Tag: business maths

Questions Related to business maths

The statement $\sim (p\rightarrow \sim q)$ is equivalence to ___________.

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

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

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

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

Show answer
Correct Option: C
Explanation:

$\sim\left({p} \rightarrow \sim{q} \right)$

We know that,
               $\sim\left({p} \rightarrow {q} \right)={p}\wedge\sim{q}$
          $\Rightarrow\sim\left({p}\rightarrow\sim{q}\right)=\sim{p}\wedge\sim\left(\sim{q}\right)$
                                    $=\sim{p}\wedge{q}$
Hence, $\left(\sim{p}\wedge{q}\right)$ is the correct answer.


Which of the following is always true?

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. $\sim(p\rightarrow q) \equiv \sim p \wedge q$

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

  3. $\sim (p \implies q ) \equiv (p \land \sim q )$

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

Show answer
Correct Option: C
Explanation:

$p \implies q \equiv \sim p \lor q  $
$\therefore \sim (p \implies q ) \equiv \sim (\sim p \lor q )$
$\therefore \sim (p \implies q ) \equiv (p \land \sim q )$

Which of the following is/are false?

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. $p\rightarrow q\equiv\sim p\rightarrow\sim q$

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

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

  4. $\sim (p\leftrightarrow q) \equiv(\sim(p\rightarrow q))\wedge\sim(q\rightarrow p)$

Show answer
Correct Option: A,B,D
Explanation:

We know that:
$p\rightarrow q \equiv \sim q\rightarrow \sim p$    {By logical equivalences }    
Hence $A$ is false


Now for option $B$
$\sim (p \ \rightarrow \ \sim q)$ $\equiv$ $\sim (\sim p\vee \sim q)=p\wedge q$   [By logical Equivalences ]
Hence $B$ is false

Now for option $C$
$\sim (\sim p\rightarrow \sim q)$ $\equiv \sim (p  \vee \sim q) $  $\equiv \sim p\wedge q$  [By Logical Equivalences]
Hence $C$ is true

Now for option $D$
$\sim (p\leftrightarrow q)$ $\equiv \sim ((p\rightarrow q)\wedge (q\rightarrow p))$ $\equiv \sim (p\rightarrow q)\vee \sim (q\rightarrow p)$
Hence $D$ is false                        [By logical Equivalences]

Which of the following is logically equivalent to $\displaystyle \sim \left (\sim p\rightarrow q\right )$?

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. $\displaystyle p\wedge q$

  2. $\displaystyle p\wedge \sim q$

  3. $\displaystyle \sim p\wedge q$

  4. $\displaystyle \sim p\wedge \sim q$

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

The values in column 6 and column 10 are same.

Hence, option D is correct.

The dual of the following statement "Reena is healthy and Meena is beautiful" is

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. Reena is not beaufiful and Meena is not healthy.

  2. Reena is not beautiful or Meena is not healthy.

  3. Reena is not healthy or Meena is not beautiful.

  4. None of these.

Show answer
Correct Option: C
Explanation:

Let $p$ denote the statement "Reena is healthy" 


and $q$ denote the statement "Meena is beautiful"

Now the given statement is $p\wedge q$

Now the Dual of this statement will be obtained by replacing $\vee$ by 

$\wedge$ and $\wedge$ by $\vee$ and inversing the true value of the statement.

So the Dual of $p\wedge q$ will be $\sim p\vee \sim q$

The statement $\sim p$ will be "Reena is not healthy"

The statement $\sim q$ will be "Meena is not beautiful"

So the dual statement will be $\sim p\vee \sim q$ or "Reena is not healthy or Meena is not beautiful."

The statement "If $2^2 = 5$ then I get first class" is logically equivalent to

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. $2^2 = 5$ and I do not get first class

  2. $2^2 = 5$ or I do not get first class

  3. $2^2 \neq 5$ or I get first class

  4. None of these.

Show answer
Correct Option: C
Explanation:

There can be two cases
$2^{2}=5$ $\rightarrow$ first class.
$2^{2}\neq 5$\rightarrow not a first class.
Hence logically equivalent statement will be 
$2^{2}=5$ or $2^{2}\neq 5$ but $2^{2}=5$ statement is equivalent to getting first class.
Hence
First class or $2^{2}\neq 5$.

The statement "If $2^2 = 5$ then I get first class" is logically equivalent to

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

  1. $2^2 = 5$ and I donot get first class

  2. $2^2 = 5$ or I do not get first class

  3. $2^2 \neq 5$ or I get first class

  4. None of these

Show answer
Correct Option: C
Explanation:

Obviously, ${ 2 }^{ 2 }\neq 5$, then the statement will be ${ 2 }^{ 2 }\neq 5$ or $I$ get first class.

Logically equivalent statement to $p \leftrightarrow  q$ is

business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

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

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

  4. none of these

Show answer
Correct Option: A
Explanation:
 $p$  $q$  $p\leftrightarrow q$
 T  T  T
 T  F  F
 F  T  F
 F  F  T
 $p$  $q$  $p\rightarrow q$  $q\rightarrow p$ $\left( p\longrightarrow q \right) \wedge \left( q\longrightarrow p \right) $ $p\wedge q$  $\left( p\wedge q \right) \vee \left( q\longrightarrow p \right) $ $q\vee p$  $\left( p\wedge q \right) \longrightarrow \left( q\vee p \right) $ 
 T  T  T  T  T  T  T  T  T
T   F  F  T  F  F  T  T  T
 F  T  T  F  F  F  F  T  T
 F F   T  T  T  F  T  F  T

In a poisson distribution, the variance is $m$ . The sum of terms in odd places in the distribution is 

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $e^{-m}$

  2. $e^{-m} \cos \, h \, (m)$

  3. $e^{-m} \sin \, h \, (m)$

  4. $e^{-m} \cot \, h \, (m)$

Show answer
Correct Option: B
Explanation:

$e^{-m}\cos h(m)$


 lf the mean is $\lambda$ and the variance is $\sigma^{2}$ in a Poisson distribution, then

business maths random variables and probability distribution poisson distribution poisson distrubution probability distributions

  1. $\displaystyle \lambda=\frac{1}{2}\sigma^{2}$

  2. $\displaystyle \sigma^{2}=\frac{1}{2}\lambda$

  3. $\lambda=\sigma^{2}$

  4. $\sigma^{2}=\lambda^{2}$

Show answer
Correct Option: C
Explanation:

For a Poisson distribution, mean and variance are equal .
i.e.$\lambda = \sigma^2$