The absorption law states that:
a) A + AB = A
b) A(A + B) = A
c) Both a and b
d) A + A’B = A + B
Answer: c) Both a and b

The involution law in Boolean algebra is:
a) (A’)’ = A
b) (A’)’ = A’
c) A + A’ = 1
d) A·A’ = 0
Answer: a) (A’)’ = A

The idempotent law is represented by:
a) A + A = A
b) A·A = A
c) Both a and b
d) A + A’ = 1
Answer: c) Both a and b

The annulment law states that:
a) A·0 = 0
b) A + 1 = 1
c) A + 0 = A
d) A·1 = A
Answer: a) A·0 = 0

Which law justifies the simplification A+A′B=A+BA + A’B = A + BA+A′B=A+B?
a) Absorption law
b) Distributive law
c) Consensus theorem
d) De Morgan’s law
Answer: a) Absorption law

The commutative law is expressed as:
a) A + B = B + A and A·B = B·A
b) A + A’ = 1
c) A + 0 = A
d) A + 1 = 1
Answer: a) A + B = B + A and A·B = B·A

The associative law is represented as:
a) (A + B) + C = A + (B + C)
b) (A·B)·C = A·(B·C)
c) Both a and b
d) A + A’ = 1
Answer: c) Both a and b

Simplify: $A(B+C)$.
a) AB + AC
b) A + BC
c) A + B + C
d) AB + C
Answer: a) AB + AC

Simplify: $A+BC=(A+B)(A+C)$.
a) True
b) False
Answer: a) True

Simplify A′B+AB′A’B + AB’A′B+AB′:
a) A ⊕ B
b) A XNOR B
c) A + B
d) A · B
Answer: a) A ⊕ B

Simplify A+A′BA + A’BA+A′B:
a) A + B
b) A’ + B
c) AB
d) A + AB
Answer: a) A + B

Simplify $A+AB$:
a) A
b) AB
c) A + B
d) B
Answer: a) A

Simplify A+A′=?A + A’ = ?A+A′=?
a) 1
b) 0
c) A
d) A’
Answer: a) 1

Simplify A⋅A′=?A·A’ = ?A⋅A′=?
a) 0
b) 1
c) A
d) A’
Answer: a) 0

Simplify (A+B′)(A′+B)(A + B’)(A’ + B)(A+B′)(A′+B):
a) A XNOR B
b) A ⊕ B
c) A · B
d) A + B
Answer: a) A XNOR B

Simplify (A+B)(A′+C)(B+C)(A + B)(A’ + C)(B + C)(A+B)(A′+C)(B+C):
a) AB + A’C
b) A’ + BC
c) A + B
d) A + C
Answer: b) A’ + BC

Simplify (A+B)(A′+B′)(A′+C)(A + B)(A’ + B’)(A’ + C)(A+B)(A′+B′)(A′+C):
a) AB’ + A’C
b) A + BC
c) A’ + B’C
d) A + B + C
Answer: a) AB’ + A’C

Simplify (A+B′)(B+C′)(A + B’)(B + C’)(A+B′)(B+C′):
a) A·B + B’·C’
b) A + C’
c) A + B + C’
d) A’ + C
Answer: b) A + C’

Simplify A′B+A′C+BCA’B + A’C + BCA′B+A′C+BC:
a) A’ + BC
b) A’ + B + C
c) A + C
d) B + C
Answer: a) A’ + BC

Simplify (A+B)(A+B′)(A′+C′)(A + B)(A + B’)(A’ + C’)(A+B)(A+B′)(A′+C′):
a) AB’ + A’C’
b) A’ + BC
c) A + C’
d) A’ + B’C
Answer: a) AB’ + A’C’

How many cells are in a 5-variable K-map?
a) 16
b) 32
c) 64
d) 128
Answer: b) 32

In a 4-variable K-map, a group of 8 adjacent 1s represents:
a) One variable term
b) Two variable term
c) Three variable term
d) No simplification
Answer: a) One variable term

In K-map simplification, grouping 1s in powers of:
a) 2
b) 3
c) 4
d) 5
Answer: a) 2

A K-map grouping of 2 adjacent cells eliminates:
a) 1 variable
b) 2 variables
c) 3 variables
d) None
Answer: a) 1 variable

In a 3-variable K-map, the corner cells are:
a) Adjacent
b) Not adjacent
c) Opposite
d) Don’t-care cells
Answer: a) Adjacent

Don’t-care terms in a K-map are used to:
a) Simplify expressions
b) Increase variables
c) Add redundancy
d) Remove maxterms
Answer: a) Simplify expressions

A group of 4 cells in a K-map eliminates:
a) 2 variables
b) 3 variables
c) 1 variable
d) 4 variables
Answer: a) 2 variables

In a K-map, adjacent cells differ in:
a) One variable
b) Two variables
c) Three variables
d) None
Answer: a) One variable

The Gray code sequence used in a 4-variable K-map ensures:
a) Only one bit changes between adjacent cells
b) Two bits change
c) Random order
d) Lexicographical order
Answer: a) Only one bit changes between adjacent cells

A cell represents a minterm if the output is:
a) 1
b) 0
c) Don’t-care
d) Undefined
Answer: a) 1

NAND gate is called universal because:
a) All other gates can be derived from it
b) It produces a universal output
c) It has multiple inputs
d) It uses De Morgan’s law
Answer: a) All other gates can be derived from it

NOR gate is also universal because:
a) Any Boolean function can be implemented using NOR gates
b) It produces a complement output
c) It is faster
d) It consumes less power
Answer: a) Any Boolean function can be implemented using NOR gates

NOT gate using NAND gate:
a) Connect both inputs together
b) Invert inputs
c) Connect one input to 1
d) Connect one input to 0
Answer: a) Connect both inputs together

AND gate using NAND gates requires:
a) Two NAND gates
b) Three NAND gates
c) Four NAND gates
d) One NAND gate
Answer: b) Three NAND gates

OR gate using NOR gates requires:
a) Three NOR gates
b) Two NOR gates
c) Four NOR gates
d) One NOR gate
Answer: a) Three NOR gates

XOR gate using NAND gates requires:
a) Four NAND gates
b) Two NAND gates
c) Three NAND gates
d) One NAND gate
Answer: a) Four NAND gates

The output of a NOR gate is 1 only when:
a) All inputs are 0
b) All inputs are 1
c) Any input is 1
d) Inputs are unequal
Answer: a) All inputs are 0

The output of a NAND gate is 0 only when:
a) All inputs are 1
b) All inputs are 0
c) One input is 1
d) Inputs are opposite
Answer: a) All inputs are 1

A NAND gate is equivalent to:
a) (A·B)’
b) (A + B)’
c) A·B
d) A + B
Answer: a) (A·B)’

A NOR gate is equivalent to:
a) (A + B)’
b) (A·B)’
c) A + B
d) A·B
Answer: a) (A + B)’

A decoder converts:
a) Binary code → one active output
b) One input → binary code
c) Decimal → binary
d) Binary → Gray code
Answer: a) Binary code → one active output

A multiplexer (MUX) selects:
a) One of many inputs to one output
b) One output to many inputs
c) One input to multiple outputs
d) None of these
Answer: a) One of many inputs to one output

A demultiplexer (DEMUX) performs:
a) One input → many outputs
b) Many inputs → one output
c) Parallel addition
d) Encoding
Answer: a) One input → many outputs

A 2-to-4 decoder requires:
a) 2 inputs and 4 outputs
b) 4 inputs and 2 outputs
c) 4 inputs and 4 outputs
d) 2 outputs and 4 inputs
Answer: a) 2 inputs and 4 outputs

A parity generator uses which gate?
a) XOR
b) XNOR
c) AND
d) NOR
Answer: a) XOR

A parity checker uses:
a) XNOR
b) XOR
c) NOR
d) NAND
Answer: a) XNOR

In logic circuits, fan-in refers to:
a) Number of inputs to a gate
b) Number of outputs from a gate
c) Power rating of gate
d) Speed of logic gate
Answer: a) Number of inputs to a gate

In logic circuits, fan-out refers to:
a) Number of gates that can be driven by one output
b) Number of inputs per gate
c) Voltage output
d) Current consumption
Answer: a) Number of gates that can be driven by one output

A hazard in a logic circuit refers to:
a) Unwanted glitch during switching
b) Extra logic gate
c) Power failure
d) Incorrect truth table
Answer: a) Unwanted glitch during switching

A race condition in digital logic occurs when:
a) Two signals change simultaneously
b) Output remains stable
c) Input is constant
d) Power is off
Answer: a) Two signals change simultaneously

461. A logic circuit that outputs 1 only when all inputs are 1 is called:
a) OR gate
b) NOR gate
c) AND gate
d) XOR gate
Answer: c) AND gate

462. The truth table of a NOT gate has:
a) 2 inputs and 2 outputs
b) 1 input and 1 output
c) 2 inputs and 1 output
d) 3 inputs and 1 output
Answer: b) 1 input and 1 output

463. The Boolean expression A + AB simplifies to:
a) A
b) B
c) AB
d) A + B
Answer: a) A

464. The expression A + A’B simplifies to:
a) A + B
b) B
c) A
d) A’B
Answer: a) A + B

465. The Demorgan’s Theorem states that:
a) (A + B)’ = A’B’
b) (AB)’ = A’ + B’
c) Both a and b
d) None of the above
Answer: c) Both a and b

466. Which gate is called the “universal gate” because any logic function can be built using it?
a) AND
b) OR
c) NAND
d) XOR
Answer: c) NAND

467. A Karnaugh Map (K-Map) is used to:
a) Add binary numbers
b) Minimize Boolean expressions
c) Multiply Boolean terms
d) Generate truth tables
Answer: b) Minimize Boolean expressions

468. A 4-variable K-Map has how many cells?
a) 8
b) 12
c) 16
d) 24
Answer: c) 16

469. The maxterm of variables A, B, and C for the combination (A=0, B=1, C=1) is:
a) (A + B’ + C’)
b) (A’ + B + C)
c) (A + B + C’)
d) (A’ + B’ + C)
Answer: b) (A’ + B + C)

470. In Boolean algebra, A + A’ = ?
a) 1
b) 0
c) A
d) A’
Answer: a) 1

471. In Boolean algebra, A·A’ = ?
a) 1
b) 0
c) A
d) A’
Answer: b) 0

472. The output of a NOR gate is 1 when:
a) All inputs are 1
b) All inputs are 0
c) One input is 1
d) One input is 0
Answer: b) All inputs are 0

473. Which law of Boolean algebra states that A + AB = A?
a) Absorption Law
b) Idempotent Law
c) Distributive Law
d) DeMorgan’s Law
Answer: a) Absorption Law

474. The number of possible combinations for a 3-variable logic system is:
a) 4
b) 6
c) 8
d) 16
Answer: c) 8

475. The Boolean expression (A + B)(A + B’) simplifies to:
a) A
b) B
c) A + B
d) AB
Answer: a) A

476. In K-Maps, grouping 1s together represents:
a) SOP simplification
b) POS simplification
c) Minterm combination
d) Don’t-care condition
Answer: a) SOP simplification

477. The output of an XOR gate is 1 when:
a) Inputs are equal
b) Inputs are different
c) One input is 0
d) Both inputs are 1
Answer: b) Inputs are different

478. Which of the following is NOT a basic logic gate?
a) NAND
b) OR
c) XOR
d) NOR
Answer: c) XOR

479. The complement of (A + B)’ is:
a) A + B
b) A’ + B’
c) AB
d) A’B’
Answer: a) A + B

480. Simplify A’B + AB + A’B’.
a) A’ + B
b) A + B’
c) A + B
d) A’ + B’
Answer: a) A’ + B

The Consensus Theorem states that AB+A′C+BC=AB+A′CAB + A’C + BC = AB + A’CAB+A′C+BC=AB+A′C.
Answer: True

The redundant term in AB+A′C+BCAB + A’C + BCAB+A′C+BC is:
a) BC
b) AB
c) A’C
d) None
Answer: a) BC

The Shannon Expansion Theorem helps in:
a) Function decomposition
b) Gate minimization
c) Transistor modeling
d) Power analysis
Answer: a) Function decomposition

Functional completeness of NAND/NOR means:
a) Any Boolean function can be implemented
b) Only AND/OR can be made
c) They are irreversible
d) Only XOR functions can be made
Answer: a) Any Boolean function can be implemented

A 3-variable K-map simplifies functions with:
a) 8 minterms
b) 4 maxterms
c) 16 minterms
d) 6 variables
Answer: a) 8 minterms

The term “minterm” corresponds to:
a) Product term giving 1
b) Sum term giving 0
c) Product term giving 0
d) None
Answer: a) Product term giving 1

The term “maxterm” corresponds to:
a) Sum term giving 0
b) Product term giving 1
c) Sum term giving 1
d) Don’t-care
Answer: a) Sum term giving 0

The Distributive Law in Boolean algebra states:
a) A(B + C) = AB + AC
b) A + BC = (A + B)(A + C)
c) Both a and b
d) None
Answer: c) Both a and b

The inverse of De Morgan’s theorem is:
a) (AB)’ = A’ + B’
b) (A + B)’ = A’·B’
c) Both a and b
d) A’ + B’ = A + B
Answer: c) Both a and b

Duality principle says:
a) Every algebraic expression remains valid if + ↔ · and 0 ↔ 1 are interchanged
b) Every expression becomes invalid
c) Complement and dual are same
d) None
Answer: a) Every algebraic expression remains valid if + ↔ · and 0 ↔ 1 are interchanged

Complementation in Boolean algebra means:
a) Changing 1 → 0 and 0 → 1
b) Swapping variables
c) Multiplying by 0
d) Adding 1
Answer: a) Changing 1 → 0 and 0 → 1

Simplify: (A+B+C)(A+B+C′)(A + B + C)(A + B + C’)(A+B+C)(A+B+C′).
a) A + B
b) A + C
c) B + C
d) A + B + C
Answer: a) A + B

Simplify: A⋅(A+B′)A·(A + B’)A⋅(A+B′).
a) A
b) A + B’
c) AB’
d) A’
Answer: a) A

The Boolean variable can take how many values?
a) 2
b) 4
c) 8
d) Infinite
Answer: a) 2

The Boolean expression $(A+0)$ simplifies to:
a) A
b) 0
c) 1
d) A’
Answer: a) A

The Boolean expression $(A\cdot 1)$ simplifies to:
a) A
b) 1
c) 0
d) A’
Answer: a) A

Simplify (A+A′)B(A + A’)B(A+A′)B:
a) B
b) A
c) A’
d) 1
Answer: a) B

Simplify A+A′B′A + A’B’A+A′B′:
a) A + B’
b) A’ + B’
c) A + B
d) A’B’
Answer: a) A + B’

Simplify A+A′B′A + A’B’A+A′B′:
a) A + B’
b) A’ + B
c) A + B
d) A’B
Answer: a) A + B’

The K-map method is based on:
a) Grouping adjacent 1s to simplify Boolean expressions
b) Converting decimal to binary
c) De Morgan’s theorem
d) Logic circuit synthesis
Answer: a) Grouping adjacent 1s to simplify Boolean expressions