CS Free MCQ Bank

[ Option A ]

Given Boolean Expression:
=x(x+y)
=xx+xy
=x+xy
=x(1+y)
=x.1
=x

[ Option A ]

Given Boolean expression:
=(A+C)(AD+AD’)+AC+C
=(A+C)(AD+AD’)+C(A+1)
=(A+C)(AD+AD’)+C.1
=(A+C)(AD+AD’)+C
=AAD+AAD’+ACD+ACD’+C
=AD+AD’+ACD+ACD’+C
=AD+AD’+C(AD+AD’+1)
=AD+AD’+C.1
=AD+AD’+C
=A(D+D’)+C
=A.1+C
=A+C
Therefore, (A+C)(AD+AD’)+AC+C = A+C

[ Option A ]

The given Boolean expression:
= x’y’+xy+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
Therefore, x’y’+xy+x’y = x’+y

[ Option C ]

Truth Table Confirmation:

[ Option C ]

The number of minterms in an n-variable truth table is 2ⁿ. This is because each Boolean variable can have two possible values, 0 or 1. When there are n variables, the total number of possible input combinations is 2ⁿ. Each unique input combination corresponds to exactly one minterm. Therefore, an n-variable truth table contains 2ⁿ minterms.

[ Option C ]

• A.A’ = 0
• A+AB = A(1+B) = A.1 = A
• A+A’B = (A+A’)(A+B) = 1.(A+B) = A+B
• A(A+B) = A.A+A.B = A+A.B = A(1+B) = A.1 = A

[ Option B ]

[ Option A ]

Given Expression:
F=A’C+A’B+AB’C+BC
F=A’C+A’B+C(B+AB’)
F=A’C+A’B+C((A+B)(B+B’))
F=A’C+A’B+C(A+B)
F=A’C+A’B+AC+BC
F=A’C+AC+A’B+BC
F=C(A’+A)+A’B+BC
F=C+BC+A’B
F=C(1+B)+A’B
F=C+A’B

[ Option B ]

In the given K-map, two quads are formed. The first quad includes minterms 0, 2 and 8 and the don’t care terms 10, while the second quad includes minterms 0, 2 and 6 and the don’t care terms 4.
Quad 1 : a’b’c’d’+ab’c’d’+a’b’cd’+ab’cd’ : b’d’
Quad 2 : a’b’c’d’+a’b’c’d+ab’c’d’+ab’c’d : b’c’
Therefore, the minimal form is b’d’+b’c’.

[ Option B ]

Given boolean expression:
=X.(X+Y)
=X.X+X.Y
=X+X.Y
=X(1+Y)
=X.1
=X
So, X.(X+Y) = X

[ Option D ]

Given Boolean expression,
=((X+Y)’+Z’)’
=(X+Y)’’.Z’’
=(X+Y).Z
So, ((X+Y)’+Z’)’ = (X+Y).Z

[ Option A ]

In digital logic, a “don’t care” condition indicates that the value of a particular input does not affect the final output of the circuit. These conditions are used during simplification of Boolean expressions or Karnaugh maps (K-maps). 
Because the output does not depend on certain inputs, we are free to treat those inputs as either 0 or 1, whichever helps produce the simplest possible expression.
“Don’t care” conditions do not prioritize inputs nor ensure completeness of truth tables, instead, they are a tool for simplification.

[ Option A ]

f(x,y,z) = z(x+y)+((z’+x+y)(x’+y’))’
f(x,y,z) = z(x+y)+((x’z’+y’z’+xx’+xy’+yy’))’
f(x,y,z) = z(x+y)+(x’z’+y’z’+xy’)’
f(x,y,z) = z(x+y)+((x’z’)’.(y’z’)’.(xy’)’)
f(x,y,z) = z(x+y)+((x’’+z’’).(y’’+z’’).(x’+y’’))
f(x,y,z) = z(x+y)+((x+z).(y+z).(x’+y))
f(x,y,z) = z(x+y)+((xy+xz+yz+zz).(x’+y))
f(x,y,z) = z(x+y)+((xy+xz+yz+z).(x’+y))
f(x,y,z) = z(x+y)+((xy+z).(x’+y))
f(x,y,z) = z(x+y)+(xyx’+xyy+x’z+yz)
f(x,y,z) = z(x+y)+(xy+x’z+yz)
f(x,y,z) = xz+yz+xy+x’z+yz
f(x,y,z) = xz+x’z+xy+yz
f(x,y,z) = z(x+x’+y)+xy
f(x,y,z) = z.1+xy
f(x,y,z) = z+xy
So, there is only 2 gates required one AND and one OR gate.

[ Option D ]

If there are n Boolean variables, then:

• Number of possible input combinations = 2ⁿ.
• For each input combination, the output can be either 0 or 1. 
• Hence, the total number of Boolean expressions (Boolean functions) is = 2^(2^n).

Here the number of Boolean variable is 3, so:

• Number of input combinations = 2³ = 8
• Total Boolean expressions = 2^(2^3) = 2⁸ = 256

[ Option A ]

Principle of Duality says, every algebraic expression remains valid if you interchange: 
(1) AND (·) with OR (+)
(2) 0 with 1
🧠 In simple words, for every Boolean law or expression, there exists a dual expression that is also valid.
E.g.:
(1) A + 0 = A, its Dual is A . 1 = A
(2) A + A =A, its Dual is A . A = A
(3) A . 0 = 0, its Dual is A + 1 =1

[ Option C ]

Apply distributive law.

[ Option A ]

Given Boolean Expression:
x+x’y
To simplify the expression, we apply Boolean algebra rules:
(x+x’) (x+y) Apply Distributive Law.
(1)(x + y) Because x+x’=1
x+y Because A.1=A
So, x+x’y = x+y

[ Option B ]

[ Option A ]

[ Option D ]

In two-valued Boolean algebra, for two variables, the number of possible Boolean functions is given by the formula  2^(2ⁿ). Where n is the number of variables. Here n=2, So, 2⁽²²⁾ = 2⁴ = 16. Therefore, the maximum number of Boolean functions for two variables is 16.

[ Option D ]

[ Option B ]

The given Boolean expression is AB + AB’ + A’C + AC. To determine which variable it is independent of, we simplify the expression. 
First, we combine AB + AB’, which simplifies to A(B + B’). Since B + B’ = 1, this becomes A. Substituting this back, the expression becomes A + A’C + AC. 
Next, we combine A + AC, which simplifies to A(1 + C) = A, because anything ORed with 1 is 1. So the expression now becomes A + A’C.
In this simplified form, we can see that the expression still contains A and C, but B no longer appears. Therefore, the Boolean expression is independent of variable B.

[ Option B ]

[ Option A ]

The given boolean expression F(P,Q,R,S) = PQ + P’QR + P’QR’S can be solved by either using K-Map or boolean theorem.
= PQ+P’QR+P’QR’S
=Q(P+P’R)+P’QR’S
=Q(P+R)+P’QR’S
=PQ+QR+P’QR’S
=PQ+Q(R+P’R’S)
=PQ+Q(R+P’S)
=PQ+QR+P’QS
=QR+Q(P+P’S)
=QR+Q(P+S)
=QR+PQ+QS
=PQ+QR+QS
Therefore, PQ+P’QR+P’QR’S = PQ+QR+QS

[ Option D ]

Principle of Duality says, every algebraic expression remains valid if you interchange: 
1)  AND (·) with OR (+)
2)  0 with 1
In simple words, for every Boolean law or expression, there exists a dual expression that is also valid. E.g.: 
1)  A + 0 = A, its Dual is A . 1 = A 
2)  A + A =A, its Dual is A . A = A 
3)    A . 0 = 0, its Dual is A + 1 =1

[ Option C ]

In option (A), The value of x depends on the input. If x=0, the expression is 0 and if x=1, the expression is 1. Since it is not always true, this is not a tautology.
In option (B), (x+x′)y, we know that x+x′=1. So, the expression becomes 1.y=y. The value now depends on y. If y=0, the output is 0. Hence, this is not always true and not a tautology.
In option (C), x+y’+x’, we know that x+x′=1. So, the expression becomes 1+y’=1. This result is always 1, irrespective of the value of y. Therefore, this expression is always true and is a tautology.
In option (D), (xy)+x’, if x=1, the expression becomes y+0=y, which can be 0 or 1 and if x=0, the expression becomes 0+1=1. Since the result is not always 1, this is not a tautology.

[ Option B ]

The given Boolean expression is Y=AB+BC+AC. Here, each term (AB, BC, AC) is a product term (logical AND). These product terms are then summed together (logical OR). This format of “logical sum of several product terms” is known as Sum of Products (SOP).

[ Option D ]

F = AB’C+A’BC+ABC’+A’B’C+AB’C’
F = AB’C+AB’C’+A’BC+A’B’C+ABC’
F = AB’(C+C’)+A’C(B+B’)+ABC’
F = AB’+A’C+ABC’
F = AB’+ABC’+A’C
F = A(B’+BC’)+A’C
F = A(B’+C’)+A’C
F = AB’+AC’+A’C

[ Option A ]

The symbol Σ (Sigma) means the function is expressed as a sum of minterms.
Since there are four variables, we use a 4-variable K-map. The variables AB represent the rows and CD represent the columns, arranged in gray code order (00, 01, 11, 10). We place 1s in the cells corresponding to minterms 1, 4, 5, 9, 11, and 12.

Here we found 3 pairs, pair one using minterm 1 and 5, pair second using minterm 4 and 12 and pair third using minterm 9 and 11.
Pair 1 : A’B’C’D + A’BC’D = A’C’D
Pair 2 : A’BC’D’ + ABC’D’ = BC’D’
Pair 3 : AB’C’D + AB’CD = AB’D
Therefore, F(A,B,C,D) = A’C’D + BC’D’ + AB’D

[ Option C ]

Given boolean expression,
=((x₁+x₂)(x₃x₄)’)’
=(x₁+x₂)’+(x₃x₄)’’
=x₁’x₂’+x₃x₄
Now, construct the truth table for this simplified Boolean expression. In the truth table, output ‘1’ represents the minterms (Sum of Products), and output ‘0’ represents the maxterms (Product of Sums).

In the final output column, the total number of 1’s is 7 and the total number of 0’s is 9. Therefore, the number of Sum of Product (SOP) terms is 7, and the number of Product of Sum (POS) terms is 9.

[ Option D ]

Given Boolean expression:
=x.(x+y)
=x.x+x.y
=x+x.y
=x(1+y)
=x.1
=x
So, x.(x+y) = x

[ Option D ]

Boolean algebra is the mathematical foundation of digital electronics. It deals with binary variables and logical operations, which makes it very useful in several areas. 

• Firstly, it is directly applied in designing digital computers because computer circuits are built on binary logic. 
• Secondly, Boolean algebra is used in building logic symbols such as AND, OR, NOT gates, and in representing how these gates behave. 
• Thirdly, it plays an important role in circuit theory, where logical expressions can be simplified to optimize digital circuits.

[ Option B ]

Given Boolean Expression:
=(X+Y+XY)(X+Z)
=(Y+X(1+Y))(X+Z)
=(Y+X.1)(X+Z)
=(Y+X)(X+Z)
=XY+YZ+XX+XZ
=X+XY+XZ+YZ
=X(1+Y+Z)+YZ
=X.1+YZ
=X+YZ

[ Option A ]

According to the Consensus Theorem, the term BC is the redundant term and can be removed without changing the value of the Boolean expression.
Consensus Theorem : AB+A′C+BC = AB+A′C
We can also verify this using a Truth Table or K-Map for extra clarity.
USING K-MAP:
For a K-map, we always require either a collection of minterms or a collection of maxterms. Since the given expression is in Sum of Products (SOP) form, we first convert it into Standard SOP (SSOP), that is, a collection of minterms.
SOP : AB+A’C+BC
SSOP : AB.1+A’C.1+BC.1
=AB(C+C’)+A’C(B+B’)+BC(A+A’)
=ABC+ABC’+A’BC+A’B’C+ABC+A’BC
=ABC+ABC’+A’BC+A’B’C (Remove redundant terms)
In a minterm, each variable appears exactly once, an uncomplemented variable indicates a value of 1, whereas a complemented variable indicates a value of 0.
Therefore,
ABC=111 = 7 = m₇
ABC’=110 = 6 = m₆
A’BC=011 = 3 = m₃
A’B’C=001 = 1 = m₁
F(A,B,C) = Σ(1,3,6,7)
Since there are three variables, we use a 3-variable K-map. The variables A represent the rows and BC represent the columns, arranged in gray code order (00, 01, 11, 10). We place 1s in the cells corresponding to minterms 1, 3, 6 and 7.

In this K-map, two pairs are formed. The first pair includes minterms 1 and 3, while the second pair includes minterms 6 and 7.
Pair 1 : A’B’C+A’BC = A’C
Pair 2 : ABC+ABC’ = AB
F(A,B,C) = A’C+AB
Therefore, BC is the redundant (consensus) term.
USING TRUTH TABLE:
F=AB+A′C+BC

The output column AB+A′C and AB+A′C+BC are identical for all input combinations. So, adding or removing the term BC does not change the final output. Therefore, BC is the redundant (consensus) term.

[ Option C ]

In two-valued Boolean algebra, each variable can take only two values: 0 or 1. When we have two Boolean variables, there are four possible input combinations, (0,0), (0,1), (1,0) and (1,1). 
For each of these combinations, a Boolean function can output either 0 or 1. Therefore, the total number of possible Boolean functions is calculated by raising 2 to the power of the number of input combinations. Since there are four combinations, the number of functions becomes 2⁴=16.

[ Option A ]

[ Option B ]

The symbol Σ (Sigma) means the function is expressed as a sum of minterms.
Since there are four variables, we use a 4-variable K-map. The variables PQ represent the rows and RS represent the columns, arranged in gray code order (00, 01, 11, 10). We place 1s in the cells corresponding to minterms 0, 2, 5, 7, 8, 10, 13 and15.

In this K-map, two quads are formed. The first quad includes minterms 0 and 10 and the don’t care terms 2 and 8, while the second quad includes minterms 5 and 15 and the don’t care terms 7 and 13.
Quad 1 : P’Q’R’S’ + P’Q’RS’ + PQ’R’S’ + PQ’RS’ : Q’S’
Quad 2 : P’QR’S + P’QRS + PQR’S + PQRS : QS
Therefore, F(P,Q,R,S) = Q’S’ + QS

[ Option B ]

Given Boolean Expression:
=AB+AB’+A’C+AC
=A(B+B’)+C(A’+A)
=A.1+C.1
=A+C
The simplified expression A+C contains variables A and C, but does not contain B. Therefore, the Boolean expression is independent of variable B.

[ Option A ]

[ Option C ]

[ Option B ]

• The rule A+1=1 is called the domination law.
• The rule A.A=A is known as the idempotent law.
• The A+0=A is the identity law.

The statement A=A’ means a variable is equal to its complement. This is not possible in Boolean algebra because a variable and its complement always have opposite values.

Truth Table Verification:

[ Option A ]

Given Boolean Expression:
=x+xy
=x(1+y)
=x.1
=x

[ Option B ]

The symbol Σ (Sigma) means the function is expressed as a sum of minterms.
Since there are three variables, we use a 3-variable K-map. The variables X represent the rows and YZ represent the columns, arranged in gray code order (00, 01, 11, 10). We place 1s in the cells corresponding to minterms 0, 2, 4, 5 and 6.

Here we found 1 quad using minterms 0, 2, 4 and 6 and 1 pair using minterms 4 and 5.
Quad : X’Y’Z’ + XY’Z’ + X’YZ’ + XYZ’ = Z’
Pair : XY’Z’ + XY’Z = XY’
Therefore, F(X,Y,Z) = Z’+XY’

[ Option A ]

Given Boolean expression:
=(X+Y)(X+Y’)(X’+Y)
=(XX+XY’+XY+YY’)(X’+Y)
=(X+XY’+XY+0)(X’+Y)
=(X+XY’+XY)(X’+Y)
=X(1+Y’+Y)(X’+Y)
=(X.1)(X’+Y)
=X(X’+Y)
=XX’+XY
=0+XY
=XY
Therefore, (X+Y)(X+Y’)(X’+Y) = XY

[ Option C ]

The canonical forms of Boolean expressions are:

• Sum of Minterms (SOM) — also called Standard Sum of Products (SSOP)
• Product of Maxterms (POM) — also called Standard Product of Sums (SPOS)

These forms represent Boolean functions in standard forms using all variables in each term.
Note:

• In SOP, not all terms necessarily contain all variables. SOP can be non-standard if terms are missing variables.
• SOM = Standard SOP = Canonical Form
• POM = Standard POS = Canonical Form

[ Option B ]

Evaluate the expression step by step using the double-negation (complement) rule (x’)’=x.