Mathematical Reasoning Test

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Mathematical Reasoning Test - 19

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Question 1
1 / -0

The contrapositive of $$p \rightarrow (\sim q \rightarrow \sim r)$$ is

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

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

C
$$( q \vee \sim r) \rightarrow \sim p $$

D
none of these.

Solution

The contrapositive of $$p \rightarrow (\sim q \rightarrow \sim r)$$ is
$$\equiv \sim (\sim q \rightarrow \sim r) \rightarrow \sim p$$
$$\equiv \sim(q \vee \sim r) \to \sim p$$
$$\equiv (\sim q \wedge r) \to \sim p$$
Question 2
1 / -0

Negation of $$\displaystyle q \vee \sim \left ( p\wedge r \right )$$ is

A
$$\displaystyle \sim q\wedge \sim \left ( p\wedge r \right )$$

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

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

D
None of these

Solution

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

\right ) \equiv \sim q\wedge \left ( \sim \left ( \sim (p\wedge r \right )

\right ) \equiv \sim q\wedge \left ( p\wedge r \right )$$
Question 3
1 / -0

Find the truth value of the compound statement, 4 is the first composite number and $$2+5=7$$.

A
T

B
F

C
Neither T nor F

D
cannot be determined

Solution

The given statement is True, as $$ 4$$ is the first composite number. Also $$ 2 + 5 = 7 $$
Question 4
1 / -0

The negation of the statement "$$2 + 3 = 5$$" and "$$8 < 10$$" is

A
$$2 + 3 \neq 5$$ and $$8\nless 10$$

B
$$2 + 3 $$\neq 5 or $$8 > 10$$

C
$$2 + 3 \neq 5$$ or $$8 \ngeqslant 10$$

D
None of these

Solution

Take $$p:2+3=5$$ and $$q:8<10$$

So the given conjunction is $$p\wedge q$$

Now $$\sim p:2+3\neq 5$$ and $$\sim q:8\nless 10$$

Now Negation of the given conjunction $$p\wedge q$$ is $$\sim \left( p\wedge q \right) $$

$$\sim \left( p\wedge q \right) :2+3\neq 5$$ or $$8\nless 10$$
Question 5
1 / -0

Find the quantifier which best describes the variable of the open sentence $$x^2+2\ge0$$

A
Universal.

B
Existential.

C
Neither (a) nor (b).

D
Does not exist.

Solution

An universal quantifier is a symbol or a logic to denote that the statement is true for all values under the scope.

We know that $$ x^2 $$ is always $$ \ge 0 $$
Hence, $$ x^2 + 2 \ge 0 $$ holds true for all values of $$ x $$.

Thus, the quantifier to describe the variable of the given sentence is universal.
Question 6
1 / -0

What is the truth value of the statement 'Two is an odd number iff 2 is a root of $$x^2+2=0$$'?

A
T

B
F

C
Neither T nor F

D
Cannot be determined

Solution

Both the statements are false as Two is not an odd number and $$ 2 $$ is not the root of $$ x^2 + 2 = 0 $$

We know that two False statements together make a True statement., Hence, truth value of the given statement is True.
Question 7
1 / -0

The negative of the statement "If a number is divisible by 15 then it is divisible by 5 or 3"

A
if a number is divisible by 15 then it is not divisible by 5 and 3

B
a number is divisible by 15 and it is not divisible by 5 or 3

C
a number is divisible by 15 or it is not divisible by 5 and 3

D
a number is divisible by 15 and it is not divisible by 5 and 3

Solution

Let $$p, q, r$$ be three statements defined as

$$p$$ : a number N is divisible by 15

$$q$$ : a number N is divisible by 5

$$r$$ :a number N is divisible by 3

Here given statement is $$p\rightarrow (q\vee r)$$

Here negative of above statement is

$$\sim (p \rightarrow (q \vee r)) $$

$$=\sim [(\sim p) \vee (q \vee r)] $$ ($$\because p\rightarrow q=\sim p \vee q$$)

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

$$= p \wedge (\sim q \wedge \sim r)$$ ($$\because \sim (\sim p)=p, \sim (p\vee q=\sim p \wedge \sim q$$)

i.e. a number is divisible by 15 and it is not divisible by 5 and 3.
Question 8
1 / -0

The inverse of the statement $$(p \wedge \sim q) \rightarrow r$$ is

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

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

C
$$(\sim p \vee q) \rightarrow \sim r$$

D
None of these.

Solution

$$\textbf{Step-1: Apply the concept of logical reasoning. }$$
$$\text{We have,}$$
$$(p \wedge \sim q) \rightarrow r$$
$$\therefore$$ $$\text{The inverse of the statement}$$ $$(p \wedge \sim q) \rightarrow r$$ $$\text{is,}$$
$$\equiv \sim (p \wedge \sim q) \rightarrow \sim r \equiv (\sim p \vee \sim (\sim q)) \to \sim r$$
$$\equiv (\sim p \vee q) \to \sim r$$

$$\textbf{Hence, option C}$$
Question 9
1 / -0

The contrapositive of $$\sim p \rightarrow ( q \rightarrow \sim r)$$ is

A
$$(~q \wedge r) \rightarrow ~p$$

B
$$(q \rightarrow r) \rightarrow~p$$

C
$$(q \vee ~r) \rightarrow ~ p$$

D
None of these.

Solution

The contrapositive of $$\sim p \rightarrow ( q \rightarrow \sim r)$$ is
$$\equiv \sim ( q \rightarrow \sim r) \rightarrow \sim (\sim p )$$
$$\equiv \sim (\sim q \vee \sim r) \rightarrow p \equiv (q \wedge r) \rightarrow p $$
Question 10
1 / -0

Let $$S$$ be non-empty subset of $$R$$ then consider the following statement
"Every number $$\displaystyle x\: \epsilon \: S $$ is an even number."
Negation of the statement will be

A
There is no number $$\displaystyle x\: \epsilon \: S $$ which is even

B
There exists a number $$\displaystyle x\: \epsilon \: S $$ which is not even

C
There exists a number $$\displaystyle x\: \epsilon \: S $$ which is odd

D
($$B$$) and ($$C$$) both

Solution

The given statement implies $$x$$ in the set $$S$$ is always even. The negation of this means there exists $$x$$ in $$S$$ which is not even, or which is odd.
So, option $$D$$ is correct.

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Mathematical Reasoning Test - 19

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Mathematical Reasoning Test - 19

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Question 1

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

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

$$( q \vee \sim r) \rightarrow \sim p $$

none of these.

Solution

Question 2

$$\displaystyle \sim q\wedge \sim \left ( p\wedge r \right )$$

$$\displaystyle \sim q\wedge \left ( p\wedge r \right )$$

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

None of these

Solution

Question 3

T

F

Neither T nor F

cannot be determined

Solution

Question 4

$$2 + 3 \neq 5$$ and $$8\nless 10$$

$$2 + 3 $$\neq 5 or $$8 > 10$$

$$2 + 3 \neq 5$$ or $$8 \ngeqslant 10$$

None of these

Solution

Question 5

Universal.

Existential.

Neither (a) nor (b).

Does not exist.

Solution

Question 6

T

F

Neither T nor F

Cannot be determined

Solution

Question 7

if a number is divisible by 15 then it is not divisible by 5 and 3

a number is divisible by 15 and it is not divisible by 5 or 3

a number is divisible by 15 or it is not divisible by 5 and 3

a number is divisible by 15 and it is not divisible by 5 and 3

Solution

Question 8

$$\sim (p \vee \sim q) \rightarrow \sim r$$

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

$$(\sim p \vee q) \rightarrow \sim r$$

None of these.

Solution

Question 9

$$(~q \wedge r) \rightarrow ~p$$

$$(q \rightarrow r) \rightarrow~p$$

$$(q \vee ~r) \rightarrow ~ p$$

None of these.

Solution

Question 10

There is no number $$\displaystyle x\: \epsilon \: S $$ which is even

There exists a number $$\displaystyle x\: \epsilon \: S $$ which is not even

There exists a number $$\displaystyle x\: \epsilon \: S $$ which is odd

($$B$$) and ($$C$$) both

Solution

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