Digital Electronics

Boolean Algebra — Practice MCQs for CCAT

25 Questions Section C: Hardware Digital Electronics

Practice 25 Boolean Algebra multiple-choice questions designed for CDAC CCAT exam preparation. Click "Show Answer" to reveal the correct option with detailed explanation.

Q1.
Simplify: A + A'B
AA
BA + B
CB
DA'B
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Correct Answer: B — A + B

Using absorption law: A + A'B = A + B.

Q2.
A + A = ?
A0
B1
CA
D2A
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Correct Answer: C — A

Idempotent law: A + A = A.

Q3.
A · A' = ?
A0
B1
CA
DA'
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Correct Answer: A — 0

Complement law: A AND NOT A = 0.

Q4.
A + A' = ?
A0
B1
CA
DA'
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Correct Answer: B — 1

Complement law: A OR NOT A = 1.

Q5.
De Morgan's theorem: (A·B)' = ?
AA' · B'
BA' + B'
CA + B
DA · B
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Correct Answer: B — A' + B'

De Morgan: NOT(A AND B) = NOT A OR NOT B.

Q6.
De Morgan's theorem: (A+B)' = ?
AA' + B'
BA' · B'
CA + B
DA · B
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Correct Answer: B — A' · B'

De Morgan: NOT(A OR B) = NOT A AND NOT B.

Q7.
Simplify: A·B + A·B'
AA
BB
C0
D1
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Correct Answer: A — A

A·B + A·B' = A(B + B') = A·1 = A.

Q8.
A + 0 = ?
A0
B1
CA
DA'
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Correct Answer: C — A

Identity law: A + 0 = A.

Q9.
A · 1 = ?
A0
B1
CA
DA'
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Correct Answer: C — A

Identity law: A · 1 = A.

Q10.
A + 1 = ?
A0
B1
CA
DA'
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Correct Answer: B — 1

Domination law: A + 1 = 1.

Q11.
What is the result of A + A'?
A0
B1
CA
DA'
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Correct Answer: B — 1

A + A' = 1. This is the complement law - a variable ORed with its complement always equals 1.

Q12.
Which law states that A + (B · C) = (A + B) · (A + C)?
AAssociative Law
BDistributive Law
CCommutative Law
DAbsorption Law
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Correct Answer: B — Distributive Law

This is the distributive law of addition over multiplication in Boolean algebra.

Q13.
What is the absorption law A + (A · B)?
AA + B
BA · B
CA
DB
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Correct Answer: C — A

A + (A · B) = A. This is the absorption law - A absorbs the term (A · B).

Q14.
What is the idempotent law for OR operation?
AA + 0 = A
BA + 1 = 1
CA + A = A
DA + A' = 1
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Correct Answer: C — A + A = A

A + A = A is the idempotent law for OR. ORing a variable with itself gives the same variable.

Q15.
Simplify: A · (A + B)
AA + B
BA · B
CA
DB
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Correct Answer: C — A

A · (A + B) = A. This is the absorption law - A absorbs the term (A + B).

Q16.
What is a minterm?
AA product term with all variables in complemented form
BA product term with each variable appearing once
CA sum term with all variables
DMinimum number of terms
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Correct Answer: B — A product term with each variable appearing once

A minterm is a product (AND) term that contains each variable of the function exactly once, either in true or complemented form.

Q17.
What is a maxterm?
AA product term with all variables
BA sum term with each variable appearing once
CMaximum product term
DTerm with most variables
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Correct Answer: B — A sum term with each variable appearing once

A maxterm is a sum (OR) term that contains each variable of the function exactly once, either in true or complemented form.

Q18.
Simplify: A' · B + A · B
AA
BB
CA + B
D0
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Correct Answer: B — B

A' · B + A · B = B · (A' + A) = B · 1 = B. Factor out B and apply complement law.

Q19.
What is the dual of the expression A + B · C?
AA · B + C
BA · (B + C)
CA' + B' · C'
D(A + B) · C
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Correct Answer: B — A · (B + C)

The dual is obtained by swapping AND with OR and vice versa. Dual of A + B · C is A · (B + C).

Q20.
What is the identity law for AND operation?
AA · 0 = 0
BA · 1 = A
CA · A = A
DA · A' = 0
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Correct Answer: B — A · 1 = A

A · 1 = A is the identity law for AND. ANDing any variable with 1 gives the variable itself.

Q21.
What is the null law for OR operation?
AA + 0 = A
BA + 1 = 1
CA + A = A
DA + A' = 1
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Correct Answer: B — A + 1 = 1

A + 1 = 1 is the null law for OR. ORing any variable with 1 always gives 1.

Q22.
What is SOP (Sum of Products)?
ASum of all minterms
BOR of AND terms
CAND of OR terms
DProduct of sums
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Correct Answer: B — OR of AND terms

SOP is a Boolean expression written as OR of AND terms (product terms), e.g., AB + CD + EF.

Q23.
What is POS (Product of Sums)?
ASum of products
BOR of AND terms
CAND of OR terms
DOnly minterms
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Correct Answer: C — AND of OR terms

POS is a Boolean expression written as AND of OR terms (sum terms), e.g., (A+B)(C+D)(E+F).

Q24.
K-map (Karnaugh map) is used for:
ACircuit layout
BBoolean expression simplification
CMemory mapping
DTiming analysis
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Correct Answer: B — Boolean expression simplification

Karnaugh maps provide a visual method to simplify Boolean expressions by grouping adjacent cells.

Q25.
In K-map, adjacent cells differ by:
ATwo variables
BOne variable only
CAll variables
DNo variables
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Correct Answer: B — One variable only

Adjacent cells in a K-map differ by exactly one variable, allowing simplification when grouped.