Mathematical Reasoning Test

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Mathematical Reasoning Test - 26

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Question 1
1 / -0

$$\sim \left[ p\wedge (\sim q) \right] =$$

A
$$\sim p \ \wedge \sim q$$

B
$$\sim p \ \vee \sim q$$

C
$$\sim p \ \wedge \ q$$

D
$$\sim p\vee q$$

Solution

$$\sim \left [p\wedge(\sim q)\right]=\sim p\vee\sim (\sim q)$$ (by demorgan's law)
$$=\sim p\vee q$$ ($$\because \sim(\sim p)=p$$)

Option $$D$$ is correct.
Question 2
1 / -0

If $$p, q, r$$ are simple proportions with truth values $$T, F, T$$, then the truth value of $$(\sim p\vee q)\wedge \sim r \Rightarrow p$$ is

A
True

B
False

C
True, if $$r$$ is false

D
True, if $$q$$ is true

Solution

$$\sim p\vee q$$ means $$F\vee F = F, \sim r$$ means $$F$$

$$\therefore [(\sim p\vee q)\wedge \sim r)=F\wedge F = F$$ which imply $$ p = T$$ means $$F\Rightarrow T = T$$.

Hence, the truth value of the given statement is True.
Question 3
1 / -0

$$p \leftrightarrow q$$ is equivalent to

A
$$p \rightarrow q$$

B
$$q \rightarrow p$$

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

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

Solution

Consider truth table of $$p\leftrightarrow q$$ $$($$ Let $$T$$ denotes true ans $$F$$ denotes false$$)$$
If $$p=T$$ and $$q=T$$ then $$p\leftrightarrow q = T$$
If $$p=T$$ and $$q=F$$ then $$p\leftrightarrow q = F$$
If $$p=F$$ and $$q=T$$ then $$p\leftrightarrow q = F$$
If $$p=F$$ and $$q=F$$ then $$p\leftrightarrow q = T$$
Now consider truth table of $$(p\rightarrow q)\wedge (q\rightarrow p)$$
If $$p=T$$ and $$q=T$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = T$$
If $$p=T$$ and $$q=F$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = F$$
If $$p=F$$ and $$q=T$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = F$$
If $$p=F$$ and $$q=F$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = T$$
Observe that both truth tables are equal , therefore the correct option is $$D$$
Question 4
1 / -0

The inverse of the propositions $$(p \wedge \sim q) \rightarrow r$$ is____.

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

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

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

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

Solution

In logic, an inverse is a type of conditional sentence which is an immediate interference made from another conditional sentence. Any conditional sentence has an inverse. The inverse of $$P\rightarrow Q\quad is\quad \neg P\rightarrow \neg Q$$
Hence, answer is (∼r) $$\rightarrow$$ (∼p) ∨ q
Question 5
1 / -0

If $$p$$ is $$T$$ and $$q$$ is $$F$$, then which of the following have the truth value $$T$$?
$$(i)p\vee q$$
$$(ii)\sim p\vee q$$
$$(iii)p\vee (\sim q)$$
$$(iv)p\wedge (\sim q)$$

A
$$(i),(ii),(iii)$$

B
$$(i),(ii),(iv)$$

C
$$(i),(iii),(iv)$$

D
$$(ii),(iii),(iv)$$

Solution

$$(i)p\vee q = T\vee F =T$$
$$(ii)\sim p\vee q=(\sim T)\vee F = F\vee F=F$$
$$(iii)p\vee (\sim q)=T\vee (\sim F)=T\vee T= T$$
$$(iv)p\wedge (\sim q)=T\wedge (\sim F)=T\wedge T=T$$

Hence option 'C' is correct
Question 6
1 / -0

If $$p:3$$ is a prime number and $$q:$$ one plus one is three, then the compound statement "It is not that $$3$$ is a prime number or it is not that one plus one is three" is

A
$$\sim p\vee q$$

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

C
$$p\wedge \sim q$$

D
$$\sim p\vee \sim q$$

E
$$p\vee \sim q$$

Solution

$$p:3$$ is a prime number
$$q:$$ one plus one is three

Compound statement: "It is not that $$3$$ is a prime number or it is not that one plus one is three" is
$$\sim p\vee \sim q$$
Question 7
1 / -0

If $$p$$'s truth value is $$T$$ and $$q$$'s truth value is $$F$$, then which of the following have the truth value $$T$$?
(i) $$p\vee q$$
(ii) $$\sim p\vee q$$
(iii) $$p\vee (\sim q)$$
(iv) $$p\wedge (\sim q)$$

A
(i), (ii), (iii)

B
(i), (iii), (iv)

C
(i), (ii), (iv)

D
(ii), (iii), (iv)

Solution

Given $$p's$$ truth value is $$T$$ and $$q's$$ truth value is $$F$$.

We have to determine which of the given have the truth value $$T$$.

$$(i) p\vee q$$

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

$$p$$ $$\sim p$$ $$q$$ $$\sim p \vee q$$
$$T$$ $$F$$ $$F$$ $$F$$

$$(iii) p\vee (\sim q)$$

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

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

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

From the above tables we see that, $$(i), (iii), (iv)$$ have truth value $$T$$.
Question 8
1 / -0

Which of the following is not a logical statement?

A
$$8$$ is less than $$6$$

B
Every set is a finite set

C
Kashmir is far from here

D
The sun is a star

Solution

A) $$8$$ is less than $$6$$ $$ \rightarrow $$ logical
B) Every set is a finite set $$ \rightarrow $$ logical
C) Kashmir is far from here $$ \rightarrow $$ not a logical statement
D) The sun is a star $$ \rightarrow $$ logical

Question 9

1 / -0

The truth values of $$p, q$$ and $$r$$ for which $$(p \wedge q) \vee (\sim r)$$ has truth value F are respectively

F, T, F

F, F, F

T, T, T

T, F, F

F, F, T

Solution

option	$$p$$	$$q$$	$$r$$	$$p\wedge q$$	$$\sim r$$	$$(p\wedge q)\vee (\sim r)$$
A	$$ F$$	$$T$$	$$ F$$	$$ F$$	$$T$$	$$T$$
B	$$ F$$	$$ F$$	$$ F$$	$$ F$$	$$T$$	$$T$$
C	$$T$$	$$T$$	$$T$$	$$T$$	$$ F$$	$$T$$
D	$$T$$	$$ F$$	$$ F$$	$$ F$$	$$T$$	$$T$$
E	$$ F$$	$$ F$$	$$T$$	$$ F$$	$$ F$$	$$ F$$

Option $$E$$ gives truth value $$F$$ .

So option $$E$$ is correct.

Question 10
1 / -0

$$\left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right) $$ is a :

A
A tautology

B
A contradiction

C
Both a tautology and a contradiction

D
Neither a tautology nor a contradiction

Solution

$$\left( p\wedge \sim q \right) \left( \sim p\wedge q \right) =\left( p\wedge \sim p \right) \wedge \left( \sim q\wedge q \right) $$

$$=f\wedge f=f$$
(By using associative laws and commutative laws)

Therefore, $$ \left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right) $$ is a contradiction.

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

Mathematical Reasoning Test - 26

Result

Mathematical Reasoning Test - 26

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SHARING IS CARING

Question 1

$$\sim p \ \wedge \sim q$$

$$\sim p \ \vee \sim q$$

$$\sim p \ \wedge \ q$$

$$\sim p\vee q$$

Solution

Question 2

True

False

True, if $$r$$ is false

True, if $$q$$ is true

Solution

Question 3

$$p \rightarrow q$$

$$q \rightarrow p$$

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

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

Solution

Question 4

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

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

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

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

Solution

Question 5

$$(i),(ii),(iii)$$

$$(i),(ii),(iv)$$

$$(i),(iii),(iv)$$

$$(ii),(iii),(iv)$$

Solution

Question 6

$$\sim p\vee q$$

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

$$p\wedge \sim q$$

$$\sim p\vee \sim q$$

$$p\vee \sim q$$

Solution

Question 7

(i), (ii), (iii)

(i), (iii), (iv)

(i), (ii), (iv)

(ii), (iii), (iv)

Solution

Question 8

$$8$$ is less than $$6$$

Every set is a finite set

Kashmir is far from here

The sun is a star

Solution

Question 9

F, T, F

F, F, F

T, T, T

T, F, F

F, F, T

Solution

Question 10

A tautology

A contradiction

Both a tautology and a contradiction

Neither a tautology nor a contradiction

Solution

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