Mathematical Reasoning Test

Result

Mathematical Reasoning Test - 30

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Question 1
1 / -0

If $$(p\wedge \sim q)\wedge (p\wedge r)\rightarrow \vee \sim p \vee q$$ is false then the truth values of $$p,q$$ and $$r$$ are respectively

A
$$F,T,F$$

B
$$T,F,T$$

C
$$F,F,F$$

D
$$T,T,T$$

Solution
Question 2
1 / -0

Let $$p\rightarrow (\sim q\vee r)$$ is false, then truth values of p, q, r are respectively.

A
$$F, T, T$$

B
$$T, F, T$$

C
$$T, T, F$$

D
$$F, F, T$$

Solution
Question 3
1 / -0

Which of the following is not true?

A
If the last column of its truth table contains only $$F$$ then it is a contradiction

B
Negation of a negation of a statement is the statement itself

C
If $$p$$ and $$q$$ are two statements then $$p \leftrightarrow q$$ is a tautology

D
If the last column of its truth table contains only $$T$$ then it is tautology

Solution

Truth table
$$p$$ $$q$$ $$ p\leftrightarrow q$$
$$ T$$ $$T$$ $$T$$
$$T$$ $$F$$ $$F$$
$$F$$ $$T$$ $$F$$
$$F$$ $$F$$ $$T$$
Since the last column of the truth table has False and Truth.It cannot be tautology.
Question 4
1 / -0

Which of the following is NOT equivalent to $$p\rightarrow q$$?

A
p only if q

B
q is necessary for p

C
q only if p

D
p is sufficient for q

Solution

$$q$$ only if $$p$$ is a statement that says if $$q$$ is true then $$p$$ is true.
This is not equivalent to $$p \rightarrow{} q$$. This non-equivalency can be checked by letting $$p$$ be a true statement and $$q$$ be a false statement. With these values, the statement $$p \rightarrow{} q$$ is false but the statement $$q$$ only if $$p$$ is true.
Question 5
1 / -0

An implication or conditional "if p then q "is denoted by

A
$$p \vee q$$

B
$$p \rightarrow q$$

C
$$p \leftarrow q$$

D
None of these

Solution
Question 6
1 / -0

The statement pattern $$(p\wedge q)\wedge [\sim r\vee (p\wedge q)]\vee (\sim p\wedge q)$$ is equivalent to _________.

A
r

B
q

C
$$p\wedge q$$

D
p

Solution

$$(p\wedge q)\wedge [\sim r\vee (p\wedge q)]\vee (\sim p\wedge q)$$
We will now use absorption law $$ [(a\vee b)\wedge a=a]$$ for first 2 bracket.
$$(p\wedge q)\vee (\sim p \wedge q)$$
Now by using disrtibutive law we get
$$( p \vee\sim p) \wedge q$$
$$ 1\wedge q$$
$$q$$
Question 7
1 / -0

Disjunction of two statements p and q is denoted by

A
$$p \leftrightarrow q$$

B
$$p \rightarrow q$$

C
$$p \leftarrow q$$

D
$$p \vee q$$

Solution

Question 8

1 / -0

The equivalent form of the statement $$\sim(p\rightarrow \sim q)$$ is ________.

$$p\wedge q$$

$$p\wedge \sim q$$

$$p\vee \sim q$$

$$\sim p\vee q$$

Solution

We have $$a \rightarrow b = \sim a \vee b$$

$$\therefore\ p \rightarrow \sim q = \sim p \vee \sim q $$

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

$$ = \sim (\sim (p \wedge q)) $$ .... De Morgan's Law

$$ = p \wedge q$$ .... Complement Law

_________________________ [OR] ____________________________

We can also solve it using Truth Table

$$p$$	$$q$$	$$\sim p$$	$$\sim q$$	$$p \wedge q$$	$$p \wedge \sim q$$	$$p \vee \sim q$$	$$\sim p \vee q $$	$$p \rightarrow \sim q $$	$$\sim (p \rightarrow \sim q)$$
$$0$$	$$0$$	$$1$$	$$1$$	$$0$$	$$0$$	$$1$$	$$1$$	$$1$$	$$0$$
$$0$$	$$1$$	$$1$$	$$0$$	$$0$$	$$0$$	$$0$$	$$1$$	$$1$$	$$0$$
$$1$$	$$0$$	$$0$$	$$1$$	$$0$$	$$1$$	$$1$$	$$0$$	$$1$$	$$0$$
$$1$$	$$1$$	$$0$$	$$0$$	$$1$$	$$0$$	$$1$$	$$1$$	$$0$$	$$1$$

So from column $$5$$ and $$10$$ we can write that,

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

Question 9
1 / -0

The negation of the compound proposition $$p \vee (p \vee q)$$ is

A
$$(p\wedge ∼q)\wedge ∼p$$

B
$$(p\wedge ∼q)\vee ∼p$$

C
$$(p\wedge ∼q)\vee ∼p$$

D
none of these

Solution
Question 10
1 / -0

Given, "If I have a Siberian Husky, then I have a dog." Identify the converse

A
If I do not have a Siberian Husky, then I do not have a dog.

B
If I have a dog, then I have a Siberian Husky.

C
If I do not have a dog, then I do not have a Siberian Husky.

D
If I do not have a Siberian Husky, then I have a dog.

Solution

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Re-Attempt Weekly Quiz Competition

$$p$$	$$q$$	$$ p\leftrightarrow q$$
$$ T$$	$$T$$	$$T$$
$$T$$	$$F$$	$$F$$
$$F$$	$$T$$	$$F$$
$$F$$	$$F$$	$$T$$

Mathematical Reasoning Test - 30

Result

Mathematical Reasoning Test - 30

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Rank

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SHARING IS CARING

Question 1

$$F,T,F$$

$$T,F,T$$

$$F,F,F$$

$$T,T,T$$

Solution

Question 2

$$F, T, T$$

$$T, F, T$$

$$T, T, F$$

$$F, F, T$$

Solution

Question 3

If the last column of its truth table contains only $$F$$ then it is a contradiction

Negation of a negation of a statement is the statement itself

If $$p$$ and $$q$$ are two statements then $$p \leftrightarrow q$$ is a tautology

If the last column of its truth table contains only $$T$$ then it is tautology

Solution

Question 4

p only if q

q is necessary for p

q only if p

p is sufficient for q

Solution

Question 5

$$p \vee q$$

$$p \rightarrow q$$

$$p \leftarrow q$$

None of these

Solution

Question 6

r

q

$$p\wedge q$$

p

Solution

Question 7

$$p \leftrightarrow q$$

$$p \rightarrow q$$

$$p \leftarrow q$$

$$p \vee q$$

Solution

Question 8

$$p\wedge q$$

$$p\wedge \sim q$$

$$p\vee \sim q$$

$$\sim p\vee q$$

Solution

Question 9

$$(p\wedge ∼q)\wedge ∼p$$

$$(p\wedge ∼q)\vee ∼p$$

$$(p\wedge ∼q)\vee ∼p$$

none of these

Solution

Question 10

If I do not have a Siberian Husky, then I do not have a dog.

If I have a dog, then I have a Siberian Husky.

If I do not have a dog, then I do not have a Siberian Husky.

If I do not have a Siberian Husky, then I have a dog.

Solution

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