Logic functions and Minimization
Question 1 |
Which one of the following expressions does NOT represent exclusive NOR of x and y?
xy+x'y' | |
x⊕y' | |
x'⊕y | |
x'⊕y' |
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Question 1 Explanation:
By Definition of XNOR,
So Option-A is correct.
Also by Definition of XOR,
Option-B is
So Option-B is also correct.
Option-C is
Option-C is also correct.
Option-D x'⊕y' = x''y' + x'y'' = xy' + x'y = x⊕y ≠ x⊙y
Therefore option (D) is false.
This explanation is provided by Chirag Manwani.




Question 2 |
b'd' | |
b'd' + b'c' | |
b'd' + a'b'c'd' | |
b'd' + b'c' + c'd' |
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Question 2 Explanation:
There are two prime implicants in the following K-Map-
Prime Implicant higlighted in Green =
Prime Implicant higlighted in Orange =
So the Boolean expression is-
Therefore option (B) is correct.
This explanation is provided by Chirag Manwani.




Question 3 |
A | |
B | |
C | |
D |
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Question 3 Explanation:
All options except D produce XOR as described below :

Question 4 |
The simplified SOP (Sum Of Product) form of the boolean expression (P + Q' + R') . (P + Q' + R) . (P + Q + R') is
(P'.Q + R') | |
(P + Q'.R') | |
(P'.Q + R) | |
(P.Q + R) |
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Question 4 Explanation:
See following :
(P+Q'+R').(P+Q'+R).(P+Q+R') =
From the K-map, POS form is :
P + Q'.R'


Question 5 |
The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is
m2 + m4 + m6 + m7 | |
m0 + m1 + m3 + m5 | |
m0 + m1 + m6 + m7 | |
m2 + m3 + m4 + m5 |
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Question 5 Explanation:
K-map,

= PQ + QR’ + PR’ = PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’ = PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ = PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) = m2 + m4 + m6 + m7Option (A) is correct.
Question 6 |
What is the minimum number of gates required to implement the Boolean function (AB+C)if we have to use only 2-input NOR gates?
2 | |
3 | |
4 | |
5 |
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Question 6 Explanation:
AB+C = (A+C)(B+C) = ((A+C)' + (B+C)')'
So, '3' 2-input NOR gates are required.
Question 7 |
In the Karnaugh map shown below, X denotes a don't care term. What is the minimal form of the function represented by the Karnaugh map?
A)
B)
C)
D)
![]()
A | |
B | |
C | |
D |
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Question 8 |
Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit


A)m(4, 6) B)
m(4, 8) C)
m(6, 8) D)
m(4, 6, 8)
A | |
B | |
C | |
D |
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Question 8 Explanation:
From logic diagram we have f=f1.f2+f3
f=m(4,5,6,7,8).f2+m(1,6,15)----(1)
from eq(1) we need to find such f2 so that we can get f=m(1,6,8,15)
eq(1) says we can get m(1,6,15) from f3 ,so only 8 left
now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired.
But option (C) m(4,5,6,7,8)(6,8)+(1,6,15)
- m(6,8)+m(1,6,15)
- m(1,6,8,15)an
Question 9 |
If P, Q, R are Boolean variables, then (P + Q')(PQ' + PR)(P'R' + Q') simplifies
PQ' | |
PR' | |
PQ' + R | |
PR'' + Q |
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Question 9 Explanation:
Step by step explanation :
= (P + Q’)(PQ’ + PR)(P’R’ + Q’) = (PPQ’ + PPR + PQ’Q’ + PQ’R) (P’R’ + Q’) = (PQ’ + PR + PQ’ + PQ’R) (P’R’ + Q’) = (PP’Q’R’ + PP’R’R + PP’Q’R’ + PP’Q’RR’ + PQ’Q’ + PQ’R + PQ’Q’ + PQ’Q’R) = (0 + 0 + 0 + 0 + PQ’ + PQ’R + PQ’ + PQ’R) = PQ’ + PQ’R = PQ'(1 + R) = PQ’So, option (A) is correct.
Question 10 |
Consider the following Boolean function of four variables:
f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14)
The function is:
independent of one variables. | |
independent of two variables. | |
independent of three variables. | |
dependent on all the variables. |
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There are 101 questions to complete.