BooleanAlgebra
Question 1 
Consider the Boolean operator with the following properties:
Then x#y is equivalent to
ExOR satisfies all the properties. Hence,
Question 2 
Let, x_{1}⊕x_{2}⊕x_{3}⊕x_{4} = 0 where x_{1}, x_{2}, x_{3}, x_{4} are Boolean variables, and ⊕ is the XOR operator. Which one of the following must always be TRUE?
x_{1}x_{2}x_{3}x_{4} = 0  
x_{1}x_{3}+x_{2} = 0  
x_{1} + x_{2} + x_{3} + x_{4} = 0 
x_{1} ⊕ x_{2} ⊕ x_{3} ⊕ x_{4} = 0 (1)
A) x_{1}x_{2}x_{3} x_{4} = 0
Put x_{1} = 1, x_{2} = 1, x_{3} = 1, x_{4} = 1
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
But,
x_{1}x_{2}x_{3} x_{4} ≠ 0
So, false.
B) x_{1}x_{3} + x_{2} = 0
Put x_{1} = 1, x_{2} = 1, x_{3} = 0 , x_{4} = 0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x_{1}x_{3} + x_{2} ≠ 0
So, false.
D) x_{1} + x_{2} + x_{3} + x_{4} = 0
Let x_{1}=1, x_{2}=1, x_{3}=0, x_{4}=0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x_{1} + x_{2} + x_{3} + x_{4} ≠ 0
So, false.
(i) True.
Question 3 
The number of minterms after minimizing the following Boolean expression is ______.
[D′ + AB′ + A′C + AC′D + A′C′D]′
1  
2  
3  
4 
[D' + AB' + A'C + AC'D + A'C'D]'
[D' + AB' + A'C + C'D (A + A')']' (since A+A' = 1)
[AB' + A'C + (D' + C') (D' + D)]' (since D' + D =1)
[AB' + A'C + D' + C']'
[AB' + (A' + C') (C + C') + D']'
[AB' + A' + C' + D']'
[(A + A') (A' + B') + C' + D']'
[A' + B' + C' + D']'
Apply demorgan's law,
ABCD
Question 4 
Consider the following Boolean expression for F:
F(P, Q, R, S) = PQ + P'QR + P'QR'S
The minimal sumofproducts form of F is
= Q(P+P’R) + P’QR’S
= Q(P+R) + P’QR’S
= QP + QR + P’QR’S
= QP + Q(R + P’R’S)
= QP + Q( R + P’S)
= QP + QR + QP’S
= Q(P+P’S) + QR
= Q(P+S)+ QR
= QP + QS + QR
Question 5 
The truth table
represents the Boolean function
X  
X + Y  
X ⊕ Y  
Y 
Question 6 
The simplified SOP (sum of product) form of the boolean expression
Question 7 
The minterm expansion of is
m_{2}+m_{4}+m_{6}+m_{7}  
m_{0}+m_{1}+m_{3}+m_{5}  
m_{0}+m_{1}+m_{6}+m_{7}  
m_{2}+m_{3}+m_{4}+m_{5} 
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
= m_{7} + m_{6} + m_{2} + m_{4}
Question 8 
If P, Q, R are Boolean variables, then
Simplifies to
Question 9 
What is the maximum number of different Boolean functions involving n Boolean variables?
n^{2}  
2^{n}  
2^{2n}  
2^{n2} 
Number of variables= n
Number of input combinations is 2^{n}.
Each “boolean” function has two possible outputs i.e 0 and 1.
Number of boolean functions possible is 2^{2n}.
Formula: The number of mary functions possible with n kary variables is m^{kn}.
Question 10 
The Boolean function x'y' + xy + x'y is equivalent to
x' + y'  
x + y  
x + y'  
x' + y 
= x'y' + x'y + xy
= x'(y'+y)+xy
= x'⋅1+xy
= x'+xy
= (x'+x)(x'+y)
= 1⋅(x'+y)
= x'+y
Question 11 
The simultaneous equations on the Boolean variables x, y, z and w,
x + y + z = 1 xy = 0 xz + w = 1 xy + = 0
have the following solution for x, y, z and w, respectively.
0 1 0 0  
1 1 0 1  
1 0 1 1  
1 0 0 0 
Question 12 
What values of A, B, C and D satisfy the following simultaneous Boolean equations?
A = 1, B = 0, C = 0, D = 1  
A = 1, B = 1, C = 0, D = 0  
A = 1, B = 0, C = 1, D = 1  
A = 1, B = 0, C = 0, D = 0 
Question 13 
{ AND, OR }  
{ AND, NOT }  
{ NOT, OR }  
{ NOR } 
→ NAND and NOR gates are universal gates.
→ With the help of universal gates, we can construct any boolean expressions. These gates are also called functionally complete.
→ AND+NOT=NAND→ Functionally Complete
→ OR+NOT=NOR→ Functionally Complete
→ NOR→ Functionally Complete
→ AND+OR→ Not functionally complete
Question 14 
(A + B) ∙ (A’ + C) ∙ (B + C) = (A + B) ∙ (A’ + C)  
AB + A’C + BC = AB + BC  
AB + A’C + BC = (A + B) ∙ ( A ‘+ C) ∙ (B + C)  
(A + B) ∙ (A’ + C) ∙ (B + C) = AB + A’C 
XY+XZ+YZ=XY+XZ Consensus Law
For minterm: AB + A'C +BC = AB + A'C
AND for maxterm : ( A + B ).( A' + C ).( B + C ) = ( A + B ).(A' + C )
Question 15 
X_{0}X_{1}X_{2} … X_{n} + X_{1}X_{2} … X_{n} + X_{2}X_{3} … X_{n} + ⋯ + X_{n}  
X_{0}X_{1} + X_{2}X_{3} + … X_{n1} X_{n}  
X_{0} + X_{1} + X_{2} + … + X_{n}  
X_{0}X_{1} + X_{3} … X_{n−1} + X_{2}X_{3} + X_{5} … X_{n−1} + ⋯ + X_{n−2}X_{n−1} + X_{n}  
None of the above 
=(X_{0}X_{1}X_{3}+X_{2}X_{3}+X_{4})X_{5}+⋯+X_{N}
=X_{0}X_{1}X_{3}X_{5}+X_{2}X_{3}X_{5}+X_{4}X_{5}+⋯+X_{N}
=X_{0}X_{1}X_{3}X_{5}⋯X_{N−1}+X_{2}X_{3}X_{5}⋯X_{N−1}+X_{4}X_{5}X_{7}⋯X_{N−1}+⋯+X_{N}
Question 16 
C’ + AB’  
C’ (A’ + B)  
B’C’ + AB’  
None of these. 
The following expression can be simplified as:
(A + C')(B'+ C')
= AB' + AC' + B'C' + C
'
= AB' + C'(A + B' + 1) // 1 + A = 1
= AB' + C'
Question 17 
A + AB  
AB  
A'  
A + B 
A'(A’ + B’)
This expression can be simplified as:
= A'A' + A'B'
= A' + A'B'
= A'(1 + B') // 1 + B' = 1
= A'
Question 18 
0 XOR 0 = 0  
1 XOR 1 = 1  
1 XOR 0 = 1  
B XOR B = 0 
XOR gate only returns 1 as the output when both inputs are different and in every other case, it returns 0.
So, options (A), (C) and (D) are correct.
Question 19 
Y+XZ  
X+YZ  
XY+Z  
XZ+Y 
(X+Y)(X+Z) = X + XZ + XY + YZ
= X(1 + Z + Y) + YZ // as (1 + A = A)
= X.1 + YZ
= X + YZ
Question 20 
OR gate  
AND gate  
NAND gate  
XOR gate 
Question 21 
X.X = X  
(X + Y).X = X  
X̄ + XY = Y  
(X + Y).(X + Z) = X + YZ 
ii) (X + Y).X
= X.X + X.Y
= X + X.Y
= X(1 + Y)
= X
iii) X' + XY
= (X' + X)(X' + Y)
= (1)(X' + Y)
= (X' + Y)
iv) (X + Y).(X + Z)
= X.X + X.Z + X.Y + Y.Z
= X(1 + Z + Y) + Y.Z
= X + Y.Z
Question 22 
Perform the following operation for the binary equivalent of the decimal numbers
(14)_{10} + (15)_{10}The solution in 8 bit representation is :
11100011  
00011101  
10011101  
11110011 
(29)_{10} = (11101)_{2}
(29)_{10} in 8bit representation = (00011101)_{2}
(29)_{10} = (00011101)_{2}
Question 23 
A ⊕ C=B  
B ⊕ C=A  
A ⊕ B ⊕ C=0  
Both A) and B) 
Question 24 
1 ⊕ 0=1  
1 ⊕ 1 ⊕ 0=1  
1 ⊕ 1 ⊕ 1=1  
1 ⊕ 1 =0 
Option A: Here, 1 ⊕ 0=1 true according to truth table.
Option B: 1⊕1=0⊕0=0 false
Option C: 1 ⊕ 1=0 ⊕ 1=1 true
Option D: 1 ⊕ 1 =0 true.
Question 25 
The boolean expression A'⋅B + A⋅B' + A⋅B is equivalent to
A+B
 
A⋅B  
(A+B)'  
A'⋅B

Question 26 
The relation ≤ and < on a boolean algebra are defined as :
x ≤ y and only if x ∨ y = y x < y means x ≤ y but x ≠ y x ≥ y means y ≤ x and x > y means y < x
Consider the above definitions, which of the following is not true in the boolean algebra ?

(i) If x ≤ y and y ≤ z, then x ≤ z
(ii) If x ≤ y and y ≤ x, then x = y
(iii) If x < y and y < z, then x ≤ y
(iv) If x < y and y < z, then x < y
Choose the correct answer from the code given below:
Code:(iv) only  
(iii) only  
(i) and (ii) only  
(ii) and (iii) only

Because x < y means x ≤ y but x ≠ y.
ii) If x ≤ y and y ≤ x, then x = y is true
Because
x ≤ y implies x v y = x
y ≤ x implies x v y = y
x v y = x = y
Question 27 
Consider the following boolean equations :
(i) wx + w(x + y) + x(x + y)= x + wy (ii) (wx’(y + xz’) + w’x’)y = x’y
What can you say about the above equations ?
Both (i) and (ii) are true
 
(i) is true and (ii) is false  
Both (i) and (ii) are false  
(i) is false and (ii) is true

wx + w(x + y) + x(x + y)
= (wx + wx) + wy + (x + xy)
= wx + wy + x(1 + y)
= wx + wy + x
= (w + 1)x + wy
= x + wy
(ii)
Question 28 
Find the boolean expression for the logic circuit shown below :
(1NAND gate, 2NOR gate, 3NOR gate)
AB
 
AB’
 
A’B’  
A’B 
Question 29 
Which of the following statements are true ?
 (i) Every logic network is equivalent to one using just NAND gates or just NOR gates.
(ii) Boolean expressions and logic networks correspond to labelled acyclic digraphs.
(iii) No two Boolean algebras with n atoms are isomorphic.
(iv) Nonzero elements of finite Boolean algebras are not uniquely expressible as joins of atoms.
Choose the correct answers from the code given below :
Code :(i) and (iv) only
 
(i) and (ii) only  
(i), (ii) and (iii) only  
(ii), (iii) and (iv) only

→ NAND and NOR are universal gates, by using these gates we can construct all gates.
→ An atom of a Boolean algebra is an element x such that there exist exactly two elements y satisfying y ≤ x, namely x and 0. A Boolean algebra is said to be atomic when every element is a sup of some set of atoms (the bottom element is always the empty sup).
→ So, the options (iii) and (iv) are false.
Question 30 
A ⊕ C=B  
B ⊕ C=A  
A ⊕ B ⊕ C=0  
Both A) and B)  
None of these 
XOR:
X ⊕ X= 0
X ⊕ X'= 1
X ⊕ 0 = X
X ⊕ 1 = X'
A) A ⊕ C = B
A ⊕ A ⊕ B = B
0 ⊕ B = B
B= B
B) B ⊕C = A
B ⊕ A ⊕ B = A
A ⊕ 0=A
A=A
C) A⊕B=C
A⊕B⊕ C=C ⊕C ,
A⊕B⊕ C= 0
All are correct
Question 31 
Which of the following does NOT represent the Exclusive NOR operation over the binary variables A and B?
A’ ⊕ B’
 
A ⊕ B’
 
A’ ⊕ B
 
AB + A’B’ 
Question 32 
The AND function of several AND functions  
The AND function of several OR functions  
The OR function of several AND functions  
The OR function of several OR functions 
Example: F=AB+AC+AD Product of sum: A boolean expression consisting purely of Maxterms (sum terms) is said to be in canonical product of sums form.
Example: F=(A+B).(A+C).(A+D)
Question 33 
In boolean algebra, (x ⋀ y)’ = x’ V y’ and (x V y)’ = x’ ⋀ y’ is known as ___ law.
Demorgan’s law
 
Absorption
 
Dominance
 
Idempotent

1. The negation of a disjunction is the conjunction of the negations.
2. The negation of a conjunction is the disjunction of the negations.
Question 34 
A'B+BC  
AB'  
AB+BC  
AB+A'C 
First compute one and primes
(AAB'+AAC+AB'C+ACC)(AC'+B')
to take common AC then we get
(AB'+AC+AB'C+AC)(AC'+B')
To take common AB'
(AB'(1+C)+AC)(AC'+B')
(AB'+AC)(AC'+B')
AB'AC'+AB'B'+ACAC'+ACB'
AB'C'+AB'+0+ACB'
AB'(C'+1)+ACB'
AB'+ACB'
AB'(1+C)
AB'
(or)
We can also use truth table to get same solution.
Question 35 
4  
6  
8  
10 
=A’+A’C+B’+B’C+C’+AC’+A’BC’+AB’+AB’C
=A’ (1+C)+B’(1+C)+C’(1+A)+A’BC’+AB’(1+C) [ 1+X=1]
=A’+B’+C’+A’BC’+AB’
=A’+C’+A’BC’+B’ (1+A)
=A’+C’+A’BC’+B’
=A’+B’+C’ (1+A’B)
=A’+B’+C’
Total four gates required three not gates and one OR gate.
Question 36 
1θ1=1
1θ0=0
0θ1=0
0θ0=1
What will be the truth value of the expression (XθY)θZ=Xθ(YθZ)?
Always false  
Always true  
Sometimes true  
Sometimes false 
From the given table, we can infer that θ is ExNOR operator.
ExNOR is associative. Hence (XθY)θZ=Xθ(YθZ) is always true
Question 37 
Use Kruskal’s algorithm to find a minimal spanning tree for the graph. The List of the edges of the tree in the order in which they are chosen is?
AD, AE, AG, GC, GB, BF  
GC, GB, BF, GA, AD, AE  
GC, AD, GB, GA, BF, AE  
AD, AG, GC, AE, GB, BF 
Here, some edges have same weight. This situation, we can get many number of spanning trees.
We have to connect strictly below order.
Note: DB not given edge weight. GC, AD, BG, AE, BF
Edge weight 2= AD or GC
Edge weight 3= AG or GB
Edge weight 4= AE or DG or BC or BF
Edge weight 5= ED or CF
Edge weight 6: AC
Question 38 
q  
p ∧ r  
p ∨ q  
p 
Question 39 
1 ⊕ 0 = 1  
1 ⊕ 1 ⊕ 1 = 1  
1 ⊕ 1 ⊕ 0 = 1  
1 ⊕ 1 = 0 
According to truth table,
OptionA is TRUE
OptionB is a 1 ⊕ 1 is 0.
0 ⊕ 1 is 1(TRUE)
OptionC is 1 ⊕ 1 is 0.
0 ⊕ 0 = 0 but given 1. So, FALSE
OptionD is TRUE.
Question 40 
F = yz’ + y’z  
F = xy’ + x’y  
F = x’z + xz’  
F = x’z + xz’ + xyz 
Method2: Using boolean simplification
= x’y’z+x’yz+xy’z’+xyz’
= x'z(y'+y)+ xz'(y'+y)
= x'z+xz' (Since y'+y=1)
Question 41 
AB’  
AB’C  
A’B  
ABC 
= (AB'+AC) (A'C'+B')
= AB'A'C' + AB'B' + ACA'C' + ACB'
= AB'B' + ACB'
= AB'(C+1)
= AB'
Question 42 
A+B  
A.B  
(A+B)’  
A’.B 
Question 43 
x ≤ y and only if x ∨ y = y
x < y means x ≤ y but x ≠ y
x ≥ y means y ≤ x and
x > y means y <x
Consider the above definitions, which of the following is not true in the boolean algebra ?
(i)If x ≤ y and y ≤ z, then x ≤ z
(ii)If x ≤ y and y ≤ x, then x=y
(iii)If x < y and y < z, then x ≤ y
(iv)If x < y and y < z, then x < y
(iv) only  
(iii) only  
(i) and (ii) only  
(ii) and (iii) only 
Because x < y means x ≤ y but x ≠ y.
ii)If x ≤ y and y ≤ x, then x=y is true
Because
x ≤ y implies x v y =x
y ≤ x implies x v y = y
X v y = x = y
Note: From the given definitions, x
Key point: y < z has nothing to do with x ≤ y. So, we are ignoring.
Question 44 
(i) Every logic network is equivalent to one using just NAND gates or just NOR gates.
(ii) Boolean expressions and logic networks correspond to labelled acyclic digraphs.
(iii) No two Boolean algebras with n atoms are isomorphic.
(iv) Nonzero elements of finite Boolean algebras are not uniquely expressible as joins of atoms.
(i) and (iv) only  
(i) and (ii) only  
(i), (ii) and (iii) only  
(ii), (iii) and (iv) only 
→ NAND and NOR are universal gates, by using these gates we can construct all gates.
→ An atom of a Boolean algebra is an element x such that there exist exactly two elements y satisfying y ≤ x, namely x and 0. A Boolean algebra is said to be atomic when every element is a sup of some set of atoms (the bottom element is always the empty sup).
→ So the options (iii) and (iv) are false
Question 45 
aiii, biv, ci, dii  
aiv, biii, ci, dii  
aiv, biii, cii, di  
aiii, biv, cii, di 
Identity laws: x+0=x and x.1=x
Dominance laws: x+1=1 and x.0=0
Absorption laws: x.(x+y)=x
Question 46 
X + X = X  
X . X = X  
x + x . y = x  
None of the above 
X • (X+Y) = X
X + X•Y = X
Question 47 
~(~x)=x  
x+x=x  
x+xy=x  
x(x+y)=x 
The idempotence in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means "(the quality of having) the same power", from idem + potence (same + power).
(i). X + X=X
(ii). X ∧ X=X
According to boolean algebra
Question 48 
0  
1  
1  
None 