Mathematical Reasoning Test

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Mathematical Reasoning Test - 29

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Question 1

1 / -0

If statement $$(p\rightarrow q)\rightarrow (q\rightarrow r)$$ is false, then truth values of statements $$p,q,r$$ respectively, can be :

$$FTF$$

$$TTT$$

$$FFF$$

$$FTT$$

Solution

p	q	r	$$p\rightarrow q$$	$$q\rightarrow r$$	$$(p\rightarrow q) \rightarrow (q\rightarrow r)$$
T	T	T	T	T	T
T	T	F	T	F	F
T	F	T	F	T	T
T	F	F	F	T	T
F	T	T	T	T	T
F	T	F	T	F	F
F	F	T	T	T	T
F	F	F	T	T	T

$$\therefore $$ If p=T or p=F

q=T q=T

r=F r=F

Then the statement $$(p\rightarrow q)\rightarrow (q\rightarrow r)$$ is false.

Question 2

1 / -0

$$p \wedge ( q \vee \sim p ) =?$$

$$p \vee q$$

$$p \wedge q$$

$$p \rightarrow q$$

none of these

Solution

We have,

$$P \wedge (q \, \vee \sim p) = ?$$

From truth table,

$$p$$	$$q$$	$$p \vee q$$	$$p \wedge q$$	$$\sim p$$	$$q \vee \sim p$$	$$p \wedge (q \vee \sim p)$$
$$T$$ $$T$$ $$F$$ $$F$$	$$T$$ $$F$$ $$T$$ $$F$$	$$T$$ $$T$$ $$T$$ $$F$$	$$T$$ $$F$$ $$F$$ $$F$$	$$F$$ $$F$$ $$T$$ $$T$$	$$T$$ $$F$$ $$T$$ $$T$$	$$T$$ $$F$$ $$F$$ $$F$$

From above truth table, we get

$$p \wedge (q \vee \sim p) \equiv p \wedge q$$

Hence, option B is correct answer.

Question 3
1 / -0

Negation of $$(\sim p\rightarrow q)$$ is ________________.

A
$$\sim { p }{ \wedge }\sim q$$

B
$$\sim \left( p\vee q \right) \vee \left( p\vee \left( \sim p \right) \right) $$

C
$$\sim \left( p\vee q \right) \wedge \left( p\vee \left( \sim p \right) \right) $$

D
$$\left( \sim p\vee q \right) \wedge \left( p\vee \sim q \right) $$
Question 4
1 / -0

Consider the statement "if it is raining, then sky is not filled with clouds which among the Q 7 following is true

A
contrapositive of the given statement "if sky is filled with clouds then it is not raining"

B
converse of the given statement is "If sky is filled with clouds then it is raining"

C
"If it is not raining then sky is filled with clouds" is the inverse of the given statement

D
Both 1 , 2 and 3 are true

Solution
Question 5
1 / -0

The statement $$P \rightarrow ( q \vee r )$$ is not equivalent to

A
$$( p \rightarrow q ) \vee ( q - r )$$

B
$$p \wedge ( \sim q ) \rightarrow r$$

C
$$P \wedge ( \sim r ) \rightarrow q$$

D
$$p \wedge q \rightarrow ( p \wedge r ) \vee ( q \wedge r )$$

Solution
Question 6
1 / -0

The false statement in the following is

A
$$p \wedge ( \sim p )$$ is a contradiction.

B
$$( p \rightarrow q ) \leftrightarrow ( \sim q \rightarrow \sim p )$$ is a contradiction

C
$$\sim ( \sim p ) \leftrightarrow p$$ is a tautology

D
$$p \vee ( \sim p )$$ is a tautology

Solution
Question 7
1 / -0

Consider the following statements
$$P:$$Suman is brilliant
$$Q:$$Suman is rich
$$R:$$Suman is honest.
The negative of the statement."Suman is brilliant and dishonest, if and only if Suman is rich" can be expressed as

A
$$\sim [Q \leftrightarrow (P \wedge \sim R)]$$

B
$$\sim Q \leftrightarrow P \wedge R$$

C
$$\sim (P \wedge \sim R) \leftrightarrow Q $$

D
$$\sim P \wedge (Q \leftrightarrow \sim R)$$

Solution

Suman is brilliant and dishonest $$\Rightarrow P \wedge \sim~ R$$
Suman is brilliant and dishonest if and only if suman is rich $$\Rightarrow Q \leftrightarrow \left( P \wedge \sim R \right)$$
Negative of the above statement will be-
$$\sim \left[ Q \leftrightarrow \left( P \wedge \sim R \right) \right]$$
Question 8
1 / -0

The negation of $$p\wedge \left( q \rightarrow\ r \right)$$ is

A
$$p\vee \left( \sim q\vee r \right)$$

B
$$\sim p\wedge \left( q\rightarrow r \right)$$

C
$$\sim p\wedge \left( \sim q\rightarrow\sim r \right)$$

D
$$\sim p\vee \left( q\wedge \sim r \right)$$

Solution
Question 9
1 / -0

Consider the following two statements: $$P:$$ if $$7$$ is an odd number, then $$7$$ is divisible by $$2$$.$$Q:$$ if $$7$$ is a prime number, then $$7$$ is an odd number If $$V_{1}$$ is the truth value of the contrapositive of $$P$$ and $$V_{2}$$ is the truth value of contrapositive of $$Q$$. Then the ordered pair $$\left(V_{1},V_{2}\right)$$ equals:

A
$$\left(F,F\right)$$

B
$$\left(T,T\right)$$

C
$$\left(F,T\right)$$

D
$$\left(T,F\right)$$

Solution
Question 10
1 / -0

Negation of $$p \to \left( {p \vee \sim q} \right)$$ is

A
$$( \sim p) \to ( \sim p \vee q)$$

B
$$p \wedge ( \sim p \wedge q)$$

C
$$\left( { \sim p} \right) \vee \left( { \sim p \vee \sim q} \right)$$

D
$$\left( { \sim p} \right) \leftarrow \left( { \sim p \to q} \right)$$

Solution