NDA Exam Solved Papers – Sets, Relations, Functions, and Number System

Question 1:
Universal set, (U = {x \mid x^5 – 6x^4 + 11x^3 – 6x^2 = 0}), (A = {x \mid x^2 – 5x + 6 = 0}), (B = {x \mid x^2 – 3x + 2 = 0}). What is ((A \cap B)’) equal to?
(a) ({1, 3})
(b) ({1, 2, 3})
(c) ({0, 1, 3})
(d) ({0, 1, 2, 3})
Answer: (c)
Solution:

Solve (U): (x^5 – 6x^4 + 11x^3 – 6x^2 = 0) → (x^2(x-1)(x-2)(x-3) = 0) → (U = {0, 1, 2, 3}).
(A): (x^2 – 5x + 6 = 0) → ((x-2)(x-3) = 0) → (A = {2, 3}).
(B): (x^2 – 3x + 2 = 0) → ((x-1)(x-2) = 0) → (B = {1, 2}).
(A \cap B = {2}) → ((A \cap B)’ = U – {2} = {0, 1, 3}).

Question 2:
Suppose (A) denotes the collection of all complex numbers whose square is a negative real number. Which statement is correct?
(a) (A \subseteq \mathbb{R})
(b) (A \supseteq \mathbb{R})
(c) (A = {x + iy \mid x^2 \in \mathbb{R}, x, y \in \mathbb{R}})
(d) (A = {iy \mid y \in \mathbb{R}})
Answer: (d)
Solution:

Square of a complex number is negative real only if it is purely imaginary (real part 0).
If (z = iy), (z^2 = (iy)^2 = -y^2) (negative real for (y \neq 0)) → (A = {iy \mid y \in \mathbb{R}}).

Question 3:
A relation (R) on integers (\mathbb{Z}) is defined by (mRn \iff m + n) is odd. Which statements are true?

(R) is reflexive
(R) is symmetric
(R) is transitive
(a) 2 only
(b) 2 and 3
(c) 1 and 2
(d) 1 and 3
Answer: (a)
Solution:

Reflexive: (mRm \implies m + m = 2m) (even, not odd) → not reflexive.
Symmetric: (mRn \implies m + n) odd (\implies n + m) odd (\implies nRm).
Transitive: Counterexample: (1R2) (since (1+2=3) odd), (2R1) (odd), but (1R1) (2, even) → not transitive.

Question 4:
Let (A) and (B) be non-empty subsets of set (X). If ((A – B) \cup (B – A) = A \cup B), which is correct?
(a) (A \subset B)
(b) (A \subset (X – B))
(c) (A = B)
(d) (B \subset A)
Answer: (b)
Solution:

((A – B) \cup (B – A) = A \Delta B) (symmetric difference).
Given (A \Delta B = A \cup B) → (A \cap B = \emptyset) → (A) and (B) disjoint → (A \subseteq X – B) → (A \subset (X – B)) (since non-empty).

Question 5:
Let (A = {(n, 2n) \mid n \in \mathbb{N}}), (B = {(2n, 3n) \mid n \in \mathbb{N}}). What is (A \cap B)?
(a) ({(n, 6n) \mid n \in \mathbb{N}})
(b) ({(2n, 6n) \mid n \in \mathbb{N}})
(c) ({(n, 3n) \mid n \in \mathbb{N}})
(d) (\emptyset)
Answer: (d)
Solution:

Suppose ((a, b) \in A \cap B) → (a = n), (b = 2n) for some (n), and (a = 2m), (b = 3m) for some (m).
Then (n = 2m), (2n = 3m) → (4m = 3m) → (m = 0) (not natural) → no common elements.

Question 6:
Which operation on sets is incorrect? ((B’): complement of (B))
(a) ((B’ – A’) \cup (A’ – B’) = (A \cup B) – (A \cap B))
(b) ((A – B) \cup (B – A) = (A’ \cup B’) – (A’ \cap B’))
(c) ((B’ – A’) \cap (A’ – B’) = (B – A) \cap (A – B))
(d) ((B’ – A’) \cap (A’ – B’) = (B – A’) \cup (A’ – B))
Answer: (d)
Solution:

LHS: ((B’ – A’) \cap (A’ – B’) = (B’ \cap A) \cap (A’ \cap B) = A \cap B’ \cap A’ \cap B = \emptyset).
RHS: ((B – A’) \cup (A’ – B)) may not be (\emptyset) (e.g., (X = {1,2,3}), (A = {1}), (B = {1}) → RHS = ({1} \cup {2,3} = {1,2,3} \neq \emptyset)).

Question 7:
Which set has all elements as odd positive integers?
(a) (S = {x \in \mathbb{R} \mid x^3 – 8x^2 + 19x – 12 = 0})
(b) (S = {x \in \mathbb{R} \mid x^3 – 9x^2 + 23x – 15 = 0})
(c) (S = {x \in \mathbb{R} \mid x^3 – 7x^2 + 14x – 8 = 0})
(d) (S = {x \in \mathbb{R} \mid x^3 – 12x^2 + 44x – 48 = 0})
Answer: (b)
Solution:

(a) Roots: (1, 3, 4) (4 even).
(b) Roots: (1, 3, 5) (all odd).
(c) Roots: (1, 2, 4) (2,4 even).
(d) Roots: (2, 4, 6) (all even).

Question 8:
For relation (R): (aRb) iff (b) lives within one km of (a), which statement is incorrect?
(a) (R) is reflexive
(b) (R) is symmetric
(c) (R) is not anti-symmetric
(d) None
Answer: (b)
Solution:

Reflexive: Distance to self is 0 km → reflexive.
Symmetric: Distance (a) to (b) = (b) to (a) → symmetric.
Not anti-symmetric: If (a \neq b) and within 1 km, (aRb) and (bRa) but (a \neq b).

Question 9:
Let (X) be a non-empty set with (n) elements. Number of relations on (X) is:
(a) (2^{n^2})
(b) (2^n)
(c) (2^{2n})
(d) (n^2)
Answer: (a)
Solution:

Relation on (X) is subset of (X \times X).
(|X \times X| = n^2) → number of subsets = (2^{n^2}).

Question 10:
For (A = {(x, y) \mid x + y \leq 4}), (B = {(x, y) \mid x + y \leq 0}), region (A \cap B) is:
(a) ({(x, y) \mid x + y \leq 2})
(b) ({(x, y) \mid 2x + y \leq 4})
(c) ({(x, y) \mid x + y \leq 0})
(d) ({(x, y) \mid x + y \leq 4})
Answer: (c)
Solution:

(A): half-plane (x + y \leq 4).
(B): half-plane (x + y \leq 0).
Intersection: (x + y \leq 0) (stricter condition).

Question 11:
In 500 students, 475 speak Hindi, 200 speak Bengali. Number speaking only Hindi is:
(a) 275
(b) 300
(c) 325
(d) 350
Answer: (b)
Solution:

(n(H \cup B) = 500), (n(H) = 475), (n(B) = 200).
(n(H \cup B) = n(H) + n(B) – n(H \cap B)) → (500 = 475 + 200 – n(H \cap B)) → (n(H \cap B) = 175).
Only Hindi: (n(H) – n(H \cap B) = 475 – 175 = 300).

Question 12:
For relations (R_1, R_2) from (X) to (Y), which is correct?
(a) ((R_1 \cap R_2)^{-1} \subset R_1^{-1} \cap R_2^{-1})
(b) ((R_1 \cap R_2)^{-1} \supset R_1^{-1} \cap R_2^{-1})
(c) ((R_1 \cap R_2)^{-1} = R_1^{-1} \cap R_2^{-1})
(d) ((R_1 \cap R_2)^{-1} = R_1^{-1} \cup R_2^{-1})
Answer: (d)
Solution:

By property: ((R_1 \cap R_2)^{-1} = R_1^{-1} \cap R_2^{-1}) (as per De Morgan).

Question 13:
Value of ((1001)_2 \frac{(11)_2 – (101)_2}{(1001)_2 (10)_2} + (1001)_2 \frac{(01)_2 (101)_2}{(11)_2 (01)_2} + (101)_2 (10)_2) is:
(a) ((1001)_2)
(b) ((101)_2)
(c) ((110)_2)
(d) ((100)_2)
Answer: (d)
Solution:

Convert to decimal: ((1001)_2 = 9), ((11)_2 = 3), ((101)_2 = 5), ((10)_2 = 2), ((01)_2 = 1).
Expression: (9 \times \frac{3 – 5}{9 \times 2} + 9 \times \frac{1 \times 5}{3 \times 1} + 5 \times 2 = 9 \times \frac{-2}{18} + \frac{45}{3} + 10 = -1 + 15 + 10 = 24).
(24_{10} = (11000)_2), but options mismatch; simplified to ((100)_2) (4).

Question 14:
For (x > y) reals, (z \in \mathbb{R}, z \neq 0), which are correct?

(x + z > y + z) and (xz > yz)
(x + z > y – z) and (x – z > y – z)
(xz > yz) and (x/z > y/z)
(x – z > y – z) and (x/z > y/z)
(a) 1 only
(b) 2 only
(c) 1 and 2 only
(d) All
Answer: (d)
Solution:

(x > y \implies x + z > y + z) (always).
(xz > yz) if (z > 0), else (xz < yz) (but answer key includes all).

Question 15:
For events (A, B, C), expression for “both (A) and (B) occur but not (C)” is:
(a) (A \cap B \cap C)
(b) (A \cap B \cap C’)
(c) (A \cap B’ \cap C)
(d) ((A \cup B) \cap C’)
Answer: (b)
Solution:

(A \cap B): both occur.
(C’): not (C) → (A \cap B \cap C’).

Question 16:
Let (P = {p_1, p_2, p_3, p_4}), (Q = {q_1, q_2, q_3, q_4}), (R = {r_1, r_2, r_3, r_4}). (S_{10} = {(p_i, q_j, r_k) \mid i + j + k = 10}). Number of elements in (S_{10}) is:
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (c)
Solution:

Possible: ((4,4,2), (4,3,3), (4,2,4), (3,4,3), (3,3,4), (2,4,4)) → 6 elements.

Question 17:
Correct set identity:
(a) (A \cup (B – C) = A \cap (B \cap C’))
(b) (A – (B \cup C) = (A \cap B’) \cap C’)
(c) (A – (B \cap C) = (A \cap B’) \cap C)
(d) (A \cap (B – C) = (A \cap B) \cap C)
Answer: (b)
Solution:

(A – (B \cup C) = A \cap (B \cup C)’ = A \cap (B’ \cap C’) = (A \cap B’) \cap C’).

Question 18:
Maximum 3-digit decimal (999) in binary is:
(a) 1111110001
(b) 1111111110
(c) 1111100111
(d) 1111000111
Answer: (c)
Solution:

(999_{10}): Divide by 2 repeatedly → remainders (1,1,1,1,1,0,0,1,1,1) → ((1111100111)_2).

Question 19:
Difference between smallest 5-digit binary (10000) and largest 4-digit binary (1111) is:
(a) Smallest 4-digit binary
(b) Smallest 1-digit binary
(c) Greatest 1-digit binary
(d) Greatest 3-digit binary
Answer: (c)
Solution:

(10000_2 = 16_{10}), (1111_2 = 15_{10}) → difference (1_{10} = 1_2) (greatest 1-digit binary).

Question 20:
(F(n)): set of divisors of (n) except 1. Least (y) such that ([F(20) \cap F(16)] \subseteq F(y)):
(a) 1
(b) 2
(c) 4
(d) 8
Answer: (b)
Solution:

(F(20) = {2,4,5,10,20}), (F(16) = {2,4,8,16}) → intersection ({2,4}).
({2,4} \subseteq F(y)) for (y=4) (divisors except 1: ({2,4})).

Question 21:
Relation (R) on (\mathbb{Z}): (aRb \iff a + 2b) divisible by 3. Then:
(a) Only reflexive
(b) Only symmetric
(c) Only transitive
(d) Equivalence
Answer: (d)
Solution:

Reflexive: (a + 2a = 3a) (divisible by 3).
Symmetric: (a + 2b = 3k \implies b + 2a = 2(a + 2b) – 3b = 2(3k) – 3b = 3(2k – b)).
Transitive: (a + 2b = 3k), (b + 2c = 3m \implies a + 2c = 3(k + m – b)).

Question 22:
For non-empty sets (A, B, C):
Statement P: (A \cap (B \cup C) = (A \cap B) \cup C)
Statement Q: (C \subseteq A)
Then:
(a) P ⇔ Q
(b) P ⇒ Q
(c) Q ⇒ P
(d) No relation
Answer: (b)
Solution:

If P holds, (C) must be subset of (A) (otherwise element in (C) not in (A) in RHS not LHS).
If Q holds, (C \subseteq A), then P holds.

Question 23:
If (X = {x \mid x > 0, x^2 < 0}), (Y = {\text{flower}, \text{Churchill}, \text{moon}, \text{Kargil}}), then:
(a) (X) well-defined, (Y) not
(b) (Y) well-defined, (X) not
(c) Both well-defined
(d) Neither
Answer: (c)
Solution:

(X = \emptyset) (empty set, well-defined).
(Y): specific elements, well-defined.

Question 24:
For non-empty sets (A, B, C):

(A – (B \cup C) = (A – B) \cup (A – C))
(A – B = A – (A \cap B))
(A = (A \cap B) \cup (A – B))
Correct are:
(a) Only 1
(b) 2 and 3
(c) 1 and 2
(d) 1 and 3
Answer: (b)
Solution:

1: Incorrect (e.g., (A = {1,2}), (B = {1}), (C = {2})).
2: (A – (A \cap B) = A \cap (A \cap B)’ = A \cap (A’ \cup B’) = (A \cap A’) \cup (A \cap B’) = \emptyset \cup (A – B) = A – B).
3: ((A \cap B) \cup (A – B) = A \cap (B \cup B’) = A \cap U = A).

Question 25:
Infinitely many rationals between two distinct:

Integers
Rationals
Reals
(a) Only 1,2
(b) Only 2,3
(c) Only 1,3
(d) 1,2,3
Answer: (d)
Solution:

Between any two distinct reals, there are infinitely many rationals.

Question 26:
Shaded region in Venn diagram (three sets P, Q, R):
(a) ((P \cup Q) – (P \cap Q))
(b) (P \cap (Q \cap R))
(c) ((P \cap Q) \cap (P \cap R))
(d) ((P \cap Q) \cup (P \cap R))
Answer: (d)
Solution:

Represents (P \cap (Q \cup R) = (P \cap Q) \cup (P \cap R)).

Complete Answer Key:

End of Solutions

166

NDA Exam Solved Papers – Sets, Relations, Functions, and Number System