GATE Questions & Answers of Boolean Algebra

What is the Weightage of Boolean Algebra in GATE Exam?

Total 11 Questions have been asked from Boolean Algebra topic of Digital circuits subject in previous GATE papers. Average marks 1.64.

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

 

The Boolean expression $ \begin{array}{l}F(X,Y,Z)=(\overline XY\overline Z)+X\overline Y\overline Z+XY\overline Z+XYZ\\\end{array} $. converted into the canonical product of sum (POS) form is

A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as

F(X,Y,Z) = (1, 2, 5, 6, 7)

Which one of the product of sums given below is equal to the function F(X, Y, Z)?

The Boolean expression X+YX+Y+XY¯+X¯¯ simplifies to

Consider the Boolean function, F=w,x,y,z=wy+xy+w¯xyz+w¯ x¯y+xz+x¯y¯z¯   Which one of the following is the complete set of essential prime implicants?

For an n-variable Boolean function, the maximum number of prime implicants is

In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is +5 V, X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is

In the sum of products function f(X,Y,Z)=2,3,4,5,the prime implicants are

If X = 1 in the logic equation X+ZY¯+Z¯+XY¯X¯+Z¯X+Y=1, then

The Boolean function Y = AB + CD is to be realized using only 2-input NAND gates. The minimum number of gates required is:

The Boolean expression Y=A¯B¯C¯D+A¯BCD¯+AB¯C¯D+ABC¯D¯ can be minimized to