Mathematical Reasoning

More Topic from Mathematics

Question 1

Consider :
Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.

  • Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I 

  • Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

  • Statement -I is True; Statement -II is False.

  • Statement -I is False; Statement -II is True

Solution

B.

Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I

p q ~p  ~q p^~q ~p^q (p^~q)^(~p^q)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
F
T
F
F
F
F
T
F
F
F
F
F
Fallacy
p q ~p  ~q p ⇒ q ~ q ⇒ ~ p (p ⇒ q) ⇔ (~ q ⇒ ~ p)
T
T
F
F
T
F
T
F
F
F
T
T
F
T
F
T
T
F
T
T
T
F
T
T
T
T
T
T
Tautology
S2 is not an explanation of S1
Question 2

Consider the following statements:
(a) Mode can be computed from histogram
(b) Median is not independent of change of scale
(c) Variance is independent of the change of origin and scale. Which of these is/are correct?

  • only (a)

  • only (b)

  • only (a) and (b)

  • (a), (b) and (c)

Solution

C.

only (a) and (b)

Question 3

Consider the following statements
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest. The negation of the statement ì Suman is brilliant and dishonest if and only if Suman is richî can be ex- pressed as

  •  ~ P ^ (Q ↔ ~ R)

  • ~ (Q ↔ (P ^ ~R)

  • ~ Q ↔ ~ P ^ R

  • ~ (P ^ ~ R)↔ Q

Solution

B.

~ (Q ↔ (P ^ ~R)

Negation of (PΛ~ R) ↔ Q is ~ ↔(PΛ ~ R)↔Q)
It may also be written as ~ (Q ↔ (PΛ ~ R))

Question 4

Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental number”, and r be the statement “x is a rational number iff y is a transcendental number”.
Statement –1: r is equivalent to either q or p
Statement –2: r is equivalent to ∼ (p ↔ ∼ q).

  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement − 1 is true, Statement − 2 is false. 

Solution

D.

Statement − 1 is true, Statement − 2 is false. 

Given statement r = ∼ p ↔ q
Statement −1 : r1 = (p ∧ ∼ q) ∨ (∼ p ∧ q)
Statement −2 : r2 = ∼ (p ↔ ∼ q) = (p ∧ q) ∨ (∼ q ∧ ∼ p)
From the truth table of r, r1 and r2,
r = r1.
Hence Statement − 1 is true and Statement −2 is false.

Question 5

Let S be a non empty subset of R. Consider the
following statement:
P: There is a rational number x∈S such that x > 0.
Which of the following statements is the negation of the statement P?

  • There is a rational number x∈S such that x ≤ 0.

  • There is no rational number x∈ S such that x≤0.

  • Every rational number x∈S satisfies x ≤ 0.

  • x∈S and x ≤ 0 ⇒ x is not rational.

Solution

C.

Every rational number x∈S satisfies x ≤ 0.