# Topic: Mathematical Reasoning (Test 3)

Topic: Mathematical Reasoning
Q.1
p∧(q∧r) is logically equivalent to
A. (p∨q)∨r
B. p→(q∧r)
C. (p∨q)∧r
D. p∧(q∧r)≡(p∧q)∧r
Explaination / Solution:

Associative law

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Q.2
The negation of the compound statement p∨(∼p∨q) is
A. (p∧∼q)∧∼p
B. (p∧∼q)∨p
C. (p∧q)∨p
D. (p∧∼q)∨∼p
Explaination / Solution:

∼(p∨(∼p∨q))≡∼p∧(p∧∼q) Applying De morgan's law

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Q.3
Let p and q be two prepositions given by p : I take only bread and butter in breakfast. q : I do not take anything in breakfast. Then , the compound proposition “ I take only bread and butter in breakfast or I do not take anything “ is represented by
A. p↔q
B. p→q
C. p ∧q
D. p ∨q
Explaination / Solution:

and is replaced by ∧ or is replaced by ∨

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Q.4
Which of the following is true for the propositions p and q ?
A. p∧q is true when atleast one of p and q is true.
B. p↔q is true only when both p and q are true
C. p→q is true when p is true and q is false
D. ∼(p∨q) is true only when both p and q are false.
Explaination / Solution:

only F∨F≡F Hence ∼F≡T

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Q.5
Let p and q be two propositions. Then, the contrapositive of the implication p→q is
A. ∼p→∼q
B. ∼q→∼p
C. p↔q
D. p→q
Explaination / Solution:

the contrapositve of p→q is∼q→∼p

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

= F

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Q.7
The proposition (p→∼p)∧(∼p→p) is
A. a tautology
B. both a tautology and a contradiction
C. neither a tautology nor a contradiction
Explaination / Solution:

definition of

Hence F

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Q.8
Which of the following is always true ?
A. (p→q)≅(∼q→∼p)
B. ∼(p∧q)≅(∼p∧∼q)
C. ∼(p→q)≅(p∨∼q)
D. ∼(p∨q)≅(∼p∨∼q)
Explaination / Solution:

Since

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Q.9
If x = 5 and y = - 2 , then x – 2y = 9 . The contrapositive of this proposition is
A. x – 2y = 9 iff x = 5 and y = - 2
B. If x – 2y = 9 , x ≠ 5 and y ≠ - 2
C. None of these
D. If x – 2y ≠ 9 , then x ≠ 5 or y ≠ - 2
Explaination / Solution:

p: x = 5 and y = - 2 , q : x – 2y = 9

The contrapositive of

Hence If x – 2y  9 , then x  5 or y  - 2

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Q.10
Logical equivalent proposition to the proposition ∼(p∨q) is
A. ∼p↔∼q
B. ∼p∧∼q
C. ∼p→∼q
D. ∼p∨∼q