GATE CSE 2014 Set 1 | Question: 53

Question

Which one of the following propositional logic formulas is TRUE when exactly two of $p,q$ and $r$ are TRUE?

• A. $(( p  \leftrightarrow q)  \wedge  r)  \vee  (p \wedge q \wedge  \sim r)$
• B. $( \sim (p \leftrightarrow q) \wedge r)\vee (p \wedge q \wedge \sim r)$
• C. $( (p \to q) \wedge r) \vee (p \wedge q \wedge \sim r)$
• D. $(\sim (p \leftrightarrow q) \wedge r) \wedge (p \wedge q \wedge \sim r) $

Comments

https://gateoverflow.in/1933/gate2014-1-53

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B is correct one ! Here is better clarification:

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Combination which gives False on exactly two of p,q,r are true.

Option A ---  (P,Q,R) = (F,T,T)

Option C---  (P,Q,R) = (T,F,T)

Option D---  (P,Q,R) = (T,T,F)  or (T,F,T) or (F,T,T)

Now, option B these combination give true..

Option B ---  (P,Q,R) = (T,T,F)  or (T,F,T) or (F,T,T)

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Ambiguous Question

Answer 1

Correct answer is option B. 

Answer 2

  Consider 4,6,7 cases and check options.

$B)\\ (\sim (p\leftrightarrow q)\wedge r) \vee (p \wedge q \wedge \sim r)\\$

$4^{th}$ case

$=(\sim (F\leftrightarrow T)\wedge T) \vee (F \wedge T \wedge \sim T)\\ =(\sim F\wedge T) \vee F\\ =(T\wedge T) \vee F\\ =T \vee F\\ =T$

 

  Similarly Check $6^{th} \&\ 7^{th}$ cases.

  Correct Answer: B

Answer 3

In option (B) : 
We assume ‘p’ and ‘q’ to be TRUE. Therefore, ‘r’ is false because exactly two of ‘p’, ‘q’, and ‘r’ can be TRUE. 
p <--> q means if both ‘p’ and ‘q’ have same values then p <--> q is TRUE. So, ((p <--> r ) and r) evaluates to be FALSE. Therefore, ~((p <--> r ) and r) is TRUE. 
Here, two sub- expression are connected via OR operator and one of the sub-expression is TRUE. So, the complete expression becomes TRUE. 
 
Therefore, option (B) evaluates to be TRUE. 

Answer 4

If you check for 110

A,B, C all are true.

If you check for 111,

A, C are true, while B is false.

So, B is the answer.