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Questions Related to business maths

Consider :
Statement - I :$(p\wedge \sim q)\wedge (\sim p\wedge q)$ is a fallacy.
Statement - II :$(p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)$ is a tautology.

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Statement - I is true: Statement - II is true: Statement - II is a correct explanation for Statement - I.

  2. Statement - I is true: Statement - II is true: Statement - II is not a correct explanation for Statement - I.

  3. Statement - I is true; Statement - II is false.

  4. Statement - I is false; Statement - II is true.

Show answer
Correct Option: A

The statement (p ^ q) ^ (-pv - q) is _______________.

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. a contingency

  4. neither a tautology nor a contradiction

Show answer
Correct Option: A

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

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. none of these.

Show answer
Correct Option: A

The statement $\sim (p \rightarrow q) \leftrightarrow  (\sim p \vee \sim q)$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. None of these

Show answer
Correct Option: C
Explanation:

When $p$ and $q$ both are true then 

$(p \rightarrow  q) and (\sim p \vee \sim q)$ both are false

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is true when $p$ and $q$ both are false then  

$\sim (p \rightarrow  q)$ is false and $(\sim p \vee \sim q)$ is true

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is false

Hence $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is neither tautology nor contradiction

Which of the following statement is a contradiction ?

logical equivalence mathematical logic discrete mathematics business maths maths

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

  2. $(p \rightarrow q) \vee (p \wedge \sim q)$

  3. $(\sim p \wedge q) \wedge (\sim q)$

  4. $(\sim p \wedge q) \vee (\sim q)$

Show answer
Correct Option: C

If $p$ is any statement, $t$ is a tautology and $c$ is a contradiction, then which for the following is NOT correct?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p \wedge (\sim c) \equiv p$

  2. $p \vee (\sim t) \equiv p$

  3. $t \vee c \equiv p \vee t$

  4. $(p\wedge t) \vee (p \vee c) \equiv (t \wedge c)$

Show answer
Correct Option: D
Explanation:

A.   $p \wedge (\sim c )\equiv p \wedge t  \equiv p$

B.   $p \vee (\sim t )\equiv p \vee c \equiv p$

C.   $ t \vee c \equiv t \equiv p \vee t$

Clearly option 'A', 'B', 'C' are correct.

Hence option 'D' is the correct choice.

If $p$ is any statement, $t$ and $c$ are a tautology and a contradiction respectively, then which of the following is INCORRECT?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p \wedge t \equiv p $

  2. $ p \wedge c \equiv c$

  3. $p \vee t \equiv p $

  4. $ p \vee c \equiv p$

Show answer
Correct Option: C
Explanation:

Truth table,

$p$             $p\wedge t$        $p\wedge c$          $p\vee  t$      $p\vee c$
T T            F               T          T
F F            F            T         F

Hence option 'C' is the correct choice.

The statement $(p \rightarrow ~p) \wedge (~ p \rightarrow p)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. None of these.

Show answer
Correct Option: A
Explanation:

If $p$ is a true statement, then $p \rightarrow p$ is true.

Also, if $p$ is a false statement, then $p \rightarrow p$ is true.
Then, $(p \rightarrow p) \wedge (p \rightarrow p)$ is always true.
Hence, the given statement is a tautology.

Which of the following statement is a contradiction?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $(p \wedge q) \wedge (\sim(p \vee q))$

  2. $p \vee (\sim p \wedge q)$

  3. $(p \rightarrow q) \rightarrow p$

  4. $\sim p \vee \sim q$

Show answer
Correct Option: A
Explanation:

We check for contradiction for all the given options.

A. $\left( p\wedge q \right) \wedge \left( \sim \left( p\vee q \right)  \right) $

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

All F so this is a contradiction.


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

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

So not a contradiction.


C. $\left( p\longrightarrow q \right) \rightarrow p$

 $p$  $q$  $\left( p\longrightarrow q \right) $  $\left( p\longrightarrow q \right) \rightarrow p$
 T  T  T  T
 T  F  F  T
 F  T  T  F
 F  F  T  F

So it is also not a contradiction.


D. $\sim p\vee \sim q$

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

So it is also not a contradiction.

The statement $\sim (p \rightarrow q )\leftrightarrow (\sim p \vee \sim q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology.

  2. a contradiction.

  3. neither a tautology nor a contradiction.

  4. None of these.

Show answer
Correct Option: C
Explanation:

When $p$ and $q$ both are true then

$\sim (p \rightarrow q ) $ and $(\sim p \vee \sim q) $ both are false

i.e. $\sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q ) $ is true when $p$ and $q$ both are false then $\sim (p \rightarrow q ) $ is false and $(\sim p \vee \sim q)$ is true 

i.e. $\sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q)$ is false 

Hence $ \sim (p \rightarrow q ) \leftrightarrow (\sim p \vee \sim q) $ is neither tautology nor contradiction.