# Topic: Mathematical Reasoning (Test 4)

Topic: Mathematical Reasoning
Q.1
∼(∼p)↔p is
A. a tautology
C. neither a contradiction nor a tautology
D. none of these
Explaination / Solution:

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Q.2
Which of the following is a preposition?
A. Delhi is on the Jupiter
C. None of these
D. A half open door is half closed
Explaination / Solution:

It is a false statement other options are open sentences

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Q.3
Let p and q be two prepositions given by p : I take medicine, q : I can sleep then ,the compound statement ∼p→q means
A. If I do not take medicine , then I can sleep.
B. I take medicine iff I can sleep.
C. I take medicine if I can sleep.
D. If I do not take medicine , then I cannot sleep.
Explaination / Solution:
No Explaination.

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Q.4
The logically equivalent propositions of p↔q is
A. (p∧q)→(q∨p)
B. (p→q)∨(q→p)
C. (p→q)∧(q→p)
D. (p∧q)∧(q∨p)
Explaination / Solution:

p↔q≡(p→q)∧(q→p)

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Q.5
Negation of the statement ∼p→(q∨r) is
A. ∼p∧(∼q∧∼r)
B. p∧(q∨r)
C. p∨(q∧r)
D. p→∼(q∨r)
Explaination / Solution:

rules of negation ∼(p→q)≡p∧∼q Hence ∼p∧(∼q∧∼r)

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Q.6
The compound statement p→(∼p∨q) is false , then the truth values of p and q are respectively
A. T,T
B. T,F
C. F,T
D. F,F
Explaination / Solution:

since T->F is false.

Hence q=F.

So p=T and q=F

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Q.7
Which of the following statement is a tautology
A. (∼p∨∼q)→(p∨q)
B. (∼p∨∼q)∨(p∨q)
C. (∼p∨q)∼(p∨∼q)
D. (p∨∼q)∧(p∨q)
Explaination / Solution:

Since

Hence tautology

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Q.8
The negation of the proposition “if a quadrilateral is a square, then it is a rhombus “ is
A. if a quadrilateral is not a square , then it is a rhombus
B. if a quadrilateral is a square , then it is not a rhombus
C. a quadrilateral is a square and it is not a rhombus
D. a quadrilateral is not a square and it is a rhombus
Explaination / Solution:

rules of negation ∼(p→q)≡p∧∼q

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Q.9
“If the figure is a rhombus then the diagonals are perpendicular “. The contrapositive of the above statement is
A. If the diagonals are perpendicular, then the figure is a rhombus
B. If the diagonals are not perpendicular, then the figure is not a rhombus
C. If the figure is not a rhombus, then its diagonals are not perpendicular
D. If the diagonals are not perpendicular, then the figure is a rhombus
Explaination / Solution:

p: the figure is a rhombus  q: the diagonals  are perpendicular

Contrapositive of

hence If the diagonals are not perpendicular, then the figure is not a rhombus

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Q.10
Let p and q be two propositions. Then the inverse of the implication p→q is
A. ∼q→p
B. ∼p→q
C. p→∼q
D. ∼p→∼q