Mathematical Reasoning Test

Result

Mathematical Reasoning Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

(p∧ ∼ q ) ∧ ( ∼ p ∨ q) is

A
a contradiction

B
a tautology

C
both a tautology and a contradiction

D
neither a tautology nor a contradiction

Solution

[ ( p ∧ ∼ q ) ∧ ( ∼ p ) ] ∨ [ ( p ∧ ∼ q ) ∧ q ) ] Since p∧ ( q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r )

F V F = F Since p ∧ ∼ p = F

Hence contracdiction
Question 2
1 / -0

Which of the following is the conditional p→q?

A
p only if q.

B
p is necessary of q.

C
q is sufficient for p.

D
if q, then p.

Solution

p only if q.
Question 3
1 / -0

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 )

Solution

(∼ p ∨ ∼ q ∨ p ∨ q ) Since p ∨ ( q ∨ r ) ≡ ( p ∨ q ) ∨ r

∼ p ∨ p ≡ T

Hence tautology
Question 4
1 / -0

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

A
∼ p → ( ∼ q ∨ r )

B
p ∧ ( ∼ q ∨ ∼ r )

C
p → ( ∼ q ∨ r )

D
∼ p ∧ ( ∼ q ∨ ∼ r )

Solution

∼ ( p → ( q ∧ r ))

=p ∧ ∼ ( q ∧ r ) since ∼ ( p → q ) ≡ p ∧ ∼ q

=p ∧ (∼ q ∨ ∼ r )
Question 5
1 / -0

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)

Solution

(p∧r)∧(∼r∧∼q) since ∼(p→q)≡p∧∼q
Question 6
1 / -0

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

A
∼q→∼(p∧r)

B
∼q∨(p∧r)

C
∼q∧∼(p∧r)

D
∼q∧(p∧r)

Solution

∼(q∨∼(p∧r)) Since ∼ ( q ∨ r ) ≡ ∼ q ∧ ∼ r

∼q∧(p∧r)
Question 7
1 / -0

Which of the following is always true ?

A
∼(p→q) ≅ (p∨∼q)

B
(p→q) ≅ (∼q→∼p)

C
∼(p∨q) ≅ (∼p∨∼q)

D
∼(p∧q) ≅ (∼p∧∼q)

Solution

∼q→∼p ≡ q∨∼p Since p→q ≡ ∼ p ∨ q

∼p∨q ≡ p→q
Question 8
1 / -0

Negation of the statement ∼p→(q∨r) is

A
∼p∧(∼q∧∼r)

B
p∨(q∧r)

C
p∧(q∨r)

D
p→∼(q∨r)p→∼(q∨r)

Solution

rules of negation ∼(p→q) ≡ p∧∼q

Hence ∼ p ∧ ( ∼ q ∧ ∼ r )

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Re-Attempt Weekly Quiz Competition

Mathematical Reasoning Test - 4

Result

Mathematical Reasoning Test - 4

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SHARING IS CARING

Question 1

a contradiction

a tautology

both a tautology and a contradiction

neither a tautology nor a contradiction

Solution

Question 2

p only if q.

p is necessary of q.

q is sufficient for p.

if q, then p.

Solution

Question 3

( ∼ p ∨ q )∼ ( p ∨ ∼ q )

(∼ p ∨ ∼ q ) ∨ ( p ∨ q )

(p ∨ ∼ q ) ∧ ( p ∨ q )

( ∼ p ∨ ∼ q ) → ( p ∨ q )

Solution

Question 4

∼ p → ( ∼ q ∨ r )

p ∧ ( ∼ q ∨ ∼ r )

p → ( ∼ q ∨ r )

∼ p ∧ ( ∼ q ∨ ∼ r )

Solution

Question 5

(p∧r)∨(r→q)

(p∧r)∨(∼r→∼q)

(p∧r)∧(∼r∧∼q)

(p∧r)∧(∼r→∼q)

Solution

Question 6

∼q→∼(p∧r)

∼q∨(p∧r)

∼q∧∼(p∧r)

∼q∧(p∧r)

Solution

Question 7

∼(p→q) ≅ (p∨∼q)

(p→q) ≅ (∼q→∼p)

∼(p∨q) ≅ (∼p∨∼q)

∼(p∧q) ≅ (∼p∧∼q)

Solution

Question 8

∼p∧(∼q∧∼r)

p∨(q∧r)

p∧(q∨r)

p→∼(q∨r)p→∼(q∨r)

Solution

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