GATE CSE 2025 | Set 2 | Question: 33

Question

​​​​Given the following Karnaugh Map for a Boolean function $F(w, x, y, z)$ :

Which one or more of the following Boolean expression(s) represent(s) $F$ ?

• A. $\bar{w} \bar{x} \bar{y} \bar{z}+w \bar{x} \bar{y} \bar{z}+\bar{w} \bar{x} y \bar{z}+w \bar{x} y \bar{z}+x z$
• B. $\bar{w} \bar{x} \bar{y} \bar{z}+\bar{w} \bar{x} y \bar{z}+w \bar{x} y z+x z$
• C. $\bar{w} \bar{x} \bar{y} \bar{z}+w \bar{x} \bar{y} \bar{z}+w \bar{x} \bar{y} z+x z$
• D. $\bar{x} \bar{z}+x z$

Comments

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$$\textbf{Hint}$$

• $\text{Using K-map minimization we get }\overline{x}\text{ }\overline{z}\text{ + }xz.$

 

• $\text{B and option C can be eliminated since you need atleast 4, four variable terms to cover a four varibale cube using 1 variable cubes.}$

 

• $\text{Option A can be evaluated as a correct option.}$

 

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by first see on option then option D is correct

like xz + x'z'

but its multiple then see again in options
the term xz is as it is but x'z' is explore with w,y in option A

option A,D are correct

 

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firstly we will find the mininmal sop by grouping the vsible quads in the k-map which will give us optD as an correct answer and next to check other option we can use two methods :

method 1- calculate min term from the map it will be 8 min terms hence option b&d got eleminated as both have 3 direct minterms + XZ a product term that will result in 4 other min terms i.e a total of 7 so we need to only map opt A min term

method 2 - its visible directly from the k-map and again after finding MSOP that 8 minterm so optB&C got eliminated and we can minimize optA by using boolean algebra like this

W'X'Y'Z'+WX'Y'Z'+W'X'YZ' +XZ

X'Y'Z'(W'+W) + W'X'YZ' + XZ

X'Z'(Y'+Y) + XZ

X'Z' + XZ

 

Answer 1

You Can Solve these type of Questions while Expending in Canonical form and Find the SOP 🫵👇

Comments

tnx bro

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in option c , it must be (0,8,9,5,7,13,15) in place of 10 it must be 9

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Done ✅ Sais

Answer 2

By Solvjng the K-map

               Y= x'z' + xz (Minimized equation) Option D matches

Now every other option has xz so some alteration with x'z' so expand it.

Y= x'z'(y'+y) + xz

= x'z'y' + x'z'y + xz

No option matches so  expand it further

= x'z'y (w'+ w) +x'z'y'(w' + w) + xz

=x'z'yw'+ x'z'yw + x'z'y'w' + x'z'y'w + xz

Option A matches .

 Hence Answer is : A;D

 

Answer 3

 Option A and D are correct.

 

Comments

This is 2 mark q??

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Do not know. Seems to be.

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option A & D is correct

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Option B and C can be easily eliminated as XZ covers 4 1's there are remaining 4 1's for which 4 min terms are required and in both B and C only 3 min terms is given apart from XZ.

Answer 4

F(w, x, y, z) = xz + x̄z̄ so option D is one of the correct option

Now, let's check other options (Term xz is given in each options so let's see other minterms)
Apart from "xz" we require four '4-literals' minterms and in option B and C we have 3 minterms apart from xz so option B and C is not possible

Now, let's check option A
F(w, x, y, z) = w'x'y'z' + wx'y'z' + w'x'yz' + wx'yz' + xz
So, option A is also correct option

Hence, answer = A, D

Answer 5

Solution by Deepak sir.
https://www.youtube.com/live/T_LcdP0XjHo?si=7jwYmvb_7pWIs5_N&t=4222

Answer 6

I think This method is more suitable for exam POV
Option B and C can be easily eliminated coz, 4 lateral minterm corresponds to just 1 cell (Opt B and C have 3), and 2 lateral minterm corresponds to 4 cell filling in a 4 Variable ftn Kmap, which adds upto 7 cells, but obv given Kmap has 8 cells filled.
Remaining option D, can simply be derived from the given Map, and as for Option A, just make a rough Kmap, and and fill it accrding to given Minterms, which comes out to be identical as the given Question