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

↔ is equal to EX-NOR = xy+x’y’

now manipulate formulas and get

pqr’+pq’r+p’qr

as this expression has exactly two among p,q,r true

you will arrive at

answer (B)

Answer 2

I feel the answer is right but explanation for option (A) is wrong in selected answer.

Just think what does ”q if p” sentence format say in propositional logic. It means p->q. If you see it’s truth table,it means when ‘p’ is true , ‘q’ should be compulsorily TRUE.But when ‘p’ is false , ‘q’ can assume any value.

 

→ Now here ‘p’=”EXACTLY TWO OF p,q,r are TRUE.”

→ ‘q’ = “Given option is right”.

(A) is wrong bcoz when ‘q’ and ‘r’ are TRUE and ‘p’ is false , it assumes FALSE value, which shouldn’t happen as discussed above.

(B) is right .

(C) and (D) are false for same reason as (A)

Answer 3

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Answer 4

from all combinations of o's and 1's

try out 011

101

110

and b satisfies