# GATE Questions & Answers of Boolean Algebra

## What is the Weightage of Boolean Algebra in GATE Exam?

Total 12 Questions have been asked from Boolean Algebra topic of Digital circuits subject in previous GATE papers. Average marks 1.67.

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 $m_0,\;m_2,\;m_3$ and $m_5$ are minterms corresponding to the inputs A,B and C with A as the MSB and C as the LSB?

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) = $\sum$(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 $\left(\mathrm{X}+\mathrm{Y}\right)\left(\mathrm{X}+\overline{)\mathrm{Y}}\right)+\overline{\left(\mathrm{X}\overline{\mathrm{Y}}\right)+\overline{\mathrm{X}}}$ simplifies to

Consider the Boolean function, 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)=$\sum \left(2,3,4,5\right)$,the prime implicants are

If X = 1 in the logic equation $\left[X+Z\left\{\overline{Y}+\left(\overline{Z}+X\overline{Y}\right)\right\}\right]\left\{\overline{X}+\overline{Z}\left(X+Y\right)\right\}=1$, then

The Boolean expression $Y=\overline{A}\overline{B}\overline{C}D+\overline{A}BC\overline{D}+A\overline{B}\overline{C}D+AB\overline{C}\overline{D}$ can be minimized to