in Digital Logic
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27 votes

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

in Digital Logic
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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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4 Answers

41 votes
 
Best answer
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 $
edited by

6 Comments

Sir how you get PQ'R'  in third step?
1
For those who don't like POS form to do simplification
Take complement->simplify in sop->take complement again to get the answer.
2

How to solve this using k-map?

1
Mistake in third step
1
Fixed now
1
Thanks a lot sir.
1
6 votes

(p+q')(p.q'+p.r)(p'r'+q')

=( p'(q')' )'.{( (pr)'.(pq')' )'.( (p'r')'(q')' )'}

={(p'q)'.( (pr)'.(pq') )'} + (p'r')'q )'

=( p'q+(pr)'.(pq')' + (p'r')'q )'

=( p'q+ (p'+r')(p'+q)+(p+r)q )'

=(p'q +p'+ p'r'+qp'+qr'+pq+qr)'

=( p'(q+1) +p'r+ q(p'+p)+q(r'+r) )'

=(p'(1+r)+q+q)'

=(p'+q)'

=p.q'

 

5 Comments

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
1
can you please elaborate it
1
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
2
No it will not eliminate @set2018 there will be conflict btw option (a) and (b)
2
1,1,0 can eliminate the conflict b/w a and b
0
4 votes

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

 

Map these POS in K-Map

$P$     $QR\rightarrow$

$\downarrow$ 

00 01 11 10
0 $0$ $0$ $0$ $0$
1 $1$ $1$ $0$ $0$

$SOP : PQ'$

0 votes
Just Multiply and put PP’  = 0.
Answer:

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