Boolean Algebra and Minimization

Question 1
The output expression for the Karnaugh map shown below is
A
Q\bar{R}+S
B
Q\bar{R}+\bar{S}
C
QR+S
D
QR+\bar{S}
GATE EE 2019   Digital Electronics
Question 1 Explanation: 


Output=Q\bar{R}+S
Question 2
Digital input signals A,B,C with A as the MSB and C as the LSB are used to realize the Boolean function
F=m_{0}+m_{2}+m_{3}+m_{5}+m_{7}, \; where \; m_{i}
denotes the i^{th} minterm. In addition, F has a don't care for m_1. The simplified expression for F is given by
A
\bar{A}\bar{C}+\bar{B}C+AC
B
\bar{A}+C
C
\bar{C}+A
D
\bar{A}C+BC+A\bar{C}
GATE EE 2018   Digital Electronics
Question 2 Explanation: 
Given, f=m_0+m_2+m_3+m_5+m_7 and m_1= don't care
Question 3
The output expression for the Karnaugh map shown below is
A
B\bar{D}+BCD
B
B\bar{D}+AB
C
B\bar{D}+ABC
D
B\bar{D}+ABC
GATE EE 2017-SET-1   Digital Electronics
Question 3 Explanation: 


ABC+B\bar{D}
Question 4
The Boolean expression AB + A\bar{C} + BC simplifies to
A
BC+A\bar{C}
B
AB+A\bar{C}+B
C
AB+A\bar{C}
D
AB+BC
GATE EE 2017-SET-1   Digital Electronics
Question 4 Explanation: 


BC+A\bar{C}
Question 5
The Boolean expression \overline{(a+\bar{b}+c+\bar{d})+(b+\bar{c})} simplifies to
A
1
B
\overline{a.b}
C
a.b
D
0
GATE EE 2016-SET-2   Digital Electronics
Question 5 Explanation: 
F=\overline{(a+\bar{b}+c+d)+(b+\bar{c})}
\;\;=\overline{(a+\bar{b}+c+d)}\cdot \overline{(b+\bar{c})}
\;\;=\bar{a}\cdot b\cdot \bar{c}\cdot d \cdot \bar{b} \cdot c
F=0
Question 6
The output expression for the Karnaugh map shown below is
A
A+\bar{B}
B
A+\bar{C}
C
\bar{A}+\bar{C}
D
\bar{A}+C
GATE EE 2016-SET-2   Digital Electronics
Question 6 Explanation: 


F=A+\bar{C}
Question 7
Consider the following Sum of Products expression, F.
F=ABC+\bar{A}\bar{B}C+A\bar{B}C+\bar{A}BC+\bar{A}\bar{B}\bar{C}
The equivalent Product of Sums expression is
A
F=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C)
B
F=(A+\bar{B}+\bar{C})(A+B+C)(\bar{A}+\bar{B}+\bar{C})
C
F=(A+\bar{B}+\bar{C})(A+\bar{B}+\bar{C})(A+B+C)
D
F=(A+\bar{B}+\bar{C})(A+B+\bar{C})(A+B+C)
GATE EE 2015-SET-2   Digital Electronics
Question 7 Explanation: 
The SOP form of F is ( shown in K-map)

So,POS form can be formed using '0' from the K-map.
POS=(A+\bar{B}+C)(\bar{A}+B+C)(\bar{A}+\bar{B}+C)
Question 8
f(A,B,C,D)=\PiM(0,1,3,4,5,7,9,11,12,13,14,15) is a maxterm representation of a Boolean function f(A,B,C,D) where A is the MSB and D is the LSB. The equivalent minimized representation of this function is
A
(A+\bar{C}+D)(\bar{A}+B+D)
B
(A\bar{C}D)+(\bar{A}BD)
C
\bar{A}C\bar{D}+A\bar{B}C\bar{D}+A\bar{B}\bar{C}\bar{D}
D
(B+\bar{C}+D)(A+\bar{B}+\bar{C}+D)(\bar{A}+B+C+D)
GATE EE 2015-SET-1   Digital Electronics
Question 9
The SOP (sum of products) form of a Boolean function is \Sigma (0,1,3,7,11), where inputs are A, B, C , D (A is MSB, and D is LSB). The equivalent minimized expression of the function is
A
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+D)
B
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{C})(\bar{C}+D)
C
(\bar{B}+C)(\bar{A}+C)(\bar{A}+\bar{B})(\bar{C}+\bar{D})
D
(\bar{B}+C)(A+\bar{B})(\bar{A}+\bar{B})(\bar{C}+D)
GATE EE 2014-SET-2   Digital Electronics
Question 9 Explanation: 
The 4 variable Boolean function is given in canonical sum of product form as,
f(A,B,C,D)=\Sigma (0,1,3,7,11)
As the options are given in the simplified product of sum form, we first convert the given function in canonical product of sum form, as under
f(A,B,C,D)=\Pi (2,4,5,6,8,9,10,12,13,14,15)
Now by plotting the above function on a 4 variable K-map (Maxterm map), we obtain the simplified expression of the function

f=(\bar{B}+C)\cdot (\bar{A}+C)\cdot (\bar{A}+\bar{B})\cdot (\bar{C}+D)
Question 10
Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter
A
1 0 0 0
B
1 0 0 1
C
0 0 1 1
D
1 1 0 0
GATE EE 2014-SET-2   Digital Electronics
Question 10 Explanation: 
Binary coded decimal counter counts from 0 to 9. So, 1100 is an initial state i.e. 12.
There are 10 questions to complete.