Mathematical Reasoning Test

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Mathematical Reasoning Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The negation of the proposition “if a quadrilateral is a square, then it is a rhombus “ is

A
if a quadrilateral is a square , then it is not a rhombus

B
a quadrilateral is a square and it is not a rhombus

C
if a quadrilateral is not a square , then it is a rhombus

D
a quadrilateral is not a square and it is a rhombus

Solution

rules of negation ∼ ( p → q ) ≡ p ∧ ∼ q
Question 2
1 / -0

The contrapositive of (p ∨ q) → r is

A
∼ r → ∼ (p ∨ q)

B
∼r → ( ∼ p ∧ ∼ q)

C
∼ r → (p ∧ q)

D
p → (p ∧ q)

Solution

the contrapositive of p → q is ∼ q → ∼ p
Question 3
1 / -0

The contrapositive of p → (∼ q → ∼ r) is

A
(∼ q ∧ r) → ∼ p

B
(q ∨ ∼ r) ∨ p

C
(q ∧ ∼ r ) → ∼ p

D
(q ∧ ∼ r ) → p

Solution

The contrapositive of p → q is ∼ q → ∼ p
Question 4
1 / -0

The contrapositive of the statement “ if 2² = 5, then I get first class” is

A
If I do not get a first class , then 2² ≠ 5,

B
If I do not get a first class , then 2² = 5,

C
If I get a first class , then 2²= 5,

D
none of these

Solution

p:2² = 5

q:I get first class

the contrapositive of p→ q is ∼ q → ∼ p. hence the answer is If I do not get a first class , then 2²≠ 5
Question 5
1 / -0

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

D
none of these

Solution

p: x = 5 and y = - 2 , q : x – 2y = 9

The contrapositive of p → q is ∼ q → ∼ p

Hence If x – 2y ≠ 9 , then x ≠ 5 or y ≠ - 2
Question 6
1 / -0

“If the figure is a rhombus then the diagonals are perpendicular “. The contrapositive of the above statement is

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

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

C
If the diagonals are perpendicular, then the figure is a rhombus

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

Solution

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

Contrapositive of p → q is ∼ q → ∼ p

hence If the diagonals are not perpendicular, then the figure is not a rhombus
Question 7
1 / -0

Which of the following statement is a tautology ?

A
(p ∧ q) ∧ ( ∼ (p ∧ q))

B
(∼ q ∧ p) ∨ (p ∨ ∼ p)

C
(∼ q ∧ p) ∧ q

D
(∼ q ∧ p) ∧ (p ∧ ∼ p)

Solution

p ∨ ∼ p ≡ T

T ∨ p ≡ T

Question 8

1 / -0

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

p→ (p ∧ q)

p→ (p → q)

p → ( p ↔ q)

p → (p ∨ q)

Solution

p	q	q→p	p∨q	p→(q→p)	p→(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

Question 9
1 / -0

The inverse of the proposition (p ∧ ∼ q ) → r is

A
∼ r → ∼ p ∨ q

B
r → p ∧ ∼ q

C
(∼ p ∨ q ) → ∼ r

D
none of these

Solution

∼(p ∧ ∼ q ) → ∼ r inverse of p → q ≡ ∼ p → ∼ q

( ∼ p ∨ q) → ∼ r Since ∼ (p ∧ q) ≡ ∼ p ∨ ∼ q
Question 10
1 / -0

Logical equivalent proposition to the proposition ∼ (p ∨ q) is

A
∼ p ∧ ∼ q

B
∼ p ∨ ∼ q

C
∼ p → ∼ q

D
∼ p ↔ ∼ q

Solution

∼ (p ∨ q) ≡ ∼ p∧ ∼ q De Morgan's law

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Mathematical Reasoning Test - 5

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Mathematical Reasoning Test - 5

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Question 1

if a quadrilateral is a square , then it is not a rhombus

a quadrilateral is a square and it is not a rhombus

if a quadrilateral is not a square , then it is a rhombus

a quadrilateral is not a square and it is a rhombus

Solution

Question 2

∼ r → ∼ (p ∨ q)

∼r → ( ∼ p ∧ ∼ q)

∼ r → (p ∧ q)

p → (p ∧ q)

Solution

Question 3

(∼ q ∧ r) → ∼ p

(q ∨ ∼ r) ∨ p

(q ∧ ∼ r ) → ∼ p

(q ∧ ∼ r ) → p

Solution

Question 4

If I do not get a first class , then 22 ≠ 5,

If I do not get a first class , then 22 = 5,

If I get a first class , then 22= 5,

none of these

Solution

Question 5

x – 2y = 9 iff x = 5 and y = - 2

If x – 2y = 9 , x ≠ 5 and y ≠ - 2

If x – 2y ≠ 9 , then x ≠ 5 or y ≠ - 2

none of these

Solution

Question 6

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

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

If the diagonals are perpendicular, then the figure is a rhombus

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

Solution

Question 7

(p ∧ q) ∧ ( ∼ (p ∧ q))

(∼ q ∧ p) ∨ (p ∨ ∼ p)

(∼ q ∧ p) ∧ q

(∼ q ∧ p) ∧ (p ∧ ∼ p)

Solution

Question 8

p→ (p ∧ q)

p→ (p → q)

p → ( p ↔ q)

p → (p ∨ q)

Solution

Question 9

∼ r → ∼ p ∨ q

r → p ∧ ∼ q

(∼ p ∨ q ) → ∼ r

none of these

Solution

Question 10

∼ p ∧ ∼ q

∼ p ∨ ∼ q

∼ p → ∼ q

∼ p ↔ ∼ q

Solution

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If I do not get a first class , then 2² ≠ 5,

If I do not get a first class , then 2² = 5,

If I get a first class , then 2²= 5,