Boolean Algebra · Digital Circuits · GATE ECE

Start Practice

Marks 1

1

For the Boolean function

$F(A, B, C, D) = \sum m(0,2,5,7,8,10,12,13,14,15)$,

the essential prime implicants are _________.

GATE ECE 2024
2

Select the Boolean function(s) equivalent to x + yz, where x, y, and z are Boolean variables, and + denotes logical OR operation.

GATE ECE 2022
3
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 = $$\overline A .\overline B .\overline C + \overline A .B.\overline C + A.\overline B .\overline C $$

where, $$\overline A $$, $$\overline B $$, and $$\overline C $$ are the complements of the respective variables. The product of sums (POS) form of the function F is
GATE ECE 2018
4
The Boolean expression (X+Y)(X+$$\overline Y $$)+($$\overline {(X\overline Y ) + \overline X } $$ simplifies to
GATE ECE 2014 Set 1
5
For an n - variable Boolean function maximum number of prime implicants is
GATE ECE 2014 Set 2
6
In the sum of products function f (x,y,z) = $$\sum {} $$m (2,3,4,5), the prime implicants are
GATE ECE 2013
7
The Boolean function Y=AB+CD is to be realized using only 2-input NAND gates. The minimum number of gates required is
GATE ECE 2007
8
The number of product terms in the minimized sum-of-product expression obtained through the following k-map is ( where , "d" denotes don't care states) GATE ECE 2006 Digital Circuits - Boolean Algebra Question 26 English
GATE ECE 2006
9
The Boolean expression for the truth table shown is
A B C D
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
GATE ECE 2005
10
The number of distinct Boolean expressions of 4 variables is
GATE ECE 2003
11
The Logical expression $$Y = A + \overline A B$$ is equivalent to
GATE ECE 1999
12
Two 2' s complement numbers having sign bits x and y added and the sign bit of the result is z. Then, the occurrence of overflow is indicated by the Boolean function.
GATE ECE 1998
13
The K-map for a Boolean function is shown in figure. The number of essential prime implicants for this function is GATE ECE 1998 Digital Circuits - Boolean Algebra Question 30 English
GATE ECE 1998
14
The number of Boolean functions that can be generated by n variable is equal to:
GATE ECE 1990

Marks 2

1
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 $$F = {m_0} + {m_2} + {m_3} + {m_5},$$ are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
GATE ECE 2017 Set 1
2
Following is the k-map of a Boolean function of five variable P, Q, R, S and X. The minimum for the function is GATE ECE 2016 Set 3 Digital Circuits - Boolean Algebra Question 10 English
GATE ECE 2016 Set 3
3
A function of Boolean variables X,Y and Z is expressed in terms of the min-terms as F(X, Y, Z)=$$\sum\limits_{}^{} {} $$m(1,2,5,6,7) Which one of the product of sums given below is equal to the funtion F(X, Y, Z)?
GATE ECE 2015 Set 2
4
The Boolean expression F(X, Y, Z)= $$\overline X Y\overline Z + X\overline {Y\,} \overline Z + XY\overline Z + XYZ$$ converted into the canonical product of sum (POS)from is
GATE ECE 2015 Set 1
5
Consider the Boolean function, F(w,z,y,z)=wy+ xy +$$\overline w \,xyz + \overline w \,\overline x y\, + xz + \,\overline {x\,} \,\overline y \,$$ $$\overline z $$ Which one of the following is the complete set of essential prime implicants?
GATE ECE 2014 Set 1
6
If X=1 in the logic equation $$\left[ {X + Z\left\{ {\overline Y + (\overline Z + X\overline {Y)} } \right\}} \right]$$ $$\left\{ {\overline X + \overline Z (X + Y)} \right\} = 1,$$ then
GATE ECE 2009
7
The Boolean expression Y= $$\overline A \,\overline B \,\overline C \,D + \overline A BC\overline D + A\overline {B\,} \overline C \,D + AB\overline C \,\overline D $$
GATE ECE 2007
8
The point p in the following figure is stuck- at-1. The output f will be GATE ECE 2006 Digital Circuits - Boolean Algebra Question 16 English
GATE ECE 2006
9
A Boolean function 'f' of two variables x and y is defined as follows: f(0,0)=f(0,1)=f(1,1)=1;f(1,0)=0 Assuming complements of x and y are not available, a minimum cost solution for realizing F using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost
GATE ECE 2004
10
The Boolean expression AC + B$$\overline C $$ is equivalent to
GATE ECE 2004
11
If the functions W, X, Y and Z are as follows

W= R+$$\overline P Q + \overline R $$ S
X = $$X = PQ\overline R \,\overline S + \overline P \,\overline Q \,\overline R \,\overline S + P\overline Q \,\overline R \,\overline S $$
Y = $$RS + \overline {OR + P\overline Q + \overline {PQ} } $$
Z = $$R + S + \overline {PQ + \overline {PQR} + P\overline {QS} } $$


GATE ECE 2003
12
For a binary half-subtractor having two inputs A and B, the correct set of Logical expressions for the output D(=Aminus B) and X(=Borrow) are
GATE ECE 1999
13
The minimized form of the logical expression ($$\overline A \,\overline B \,\overline C + B\overline C + \overline A B\overline C + \overline A BC + AB\overline C )$$
GATE ECE 1999

Marks 4

Marks 5

Marks 10

EXAM MAP
Medical
NEETAIIMS
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
Defence
NDA
Staff Selection Commission
SSC CGL Tier I
CBSE
Class 12