# Boolean Algebra

 Question 1
A function F(A,B,C) defined by three Boolean variables A, B and C when expressed as sum of products is given by

$F=\bar{A}\cdot \bar{B} \cdot \bar{C} + \bar{A}\cdot B \cdot \bar{C} + A\cdot \bar{B} \cdot \bar{C}$

where,$\bar{A},\bar{B} \;and \; \bar{C}$ are complements of the respective variable. The product of sums (POS) form of the function F is
 A $F=(A+B+C)\cdot (A+\tilde{B}+C)\cdot (\bar{A}+B+C)$ B $F=(\bar{A}+\bar{B}+\bar{C})\cdot (\bar{A}+B+\bar{C})\cdot (A+\bar{B}+\bar{C})$ C $F=(A + B + \bar{C}) \cdot (A + \bar{B} + \bar{C} ) \cdot (\bar{A} + B + \bar{C}) \cdot$ $(\bar{A}+\bar{B}+C) \cdot (\bar{A}+\bar{B}+\bar{C})$ D $F=(\bar{A} + \bar{B} + C) \cdot (\bar{A} + B + C) \cdot$ $(A + B + \bar{C}) \cdot (A+B+C)$
GATE EC 2018   Digital Circuits
Question 1 Explanation:
\begin{aligned} F(A, B, C, D) &=\bar{A} \bar{B} \bar{C}+\bar{A} B \bar{C}+A \bar{B} \bar{C} \\ &=\Sigma m(0,2,4)=\Pi M(1,3,5,6,7) \\ =&(A+B+\bar{C})(A+\bar{B}+\bar{C})(\bar{A}+B+\bar{C}) \\ &(\bar{A}+\bar{B}+C)(\bar{A}+\bar{B}+\bar{C}) & \end{aligned}
 Question 2
Which one of the following gives the simplified sum of products expression for the Boolean function $F=m_{0}+m_{2}+m_{3}+m_{5}$, where $m_{0},m_{2},m_{3},m_{5}$, are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
 A $\bar{A}B+\bar{A}\bar{B}\bar{C}+A\bar{B}C$ B $\bar{A}\bar{C}+\bar{A}B+A\bar{B}C$ C $\bar{A}\bar{C}+A\bar{B}+A\bar{B}C$ D $\bar{A}BC+\bar{A}\bar{C}+A\bar{B}C$
GATE EC 2017-SET-1   Digital Circuits
Question 2 Explanation:
Given Boolean function is,
$F=m_{0}+m_{2}+m_{3}+m_{5}$
It can be minimized by using K-map as given below.

$F=\bar{A} \bar{C}+\bar{A} B+A \bar{B} C$
 Question 3
Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sum-of-product (SOP) expression for the function is
 A $\bar{P}\bar{Q}S\bar{X}+P\bar{Q}S\bar{X}+Q\bar{R}\bar{S}X+QR\bar{S}X$ B $\bar{Q}S\bar{X}+Q\bar{S}X$ C $\bar{Q}SX+Q\bar{S}\bar{X}$ D $\bar{Q}S+Q\bar{S}$
GATE EC 2016-SET-3   Digital Circuits
Question 3 Explanation:

$\therefore$ minimum sum of product expression of the function is
$=\bar{Q}S\bar{X}+Q\bar{S}X$
 Question 4
A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as
$F(X,Y,Z)=\sum (1,2,5,6,7)$
Which one of the product of sums given below is equal to the function F(X,Y,Z)?
 A $(\bar{X}+\bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (X+\bar{Y}+\bar{Z})$ B $(X+Y+Z)\cdot (X+ \bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)$ C $(\bar{X}+\bar{Y}+Z)\cdot (\bar{X}+Y+\bar{Z})\cdot (X+\bar{Y}+Z)$ $\cdot (X+Y+\bar{Z})\cdot (X+Y+Z)$ D $(X+Y+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (\bar{X}+Y+\bar{Z})$ $\cdot (\bar{X}+\bar{Y}+Z)\cdot (\bar{X}+\bar{Y}+\bar{Z})$
GATE EC 2015-SET-2   Digital Circuits
Question 4 Explanation:
\begin{aligned} F(X, Y, Z) &=\Sigma(1,2,5,6,7) \\ &=\pi[0,3,4] \\ =(X+Y+Z)&(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z) \end{aligned}
 Question 5
The Boolean expression $f(X,Y,Z)=\bar{X}Y\bar{Z}+X\bar{Y}\bar{Z}+XY\bar{Z}+XYZ$ converted into the canonical product of sum (POS) form is
 A $(X+Y+Z)(X+Y+\bar{Z})(X+\bar{Y}+\bar{Z})(\bar{X}+Y+\bar{Z})$ B $(X+\bar{Y}+Z)(\bar{X}+Y+\bar{Z})(\bar{X}+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z})$ C $(X+Y+Z)(\bar{X}+Y+\bar{Z})(X+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z})$ D $(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z)(\bar{X}+\bar{Y}+Z)(X+Y+Z)$
GATE EC 2015-SET-1   Digital Circuits
Question 5 Explanation:
$F(X, Y, Z)=\bar{X} Y \bar{Z}+X \overline{Y Z}+X Y \bar{Z}+X Y Z$

$F=(X+Y+Z)(X+Y+\bar{Z})(X+\bar{Y}+\bar{Z})(\bar{X}+Y+\bar{Z})$
 Question 6
For an n-variable Boolean function, the maximum number of prime implicants is
 A 2(n-1) B n/2 C $2^{n}$ D $2^{n-1}$
GATE EC 2014-SET-2   Digital Circuits
Question 6 Explanation:
For a variable Boolean function the maximum number of prime implicants is $2^{(n- l)}.$
 Question 7
Consider the Boolean function, $F(w,x,y,z)=wy+xy+\bar{w}xyz+\bar{w}\bar{x}y+xz+\bar{x}\bar{y}\bar{z}$. Which one of the following is the complete set of essential prime implicants ?
 A $w,y,xz,\bar{x}\bar{z}$ B w,y,xz C $y,\bar{x}\bar{y}\bar{z}$ D $y,xz,\bar{x}\bar{z}$
GATE EC 2014-SET-1   Digital Circuits
Question 7 Explanation:
$\begin{array}{l} f(w, x, y, z) \\ =w+x y+\bar{w} x y z+\bar{w} \bar{x} y+x z+\bar{x} \bar{y} \bar{z} \end{array}$

$y, x z, \bar{x} \bar{z}$
From the above K-map, it can be seen that, the set of essential prime-implicants is
$y, x z, \bar{x} \bar{z}$
 Question 8
The Boolean expression $(X+Y)(X+\bar{Y})+\overline{(X\bar{Y}+\bar{X})}$ simplifies to
 A X B Y C XY D X+Y
GATE EC 2014-SET-1   Digital Circuits
Question 8 Explanation:
\begin{aligned} \text{Let}\quad F &=(X+Y)(X+\bar{Y})+(\overline{X \bar{Y}+\bar{X}}) \\ F &=X+X \bar{Y}+X Y+(\overline{X \bar{Y}} \cdot \bar{X}) \\ &=X+(\bar{X}+\eta) \cdot X \\ &=X+X Y \\ &=X \end{aligned}
 Question 9
In the sum of products function f (X, Y, Z) =$\sum$(2, 3, 4, 5), the prime implicants are
 A $\bar{X}Y,X\bar{Y}$ B $\bar{X}Y,X\bar{Y}\bar{Z},X\bar{Y}Z$ C $\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}$ D $\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}\bar{Z},X\bar{Y}Z$
GATE EC 2012   Digital Circuits
Question 9 Explanation:
$f(X,Y,Z)=\sum(2,3,4,5)$

$f(X, Y, Z)=X \bar{Y}+\bar{X} Y$
So prime implicants are $X\bar{Y}$ and $\bar{X}Y .$
 Question 10
If X=1 in logic equation
$[X+Z\{\bar{Y}+(\bar{Z}+X\bar{Y})\}]\{\bar{X}+\bar{Z}(X+Y)\}=1$, then
 A Y=Z B $Y=\bar{Z}$ C Z=1 D Z=0
GATE EC 2009   Digital Circuits
Question 10 Explanation:
$[X+Z\{\bar{Y}+(\bar{Z}+X\bar{Y})\}]\{\bar{X}+\bar{Z}(X+Y)\}=1$
\begin{aligned} \Rightarrow Z(1+Y)&=1\\ \Rightarrow \quad\bar{Z}&=1\\ \Rightarrow \quad Z&=0 \end{aligned}
There are 10 questions to complete.