⁠√12⁠ ≈ 3.46 and ⁠√15⁠ ≈ 3.87; infinitely many irrationals lie between any two distinct reals. Correct.
II. Odd integers less than 1000 include 999, 997, … and all the negative odd integers, which never end. Infinite set. Incorrect.
I only.

Answer: (a) 1 only

Method. Finite or infinite sets

1. Between any two distinct real numbers there are infinitely many irrationals (⁠√2⁠ ≈ 1.414, ⁠√5⁠ ≈ 2.236). Infinite set. Correct.
2. Odd integers less than 100 include 99, 97, …, 1, -1, -3, … with no lower end. Infinite set. Incorrect.
Statement 1 only.

Answer: (c) Both 1 and 2

Method. Equivalent sets have equal cardinality; a set ignores repeated listing

1. n(A) = 3 and n(B) = 3, so A and B are equivalent (a one-to-one correspondence exists). Statement 1 is correct.
2. Repeated elements in a listing are not counted again: {1, 5, 5, 9, 9} = {1, 5, 9} = A. Statement 2 is correct.
Both are correct.

Answer: (d) an empty set

Method. solve the equation over the reals

x² + 1 = 0 gives x² = -1, which has no real solution because x² ≥ 0 for every real x.
So S has no element: S = φ, the empty set, option (d).

Answer: (b) 2

Method. Power set of a set with one element (which is itself a set)

A = {{1, 2, 3}} has exactly one element, namely the set {1, 2, 3}. So n(A) = 1.
Power set of A has 2¹ = 2 elements: φ and {{1, 2, 3}}.

Answer: (c) 8

Method. Subsets containing a fixed element

A superset of {4} inside A is a subset of A that contains 4.
For each of 1, 2, 3 choose in or out: 2³ = 8 subsets (the sets {4}, {1,4}, …, {1,2,3,4}).

Shortcut. Fix 4 in; the other n − 1 = 3 elements are free: 2³ = 8.

Answer: (c) 1023

Method. Count elements, then 2ⁿ - 1 proper subsets

The 15 elements are distinct, so X and Y have no common element and X − Y = X, with 6 elements.
Z is disjoint from X, so n(S) = n(X) + n(Z) = 6 + 4 = 10.
Number of subsets = 2¹⁰ = 1024; proper subsets exclude S itself: 1024 − 1 = 1023.

Answer: (d) 1, 2 and 3

Method. Basic facts on subsets and power sets

1. φ ⊆ every set: true (no element of φ can fail to lie in the set).
3. A set with n elements has 2ⁿ subsets; 2¹⁰ = 1024. True.
All three are correct.

Answer: (c) 7

Method. count subsets containing 3, then remove the set itself

A subset containing 3 is fixed by the choice among 1, 2, 4: each is in or out, giving 2³ = 8 subsets, including {3} itself and {1, 2, 3, 4}.
A proper subset excludes {1, 2, 3, 4} itself, so 8 − 1 = 7.
Option (c).

Answer: (d) 1024

Method. a set with n elements has 2ⁿ subsets

S = {2, 4, 6, …, 20} has 10 elements (the even numbers 2 to 20).
Number of subsets = 2¹⁰ = 1024, option (d).
('Maximum' adds nothing: the number of subsets of S is fixed at 1024.)

Answer: (b) {φ, {λ}, {{λ, μ}}, {λ, {λ, μ}}}

Method. the power set of a 2-element set has 4 members; the elements of A are λ and {λ, μ}

A has exactly two elements: λ and the set {λ, μ}. The power set lists all subsets: φ, {λ}, {{λ, μ}}, and {λ, {λ, μ}} (4 = 2² of them).
In option (b) the subsets are φ, {λ}, {{λ, μ}} and {λ, {λ, μ}}. This matches, so (b).
Option (c) is wrong because {λ, μ} is not a subset of A (μ is not an element of A), only an element of A.

Answer: (b) A

Method. Simplify the inner brackets using X = (X ∩ Y) ∪ (X − Y)

(A ∩ B) ∪ (A − B) = A (A split into the part inside B and the part outside B).
(A ∩ B) ∪ (B − A) = B similarly.
So the expression is (A − B) ∪ A, and A − B ⊆ A gives A.

Answer: (b) II only

Method. Remove C from both sides using disjointness

Since A ∩ C = φ, (A ∪ C) − C = A. Likewise (B ∪ C) − C = B (using B ∩ C = φ).
A ∪ C = B ∪ C, so A = B. II is true.
I: C need not be φ. A = B = {1}, C = {2} satisfies every given condition. False.
III: in that example A ∪ B = {1} ≠ C = {2}. False.
II only. (Unlike the general cancellation law, the disjointness condition makes cancelling C valid.)

Answer: (a) I only

Method. Rewrite A − B as A ∩ B^c

I. A − B = A ∩ B^c. B^c - A^c = B^c ∩ (A^c)^c = B^c ∩ A. Same set. Correct.
II. A − B^c = A ∩ B, while A^c - B = A^c ∩ B^c. These differ: A = B = {1} in the universe {1, 2} gives {1} versus {2}. Incorrect.
I only.

Answer: (a) Null set

Method. The three pieces A − B, B − A, A ∩ B partition A ∪ B

Every element of A ∪ B is in A only (A − B), in B only (B − A), or in both (A ∩ B). So (A − B) ∪ (B − A) ∪ (A ∩ B) = A ∪ B.
Therefore (A ∪ B) − (A ∪ B) = φ, the null set.

Answer: (b) 2 and 3 only

Method. Test each identity by element-chasing

1. (A ∪ B) ∪ (A − B) = A ∪ B, since A − B ⊆ A. This equals A only if B ⊆ A. Fails for A = {1}, B = {2}. Incorrect.
2. B − A ⊆ B, so A ∪ (B − A) = A ∪ B. Correct.
3. A ∪ B splits as (A − B) ∪ B with A − B disjoint from B; removing A − B leaves exactly B. Correct.
Statements 2 and 3 only.

Answer: (d) Neither 1 nor 2

Method. Cancellation laws fail for ∩ and ∪; find counterexamples

1. Take A = φ, B = {1}, C = φ. Then A ∩ B = φ = A ∩ C, but B ≠ C. Not correct.
2. Take A = {1}, B = {1}, C = φ. Then A ∪ B = {1} = A ∪ C, but B ≠ C. Not correct.
(Both conclusions would hold only if both A ∩ B = A ∩ C and A ∪ B = A ∪ C were given together.)
Neither 1 nor 2.

Answer: (a) 1 only

Method. Element-wise reasoning with De Morgan's laws

1. x ∉ (A ∪ B) means x ∉ A and x ∉ B. If x is outside both, it is certainly outside at least one, so 'x ∉ A or x ∉ B′ follows. Correct.
2. x ∉ (A ∩ B) means x ∉ A or x ∉ B, not 'and'. Example: A = {1, 2}, B = {2, 3}, x = 1: x ∉ A ∩ B = {2}, but x ∈ A. Incorrect.
Statement 1 only.

Answer: (c) Both 1 and 2

Method. A ∪ B = A ∩ B forces equality; symmetric difference is empty iff the sets agree

1. A ∩ B ⊆ A ⊆ A ∪ B and A ∩ B ⊆ B ⊆ A ∪ B. If A ∪ B = A ∩ B, then A and B are both squeezed to the same set, so A = B. Conversely, if A = B then A ∪ B = A = A ∩ B. Correct.
2. A Δ B = (A − B) ∪ (B − A) = φ iff A − B = φ and B − A = φ iff A ⊆ B and B ⊆ A iff A = B. Correct.
Both 1 and 2.

Answer: (a) 1 and 2 only

Method. Distributive law, De Morgan's law, and A − B = φ

1. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C): union is distributive over intersection. Correct.
2. (A ∪ B)^c = A^c ∩ B^c: De Morgan's law. Correct.
3. A − B = φ only says A ⊆ B, not A = B. Example: A = {1}, B = {1, 2}: A − B = φ but A ≠ B. Incorrect.
Statements 1 and 2 only.

Answer: (a) A ∪ (A ∩ B) = A ∪ B

Method. test each identity

(a) A ∪ (A ∩ B) = A by the absorption law, not A ∪ B. Counter-example: A = {1}, B = {2}: left side = {1}, A ∪ B = {1, 2}. So (a) is not correct.
(b) A ∩ (A ∪ B) = A, absorption law. True.
(c) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C), distributive law. True.
(d) (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C), distributive law. True.
The relation that is not correct is (a).

Answer: (c) {2}

Method. find the primes inside the interval [0, 2]

A = {x: 0 ≤ x ≤ 2} is the interval from 0 to 2, and B is the set of primes.
The only prime in [0, 2] is 2 (1 is not a prime; 0 is not a prime).
A ∩ B = {2}, option (c).

Answer: (a) 1 only

Method. simplify each statement with distributive and complement laws

Statement 1: (A ∩ B) ∪ (A ∩ B′) = A ∩ (B ∪ B′) = A. Then A ∪ (A′ ∩ B) = (A ∪ A′) ∩ (A ∪ B) = A ∪ B. True.
Statement 2: A ∪ (A′ ∩ B′) = (A ∪ A′) ∩ (A ∪ B′) = A ∪ B′, not A ∪ B. Counter-example: universal set {1, 2, 3}, A = {1}, B = {2}: left side = {1} ∪ {3} = {1, 3}, right side A ∪ B = {1, 2}. False.
Only statement 1 is correct, option (a).

Answer: (d) 1, 2 and 3

Method. use the definition of subset, and choose B = X or B = φ to extract A ⊂ C(⊂ read as 'is a subset of', as in NCERT)

Statement 1: if A ⊂ C, any x in A ∩ B lies in A, hence in C, and in B, so x is in C ∩ B; any x in A ∪ B lies in C or in B, so x is in C ∪ B. True.
Statement 2: the condition holds for all B, so take B = X: A ∩ X = A and C ∩ X = C, so A ⊂ C. True.
Statement 3: take B = φ: A ∪ φ = A and C ∪ φ = C, so A ⊂ C. True.
All three are correct, option (d).

Answer: (c) A′ ∪ (B ∪ C) = (C′ ∩ B)' ∩ A′

Method. check each identity (distributive laws, De Morgan); the false one is the answer

(a) is the distributive law of ∪ over ∩. True.
(b) A′ ∪ (A ∪ B) = (A′ ∪ A) ∪ B = the universal set. And (B′ ∩ A)' ∪ A = (B ∪ A′) ∪ A = the universal set. Equal. True.
(c) A′ ∪ (B ∪ C) = A′ ∪ B ∪ C. But (C′ ∩ B)' ∩ A′ = (C ∪ B′) ∩ A′. Take A = the universal set: left side = B ∪ C, right side = φ. Not equal in general. False.
(d) (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C), distributive law. True.
The relation that is not correct is (c).

Shortcut. Only (c) mixes ∪ on the left with ∩ A′ on the right; the other three are standard laws.

No correct option. This question is defective as printed in UPSC's paper.

Method. C is the symmetric difference A Δ B; compare each option by De Morgan

C = (A ∩ B′) ∪ (A′ ∩ B) is the set of elements in exactly one of A, B, which equals (A ∪ B) − (A ∩ B), so option (c) is correct.
Option (d): (A′ ∪ B′) − (A′ ∩ B′) = (A ∩ B)' ∩ (A ∪ B) [since A′ ∪ B′ = (A ∩ B)' and A′ ∩ B′ = (A ∪ B)', and X − Y = X ∩ Y′] = (A ∪ B) − (A ∩ B) as well. So (d) is the same set as (c).
Checked by brute force over all subsets of a 4-element set: (c) and (d) both equal C; (a) and (b) do not (they give the complement of C).
Two options are correct.

Options (c) and (d) both equal (A ∪ B) − (A ∩ B), because (A′ ∪ B′) − (A′ ∩ B′) = (A ∩ B)' ∩ (A ∪ B); repair: change (d) to a set that differs from the symmetric difference, which would leave (c) as the answer.

Answer: (c) B′ ∩ C′

Method. E − (E − X) = X, so nested differences cancel in pairs; De Morgan for A′

For any subset X of E, E − X = X′, and E − X′ = X.
The expression has five subtractions from E, innermost first: E − A = A′; E − A′ = A; E − A = A′; E − A′ = A; E − A = A′.
Odd number of complements, so the whole expression equals A′.
A = B ∪ C, so A′ = (B ∪ C)' = B′ ∩ C′ (De Morgan), option (c).

Shortcut. Each E − ( ) is one complement; five complements = one complement = A′ = B′ ∩ C′.

Answer: (c) C

Method. B ∩ C where C ⊆ B, then C ⊆ A

Every multiple of 10 is a multiple of 5, so C ⊆ B and B ∩ C = C.
Every multiple of 10 is a multiple of 2, so C ⊆ A and A ∩ C = C.
Hence A ∩ (B ∩ C) = A ∩ C = C, option (c).
Option (d) is wrong: 10 is in C but is not a multiple of 100.

Answer: (b) 2, 3 and 4

Method. test each identity: (A − B) is the part of A outside B

(1) (A − B) ∪ B = A ∪ B, which is bigger than A whenever B has an element outside A. Take A = {1}, B = {2}: (A − B) ∪ B = {1, 2} ≠ A. So (1) is false.
(2) A − B ⊆ A, so (A − B) ∪ A = A. True.
(3) A − B has no element of B, so (A − B) ∩ B = φ. True.
(4) If A ⊆ B then every element of A is already in B, so A ∪ B = B. True.
Statements 2, 3 and 4 are correct, so the answer is (b).

Answer: (c) 110

Method. Inclusion-exclusion: n(X ∩ Y) ≥ n(X) + n(Y) − total

n(X ∪ Y) ≤ 500, so n(X ∩ Y) = n(X) + n(Y) − n(X ∪ Y) ≥ 370 + 240 − 500 = 110.
The value 110 occurs when every consumer prefers at least one product, so the least number is 110.

Answer: (b) 30

Method. Inclusion-exclusion, then exactly one = total − both (everyone likes at least one)

n(C ∪ F) = 45 (everyone likes at least one game). n(C ∩ F) = 34 + 26 − 45 = 15.
Exactly one game = 45 − 15 = 30.

Shortcut. Exactly one = 34 + 26 − 2(15) = 30.

Answer: (c) 384

Method. Inclusion-exclusion with the common multiples of 6

A = {3, 6, …, 750} (250 terms); B = {2, 4, …, 400} (200 terms).
A ∩ B = numbers that are multiples of both 3 and 2, i.e. multiples of 6, lying in both lists, so at most 400: 6, 12, …, 396, which is ⌊400/⁠6⌋ = 66 numbers. (Each is a multiple of 3 within the first 250, since 396 < 750.)
n(A ∪ B) = 250 + 200 − 66 = 384.

Answer: (b) 6

Method. n(A ∪ B) ≥ n(B)

A ∪ B contains B, so it has at least 6 elements. The value 6 is reached when A ⊆ B.
Minimum = 6, option (b).

Answer: (c) 20%

Method. n(C ∪ F) = n(C) + n(F) − n(C ∩ F)

Playing cricket or football = 50 + 40 − 10 = 80%.
Neither = 100% - 80% = 20%. Option (c).

Answer: (b) 45

Method. n(M ∪ D) ≤ 850, so n(M ∩ D) = n(M) + n(D) − n(M ∪ D)

n(M ∩ D) = 680 + 215 − n(M ∪ D).
n(M ∪ D) cannot exceed the 850 students, so n(M ∩ D) ≥ 680 + 215 − 850 = 45.
This is reached when every student likes at least one of the two. Least number = 45, option (b).

Answer: (a) 60

Method. Count students in 'at least one' by regions and by subject totals

At least one subject: 240 − 10 = 230. Let t = passed all three.
Only one + only two + all three = 230: 60 + 110 + t = 230, so t = 60.
Check with subject totals: 180 + 130 + 150 = 460 = (only one) + 2(only two) + 3t = 60 + 220 + 3(60) = 460. (holds).

Answer: (c) 40 − 7x

Method. Three-set inclusion-exclusion with exactly-two and exactly-one regions

Let a = number with exactly two medals, e = exactly one. Exactly three: x.
Students: e + a + x = 15x, so e + a = 14x. … (i)
Medals: e + 2a + 3x = 5x + (4x + 15) + (x + 25) = 10x + 40, so e + 2a = 7x + 40. … (ii)
(ii) − (i): a = 7x + 40 − 14x = 40 − 7x.
Hence exactly two = 40 − 7x.
(Consistent for x = 2, 3, 4, 5; checked by enumeration that each such x gives non-negative region sizes.)

Answer: (d) 40 − 6x

Method. Same equations as the previous item; add the all-three group

From the common stem: exactly two = a = 40 − 7x (item 34).
At least two = exactly two + exactly three = (40 − 7x) + x = 40 − 6x.

Answer: (a) 21x − 40

Method. Same equations as the earlier items; solve for the exactly-one group

Exactly one = e, with e + a = 14x (students) and a = 40 − 7x.
e = 14x − (40 − 7x) = 21x − 40.
(Check: x = 3 gives e = 23, a = 19, three = 3: total 45 = 15 · 3.)

Answer: (c) 21

Method. read n(Z) and n(Y) off the regions of the diagram

Z consists of: Z only (c), X ∩ Z only (12), the centre (18) and Y ∩ Z only (17).
n(Z) = c + 12 + 18 + 17 = c + 47 = 90, so c = 43.
Y consists of: Y only (b), X ∩ Y only (16), the centre (18) and Y ∩ Z only (17).
n(Y) = b + 16 + 18 + 17 = b + 51.
n(Y): n(Z) = 4 : 5, so n(Y) = (4/⁠5)(90) = 72.
b + 51 = 72, so b = 21. Option (c).

Answer: (d) a + b + 106

Method. the expression is n(X ∪ Y ∪ Z) by inclusion-exclusion; add up all seven regions

n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z) = n(X ∪ Y ∪ Z).
Add the seven regions: a + 16 + b + 18 + 12 + 17 + c = a + b + 63 + c.
From the first item, c = 90 − (12 + 18 + 17) = 43.
So the value = a + b + 63 + 43 = a + b + 106. Option (d).

Shortcut. The expression is the union; the seven region labels sum to a + b + c + 63, and c = 43.

Answer: (a) p + b + 60

Method. complement of X = everything outside circle X, including the p outside all circles

Outside X: Y only (b), Y ∩ Z only (17), Z only (c), and the p elements in none of the sets.
n(X′) = p + b + 17 + c.
c = 90 − 12 − 18 − 17 = 43 (from n(Z) = 90).
n(X′) = p + b + 17 + 43 = p + b + 60. Option (a).

Answer: (b) 111

Method. inclusion-exclusion with the triple overlap t unknown

Let t = number playing all three games. All students play at least one game, so
n(total) = 60 + 50 + 48 − 12 − 15 − 20 + t = 111 + t.
For the region counts to stay non-negative, t ≤ the smallest pair overlap: t ≤ 12 (and t ≥ 0).
Minimum is at t = 0: 111 students. Option (b).
(Check at t = 0: chess only 28, table tennis only 15, carrom only 21, no region negative.)

Answer: (b) 123

Method. same formula, largest allowed t

Total = 111 + t, with 0 ≤ t ≤ 12 (t cannot exceed the chess-and-carrom overlap of 12).
Maximum at t = 12: 111 + 12 = 123. Option (b).
(Check at t = 12: chess and carrom only = 0, carrom and table tennis only = 3, table tennis and chess only = 8, all regions non-negative.)

Answer: (a) 14

Method. inclusion-exclusion with 'exactly two' and 'all three' counted separately; every one of the 300 students likes at least one game

Let x = number liking all three, and N1 = number liking exactly one game.
Counting the 'likes' of the three games: 125 + 145 + 90 = 360 = N1 + 2(32) + 3x.
So N1 = 296 − 3x.
Working, that line: 'The survey population is taken to be the union (every one of the 300 students likes at least one game; the stem does not say so, but it is the only reading under which both items have an option): N1 + 32 + x = 300, so 296 − 3x + 32 + x = 300, giving 2x = 28 and x = 14.'
Option (a). Check: N1 = 296 − 42 = 254, and 254 + 32 + 14 = 300.

Answer: (c) 254

Method. use the value found for the previous item

From the previous item, x = 14 like all three games (every student is taken to like at least one game, as in the previous item).
Exactly one game: N1 = 296 − 3x = 296 − 42 = 254, option (c).
Check: 254 + 32 + 14 = 300.

Answer: (c) 125

Method. Venn regions: the 18 who are good in both Hindi and Mathematics include the 10 who are good in all three

Students good in Hindi and Mathematics: 18, and 10 of them are also good in English.
So Hindi and Mathematics but not English = 18 − 10 = 8.
Hindi or Mathematics but not English = (Hindi only) + (Mathematics only) + (Hindi and Mathematics only) = 54 + 63 + 8 = 125.
Option (c). (Reading 18 as 'exactly Hindi and Mathematics' would give 135, which is not an option.)

Answer: (d) 8

Method. same regions as the previous item

The 18 students good in both Hindi and Mathematics include the 10 who are good in all three subjects.
Good in Hindi and Mathematics but not in English = 18 − 10 = 8, option (d).

Answer: (a) I only

Method. |A × A| = |A|², and the listed pairs reveal the members of A

|A|² = 25, so |A| = 5.
From (3, 1), (6, 2), (5, 3): 1, 2, 3, 5, 6 are in A. That is five elements, so A = {1, 2, 3, 5, 6} exactly. Hence every pair of A × A can be listed. I is correct.
II: (5, 5) ∈ A × A, but (1, 3) ∈ A × A too (1 and 3 are in A). So '(1, 3) ∉ A × A′ is false. Incorrect.
I only.