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13. Mathematical Reasoning

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Question 1/137
4 / -1

Mark Review

The compound statement (∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q)) is equivalent to

[24-Jan-2023 Shift 1]

The compound statement (∼(P∧Q))∨((∼P)∧Q)⇒((∼P)∧(∼Q)) is equivalent to

[24-Jan-2023 Shift 1]

A
((∼P)∨Q)∧((∼Q)∨P)
((∼P)∨Q)∧((∼Q)∨P)

B
(∼Q)∨P
(∼Q)∨P

C
((∼P)∨Q)∧(∼Q)
((∼P)∨Q)∧(∼Q)

D
(∼P)∨Q
(∼P)∨Q
Question 2/137
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Mark Review

Let p and q be two statements.
Then ∼(p∧(p⇒∼q)) is equivalent to

[24-Jan-2023 Shift 2]

Let p and q be two statements.
Then ∼(p∧(p⇒∼q)) is equivalent to

[24-Jan-2023 Shift 2]

A
p∨(p∧(∼q))
p∨(p∧(∼q))

B
p∨((∼p)∧q)
p∨((∼p)∧q)

C
(∼p)∨q
(∼p)∨q

D
p∨(p∧q)
p∨(p∧q)
Question 3/137
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Mark Review

The statement (p∧(∼q))⇒(p⇒(∼q)) is

[25-Jan-2023 Shift 1]

The statement (p∧(∼q))⇒(p⇒(∼q)) is

[25-Jan-2023 Shift 1]

A
equivalent to (∼p)∨(∼q)
equivalent to (∼p)∨(∼q)

B
a tautology
a tautology

C
equivalent to p∨q
equivalent to p∨q

D
a contradiction
a contradiction
Question 4/137
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Mark Review

The statement (p∧(∼q))⇒(p⇒(∼q)) is

[25-Jan-2023 Shift 1]

The statement (p∧(∼q))⇒(p⇒(∼q)) is

[25-Jan-2023 Shift 1]

A
equivalent to (∼p)∨(∼q)
equivalent to (∼p)∨(∼q)

B
a tautology
a tautology

C
equivalent to p∨q
equivalent to p∨q

D
a contradiction
a contradiction
Question 5/137
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Mark Review

Let ∆,∇∈{∧,∨} be such that (p→q)∆(p∇q) is a tautology. Then

[25-Jan-2023 Shift 2]

Let ∆,∇∈{∧,∨} be such that (p→q)∆(p∇q) is a tautology. Then

[25-Jan-2023 Shift 2]

A
∆=∧,∇=∨
∆=∧,∇=∨

B
∆=∨,∇=∧
∆=∨,∇=∧

C
∆=∨,∇=∨
∆=∨,∇=∨

D
∆=∧,∇=∧
∆=∧,∇=∧
Question 6/137
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Mark Review

If p,q and r are three propositions, then which of the following combination of truth values of p,q and r makes the logical expression {(p∨q)∧((∼p)∨r)}→((∼q)∨r) false ?

[29-Jan-2023 Shift 1]

If p,q and r are three propositions, then which of the following combination of truth values of p,q and r makes the logical expression {(p∨q)∧((∼p)∨r)}→((∼q)∨r) false ?

[29-Jan-2023 Shift 1]

A
p=T,q=F,r=T
p=T,q=F,r=T

B
p=T,q=T,r=F
p=T,q=T,r=F

C
p=F,q=T,r=F‌‌
p=F,q=T,r=F‌‌

D
p=T,q=F,r=F
p=T,q=F,r=F
Question 7/137
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Mark Review

The statement B⇒((∼A)∨B) is equivalent to :

[29-Jan-2023 Shift 2]

The statement B⇒((∼A)∨B) is equivalent to :

[29-Jan-2023 Shift 2]

*This question may have multiple correct answers

A
B⇒(A⇒B)
B⇒(A⇒B)

B
A⇒(A⇔B)
A⇒(A⇔B)

C
A⇒((∼A)⇒B)
A⇒((∼A)⇒B)

D
B⇒((∼A)⇒B)
B⇒((∼A)⇒B)
Question 8/137
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Mark Review

Among the statements:
(S1) ‌‌((p∨q)⇒r)⇔(p⇒r)
(S2) ‌‌((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))

[30-Jan-2023 Shift 1]

Among the statements:
(S1) ‌‌((p∨q)⇒r)⇔(p⇒r)
(S2) ‌‌((p∨q)⇒r)⇔((p⇒r)∨(q⇒r))

[30-Jan-2023 Shift 1]

A
Only ( S1) is a tautology
Only ( S1) is a tautology

B
Neither (S1) nor (S2) is a tautology
Neither (S1) nor (S2) is a tautology

C
Only ( S2) is a tautology
Only ( S2) is a tautology

D
Both (S1) and (S2) are tautologies
Both (S1) and (S2) are tautologies
Question 9/137
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Mark Review

Consider the following statements:
P : I have fever
Q : I will not take medicine
R : I will take rest
The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:

[30-Jan-2023 Shift 2]

Consider the following statements:
P : I have fever
Q : I will not take medicine
R : I will take rest
The statement "If I have fever, then I will take medicine and I will take rest" is equivalent to:

[30-Jan-2023 Shift 2]

A
((∼P)∨∼Q)∧((∼P)∨R)
((∼P)∨∼Q)∧((∼P)∨R)

B
((∼P)∨∼Q)∧((∼P)∨∼R)
((∼P)∨∼Q)∧((∼P)∨∼R)

C
(P∨Q)∧((∼P)∨R)
(P∨Q)∧((∼P)∨R)

D
(P∨∼Q)∧(P∨∼R)
(P∨∼Q)∧(P∨∼R)
Question 10/137
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Mark Review

(S1)(p⇒q)∨(p∧(∼q)) is a tautology
(S2)((∼p)⇒(∼q))∧((∼p)∨q) is a
Contradiction. Then

[31-Jan-2023 Shift 1]

(S1)(p⇒q)∨(p∧(∼q)) is a tautology
(S2)((∼p)⇒(∼q))∧((∼p)∨q) is a
Contradiction. Then

[31-Jan-2023 Shift 1]

A
only (S2) is correct
only (S2) is correct

B
both (S1) and (S2) are correct
both (S1) and (S2) are correct

C
both (S1) and (S2) are wrong
both (S1) and (S2) are wrong

D
only (S1) is correct
only (S1) is correct
Question 11/137
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Mark Review

The number of values of r∈{p,q,∼p,∼q} for which ((p∧q)⇒(r∨q))∧((p∧r)⇒q) is a tautology, is:

[31-Jan-2023 Shift 2]

The number of values of r∈{p,q,∼p,∼q} for which ((p∧q)⇒(r∨q))∧((p∧r)⇒q) is a tautology, is:

[31-Jan-2023 Shift 2]

A
3
3

B
2
2

C
1
1

D
4
4
Question 12/137
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Mark Review

The negation of the expression q∨((∼q)∧p) is equivalent to

[1-Feb-2023 Shift 1]

The negation of the expression q∨((∼q)∧p) is equivalent to

[1-Feb-2023 Shift 1]

A
(∼p)∧(∼q)
(∼p)∧(∼q)

B
p∧(∼q)
p∧(∼q)

C
(∼p)∨(∼q)
(∼p)∨(∼q)

D
(∼p)∨q
(∼p)∨q
Question 13/137
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Mark Review

Which of the following statements is a tautology ?

[1-Feb-2023 Shift 2]

Which of the following statements is a tautology ?

[1-Feb-2023 Shift 2]

A
p→(pΛ(p→q))
p→(pΛ(p→q))

B
(pΛq)→(∼(p)→q))
(pΛq)→(∼(p)→q))

C
(pΛ(p→q))→∼q
(pΛ(p→q))→∼q

D
pV(pΛq)
pV(pΛq)
Question 14/137
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Mark Review

Statement (P⇒Q)∧(R⇒Q) is logically equivalent to:

[6-Apr-2023 shift 1]

Statement (P⇒Q)∧(R⇒Q) is logically equivalent to:

[6-Apr-2023 shift 1]

A
(P∨R)⇒Q
(P∨R)⇒Q

B
(P⇒R)∨(Q⇒R)
(P⇒R)∨(Q⇒R)

C
(P⇒R)∧(Q⇒R)
(P⇒R)∧(Q⇒R)

D
(P∧R)⇒Q
(P∧R)⇒Q
Question 15/137
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Mark Review

Among the statements :
(S1):(p⇒q)∨((∼p)∧q) is a tautology
(S2):(q⇒p)⇒((∼p)∧q) is a contradiction

[6-Apr-2023 shift 2]

Among the statements :
(S1):(p⇒q)∨((∼p)∧q) is a tautology
(S2):(q⇒p)⇒((∼p)∧q) is a contradiction

[6-Apr-2023 shift 2]

A
only (S2) is True
only (S2) is True

B
only (S1) is True
only (S1) is True

C
neigher (S1) and (S2) is True
neigher (S1) and (S2) is True

D
both (S1) and (S2) are True
both (S1) and (S2) are True
Question 16/137
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Mark Review

Negation of (p→q)→(q→p) is

[8-Apr-2023 shift 1]

Negation of (p→q)→(q→p) is

[8-Apr-2023 shift 1]

A
(−q)∧p
(−q)∧p

B
p∨(∼q)
p∨(∼q)

C
(∼p)∨q
(∼p)∨q

D
q∧(∼p)
q∧(∼p)
Question 17/137
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Mark Review

The negation of (p∧(∼q))∨(∼p) is equivalent to

[8-Apr-2023 shift 2]

The negation of (p∧(∼q))∨(∼p) is equivalent to

[8-Apr-2023 shift 2]

A
p∧(∼q)
p∧(∼q)

B
p∧(q∧(∼p))
p∧(q∧(∼p))

C
p∨(q∨(∼p))
p∨(q∨(∼p))

D
p∧q
p∧q
Question 18/137
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Mark Review

The negation of the statement :
(p∨q)∧(q∨(∼r)) is

[10-Apr-2023 shift 1]

The negation of the statement :
(p∨q)∧(q∨(∼r)) is

[10-Apr-2023 shift 1]

A
((∼p)∨r))∧(∼q)
((∼p)∨r))∧(∼q)

B
((∼ p) ∨(∼q))∧(∼r)
((∼ p) ∨(∼q))∧(∼r)

C
((∼p)∨(∼q))∨(∼r)
((∼p)∨(∼q))∨(∼r)

D
(p∨r)∧(∼q)
(p∨r)∧(∼q)
Question 19/137
4 / -1

Mark Review

The statement ∼[pV(∼(p∧q))] is equivalent to

[10-Apr-2023 shift 2]

The statement ∼[pV(∼(p∧q))] is equivalent to

[10-Apr-2023 shift 2]

A
(∼(p∧q))∧q
(∼(p∧q))∧q

B
∼(p∨q)
∼(p∨q)

C
∼(p∧q)
∼(p∧q)

D
(p∧q)∧(∼p)
(p∧q)∧(∼p)
Question 20/137
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Mark Review

The number of ordered triplets of the truth values of p,q and r such that the truth value of the statement (p ∨q)∧(p∨r)⇒(q∨r) is True, is equal to ________.
[11-Apr-2023 shift 1]

The number of ordered triplets of the truth values of p,q and r such that the truth value of the statement (p ∨q)∧(p∨r)⇒(q∨r) is True, is equal to ________.
[11-Apr-2023 shift 1]
Question 21/137
4 / -1

Mark Review

The converse of ((∼p)∧q)⇒r is

[11-Apr-2023 shift 2]

The converse of ((∼p)∧q)⇒r is

[11-Apr-2023 shift 2]

A
(pv(∼q))⇒(∼r)
(pv(∼q))⇒(∼r)

B
((∼p)vq)⇒r
((∼p)vq)⇒r

C
(∼r)⇒((∼p)∧q)
(∼r)⇒((∼p)∧q)

D
(∼r)⇒p∧q
(∼r)⇒p∧q
Question 22/137
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Mark Review

Among the two statements
(S1): (p⇒q)∧(p∧(∼q)) is a contradiction and
(S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q)) is a tautology

[12-Apr-2023 shift 1]

Among the two statements
(S1): (p⇒q)∧(p∧(∼q)) is a contradiction and
(S2):(p∧q)∨((∼p)∧q)∨(p∧(∼q))∨((∼p)∧(∼q)) is a tautology

[12-Apr-2023 shift 1]

A
only (S2) is true
only (S2) is true

B
only (S1) is true
only (S1) is true

C
both are true
both are true

D
both are false
both are false
Question 23/137
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Mark Review

The negation of the statement ((A∧(B∨C))⇒(A∨B))⇒A is

[13-Apr-2023 shift 1]

The negation of the statement ((A∧(B∨C))⇒(A∨B))⇒A is

[13-Apr-2023 shift 1]

A
equivalent to B∨∼C
equivalent to B∨∼C

B
a fallacy
a fallacy

C
equivalent to ∼C
equivalent to ∼C

D
equivalent to ∼ A
equivalent to ∼ A
Question 24/137
4 / -1

Mark Review

The statement (p∧(∼q))∨((∼p)∧q)∨((∼p)∧(∼q)) is equivalent to

[13-Apr-2023 shift 2]

The statement (p∧(∼q))∨((∼p)∧q)∨((∼p)∧(∼q)) is equivalent to

[13-Apr-2023 shift 2]

A
(∼p)∨(∼q)
(∼p)∨(∼q)

B
(∼p)∧(∼q)
(∼p)∧(∼q)

C
p∨(∼q)
p∨(∼q)

D
p∨q
p∨q
Question 25/137
4 / -1

Mark Review

Negation of p∧(q∧∼(p∧q)) is

[15-Apr-2023 shift 1]

Negation of p∧(q∧∼(p∧q)) is

[15-Apr-2023 shift 1]

A
(∼(p∧q))∧q
(∼(p∧q))∧q

B
∼(p∨q)
∼(p∨q)

C
p∨q
p∨q

D
(∼(p∧q))∨p
(∼(p∧q))∨p
Question 26/137
4 / -1

Mark Review

The number of choices for ∆∈{∧,∨,⇒,⇔}, such that (p∆q)⇒((p∆∼q)∨((∼p)∆q)) is a tautology, is :

[24-Jun-2022-Shift-1]

The number of choices for ∆∈{∧,∨,⇒,⇔}, such that (p∆q)⇒((p∆∼q)∨((∼p)∆q)) is a tautology, is :

[24-Jun-2022-Shift-1]

A
1
1

B
2
2

C
3
3

D
4
4
Question 27/137
4 / -1

Mark Review

Consider the following statements:
A : Rishi is a judge.
B : Rishi is honest.
C : Rishi is not arrogant.
The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

[24-Jun-2022-Shift-2]

Consider the following statements:
A : Rishi is a judge.
B : Rishi is honest.
C : Rishi is not arrogant.
The negation of the statement "if Rishi is a judge and he is not arrogant, then he is honest" is

[24-Jun-2022-Shift-2]

A
B→(A∨C)
B→(A∨C)

B
(∼B)∧(A∧C)
(∼B)∧(A∧C)

C
B→((∼A)∨(∼C))
B→((∼A)∨(∼C))

D
B→(A∧C)
B→(A∧C)
Question 28/137
4 / -1

Mark Review

Consider the following two propositions:
P1:∼(p→∼q)
P2:(p∧∼q)∧((∼p)∨q)
If the proposition p→((∼p)∨q) is evaluated as FALSE, then :

[25-Jun-2022-Shift-1]

Consider the following two propositions:
P1:∼(p→∼q)
P2:(p∧∼q)∧((∼p)∨q)
If the proposition p→((∼p)∨q) is evaluated as FALSE, then :

[25-Jun-2022-Shift-1]

A
P1 is TRUE and P2 is FALSE
P1 is TRUE and P2 is FALSE

B
P1 is FALSE and P2 is TRUE
P1 is FALSE and P2 is TRUE

C
Both P1 and P2 are FALSE
Both P1 and P2 are FALSE

D
Both P1 and P2 are TRUE
Both P1 and P2 are TRUE
Question 29/137
4 / -1

Mark Review

The negation of the Boolean expression ((∼q)∧p)⇒((∼p)∨q) is logically equivalent to :

[25-Jun-2022-Shift-2]

The negation of the Boolean expression ((∼q)∧p)⇒((∼p)∨q) is logically equivalent to :

[25-Jun-2022-Shift-2]

A
p⇒q
p⇒q

B
q⇒p
q⇒p

C
∼(p⇒q)
∼(p⇒q)

D
∼(q⇒p)
∼(q⇒p)
Question 30/137
4 / -1

Mark Review

Let ∆,∇∈{∧,V} be such that p∇q⇒((p∆q)∇r) is a tautology. Then (p ∇q)∆r is logically equivalent to :

[26-Jun-2022-Shift-1]

Let ∆,∇∈{∧,V} be such that p∇q⇒((p∆q)∇r) is a tautology. Then (p ∇q)∆r is logically equivalent to :

[26-Jun-2022-Shift-1]

A
(p∆r)∨q
(p∆r)∨q

B
(p∆r)∧q
(p∆r)∧q

C
(p∧r)∆q
(p∧r)∆q

D
(p∇r)∧q
(p∇r)∧q
Question 31/137
4 / -1

Mark Review

Let r∈{p,q,∼p,∼q} be such that the logical statement
r∨(∼p)⇒(p∧q)∨r
is a tautology. Then r is equal to :

[26-Jun-2022-Shift-2]

Let r∈{p,q,∼p,∼q} be such that the logical statement
r∨(∼p)⇒(p∧q)∨r
is a tautology. Then r is equal to :

[26-Jun-2022-Shift-2]

A
p
p

B
q
q

C
∼p
∼p

D
∼q
∼q
Question 32/137
4 / -1

Mark Review

The boolean expression (∼(p∧q))∨q is equivalent to :

[27-Jun-2022-Shift-1]

The boolean expression (∼(p∧q))∨q is equivalent to :

[27-Jun-2022-Shift-1]

A
q→(p∧q)
q→(p∧q)

B
p→q
p→q

C
p→(p→q)
p→(p→q)

D
p→(p∨q)
p→(p∨q)
Question 33/137
4 / -1

Mark Review

Which of the following statement is a tautology?

[27-Jun-2022-Shift-2]

Which of the following statement is a tautology?

[27-Jun-2022-Shift-2]

A
((∼q)∧p)∧q
((∼q)∧p)∧q

B
((∼q)∧p)∧(p∧(∼p))
((∼q)∧p)∧(p∧(∼p))

C
((∼q)∧p)∨(p∨(∼p))
((∼q)∧p)∨(p∨(∼p))

D
(p∧q)∧(∼p∧q))
(p∧q)∧(∼p∧q))
Question 34/137
4 / -1

Mark Review

Let p,q,r be three logical statements. Consider the compound statements S1:((∼p)∨q)∨((∼p)∨r) and S2:p→(q∨r)
Then, which of the following is NOT true?

[28-Jun-2022-Shift-1]

Let p,q,r be three logical statements. Consider the compound statements S1:((∼p)∨q)∨((∼p)∨r) and S2:p→(q∨r)
Then, which of the following is NOT true?

[28-Jun-2022-Shift-1]

A
If S2 is True, then S1 is True
If S2 is True, then S1 is True

B
If S2 is False, then S1 is False
If S2 is False, then S1 is False

C
If S2 is False, then S1 is True
If S2 is False, then S1 is True

D
If S1 is False, then S2 is False
If S1 is False, then S2 is False
Question 35/137
4 / -1

Mark Review

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62 , and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is_____
[28-Jun-2022-Shift-2]

Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62 , and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is_____
[28-Jun-2022-Shift-2]
Question 36/137
4 / -1

Mark Review

The maximum number of compound propositions, out of p∨r∨s,p∨r⋁∼s,p∨∼q∨s,∼p∨∼r∨s,∼p∨∼r∨∼s,∼p∨q∨∼ s,q∨r∨∼s,q∨∼r∨∼s,∼p∨∼q∨∼s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to___
[28-Jun-2022-Shift-2]

The maximum number of compound propositions, out of p∨r∨s,p∨r⋁∼s,p∨∼q∨s,∼p∨∼r∨s,∼p∨∼r∨∼s,∼p∨q∨∼ s,q∨r∨∼s,q∨∼r∨∼s,∼p∨∼q∨∼s that can be made simultaneously true by an assignment of the truth values to p, q, r and s, is equal to___
[28-Jun-2022-Shift-2]
Question 37/137
4 / -1

Mark Review

Let ∆∈{∧,∨,⇒,⇔} be such that (p∧q)∆((p∨q)⇒q) is a tautology. Then ∆ is equal to :

[29-Jun-2022-Shift-1]

Let ∆∈{∧,∨,⇒,⇔} be such that (p∧q)∆((p∨q)⇒q) is a tautology. Then ∆ is equal to :

[29-Jun-2022-Shift-1]

A
∧
∧

B
V
V

C
⇒
⇒

D
⇔
⇔
Question 38/137
4 / -1

Mark Review

Negation of the Boolean statement (p∨q)⇒((∼r)∨p) is equivalent to

[29-Jun-2022-Shift-2]

Negation of the Boolean statement (p∨q)⇒((∼r)∨p) is equivalent to

[29-Jun-2022-Shift-2]

A
p∧(∼q)∧r
p∧(∼q)∧r

B
(∼p)∧(∼q)∧r
(∼p)∧(∼q)∧r

C
(∼p)∧q∧r
(∼p)∧q∧r

D
p∧q∧(∼r)
p∧q∧(∼r)
Question 39/137
4 / -1

Mark Review

Which of the following statements is a tautology?

[25-Jul-2022-Shift-1]

Which of the following statements is a tautology?

[25-Jul-2022-Shift-1]

A
((∼p)∨q)⇒p
((∼p)∨q)⇒p

B
p⇒((∼p)∨q)
p⇒((∼p)∨q)

C
((∼p)∨q)⇒q
((∼p)∨q)⇒q

D
q⇒((∼p)∨q)
q⇒((∼p)∨q)
Question 40/137
4 / -1

Mark Review

Consider the following statements:
P : Ramu is intelligent.
Q : Ramu is rich.
R : Ramu is not honest.
The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as:

[25-Jul-2022-Shift-2]

Consider the following statements:
P : Ramu is intelligent.
Q : Ramu is rich.
R : Ramu is not honest.
The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as:

[25-Jul-2022-Shift-2]

A
((P∧(∼R))∧Q)∧((∼Q)∧((∼P)∨R))
((P∧(∼R))∧Q)∧((∼Q)∧((∼P)∨R))

B
((P∧R)∧Q)∨((∼Q)∧((∼P)∨(∼R)))
((P∧R)∧Q)∨((∼Q)∧((∼P)∨(∼R)))

C
((P∧R)∧Q)∧((∼Q)∧((∼P)∨(∼R)))
((P∧R)∧Q)∧((∼Q)∧((∼P)∨(∼R)))

D
((P∧(∼R))∧Q)∨((∼Q)∧((∼P)∨R))
((P∧(∼R))∧Q)∨((∼Q)∧((∼P)∨R))
Question 41/137
4 / -1

Mark Review

The statement (∼(p⇔∼q))∧q is :

[26-Jul-2022-Shift-1]

The statement (∼(p⇔∼q))∧q is :

[26-Jul-2022-Shift-1]

A
a tautology
a tautology

B
a contradiction
a contradiction

C
equivalent to (p⇒q)∧q
equivalent to (p⇒q)∧q

D
equivalent to (p⇒q)∧p
equivalent to (p⇒q)∧p
Question 42/137
4 / -1

Mark Review

Negation of the Boolean expression p⇔(q⇒p) is
[26-Jul-2022-Shift-2]

Negation of the Boolean expression p⇔(q⇒p) is
[26-Jul-2022-Shift-2]

A
(∼p)∧q
(∼p)∧q

B
p∧(∼q)
p∧(∼q)

C
(∼p)∨(∼q)
(∼p)∨(∼q)

D
(∼p)∧(∼q)
(∼p)∧(∼q)
Question 43/137
4 / -1

Mark Review

(p∧r)⇔(p∧(∼q)) is equivalent to (∼p) when r is
[27-Jul-2022-Shift-1]

(p∧r)⇔(p∧(∼q)) is equivalent to (∼p) when r is
[27-Jul-2022-Shift-1]

A
p
p

B
∼p
∼p

C
q
q

D
∼q
∼q
Question 44/137
4 / -1

Mark Review

If the truth value of the statement (P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F ?
[27-Jul-2022-Shift-2]

If the truth value of the statement (P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F ?
[27-Jul-2022-Shift-2]

A
P∨Q→∼R
P∨Q→∼R

B
R∨Q→∼P
R∨Q→∼P

C
∼(P∨Q)→∼R
∼(P∨Q)→∼R

D
∼(R∨Q)→∼P
∼(R∨Q)→∼P
Question 45/137
4 / -1

Mark Review

Let the operations *,⊙∈{∧,∨}. If (p*q)⊙(p⊙∼q) is a tautology, then the ordered pair (*,⊙) is:

[28-Jul-2022-Shift-1]

Let the operations *,⊙∈{∧,∨}. If (p*q)⊙(p⊙∼q) is a tautology, then the ordered pair (*,⊙) is:

[28-Jul-2022-Shift-1]

A
(∨,∧)
(∨,∧)

B
(V,V)
(V,V)

C
(∧,∧)
(∧,∧)

D
(∧,∨)
(∧,∨)
Question 46/137
4 / -1

Mark Review

Let
p : Ramesh listens to music.
q : Ramesh is out of his village.
r : It is Sunday.
s : It is Saturday.
Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as
[28-Jul-2022-Shift-2]

Let
p : Ramesh listens to music.
q : Ramesh is out of his village.
r : It is Sunday.
s : It is Saturday.
Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as
[28-Jul-2022-Shift-2]

A
((∼q)∧(r∨s))⇒p
((∼q)∧(r∨s))⇒p

B
(q∧(r∨ s))⇒p
(q∧(r∨ s))⇒p

C
p⇒(q∧(r∨s))
p⇒(q∧(r∨s))

D
p⇒((∼q)∧(r∨ s))
p⇒((∼q)∧(r∨ s))
Question 47/137
4 / -1

Mark Review

The statement (p∧q)⇒(p∧r) is equivalent to :

[29-Jul-2022-Shift-1]

The statement (p∧q)⇒(p∧r) is equivalent to :

[29-Jul-2022-Shift-1]

A
q⇒(p∧r)
q⇒(p∧r)

B
p⇒(p∧r)
p⇒(p∧r)

C
(p∧r)⇒(p∧q)
(p∧r)⇒(p∧q)

D
(p∧q)⇒r
(p∧q)⇒r
Question 48/137
4 / -1

Mark Review

The statement (p⇒q)∨(p⇒r) is NOT equivalent to
[29-Jul-2022-Shift-2]

The statement (p⇒q)∨(p⇒r) is NOT equivalent to
[29-Jul-2022-Shift-2]

A
(p∧(∼r))⇒q
(p∧(∼r))⇒q

B
(∼q)⇒((∼r)∨p)
(∼q)⇒((∼r)∨p)

C
p⇒(q∨r)
p⇒(q∨r)

D
(p∧(∼q))⇒r
(p∧(∼q))⇒r
Question 49/137
4 / -1

Mark Review

The contrapositive of the statement; "If you will work, you will earn money" is
[2021, 25 Feb. Shift-II]

The contrapositive of the statement; "If you will work, you will earn money" is
[2021, 25 Feb. Shift-II]

A
to earn money, you need to work
to earn money, you need to work

B
you will earn money, if you will not work
you will earn money, if you will not work

C
if you will not earn money, you will not work
if you will not earn money, you will not work

D
if you will earn money, you will work
if you will earn money, you will work
Question 50/137
4 / -1

Mark Review

Let F1(A,B,C)=(A∧∼B)∨[∼C∧(A∨B)]∨∼A and F2(A,B)=(A∨B)∨(B→∼A) be two logical expressions. Then,
[2021, 26 Feb. Shift-II]

Let F1(A,B,C)=(A∧∼B)∨[∼C∧(A∨B)]∨∼A and F2(A,B)=(A∨B)∨(B→∼A) be two logical expressions. Then,
[2021, 26 Feb. Shift-II]

A
F1 and F2 both are tautologies
F1 and F2 both are tautologies

B
F1 is a tautology but F2 is not a tautology
F1 is a tautology but F2 is not a tautology

C
F1 is not tautology but F2 is a tautology
F1 is not tautology but F2 is a tautology

D
Both F1 and F2 are not tautologies
Both F1 and F2 are not tautologies
Question 51/137
4 / -1

Mark Review

The statement A→(B→A) is equivalent to
[2021, 25 Feb. Shift-1]

The statement A→(B→A) is equivalent to
[2021, 25 Feb. Shift-1]

A
A→(A∧B)
A→(A∧B)

B
A→(A→B)
A→(A→B)

C
A→(A↔B)
A→(A↔B)

D
A→(A∨B)
A→(A∨B)
Question 52/137
4 / -1

Mark Review

For the statements p and q, consider the following compound statements
A. [∼q∧(p→q)]→∼p
B. [(p∨q)∧∼p]→q
Then, which of the following statement(s) is/are correct?
[2021, 24 Feb. Shift-II]

For the statements p and q, consider the following compound statements
A. [∼q∧(p→q)]→∼p
B. [(p∨q)∧∼p]→q
Then, which of the following statement(s) is/are correct?
[2021, 24 Feb. Shift-II]

A
(A) and (B) both are not tautologies.
(A) and (B) both are not tautologies.

B
(A) and (B) both are tautologies.
(A) and (B) both are tautologies.

C
(A) is a tautology but not (B).
(A) is a tautology but not (B).

D
(B) is a tautology but not (A).
(B) is a tautology but not (A).
Question 53/137
4 / -1

Mark Review

The negative of the statement ∼p∧(p∨q) is
[2021, 24 Feb. Shift-II]

The negative of the statement ∼p∧(p∨q) is
[2021, 24 Feb. Shift-II]

A
p ∨ q
p ∨ q

B
p∨∼q
p∨∼q

C
p∧q
p∧q

D
p∧∼q
p∧∼q
Question 54/137
4 / -1

Mark Review

The statement among the following that is a tautology is
[2021, 24 Feb. Shift-1]

The statement among the following that is a tautology is
[2021, 24 Feb. Shift-1]

A
A∧(A∨B)
A∧(A∨B)

B
A∨(A∧B)
A∨(A∧B)

C
[A∧(A→B)]→B
[A∧(A→B)]→B

D
B→[A∧(A→B)]
B→[A∧(A→B)]
Question 55/137
4 / -1

Mark Review

The statement among the following that is a tautology is:
24 Feb 2021 Shift 1

The statement among the following that is a tautology is:
24 Feb 2021 Shift 1

A
A∨(A∧B)
A∨(A∧B)

B
A∧(A∨B)
A∧(A∨B)

C
B→[A∧(A→B)]
B→[A∧(A→B)]

D
[A∧(A→B)]→B
[A∧(A→B)]→B
Question 56/137
4 / -1

Mark Review

If p and q are two statements, then which of the following compound statement is a tautology?
[2021, 18 March Shift-II]

If p and q are two statements, then which of the following compound statement is a tautology?
[2021, 18 March Shift-II]

A
(p⇒q)∧∼q]⇒0
(p⇒q)∧∼q]⇒0

B
(p⇒q)∧∼q]⇒∼p
(p⇒q)∧∼q]⇒∼p

C
(p⇒q)∧∼q]⇒p
(p⇒q)∧∼q]⇒p

D
(p⇒q)∧∼q]⇒(p∧q)
(p⇒q)∧∼q]⇒(p∧q)
Question 57/137
4 / -1

Mark Review

If the Boolean expression (p⇒q)⇔(q*(∼p)) is a tautology, then the Boolean expression (p*(∼q)) is equivalent to
[2021, 17 March Shift-1]

If the Boolean expression (p⇒q)⇔(q*(∼p)) is a tautology, then the Boolean expression (p*(∼q)) is equivalent to
[2021, 17 March Shift-1]

A
q⇒p
q⇒p

B
q⇒p
q⇒p

C
p⇒∼q
p⇒∼q

D
p⇒q
p⇒q
Question 58/137
4 / -1

Mark Review

If the Boolean expression (p∧q)*(p⊗q) is a tautology, then ‌* and ⊗ are respectively, given by
[2021, 17 March Shift-II]

If the Boolean expression (p∧q)*(p⊗q) is a tautology, then ‌* and ⊗ are respectively, given by
[2021, 17 March Shift-II]

A
→,→
→,→

B
∧∨
∧∨

C
v1→
v1→

D
∧→
∧→
Question 59/137
4 / -1

Mark Review

Which of the following Boolean expression is a tautology?
[2021, 16 March Shift-1]

Which of the following Boolean expression is a tautology?
[2021, 16 March Shift-1]

A
(p∧q)∨(p∨q)
(p∧q)∨(p∨q)

B
(p∧q)∨(p→q)
(p∧q)∨(p→q)

C
(p∧q)∧(p→q)
(p∧q)∧(p→q)

D
(p∧q)→(p→q)
(p∧q)→(p→q)
Question 60/137
4 / -1

Mark Review

Which of the following is the negation of the statement "for all M>0, there exist x∈S such that x≥M "?
[2021, 27 July Shift-II]

Which of the following is the negation of the statement "for all M>0, there exist x∈S such that x≥M "?
[2021, 27 July Shift-II]

A
there exists M>0, such that x<M for all x∈S
there exists M>0, such that x<M for all x∈S

B
there exists M>0, there exists x∈S such that x≥M
there exists M>0, there exists x∈S such that x≥M

C
there exists M>0, there exists x∈S such that x<M
there exists M>0, there exists x∈S such that x<M

D
there exists M>0, such that x≥M for all x∈S
there exists M>0, such that x≥M for all x∈S
Question 61/137
4 / -1

Mark Review

Consider the statement "The match will be played only if the weather is good and ground is not wet".
Select the correct negation from the following
[2021,25 July Shift-II]

Consider the statement "The match will be played only if the weather is good and ground is not wet".
Select the correct negation from the following
[2021,25 July Shift-II]

A
The match will not be played andweather is not good and ground is wet.

The match will not be played andweather is not good and ground is wet.

B
If the match will not be played, then either weather is not good or ground is wet.

If the match will not be played, then either weather is not good or ground is wet.

C
The match will be played and weather is not good or ground is wet.

The match will be played and weather is not good or ground is wet.

D
The match will not be played or weather is good and ground is not wet.
The match will not be played or weather is good and ground is not wet.
Question 62/137
4 / -1

Mark Review

Consider the following three statements,
(A) If 3+3=7, then 4+3=8
(B) If 5+3=8, then earth is flat
(C) If both (A) and (B) are true, then 5+6=17
Then, which of the following statements is correct
[2021, 20 July Shift II]

Consider the following three statements,
(A) If 3+3=7, then 4+3=8
(B) If 5+3=8, then earth is flat
(C) If both (A) and (B) are true, then 5+6=17
Then, which of the following statements is correct
[2021, 20 July Shift II]

A
(A) is false, but (B) and (C) are true
(A) is false, but (B) and (C) are true

B
(A) and (C) are true while (B) is false
(A) and (C) are true while (B) is false

C
(A) is true while (B) and (C) are false
(A) is true while (B) and (C) are false

D
(A) and (B) are false while (C) is true
(A) and (B) are false while (C) is true
Question 63/137
4 / -1

Mark Review

The compound statement (P∨Q)∧(∼P)⇒Q is equivalent to
[2021, 27 July Shift-1]

The compound statement (P∨Q)∧(∼P)⇒Q is equivalent to
[2021, 27 July Shift-1]

A
P∨Q
P∨Q

B
P∧∼Q
P∧∼Q

C
~(P⇒Q)
~(P⇒Q)

D
~(P⇒Q)⇔P∧∼Q
~(P⇒Q)⇔P∧∼Q
Question 64/137
4 / -1

Mark Review

The Boolean expression (p⇒q)∧(q⇒∼p) is equivalent to
[2021, 25 July Shift-1]

The Boolean expression (p⇒q)∧(q⇒∼p) is equivalent to
[2021, 25 July Shift-1]

A
∼q
∼q

B
q
q

C
p
p

D
p
p
Question 65/137
4 / -1

Mark Review

Which of the following Boolean expression is not a tautology ?
[2021, 22 July Shift-II]

Which of the following Boolean expression is not a tautology ?
[2021, 22 July Shift-II]

A
(p⇒q)∨(∼q⇒p)
(p⇒q)∨(∼q⇒p)

B
(q⇒p)∨(∼q⇒p)
(q⇒p)∨(∼q⇒p)

C
(p⇒∼q)∨(∼q⇒p)
(p⇒∼q)∨(∼q⇒p)

D
(∼p⇒q)∨(∼q⇒p)
(∼p⇒q)∨(∼q⇒p)
Question 66/137
4 / -1

Mark Review

The Boolean expression (p∧∼q)⇒(q∨∼p) is equivalent to
[2021, 20 July Shift-1]

The Boolean expression (p∧∼q)⇒(q∨∼p) is equivalent to
[2021, 20 July Shift-1]

A
q⇒p
q⇒p

B
p⇒q
p⇒q

C
q⇒p
q⇒p

D
p⇒∼q
p⇒∼q
Question 67/137
4 / -1

Mark Review

Let *,square,in{wedge,v} be such that the Boolean expression left(p*‌sim‌q‌right)‌Rightarrow(pq) is a tautology. Then
[2021,31 Aug. Shift-I]

Let *,square,in{wedge,v} be such that the Boolean expression left(p*‌sim‌q‌right)‌Rightarrow(pq) is a tautology. Then
[2021,31 Aug. Shift-I]

A
‌*=v1=v
‌*=v1=v

B
‌*=∧1=∧
‌*=∧1=∧

C
‌*=Λ′=∨
‌*=Λ′=∨

D
‌*=v1=∧
‌*=v1=∧
Question 68/137
4 / -1

Mark Review

Negation of the statement (p∨q)⇒(q∨r) is
[2021, 31 Aug. Shift-II]

Negation of the statement (p∨q)⇒(q∨r) is
[2021, 31 Aug. Shift-II]

A
p∧∼q∧∼r
p∧∼q∧∼r

B
∼p∧q∧∼r
∼p∧q∧∼r

C
∼p∧∧r
∼p∧∧r

D
p∧q∧r
p∧q∧r
Question 69/137
4 / -1

Mark Review

The statement (p∧(p→q)∧(q→r))→r is
[2021, 27 Aug. Shift-1]

The statement (p∧(p→q)∧(q→r))→r is
[2021, 27 Aug. Shift-1]

A
a tautology
a tautology

B
equivalent to p→∼r
equivalent to p→∼r

C
a fallacy
a fallacy

D
equivalent to →→r
equivalent to →→r
Question 70/137
4 / -1

Mark Review

The Boolean expression (p∧q)⇒((r∧q)∧p) is equivalent to
[2021,27 Aug. Shift-11]

The Boolean expression (p∧q)⇒((r∧q)∧p) is equivalent to
[2021,27 Aug. Shift-11]

A
(p∧q)⇒(r∧q)
(p∧q)⇒(r∧q)

B
(q∧r)⇒(p∧q)
(q∧r)⇒(p∧q)

C
(p∧q)⇒(tr∨q)
(p∧q)⇒(tr∨q)

D
(p∧r)⇒(p∧q)
(p∧r)⇒(p∧q)
Question 71/137
4 / -1

Mark Review

If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r)→(p∧q) is false, then the truth values of the statements, p,q and r respectively can be
[2021, 26 Aug. Shift-1]

If the truth value of the Boolean expression ((p∨q)∧(q→r)∧(∼r)→(p∧q) is false, then the truth values of the statements, p,q and r respectively can be
[2021, 26 Aug. Shift-1]

A
T F T
T F T

B
FFT
FFT

C
TFF
TFF

D
FTF
FTF
Question 72/137
4 / -1

Mark Review

Consider the two statements :
(S1):(p→q)∨(∼q→p) is a tautology (S2):(p∧∼q)∧(∼p∨q) is a fallacy Then,
[2021, 26 Aug. Shift-11]

Consider the two statements :
(S1):(p→q)∨(∼q→p) is a tautology (S2):(p∧∼q)∧(∼p∨q) is a fallacy Then,
[2021, 26 Aug. Shift-11]

A
only (S1) is true
only (S1) is true

B
both (S1) and (S2) are false
both (S1) and (S2) are false

C
both (S1) and (S2) are true
both (S1) and (S2) are true

D
only (S2) is true
only (S2) is true
Question 73/137
4 / -1

Mark Review

Which of the following is equivalent to the Boolean expression p⋀∽q ?
[1 Sep 2021 Shift 2]

Which of the following is equivalent to the Boolean expression p⋀∽q ?
[1 Sep 2021 Shift 2]

A
∽ (q→p)
∽ (q→p)

B
∽p→∽q
∽p→∽q

C
∽ ( p→∽q )
∽ ( p→∽q )

D
∽ (p→q)
∽ (p→q)
Question 74/137
4 / -1

Mark Review

Let A,B,C and D be four non-empty sets. The contrapositive statement of "If A⊆B and B⊆D, then A⊆C" is:
[Jan. 7, 2020 (II)]

Let A,B,C and D be four non-empty sets. The contrapositive statement of "If A⊆B and B⊆D, then A⊆C" is:
[Jan. 7, 2020 (II)]

A
If A⊈C, then A⊆B and B⊆D
If A⊈C, then A⊆B and B⊆D

B
If A⊆C, then B⊂A or D⊂B
If A⊆C, then B⊂A or D⊂B

C
If A⊈=C, then A⊈B and B⊆D
If A⊈=C, then A⊈B and B⊆D

D
If A⊈C, then A⊈B or B⊈D
If A⊈C, then A⊈B or B⊈D
Question 75/137
4 / -1

Mark Review

Negation of the statement:
√5 is an integer of 5 is irrational is:
[Jan. 9, 2020 (I)]

Negation of the statement:
√5 is an integer of 5 is irrational is:
[Jan. 9, 2020 (I)]

A
√5 is not an integer or 5 is not irrational
√5 is not an integer or 5 is not irrational

B
√5 is not an integer and 5 is not irrational
√5 is not an integer and 5 is not irrational

C
√5 is irrational or 5 is an integer.

√5 is irrational or 5 is an integer.

D
√5 is an integer and 5 is irrational
√5 is an integer and 5 is irrational
Question 76/137
4 / -1

Mark Review

If p→(p∧∼q) is false, then the truth values of p and q are respectively:
[Jan. 9, 2020 (II)]

If p→(p∧∼q) is false, then the truth values of p and q are respectively:
[Jan. 9, 2020 (II)]

A
F, F
F, F

B
T, F
T, F

C
T, T
T, T

D
F, T
F, T
Question 77/137
4 / -1

Mark Review

Which one of the following is a tautology?
[Jan. 8, 2020 (I)]

Which one of the following is a tautology?
[Jan. 8, 2020 (I)]

A
(p∧(p→q))→q
(p∧(p→q))→q

B
q→(p∧(p→q))
q→(p∧(p→q))

C
p∧(p∨q)
p∧(p∨q)

D
p∨(p∧q)
p∨(p∧q)
Question 78/137
4 / -1

Mark Review

Which of the following statements is a tautology?
[Jan. 8, 2020 (II)]

Which of the following statements is a tautology?
[Jan. 8, 2020 (II)]

A
p∨(∼q)→p∧q
p∨(∼q)→p∧q

B
∼(p∧∼q)→p∨q
∼(p∧∼q)→p∨q

C
∼(p∨∼q)→p∧q
∼(p∨∼q)→p∧q

D
∼(p∨∼q)→p∨q
∼(p∨∼q)→p∨q
Question 79/137
4 / -1

Mark Review

The logical statement (p⇒q)∧(q⇒∼p) is equivalent to:
[Jan. 7, 2020 (I)]

The logical statement (p⇒q)∧(q⇒∼p) is equivalent to:
[Jan. 7, 2020 (I)]

A
p
p

B
q
q

C
~p
~p

D
~q
~q
Question 80/137
4 / -1

Mark Review

The negation of the Boolean expression p∨(∼p∧q) is equivalent to :
[Sep. 06, 2020 (I)]

The negation of the Boolean expression p∨(∼p∧q) is equivalent to :
[Sep. 06, 2020 (I)]

A
p∧∼q
p∧∼q

B
∼p∧∼q
∼p∧∼q

C
∼p∨∼q
∼p∨∼q

D
∼p∨q
∼p∨q
Question 81/137
4 / -1

Mark Review

The negation of the Boolean expression x↔∼y is equivalent to:
[Sep. 05, 2020 (I)]

The negation of the Boolean expression x↔∼y is equivalent to:
[Sep. 05, 2020 (I)]

A
(x∧y)∨(∼x∧∼y)
(x∧y)∨(∼x∧∼y)

B
(x∧y)∧(∼x∨∼y)
(x∧y)∧(∼x∨∼y)

C
(x∧∼y)∨(∼x∧y)
(x∧∼y)∨(∼x∧y)

D
(∼x∧y)∨(∼x∧∼y)
(∼x∧y)∨(∼x∧∼y)
Question 82/137
4 / -1

Mark Review

Given the following two statements:
(S1):(q ∨p)→(p↔∼q) is a tautology.
(S2):∼q∧(∼p↔q) is a fallacy. Then :
[Sep. 04, 2020 (I)]

Given the following two statements:
(S1):(q ∨p)→(p↔∼q) is a tautology.
(S2):∼q∧(∼p↔q) is a fallacy. Then :
[Sep. 04, 2020 (I)]

A
both (S1) and (S2) are correct
both (S1) and (S2) are correct

B
only (S1) is correct
only (S1) is correct

C
only (S2) is correct
only (S2) is correct

D
both (S1) and (S2) are not correct
both (S1) and (S2) are not correct
Question 83/137
4 / -1

Mark Review

The proposition p→∼(p∧∼q) is equivalent to :
[Sep. 03, 2020 (I)]

The proposition p→∼(p∧∼q) is equivalent to :
[Sep. 03, 2020 (I)]

A
q
q

B
(∼p)∨q
(∼p)∨q

C
(∼p)∧q
(∼p)∧q

D
(∼p)∨(∼q)
(∼p)∨(∼q)
Question 84/137
4 / -1

Mark Review

Let p, q,r be three statements such that the truth value of (p∧q)→(∼q∨r) is F. Then the truth values of p,q,r are respectively:
[Sep. 03, 2020 (II)]

Let p, q,r be three statements such that the truth value of (p∧q)→(∼q∨r) is F. Then the truth values of p,q,r are respectively:
[Sep. 03, 2020 (II)]

A
T, F, T
T, F, T

B
T, T, T
T, T, T

C
F, T, F
F, T, F

D
T, T, F
T, T, F
Question 85/137
4 / -1

Mark Review

Consider the statement: “For an integer n, if n3−1 is even, then n is odd.” The contrapositive statement of this statement is:
[Sep. 06, 2020 (II)]

Consider the statement: “For an integer n, if n3−1 is even, then n is odd.” The contrapositive statement of this statement is:
[Sep. 06, 2020 (II)]

A
For an integer n, if n is even, then n3−1 is odd.
For an integer n, if n is even, then n3−1 is odd.

B
For an intetger n, if n3−1 is not even, then n is not odd.
For an intetger n, if n3−1 is not even, then n is not odd.

C
For an integer n, if n is even, then n3−1 is even.
For an integer n, if n is even, then n3−1 is even.

D
For an integer n, if n is odd, then n3−1 is even.
For an integer n, if n is odd, then n3−1 is even.
Question 86/137
4 / -1

Mark Review

The statement (p→(q→p))→(p→(p∨q)) is :
[Sep. 05,2020 (II)]

The statement (p→(q→p))→(p→(p∨q)) is :
[Sep. 05,2020 (II)]

A
equivalent to (p∧q)∨(∼q)
equivalent to (p∧q)∨(∼q)

B
a contradiction
a contradiction

C
equivalent to (p∨q)∧(∼p)
equivalent to (p∨q)∧(∼p)

D
a tautology
a tautology
Question 87/137
4 / -1

Mark Review

Contrapositive of the statement :
'If a function f is differentiable at a, then it is also continuous at a', is :
[Sep. 04, 2020 (II)]

Contrapositive of the statement :
'If a function f is differentiable at a, then it is also continuous at a', is :
[Sep. 04, 2020 (II)]

A
If a function f is continuous at a, then it is not differentiable at a.

If a function f is continuous at a, then it is not differentiable at a.

B
If a function f is not continuous at a, then it is not differentiable at a.

If a function f is not continuous at a, then it is not differentiable at a.

C
If a function f is not continuous at a, then it is differentiable at a
If a function f is not continuous at a, then it is differentiable at a

D
If a function f is continuous at a, then it is differentiable at a
If a function f is continuous at a, then it is differentiable at a
Question 88/137
4 / -1

Mark Review

The contrapositive of the statement "If I reach the station in time, then I will catch the train" is :
[Sep. 02, 2020 (I)]

The contrapositive of the statement "If I reach the station in time, then I will catch the train" is :
[Sep. 02, 2020 (I)]

A
If I do not reach the station in time, then I will catch the train.

If I do not reach the station in time, then I will catch the train.

B
If I do not reach the station in time, then I will not catch the train.

If I do not reach the station in time, then I will not catch the train.

C
If I will catch the train, then I reach the station in time.

If I will catch the train, then I reach the station in time.

D
If I will not catch the train, then I do not reach the station in time.
If I will not catch the train, then I do not reach the station in time.
Question 89/137
4 / -1

Mark Review

The Boolean expression((p∧q)∨(p∨∼q))∧(∼p∧∼q) is equivalent to :
[Jan. 12, 2019 (I)]

The Boolean expression((p∧q)∨(p∨∼q))∧(∼p∧∼q) is equivalent to :
[Jan. 12, 2019 (I)]

A
p∧q
p∧q

B
p∧(∼q)
p∧(∼q)

C
(∼p)∧(∼q)
(∼p)∧(∼q)

D
p∨(∼q)
p∨(∼q)
Question 90/137
4 / -1

Mark Review

The expression ∼(∼p→q) is logically equivalent to:
[Jan. 12,2019 (II)]

The expression ∼(∼p→q) is logically equivalent to:
[Jan. 12,2019 (II)]

A
∼p∧∼q
∼p∧∼q

B
p∧∼q
p∧∼q

C
∼p∧q
∼p∧q

D
p∧q
p∧q
Question 91/137
4 / -1

Mark Review

If q is false and p∧q↔r is true, then which one of the following statements is a tautology?
[Jan. 11, 2019 (I)]

If q is false and p∧q↔r is true, then which one of the following statements is a tautology?
[Jan. 11, 2019 (I)]

A
(p∨r)→(p∧r)
(p∨r)→(p∧r)

B
(p∧r)→(p∨r)
(p∧r)→(p∨r)

C
p∧r
p∧r

D
p∨r
p∨r
Question 92/137
4 / -1

Mark Review

Consider the following three statements:
P:5 is a prime number.
Q:7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35 .
Then the truth value of which one of the following statements is true?
[Jan. 10, 2019 (II)]

Consider the following three statements:
P:5 is a prime number.
Q:7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35 .
Then the truth value of which one of the following statements is true?
[Jan. 10, 2019 (II)]

A
(∼P)∨(Q∧R)
(∼P)∨(Q∧R)

B
(P∧Q)∨(∼R)
(P∧Q)∨(∼R)

C
(∼P)∧(∼Q∧R)
(∼P)∧(∼Q∧R)

D
P∨(∼Q∧R)
P∨(∼Q∧R)
Question 93/137
4 / -1

Mark Review

If the Boolean expression (p⊕q)∧(∼p◉q) is equivalent to p∧q, where ⊕,◉∈{∧,∨} then the ordered pair (⊕,◉) is:
[Jan. 09, 2019 (I)]

If the Boolean expression (p⊕q)∧(∼p◉q) is equivalent to p∧q, where ⊕,◉∈{∧,∨} then the ordered pair (⊕,◉) is:
[Jan. 09, 2019 (I)]

A
(∨,∧)
(∨,∧)

B
(∨,∨)
(∨,∨)

C
(∧,∨)
(∧,∨)

D
(∧,∧)
(∧,∧)
Question 94/137
4 / -1

Mark Review

The logical statement [∼(∼p∨q)∨(p∧r)]∧(∼p∧r) is equivalent to:
[Jan. 09, 2019 (II)]

The logical statement [∼(∼p∨q)∨(p∧r)]∧(∼p∧r) is equivalent to:
[Jan. 09, 2019 (II)]

A
(∼p∧∼q)∧r
(∼p∧∼q)∧r

B
∼p∨r
∼p∨r

C
(p∧r)∧∼q
(p∧r)∧∼q

D
(p∧∼q)∨r
(p∧∼q)∨r
Question 95/137
4 / -1

Mark Review

Contrapositive of the statement “If two numbers are not equal, then their squares are not equal”. is :
[Jan. 11, 2019 (II)]

Contrapositive of the statement “If two numbers are not equal, then their squares are not equal”. is :
[Jan. 11, 2019 (II)]

A
If the squares of two numbers are not equal, then the numbers are equal.

If the squares of two numbers are not equal, then the numbers are equal.

B
If the squares of two numbers are equal, then the numbers are not equal .

If the squares of two numbers are equal, then the numbers are not equal .

C
If the squares of two numbers are equal, then the numbers are equal.

If the squares of two numbers are equal, then the numbers are equal.

D
If the squares of two numbers are not equal, then the numbers are not equal.
If the squares of two numbers are not equal, then the numbers are not equal.
Question 96/137
4 / -1

Mark Review

If the truth value of the statement p→(~q∨r) is false (F), then the truth values of the statements p, q, r arerespectively.
[April 12, 2019 (I)]

If the truth value of the statement p→(~q∨r) is false (F), then the truth values of the statements p, q, r arerespectively.
[April 12, 2019 (I)]

A
T, T, F
T, T, F

B
T, F, F
T, F, F

C
T, F, T
T, F, T

D
F, T, T
F, T, T
Question 97/137
4 / -1

Mark Review

The Boolean expression ∼(p⇒(∼q)) is equivalent to:
[April 12, 2019 (II)]

The Boolean expression ∼(p⇒(∼q)) is equivalent to:
[April 12, 2019 (II)]

A
p∧q
p∧q

B
q⇒∼p
q⇒∼p

C
p∨q
p∨q

D
(∼p)⇒q
(∼p)⇒q
Question 98/137
4 / -1

Mark Review

Which one of the following Boolean expressions is a tautology ?
[April 10, 2019 (I)]

Which one of the following Boolean expressions is a tautology ?
[April 10, 2019 (I)]

A
(p∧q)∨(p∧∼q)
(p∧q)∨(p∧∼q)

B
(p∨q)∨(p∨∼q)
(p∨q)∨(p∨∼q)

C
(p∨q)∧(p∨∼q)
(p∨q)∧(p∨∼q)

D
(p∨q)∧(∼p∨∼q)
(p∨q)∧(∼p∨∼q)
Question 99/137
4 / -1

Mark Review

If p⇒(q∨r) is false, then the truth values of p, q, r are respectively:
[April 09, 2019 (II)]

If p⇒(q∨r) is false, then the truth values of p, q, r are respectively:
[April 09, 2019 (II)]

A
F, T, T
F, T, T

B
T, F, F
T, F, F

C
T, T, F
T, T, F

D
F, F, F
F, F, F
Question 100/137
4 / -1

Mark Review

Which one of the following statements is not a tautology?
[April 08, 2019(II)]

Which one of the following statements is not a tautology?
[April 08, 2019(II)]

A
(p∨q)→(p∨(∼q))
(p∨q)→(p∨(∼q))

B
(p∧q)→(∼p)∨q
(p∧q)→(∼p)∨q

C
p→(p∨q)
p→(p∨q)

D
(p∧q)→p
(p∧q)→p
Question 101/137
4 / -1

Mark Review

The negation of the Boolean expression ∼s∨(∼r∧s) is equivalent to :
[April 10, 2019 (II)]

The negation of the Boolean expression ∼s∨(∼r∧s) is equivalent to :
[April 10, 2019 (II)]

A
∼s∧∼r
∼s∧∼r

B
r
r

C
s∨r
s∨r

D
s∧r
s∧r
Question 102/137
4 / -1

Mark Review

For any two statements p and q, the negation of the expression p∨(∼p∧q) is:
[April 9, 2019 (I)]

For any two statements p and q, the negation of the expression p∨(∼p∧q) is:
[April 9, 2019 (I)]

A
∼p∧∼q
∼p∧∼q

B
p∧q
p∧q

C
p↔q
p↔q

D
∼p∨∼q
∼p∨∼q
Question 103/137
4 / -1

Mark Review

The contrapositive of the statement "If you are born in India, then you are a citizen of India", is :
[April 8, 2019 (I)]

The contrapositive of the statement "If you are born in India, then you are a citizen of India", is :
[April 8, 2019 (I)]

A
If you are not a citizen of India, then you are not born in India.

If you are not a citizen of India, then you are not born in India.

B
If you are a citizen of India, then you are born in India.

If you are a citizen of India, then you are born in India.

C
If you are born in India, then you are not a citizen of India.

If you are born in India, then you are not a citizen of India.

D
If you are not born in India, then you are not a citizen of India.
If you are not born in India, then you are not a citizen of India.
Question 104/137
4 / -1

Mark Review

If p→(∼p∨∼q) is false, then the truth values of p and q are respectively.
[Online April 16, 2018]

If p→(∼p∨∼q) is false, then the truth values of p and q are respectively.
[Online April 16, 2018]

A
T, F
T, F

B
F, F
F, F

C
F, T
F, T

D
T,T
T,T
Question 105/137
4 / -1

Mark Review

The Boolean expression∼(p∨q)∨(∼p∧q) is equivalent to
[2018]

The Boolean expression∼(p∨q)∨(∼p∧q) is equivalent to
[2018]

A
p
p

B
q
q

C
∼q
∼q

D
∼p
∼p
Question 106/137
4 / -1

Mark Review

If (p∧∼q)∧(p∧r)→∼p ∨q is false, then the truth values of p,q and r are respectively
[Online April 15, 2018]

If (p∧∼q)∧(p∧r)→∼p ∨q is false, then the truth values of p,q and r are respectively
[Online April 15, 2018]

A
F, T, F
F, T, F

B
T, F, T
T, F, T

C
F, F, F
F, F, F

D
T, T, T
T, T, T
Question 107/137
4 / -1

Mark Review

Consider the following two statements.
Statement p:
The value of sin 120° can be divided by taking θ=240° in the equation 2‌sin‌
θ
2
=√1+sin‌θ−√1−sin‌θ.
Statement q: The angles A,B,C and D ofany quadrilateral ABCD satisfy the equation cos(
1
2
(A+C))+cos(
1
2
(B+D))=0
Then the truth values of p and q are respectively.
[Online April 15, 2018]

Consider the following two statements.
Statement p:
The value of sin 120° can be divided by taking θ=240° in the equation 2‌sin‌
θ
2
=√1+sin‌θ−√1−sin‌θ.
Statement q: The angles A,B,C and D ofany quadrilateral ABCD satisfy the equation cos(
1
2
(A+C))+cos(
1
2
(B+D))=0
Then the truth values of p and q are respectively.
[Online April 15, 2018]

A
F, T
F, T

B
T, T
T, T

C
F, F
F, F

D
T, F
T, F
Question 108/137
4 / -1

Mark Review

Which of the following is a tautology?
[2017]

Which of the following is a tautology?
[2017]

A
(∼p)∧(p∨q)→q
(∼p)∧(p∨q)→q

B
(q→p)∨∼(p→q)
(q→p)∨∼(p→q)

C
(∼q)∨(p∧q)→q
(∼q)∨(p∧q)→q

D
(p→q)∧(q→p)
(p→q)∧(q→p)
Question 109/137
4 / -1

Mark Review

The following statement (p→q)→[(∼p→q)→q] is
[2017]

The following statement (p→q)→[(∼p→q)→q] is
[2017]

A
a fallacy
a fallacy

B
a tautology
a tautology

C
equivalent to ~p→q
equivalent to ~p→q

D
equivalent to p→∼q
equivalent to p→∼q
Question 110/137
4 / -1

Mark Review

The proposition (∼p)∨(p∧∼q)
[Online April 8, 2017]

The proposition (∼p)∨(p∧∼q)
[Online April 8, 2017]

A
p→∼q
p→∼q

B
p∧(∼q)
p∧(∼q)

C
q→p
q→p

D
p∨(∼q)
p∨(∼q)
Question 111/137
4 / -1

Mark Review

Contrapositive of the statement
‘If two numbers are not equal, then their squares are not equal’, is :
[Online April 9, 2017]

Contrapositive of the statement
‘If two numbers are not equal, then their squares are not equal’, is :
[Online April 9, 2017]

A
If the squares of two numbers are equal, then the numbers are equal.

If the squares of two numbers are equal, then the numbers are equal.

B
If the squares of two numbers are equal, then the numbers are not equal.

If the squares of two numbers are equal, then the numbers are not equal.

C
If the squares of two numbers are not equal, then the numbers are not equal.

If the squares of two numbers are not equal, then the numbers are not equal.

D
If the squares of two numbers are not equal, then the numbers are equal.
If the squares of two numbers are not equal, then the numbers are equal.
Question 112/137
4 / -1

Mark Review

The Boolean Expression (p∧∼q)∨q∨(∼p∧q) is equivalent to:
[2016]

The Boolean Expression (p∧∼q)∨q∨(∼p∧q) is equivalent to:
[2016]

A
p∨q
p∨q

B
p∨∼q
p∨∼q

C
∼p∧q
∼p∧q

D
p∧q
p∧q
Question 113/137
4 / -1

Mark Review

The contrapositive of the following statement, "If the side of a square doubles, then its area increases four times", is :
[Online April 10, 2016]

The contrapositive of the following statement, "If the side of a square doubles, then its area increases four times", is :
[Online April 10, 2016]

A
If the area of a square increases four times, then its side is not doubled.

If the area of a square increases four times, then its side is not doubled.

B
If the area of a square increases four times, then its side is doubled.

If the area of a square increases four times, then its side is doubled.

C
If the area of a square does not increases four times, then its side is not doubled.

If the area of a square does not increases four times, then its side is not doubled.

D
If the side of a square is not doubled, then its area does not increase four times.
If the side of a square is not doubled, then its area does not increase four times.
Question 114/137
4 / -1

Mark Review

Consider the following two statements:
P: If 7 is an odd number, then 7 is divisible by 2 .
Q: If 7 is a prime number, then 7 is an odd number.
If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair (V1,V2) equals:
[Online April 9, 2016]

Consider the following two statements:
P: If 7 is an odd number, then 7 is divisible by 2 .
Q: If 7 is a prime number, then 7 is an odd number.
If V1 is the truth value of the contrapositive of P and V2 is the truth value of contrapositive of Q, then the ordered pair (V1,V2) equals:
[Online April 9, 2016]

A
(F,F)
(F,F)

B
(F,T)
(F,T)

C
(T,F)
(T,F)

D
(T,T)
(T,T)
Question 115/137
4 / -1

Mark Review

The negation of ∼s∨(∼r∧s) is equivalent to :
[2015]

The negation of ∼s∨(∼r∧s) is equivalent to :
[2015]

A
s∨(r∨∼s)
s∨(r∨∼s)

B
s∧r
s∧r

C
S∧∼r
S∧∼r

D
S∧(r∧∼s)
S∧(r∧∼s)
Question 116/137
4 / -1

Mark Review

Consider the following statements :
P: Suman is brilliant
Q: Suman is rich.
R: Suman is honest
the negation of the statement
"Suman is brilliant and dishonest if and only if suman is rich" can be equivalently expressed as :
[Online April 11, 2015]

Consider the following statements :
P: Suman is brilliant
Q: Suman is rich.
R: Suman is honest
the negation of the statement
"Suman is brilliant and dishonest if and only if suman is rich" can be equivalently expressed as :
[Online April 11, 2015]

A
∼Q↔∼P∨R
∼Q↔∼P∨R

B
∼Q↔∼P∧R
∼Q↔∼P∧R

C
∼Q↔P∨∼R
∼Q↔P∨∼R

D
∼Q↔P∧∼R
∼Q↔P∧∼R
Question 117/137
4 / -1

Mark Review

The contrapositive of the statement “If it is raining, then I will not come”, is :
[Online April 10, 2015]

The contrapositive of the statement “If it is raining, then I will not come”, is :
[Online April 10, 2015]

A
If I will not come, then it is raining.

If I will not come, then it is raining.

B
If I will not come, then it is not raining.

If I will not come, then it is not raining.

C
If I will come, then it is raining.

If I will come, then it is raining.

D
If I will come, then it is not raining.
If I will come, then it is not raining.
Question 118/137
4 / -1

Mark Review

The statement ∼(p↔∼q) is:
[2014]

The statement ∼(p↔∼q) is:
[2014]

A
a tautology
a tautology

B
a fallacy
a fallacy

C
eqivalent to p↔q
eqivalent to p↔q

D
equivalent to ∼p↔q
equivalent to ∼p↔q
Question 119/137
4 / -1

Mark Review

Let p,q,r denote arbitrary statements. Then the logically equivalent of the statement p⇒(q ∨r) is:
[Online April 12, 2014]

Let p,q,r denote arbitrary statements. Then the logically equivalent of the statement p⇒(q ∨r) is:
[Online April 12, 2014]

A
(p ∨q)⇒r
(p ∨q)⇒r

B
(p⇒q) ∨(p⇒r)
(p⇒q) ∨(p⇒r)

C
(p⇒∼q)∧(p⇒r)
(p⇒∼q)∧(p⇒r)

D
(p⇒q)∧(p⇒∼r)
(p⇒q)∧(p⇒∼r)
Question 120/137
4 / -1

Mark Review

The proposition ∼(p∨∼q)∨∼(p∨q) is logically equivalent to:
[Online April 11, 2014]

The proposition ∼(p∨∼q)∨∼(p∨q) is logically equivalent to:
[Online April 11, 2014]

A
p
p

B
q
q

C
∼p
∼p

D
∼q
∼q
Question 121/137
4 / -1

Mark Review

The contrapositive of the statement “if I am not feeling well, then I will go to the doctor” is
[Online April 19, 2014]

The contrapositive of the statement “if I am not feeling well, then I will go to the doctor” is
[Online April 19, 2014]

A
If I am feeling well, then I will not go to the doctor
If I am feeling well, then I will not go to the doctor

B
If I will go to the doctor, then I am feeling well
If I will go to the doctor, then I am feeling well

C
If I will not go to the doctor, then I am feeling well
If I will not go to the doctor, then I am feeling well

D
If I will go to the doctor, then I am not feeling well.
If I will go to the doctor, then I am not feeling well.
Question 122/137
4 / -1

Mark Review

The contrapositive of the statement “I go to school if it does not rain” is
[Online April 9, 2014]

The contrapositive of the statement “I go to school if it does not rain” is
[Online April 9, 2014]

A
If it rains, I do not go to school.

If it rains, I do not go to school.

B
If I do not go to school, it rains.

If I do not go to school, it rains.

C
If it rains, I go to school.

If it rains, I go to school.

D
If I go to school, it rains.
If I go to school, it rains.
Question 123/137
4 / -1

Mark Review

Consider
Statement-1 : (p∧∼q)∧(∼p∧q) is a fallacy.
Statement- 2: (p→q)↔(∼q→∼p) is a tautology.
[2013]

Consider
Statement-1 : (p∧∼q)∧(∼p∧q) is a fallacy.
Statement- 2: (p→q)↔(∼q→∼p) is a tautology.
[2013]

A
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 a correct explanation for Statement-1.

B
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 true; Statement-2 is not a correct explanation for Statement-1.

C
Statement-1 is true; Statement-2 is false.

Statement-1 is true; Statement-2 is false.

D
Statement-1 is false; Statement-2 is true.
Statement-1 is false; Statement-2 is true.
Question 124/137
4 / -1

Mark Review

Let p and q be any two logical statements andr:p→(∼p∨q). If r has a truth value F, then the truth values of p and q are respectively:
[Online April 25, 2013]

Let p and q be any two logical statements andr:p→(∼p∨q). If r has a truth value F, then the truth values of p and q are respectively:
[Online April 25, 2013]

A
F, F
F, F

B
T, T
T, T

C
T, F
T, F

D
F, T
F, T
Question 125/137
4 / -1

Mark Review

For integers m and n, both greater than 1, consider the following three statements:
P:m divides n
Q:m divides n2
R:m is prime,
then
[Online April 23, 2013]

For integers m and n, both greater than 1, consider the following three statements:
P:m divides n
Q:m divides n2
R:m is prime,
then
[Online April 23, 2013]

A
Q∧R→P
Q∧R→P

B
P∧Q→R
P∧Q→R

C
Q→R
Q→R

D
Q→P
Q→P
Question 126/137
4 / -1

Mark Review

The statement p→(q→p) is equivalent to :
[Online April 22, 2013]

The statement p→(q→p) is equivalent to :
[Online April 22, 2013]

A
p→q
p→q

B
p→(p∨q)
p→(p∨q)

C
p→(p→q)
p→(p→q)

D
p→(p∧q)
p→(p∧q)
Question 127/137
4 / -1

Mark Review

Statement-1: The statement A→(B→A) is equivalent to A→(A∨B).
Statement-2: The statement ∼[(A∧B)→(∼A∨B)] is a Tautology.
[Online April 9, 2013]

Statement-1: The statement A→(B→A) is equivalent to A→(A∨B).
Statement-2: The statement ∼[(A∧B)→(∼A∨B)] is a Tautology.
[Online April 9, 2013]

A
Statement-1 is false; Statement-2 is true.

Statement-1 is false; Statement-2 is true.

B
Statement-1 is true; Statement-2 is true; Statement- 2 is not correct explanation for Statement-1.

Statement-1 is true; Statement-2 is true; Statement- 2 is not correct explanation for Statement-1.

C
Statement-1 is true; Statement-2 is false.

Statement-1 is true; Statement-2 is false.

D
Statement-1 is true; Statement-2 is true; Statement- 2 is the correct explanation for Statement-1.
Statement-1 is true; Statement-2 is true; Statement- 2 is the correct explanation for Statement-1.
Question 128/137
4 / -1

Mark Review

Let p and q be two Statements. Amongst the following, the Statement that is equivalent to p→q is
[Online May 19, 2012]

Let p and q be two Statements. Amongst the following, the Statement that is equivalent to p→q is
[Online May 19, 2012]

A
p∧∼q
p∧∼q

B
∼p∨q
∼p∨q

C
∼p∧q
∼p∧q

D
p∨∼q
p∨∼q
Question 129/137
4 / -1

Mark Review

The logically equivalent preposition of p⇔q is
[Online May 12, 2012]

The logically equivalent preposition of p⇔q is
[Online May 12, 2012]

A
(p⇒q)∧(q⇒p)
(p⇒q)∧(q⇒p)

B
p∧q
p∧q

C
(p∧q)∨(q⇒p)
(p∧q)∨(q⇒p)

D
(p∧q)⇒(q∨p)
(p∧q)⇒(q∨p)
Question 130/137
4 / -1

Mark Review

The negation of the statement
"If I become a teacher, then I will open a school", is :
[2012]

The negation of the statement
"If I become a teacher, then I will open a school", is :
[2012]

A
I will become a teacher and I will not open a school.

I will become a teacher and I will not open a school.

B
Either I will not become a teacher or I will not open a school.

Either I will not become a teacher or I will not open a school.

C
Neither I will become a teacher nor I will open a school.

Neither I will become a teacher nor I will open a school.

D
I will not become a teacher or I will open a school.
I will not become a teacher or I will open a school.
Question 131/137
4 / -1

Mark Review

Let p and q denote the following statements
p : The sun is shining
q: I shall play tennis in the afternoon
The negation of the statement “If the sun is shining then I shall play tennis in the afternoon”, is
[Online May 26, 2012]

Let p and q denote the following statements
p : The sun is shining
q: I shall play tennis in the afternoon
The negation of the statement “If the sun is shining then I shall play tennis in the afternoon”, is
[Online May 26, 2012]

A
q⇒∼p
q⇒∼p

B
q∧∼p
q∧∼p

C
p∧∼q
p∧∼q

D
∼q⇒∼p
∼q⇒∼p
Question 132/137
4 / -1

Mark Review

The Statement that is TRUE among the following is
[Online May 7,2012 ]

The Statement that is TRUE among the following is
[Online May 7,2012 ]

A
The contrapositive of 3x+2=8⇒x=2 is x≠2 ⇒3x+2≠8
The contrapositive of 3x+2=8⇒x=2 is x≠2 ⇒3x+2≠8

B
The converse of tan‌x=0⇒x=0 is x≠0⇒tan‌x=0.
The converse of tan‌x=0⇒x=0 is x≠0⇒tan‌x=0.

C
p⇒q is equivalent to p∨∼q.
p⇒q is equivalent to p∨∼q.

D
p∨q and p∧q have the same truth table.
p∨q and p∧q have the same truth table.
Question 133/137
4 / -1

Mark Review

The only statement among the following that is a tautology is
[2011 RS]

The only statement among the following that is a tautology is
[2011 RS]

A
A∧(A∨B)
A∧(A∨B)

B
A∨(A∧B)
A∨(A∧B)

C
[A∧(A→B)]→B
[A∧(A→B)]→B

D
B→[A∧(A→B)]
B→[A∧(A→B)]
Question 134/137
4 / -1

Mark Review

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?
[2010]

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?
[2010]

A
There is no rational number x∈S such than x≤0.
There is no rational number x∈S such than x≤0.

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

Every rational number x∈S satisfies x≤0.

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

D
There is a rational number x∈S such that x≤0.
There is a rational number x∈S such that x≤0.
Question 135/137
4 / -1

Mark Review

Statement-1: ∼(p↔∼q) is equivalent to p↔q.
Statement- 2: ∼(p↔∼q) is a tantology
[2009]

Statement-1: ∼(p↔∼q) is equivalent to p↔q.
Statement- 2: ∼(p↔∼q) is a tantology
[2009]

A
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 true;
Statement-2 is not a correct explanation for Statement-1.

B
Statement-1 is true, Statement-2 is false.

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

C
Statement-1 is false, Statement-2 is true.

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

D
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 a correct explanation for statement -1
Question 136/137
4 / -1

Mark Review

The statement p→(q→p) is equivalent to
[2008]

The statement p→(q→p) is equivalent to
[2008]

A
p→(p→q)
p→(p→q)

B
p→(p∨q)
p→(p∨q)

C
p→(p∧q)
p→(p∧q)

D
p→(p↔q)
p→(p↔q)
Question 137/137
4 / -1

Mark Review

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).
[2008]

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).
[2008]

A
Statement -1 is false, Statement-2 is true
Statement -1 is false, Statement-2 is true

B
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 a correct explanation for Statement-1

C
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 true; Statement -2 is not a correct explanation for Statement-1

D
None
None

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