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43 votes

Let $f(w, x, y, z) = \sum {\left(0,4,5,7,8,9,13,15\right)}$. Which of the following expressions are NOT equivalent to $f$?

P: $x'y'z' + w'xy' + wy'z + xz$

Q: $w'y'z' + wx'y' + xz$

R: $w'y'z' + wx'y' + xyz+xy'z$

S: $x'y'z' + wx'y'+ w'y$

  1. P only
  2. Q and S
  3. R and S
  4. S only
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6 Answers

Best answer
59 votes
59 votes
K-map

 

So, minimized expression will be 

$xz  + w'y'z' + wx'y'$ which is Q. From the K-map, we can also get P and R. So, only S is NOT equivalent to $f$.

http://www.eecs.berkeley.edu/~newton/Classes/CS150sp98/lectures/week4_2/sld011.htm


Alternatively,

Go with Minterm representation of each option, (Note that order of $W,X,Y,Z$ should be preserved.)

Here, $x$ means do not care ($x$ takes value either $0$ or $1$)

$P : X' Y' Z' + W' X Y' + W Y' Z + X Z$
$\qquad = x X' Y' Z' + W' X Y' x + W x Y' Z + x X x Z$
$\qquad= x 000                  +             010x          +           1 x 01          +                        x 1 x 1$
$\qquad= 0000 + 1000        +      0100 + 0101      +       1001 + 1101      +    0101+0111+1101+1111$
$\qquad= 0 + 8                  +         4 + 5              +          9+13              +      5+7+13+15$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$Q : W' Y' Z'   + W X' Y'  + XZ$
$\qquad= W' x Y' Z'   + W X' Y' x + x X x Z$
$\qquad= 0 x 0 0             +        100 x       +           x 1 x 1$
$\qquad= 0000+0100       +    1000+1001   +    0101 + 0111+1101+1111$
$\qquad= 0+4                 +     8+9            +       5+7+13+15$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$R : W' Y' Z'  +  W X' Y'  + XYZ + X Y' Z$
$\qquad= W' x Y' Z'     +     W X' Y' x     +    x X Y Z    +    x X Y' Z$
$\qquad= 0 x 0 0           +  100 x         +    x 111   +    x 101$
$\qquad=  0000+ 0100   + 1000+1001  +  0111+1111   + 0101 + 1101$
$\qquad= 0+4              +     8+9          +   7+15         +     5+13$
$\qquad= ∑ m(0,4,5,7,8,9,13,15)$

$S : X' Y' Z'  +  W X' Y'  + W' Y$
$\qquad= x X' Y' Z'  +  W X' Y' x  + W' x Y x$
$\qquad= x 000            +     100 x          +      0 x 1 x$
$\qquad=  0000+1000   +   1000+1001    +    0010+0011+0110+0111$
$\qquad= 0+8              +     8+9             +   2+3+6+7$
$\qquad= ∑ m(0,2,3,6,7,8,9)$

Correct Answer: $D$

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40 votes
40 votes

Let me show u a very simple method 

Let w =1 ,x =1 ,y=1 ,z=1 then the value of f is 1

consider each statement

x'y'z' + w' x y' +w y' z +x z = 0.0.0 + 0.1.0 + 1.0.1 + 1.1 =1

w' y' z' +w x' z' y' +x z = 0.0.0 +1.0.0 +1.1  =1

w' y' z' +w x' y' +x y z +x y' z =0.0.0 +1.0.0 +1.1.1 +1.0.1 =1

x' y' z' +w x' y' + w' y = 0.0.0 + 1.0.0 + 0.1 =0

So statement (d) is false because w=1 x=1 y=1 z=1 the value of f is 0 .

(d) does not contain the essential Minterms .

xyz+wxy+wyz+xz

xyz+wxy+wyz+xz

4 votes
4 votes
putting x=y=z=w=1

s is false.

So option A is eliminated as S is not there.

Out of B and C ,

Q=R , so either both B and C correct or both wrong.

As this ques can have only one answer , so B and C are incorrect.

Only S ie D is the correct answer.
Answer:

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