59 59 votes If $w, x, y, z$ are Boolean variables, then which one of the following is INCORRECT? $wx+w(x+y)+x(x +y) = x+wy$ $\overline{w \bar{x}(y+\bar{z})} + \bar{w}x = \bar{w} + x + \bar{y}z$ $(w \bar{x}(y+x\bar{z}) + \bar{w} \bar{x}) y = x \bar{y}$ $(w+y)(wxy+wyz) = wxy+wyz$ Digital Logic gatecse-2017-set2 digital-logic boolean-algebra normal + – khushtak 18.0k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments Sudo_404_Div commented Jun 1, 2025 reply Follow flag The most efficient method is to apply the By-Case method.... 2 2 replyShare Deepak Poonia commented Jul 1, 2025 reply Follow flag Detailed Video Explanation with multiple ways to solve: https://youtu.be/3Wj-BXY41MU?t=4876 2 2 replyShare Raj_Dev_Verma commented Jul 5 reply Follow flag Don't Solve all the option Always put the values this type of question Like put W=1 in all the option bot side Automatic A,B,D is match in RHS and LHS Hence Option C is Correct 0 0 replyShare Please log in or register to add a comment.
Best answer 50 50 votes Let us try to simplify (minimize) the expression given in each option Option - A: $wx+w(x+y)+x(x+y)=x+wy$ $wx + wx + wy + x$ $wx + wy + x$ $x (1+w) + wy$ $x + wy$ Option - B: $\overline{w \bar{x}(y+\bar{z})} + \bar{w}x = \bar{w} + x + \bar{y}z$ $\overline{w\bar{x}} + \overline{(y+\bar{z})} + \bar{w}x$ $\bar{w} + x + \bar{y}z + \bar{w}x$ $\bar{w} + \bar{w}x + x + \bar{y}z$ $\bar{w} + x + \bar{y}z$ Option - D: $(w+y)(wxy+wyz)=wxy+wyz$ $wxy + wyz + wxy + wyz$ $wxy + wyz$ Option A, B, D are matching fine. Hence, Option - C is the answer Arunav Khare answered Mar 29, 2017 • edited May 6, 2021 by Shiva Sagar Rao Arunav Khare comment Share Follow See all 7 Comments 7 7 Comments reply Show 4 previous comments ꧁༒☬ĿọŗԀ 🆂🅷🅸🆅🅰☬༒꧂ commented Nov 5, 2024 reply Follow flag @Abhrajyoti00 but choosing these values sometimes will land up wrong result like if we put $x=1,y=1,w=0$ so first will give same for LHS and RHS but its incorrect so instead of this We can use by case method where we can choose one variable value as either 1 or 0 and check for both the cases 0 0 replyShare Rishabh_Chaudhari commented Jul 19 reply Follow flag Common Identitis to remember 3 3 replyShare rishabh_pandey 1 commented Sep 3 reply Follow flag can i do like put anything or any specific trick 0 0 replyShare Please log in or register to add a comment.
17 17 votes Option C-(wx'(y+xz')+w'.x')y= (wx'y+wx'xz'+w'x')y= xx'=0 ..so Wx'y+w'x'y=x'y(w+w')=x'y Joker answered Feb 14, 2017 Joker comment Share Follow See all 2 Comments 2 2 Comments reply sushmita commented Mar 5, 2017 reply Follow flag HOW IS A CORRECT?? 0 0 replyShare Joker commented Mar 7, 2017 reply Follow flag 1st option given here is not the one they asked..the 1 st option is: wx+w(x+y)+x(x+y)=x+wy wx+wx+wy+x+xy wx+wy+x(1+y) wx+wy+x x(w+1)+wy x+wy 9 9 replyShare Please log in or register to add a comment.
6 6 votes LHS = (wx'(y + xz') + w'x')y = x'y RHS =xy' Option C- Put y=0 and see LHS !=RHS swap_it answered Feb 14, 2017 • edited Feb 14, 2017 by vijaycs swap_it comment Share Follow See all 7 Comments 7 7 Comments reply Show 4 previous comments Sanjay Sharma commented Feb 14, 2017 reply Follow flag x(w+1) =x.1=x ans is C 1 1 replyShare sushmita commented Mar 5, 2017 reply Follow flag BOTH A AND C ARE INCORRECT. 0 0 replyShare Lakshman Bhaiya commented Apr 12, 2017 reply Follow flag No only C is Not correct 1 1 replyShare Please log in or register to add a comment.
3 3 votes Just observe option C: on the LHS, there is no y' term, but on the RHS, there is y'. Unless we fully complement the LHS, we won't get y' on the RHS. However, we are not complementing the LHS, so option C is the only false one. advaith_xyz answered Jun 23, 2025 advaith_xyz comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Here is the Explanation and correct answer is option -C ✅ Prashant-G answered Jun 12 Prashant-G comment Share Follow 0 reply Please log in or register to add a comment.