Tag: principle of mathematical induction

Questions Related to principle of mathematical induction

$(p \wedge q) \vee  \sim p$ is equivalent to 

  1. $\sim p \wedge q$

  2. $\sim p \vee q$

  3. $p \wedge q$

  4. $p \vee q$


Correct Option: A

In a certain code language, $'543'$ means 'give my water'; $'247'$ means 'water is life' and $'632'$ means 'enjoy my life'. Which of the following stands for 'enjoy' in that language?

  1. $7$

  2. $6$

  3. $2$

  4. $5$


Correct Option: A

$\sim (p \wedge q)\Rightarrow (\sim p)\vee (\sim p \vee q)$ is equal to

  1. $\sim p \vee q$

  2. $\sim p \wedge q$

  3. $p\vee \sim q$

  4. $p\wedge \sim q$


Correct Option: A

The equivalent of $(p \rightarrow \sim p) \vee (\sim p \rightarrow p)$ is 

  1. $p \vee \sim p$

  2. $T \rightarrow F$

  3. $T \leftrightarrow F$

  4. $p \wedge \sim p$


Correct Option: A

Identify which of the following statement is not equivalent to the others

  1. If $x$ is bass then $x$ is bad.

  2. Boss implies bad,

  3. Bad is necessary condition for bass.

  4. $x$ is boss iff $x$ is bad.


Correct Option: A

Either $p$ or $q$ is equivalent to:

  1. $p \vee q$

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

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

  4. none


Correct Option: A

Equivalent statement of ''If $x\in Q$, then $x\in T$'' is
$x\in Q$ is necessary for $x\in l$
$x\in l$ is sufficient for $x\in Q$
$z\in Q$ or $x\in l$
$x\in Q$ but $x\in l$ 

  1. $I$ & $II$

  2. $I,\ II$ & $IV$

  3. $III$

  4. $All$


Correct Option: B

Let  $P , Q , R$  and  $S$  be statements and suppose that  $P \rightarrow Q \rightarrow R \rightarrow P.$  If  $\sim S \rightarrow R,$  then

  1. $S \rightarrow \sim Q$

  2. $\sim Q \rightarrow S$

  3. $\sim S \rightarrow \sim Q$

  4. $Q \rightarrow \sim S$


Correct Option: B

$(p\rightarrow q)\leftrightarrow (q\vee \sim p)$ is - 

  1. Equivalent to $p\wedge q$

  2. Tautology

  3. Fallacy

  4. Neither tautology nor fallacy


Correct Option: A

Let S be a set of n persons such that:(i)any person is acquainted to exactly k other persons in s;(ii)any two persons that are acquainted have exactly $\displaystyle l $ common acquaintances in s;(iii)any two persons that are not acquainted have exactly m common acquaintances in S.Prove that $\displaystyle m\left ( n-k \right )-k\left ( k-1 \right )+k-m= 0.$

  1. $\displaystyle k\left ( k-1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

  2. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

  3. $\displaystyle k\left ( k-1+l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired zero.

  4. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k+1 \right )$ is equivalent to the desired one.


Correct Option: A
Explanation:

Let a be a fixed element os S. Let us count the triples (a, x, y) such that a, x are acquainted, x, y are acquainted and a, y are not acquainted. Because a isacquainted to exactly k other persons in S, x can be chosen in k ways and for fixeda and x, y can be chosen in $\displaystyle k-1-l $ ways. Thus the number of such triples is $\displaystyle k\left ( k-1-l  \right ).$ Let us count again, choosing y first. The number of persons not acquainted to a equals n-k-1, hence y can be chosen in n-k-1 ways. Because x is a commonacquaintance of a and y, it can be chosen in m ways, yielding a total of $\displaystyle m\left ( n-k-1 \right )$ triples. It is not difficult to see that the equality $\displaystyle k\left ( k-1-l  \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.