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If $P, Q, R$ are Boolean variables, then

$(P + \bar{Q}) (P.\bar{Q} + P.R) (\bar{P}.\bar{R} + \bar{Q})$ simplifies to

1. $P.\bar{Q}$

2. $P.\bar{R}$

3. $P.\bar{Q} + R$

4. $P.\bar{R} + Q$

### 1 comment

If we take out p common from second bracket, function will be 0 when p=0

Take cases of p=1

Now when p=1 and because of third bracket when Q =1 function will be 0.

Hence we are left only with two min terms (P,Q,R) = 100 + 101.

Try doing such examples using truth table in exam won't take much time and increases accuracy.

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Ans is (A) $P \bar Q$

$(P + \bar Q)(P \bar Q + PR)(\bar P \bar R + \bar Q)$

$= (PP \bar Q + PPR + P \bar Q + P \bar QR)(\bar P \bar R + \bar Q)$

$= (P \bar Q + PR + P \bar Q + P \bar QR)(\bar P \bar R + \bar Q)$

$= P \bar Q + P \bar QR$

$= P \bar Q$
by

Sir how you get PQ'R'  in third step?
For those who don't like POS form to do simplification
Take complement->simplify in sop->take complement again to get the answer.

How to solve this using k-map?

Mistake in third step
Fixed now
Thanks a lot sir.

## =p.q'

by

simple method

assume p= q=r=1 calculate function value

check for same given option

some options will automatically eliminate .if same put another value
take p=q=r=1

funcion will be 0

now put this p,q,r into given options c and d will not satisfy (eliminate )

for option a and b re evaluate the function by p=1 q= 0 r=1  function value= 1

put these into option a and b

it will eliminate b
No it will not eliminate @set2018 there will be conflict btw option (a) and (b)
1,1,0 can eliminate the conflict b/w a and b

$(P+Q').P.(Q'+R).(Q'+P').(Q'+R')$

Map these POS in K-Map

 $P$     $QR\rightarrow$ $\downarrow$ 00 01 11 10 0 $0$ $0$ $0$ $0$ 1 $1$ $1$ $0$ $0$

$SOP : PQ'$

Just Multiply and put PP’  = 0.