44 44 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 $P.\bar{Q}$ $P.\bar{R}$ $P.\bar{Q} + R$ $P.\bar{R} + Q$ Digital Logic gatecse-2008 easy digital-logic boolean-algebra + – Kathleen 18.0k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments anon1 commented Nov 7, 2024 reply Follow flag Sir can you please explain the P = 0 part lill bit more and how options C, D getting eliminated ? 0 0 replyShare Deepak Poonia commented Jul 1, 2025 reply Follow flag @Skyquake._, Here: Detailed Video Explanation with multiple ways to solve: https://youtu.be/3Wj-BXY41MU?t=4425 3 3 replyShare Raj_Dev_Verma commented Jul 5 reply Follow flag Put the value of P=0 options C,D get eliminated put P=1, than Only Q' is left Hence Option A 0 0 replyShare Please log in or register to add a comment.
Best answer 59 59 votes 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 $ Keith Kr answered Sep 12, 2014 • edited Jul 25, 2019 by Arjun Keith Kr comment Share Follow See all 8 Comments 8 8 Comments reply Show 5 previous comments Rishav_Bhatt commented Jul 27, 2019 reply Follow flag Thanks a lot sir. 1 1 replyShare anon1 commented Nov 7, 2024 reply Follow flag Can anyone explain lill bit more on( the shortcut method) explained by Deepak Poonia Sir . 0 0 replyShare Deepak Poonia commented Aug 5, 2025 reply Follow flag @Skyquake._, Here it is explained: https://youtu.be/3Wj-BXY41MU?t=4425 1 1 replyShare Please log in or register to add a comment.
14 14 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 $1$ $1$ $0$ $0$ $SOP : PQ'$ KUSHAGRA गुप्ता answered Jan 11, 2020 KUSHAGRA गुप्ता comment Share Follow 0 reply Please log in or register to add a comment.
7 7 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' Raghav36 answered Jan 14, 2017 Raghav36 comment Share Follow See all 5 Comments 5 5 Comments reply set2018 commented Sep 1, 2017 reply Follow flag 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 1 replyShare Tuhin Dutta commented Sep 5, 2017 reply Follow flag can you please elaborate it 1 1 replyShare set2018 commented Sep 6, 2017 reply Follow flag 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 2 replyShare Rishi yadav commented Sep 15, 2017 reply Follow flag No it will not eliminate @set2018 there will be conflict btw option (a) and (b) 2 2 replyShare himgta commented Dec 30, 2018 reply Follow flag 1,1,0 can eliminate the conflict b/w a and b 0 0 replyShare Please log in or register to add a comment.
0 0 votes Just Multiply and put PP’ = 0. Surya_Dev Chaturvedi answered Jan 15, 2021 Surya_Dev Chaturvedi comment Share Follow 0 reply Please log in or register to add a comment.
