NIMCET Boolean Algebra Previous Year Questions (PYQs) – Page 1 of 4

Given expression is

$(x+y'+z')(x+y'+z)(x+y+z')$

Using the identity

$(x+A)(x+B)=x+AB$

First take

$(x+y'+z')(x+y'+z)$

Here,

$A=y'+z'$ and $B=y'+z$

So,

$(x+y'+z')(x+y'+z)=x+(y'+z')(y'+z)$

Now,

$(y'+z')(y'+z)=y'+z'z$

Since

$z'z=0$

So,

$(y'+z')(y'+z)=y'$

Now the expression becomes

$(x+y')(x+y+z')$

Again using the identity,

$(x+A)(x+B)=x+AB$

Here,

$A=y'$ and $B=y+z'$

So,

$(x+y')(x+y+z')=x+y'(y+z')$

$=x+y'y+y'z'$

Since

$y'y=0$

So,

$=x+y'z'$

Therefore, the simplified value is

$x+y'z'$

We have $ (A \oplus B) \land (B \to C) = (A \oplus B)\land(\lnot B \lor C)$.

• If $B=0$: $(\lnot B \lor C)=1$ always, so output $=A\oplus 0=A$. Hence rows with $A=1$ work: $(A,B,C)=(1,0,0)$ and $(1,0,1)$.
• If $B=1$: need $C=1$ (for $B\to C$) and $A\oplus 1=1\Rightarrow A=0$. Hence $(A,B,C)=(0,1,1)$.

Rows with output 1: $(1,0,0)$, $(1,0,1)$, $(0,1,1)$.