Tag: business maths

Questions Related to business maths

Which of the following is a contradiction?

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\vee q$

  2. $p\wedge q$

  3. $p\vee (\sim p)$

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

Show answer
Correct Option: D
Explanation:
$p$  $q$  $p\vee q$  $p\wedge q$
$ T$ $ T$ $ T$ $ T$
$ T$ $ F$ $ T$ $ F$
$ F$ $T$ $T$ $F$
$ F$ $F$ $ F$ $ F$
So $p\vee q$ and $p\wedge q$ are not contradiction 

Now check other options:

| $p$ |  $\sim p$ |  $p\vee (\sim p)$ |  $p\wedge (\sim p)$ | | --- | --- | --- | --- | |  $T$ | $ F$ | $ T$ | $ F$ | | $ F$ | $T$ | $T$ | $F$      |
Hence $p\wedge (\sim p)$ is contradiction 

Note: If a compound statement is always False , then it is called contradiction 

The proposition $(p\rightarrow \sim p)\wedge (\sim p\rightarrow q)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. a tautology and a contradiction

Show answer
Correct Option: C
Explanation:
$p$      $q$      $\sim p$      $p\rightarrow \sim p$     $\sim p \rightarrow q$     $(p\rightarrow \sim p)\wedge(\sim p\rightarrow q)$
T T    F      F       T         F
T F    F      F       T         F
F T    T      T       T         T
F F    T      T       F         F

The statement $(p-q)\rightarrow [(\sim p \rightarrow q)\rightarrow q]$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tautology

  2. equivalent to $\sim p \rightarrow q$

  3. equivalent to $p\rightarrow \sim q$

  4. a fallacy

Show answer
Correct Option: A

Which of the following proposition 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
Explanation:

Contradiction is a preposition which is always false(F).
Here $\sim \equiv negation$ and $\wedge \equiv AND$
Let $ x=(\sim p \wedge q) \wedge (\sim q)$
If $p=F$ and $q=F$ then $x=F$
If $p=F$ and $q=T$ then $x=F$
If $p=T$ and $q=F$ then $x=F$
If $p=T$ and $q=T$ then $x=F$
Hence option (c) is correct

Which of the following is not true (where $p, q$ and $r$ take truth values and $t$ is a tautology, $c$ is a contradiction)

logical equivalence mathematical logic discrete mathematics business maths maths

  1. $p\wedge p\equiv p$

  2. $p\vee t=t$

  3. $p\wedge c= p$

  4. $p\vee (q\wedge r)=(p\vee q)\vee (p\vee r)$

Show answer
Correct Option: A

$p,q,r$ are $3$ statement such that $(p\rightarrow q)\wedge (q\rightarrow r)\Rightarrow (p\rightarrow r)$ is 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Tautology

  2. Contradiction

  3. $P\wedge q$

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

Show answer
Correct Option: A

The statement $[p \wedge (p \rightarrow q)]\rightarrow q$,is :

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a fallacy

  2. a tautology

  3. neither a fallacy nor a tautology

  4. not a compound statement

Show answer
Correct Option: B
Explanation:

The statement

$[p \wedge (p \rightarrow q)] \rightarrow q$
a tautology 

$p,q,r$ are $3$ statement such that $(p \rightarrow q)\wedge (q \rightarrow r)\Rightarrow (P \rightarrow r)$ is

logical equivalence mathematical logic discrete mathematics business maths maths

  1. Tautology

  2. Contradiction

  3. $P \wedge q$

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

Show answer
Correct Option: A

Which of the following is a tautology?

logical equivalence mathematical logic discrete mathematics business maths maths

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

  2. $p\wedge c$

  3. $p\vee t$

  4. $p\wedge p$

Show answer
Correct Option: A

The proposition $p\vee (\sim p\vee q)$ is a 

logical equivalence mathematical logic discrete mathematics business maths maths

  1. a tatutology

  2. a contradiction

  3. Logically equivalent to $p$ & $q$

  4. both $1$ & $2$

Show answer
Correct Option: A