ELEVATE SKILLS!
Q: 1 The Boolean expression x(x+y) is equal to
x
y
xy
None of the above
[ Option A ]
Given Boolean Expression:
=x(x+y)
=xx+xy
=x+xy
=x(1+y)
=x.1
=x
Q: 2 The Boolean expression A + BC equals
(A’+B)(A’+C)
(A+B)(A’+C)
(A+B)(A+C)
None of the above
[ Option C ]
Truth Table Confirmation:
| A | B | C | BC | A+B.C | (A+B) | (A+C) | (A+B).(A+C) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Q: 3 In K-Map, those groups which cover at least one minterm that can't be covered by any other prime implicant is called –
Prime Implicit
Essential Prime Implicit
Redundant Prime Implicit
Selective Prime Implicit
[ Option B ]
Q: 4 Simplified Boolean function of Boolean function F=A’C+A’B+AB’C+BC is
F=C+A’B
F=A’B+B’C
F=BC+A’B
None of the above
[ 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
Q: 5 What is the purpose of the “don’t care” condition in digital logic?
To indicate that the value of a variable does not affect the output
To prioritize certain inputs over others
To ensure that all possible input combinations are covered in truth tables
None of the above
[ 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.
Q: 6 Which of these is a dual of A+(B.C) = (A+B) . (A+C) ?
A.(B+C) = (A.B) + (A.C)
A+ (B+C) = (A+B) +C
A. (B.C) = (A.B) .C
All of the above
[ 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
Q: 7 The Boolean expression A+BC equals
(A’+B)(A’+C)
(A+B)(A’+C)
(A+B)(A+C)
More than one of the above
[ Option C ]
Apply distributive law.
Q: 8 The Boolean expression x+x’y equals
x+y
x+xy
y+yx
None of the above
[ 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
Q: 10 Derive the Boolean expression for the logic circuit shown below:
C(A+B)DE
[C(A+B)D+E’]
[[C(A+B)D]E’]
None of the above
[ Option A ]
Q: 11 In two-valued Boolean algebra, the maximum number of Boolean functions for two variables will be
4
8
12
16
[ 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.
Q: 12 For Boolean function, f(A, B, C) = Σ m(0, 1, 3, 5) + Σ d(2, 7) , the minimized sum-of-product equation is : (using K-Map)
A' + C'
A' + B' + C'
A' . B' . C'
C + A'
[ Option D ]
Q: 13 The Boolean expression AB+AB’+A’C+AC is independent of Boolean variable?
A
B
C
A'
[ 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.
Q: 14 The minimized expression for a 4-variables logic function, f(A, B, C, D) = Σm(0, 1, 2, 3, 5, 7, 8, 9, 11, 14) , using K-Map is
(Ā + B + C̅ + D) (Ā + B̅ + C) (Ā + B̅ + D̅) (A + B̅ + D)
ABCD̅ + B̅C̅ + B̅D + A̅D + A̅B̅
B̅ + A C̅ + A̅ C D
A̅B̅C̅D + BC + BD̅ + AD̅ + AB̅
[ Option B ]
Q: 15 According to the duality theorem, the dual of the Boolean expression A+0=A is:
A.0=0
A+1=A
A.A=A
A.1=A
[ 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
Q: 16 The expression Y=AB+BC+AC show the ____________ operations.
EX-OR
SOP
POS
More than one of the above
[ 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).
Q: 17 Boolean algebra can be used
for designing the digital computers
in building logic symbols
in circuit theory
More than one of the above
[ 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.
Q: 18 The simplified form of the Boolean expression (X+Y+XY)(X+Z) is
XY+YZ
X+YZ
XZ+Y
None of the above
[ 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
Q: 19 In two-valued Boolean algebra, the maximum number of Boolean functions for two variables will be
8
12
16
None of the above
[ 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.
Q: 21 The Boolean expression AB+AB’+A’C+AC is independent of Boolean variable
A
B
C
None of the above
[ 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.
Q: 24 The Boolean expression x+xy will be equal to
x
y
x+y
None of the above
[ Option A ]
Given Boolean Expression:
=x+xy
=x(1+y)
=x.1
=x
Q: 25 What are the canonical forms of Boolean expressions?
OR and XOR
NOR and XNOR
SOM and POM
More than one of the above
[ Option C ]
The canonical forms of Boolean expressions are:
These forms represent Boolean functions in standard forms using all variables in each term.
Note:
Q: 26 If x is a Boolean variable, then (((x’)’)’)’ will be
x'
x
x+x’
More than one of the above
[ Option B ]
Evaluate the expression step by step using the double-negation (complement) rule (x’)’=x.
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