For any three propositions p , q , and r , the proposition (p∧q)∧(q∧r) is true , when

**A. ** p, q are true and r is false

**B. ** p, q , r are all false

**C. ** p,q , r are all true

**D. ** p is true and q, r are false

**Answer : ****Option C**

**Explaination / Solution: **

hence p=T , q=T, r=T

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Which of the following proposition is a tautology ?

**A. ** (∼p∨∼q)∧(p∨∼q)

**B. ** (∼p∨∼q)∨(p∨∼q)

**C. ** ∼p∧(∼p∨∼q)

**D. ** ∼q∧(∼p∨∼q)

**Answer : ****Option B**

**Explaination / Solution: **

and Associative law

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Let p and q be two prepositions given by p : I play cricket during the holidays, q : I just sleep throughout the day then , the compound statement p ∧q is

**A. ** I just sleep throughout the day if and only if I play cricket during the holidays

**B. ** I play cricket during the holidays or I just sleep throughout the day

**C. ** If I play cricket during the holidays , I just sleep throughout the day

**D. ** I play cricket during the holidays and just sleep throughout the day

**Answer : ****Option D**

**Explaination / Solution: **

∧ is replaced by and in verbal form

∧ is replaced by and in verbal form

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Let p and q be two prepositions given by p : A parallelogram is a rhombus. q : The diagonals are at right angles. The compound proposition “ A parallelogram is a rhombus iff its diagonals are at right angles “ is represented by

**A. ** p→q

**B. ** p↔q

**C. ** p ∨q

**D. ** p ∨q

**Answer : ****Option B**

**Explaination / Solution: **

iff means a double implication statement so symbolic form has ↔

iff means a double implication statement so symbolic form has ↔

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If the inverse of implication p→q is defined as ∼p→∼q , then the inverse of proposition (p∧∼q)→r is

**A. ** None of these

**B. ** ∼r→∼p∨q

**C. ** (∼p∨q)→∼r

**D. ** r→p∧∼q

**Answer : ****Option C**

**Explaination / Solution: **

as given the rule of inverse in question it becomes ∼(p∧∼q)→∼r =(∼p∨q)→∼r

as given the rule of inverse in question it becomes ∼(p∧∼q)→∼r =(∼p∨q)→∼r

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∼(p∧q) is logically equivalent to

**A. ** ∼p→q

**B. ** ∼p∨∼q

**C. ** ∼p↔∼q

**D. ** ∼p→∼q

**Answer : ****Option B**

**Explaination / Solution: **

∼(p∧q)≡∼p∨∼q De Morgan's law

∼(p∧q)≡∼p∨∼q De Morgan's law

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Which of the following is not a proposition ?

**A. ** Mathematics is interesting

**B. ** 3 is a prime

**C. ** √2 is irrational

**D. ** 5 is an even integer

**Answer : ****Option A**

**Explaination / Solution: **

The above statement is a fact and the other options are mathematical statements which are proposition .

The above statement is a fact and the other options are mathematical statements which are proposition .

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Negation of the statement (p∧r)→(r∨q) is

**A. ** (p∧r)∧(∼r∧∼q)

**B. ** (p∧r)∨(∼r→∼q)

**C. ** (p∧r)∨(r→q)

**D. ** (p∧r)∧(∼r→∼q)

**Answer : ****Option A**

**Explaination / Solution: **

(p∧r)∧(∼r∧∼q) since ∼(p→q)≡p∧∼q

(p∧r)∧(∼r∧∼q) since ∼(p→q)≡p∧∼q

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The contrapositive of p→(∼q→∼r) is

**A. ** (q∧∼r)→p

**B. ** (∼q∧r)→∼p

**C. ** (q∨∼r)∨p

**D. ** (q∧∼r)→∼p

**Answer : ****Option B**

**Explaination / Solution: **

The contrapositive of p→q is ∼q→∼p

The contrapositive of p→q is ∼q→∼p

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The statement p→(q→p) is equivalent to

**A. ** p→(p∧q)

**B. ** p→(p↔q)

**C. ** p→(p→q)

**D. ** p→(p∨q)

**Answer : ****Option D**

**Explaination / Solution: **

p | q | ||||

T | T | T | T | T | T |

T | F | T | T | T | T |

F | T | F | T | T | T |

F | F | T | F | T | T |

Hence they are equivalent

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