UGC NET CSE | June 2013 | Part 2 | Question: 39

Question

Which of the following shall be a compound proposition involving the propositions p, q and r, that is true when exactly two of the p, q and r are true and is false otherwise?

• A. $(p \vee q \wedge \rceil r) \vee (p \wedge q \wedge r) \wedge (\rceil p \wedge q \vee r)$
• B. $(p \wedge q \vee r) \wedge (p \wedge q \wedge r) \vee (\rceil q \wedge \rceil q \wedge \rceil r)$
• C. $(p \wedge q \wedge \rceil r) \vee (p \wedge \rceil q \wedge r) \vee (\rceil p \wedge q \wedge r)$
• D. $(p \vee r \wedge q) \vee (p \wedge q \wedge r) \vee (\rceil p \wedge q \wedge r)$

Comments

I am unable to find difference with the options, all seems same

Answer 1

(P⋂Q⋂R')    -> P and Q are true and R is False [exactly two are true]

U

(P⋂Q'⋂R)     -> P and R are true and Q is False [exactly two are true]

U

(P'⋂Q⋂R)     -> Q and R are true and P is False [exactly two are true]

Comments

y not and in (PUQUR')    -> P and Q are true and R is False [exactly two are true]

U
 here y not and?? y or??
(PUQ'UR)     -> P and R are true and Q is False [exactly two are true]

U

(P'UQUR

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Exactly you are right. Edited the solution

Answer 2

Catching the condition given in the question that "exactly two of the p, q and r are true and is false otherwise." 

by solving option (b) like   (1 1 1) ^ (1 1 1) v (0 0 0) = (1 1 1) also gives true but,

Option (c)  is correct because it matches the condition given in the question having exactly two of p,q and r are true.

Hence, option (c)  is correct.