Chapter 8: Boolean Algebra Class 12 Computer Science NCERT Solutions

Chapter 8 – Boolean Algebra introduces students to the mathematical framework used in digital logic and computer programming. Boolean Algebra deals with binary variables (0 and 1) and logical operations such as AND, OR, and NOT. These concepts form the foundation of digital electronics, computer architecture, and logic circuit design.

The chapter explains how logical expressions can be simplified using Boolean identities, truth tables, and logic gates. It also focuses on designing logic circuits and solving real-world problems using Boolean logic. This topic is not only critical for board exams but also builds conceptual clarity for fields like electronics, programming, and software design.

What You Will Learn in Chapter 8

This chapter helps students understand how digital systems make logical decisions using Boolean expressions. It covers logical operators, Boolean laws, truth tables, and simplification techniques. The chapter also teaches how to design logic circuits based on Boolean expressions and verify them using truth tables.

Key Topics Covered

Introduction to Boolean Algebra

Boolean Algebra is a branch of mathematics that deals with binary values (0 and 1).
Used extensively in designing digital circuits and logical reasoning in computer science.

Boolean Operations

AND (·) – True only if both inputs are true.
OR (+) – True if at least one input is true.
NOT (‾ or ‘) – Reverses the truth value (0 becomes 1, and 1 becomes 0).

Truth Tables

A truth table lists all possible input combinations and their corresponding output.
Helps verify the correctness of logical expressions.

Boolean Laws and Identities

Important rules include:
- Identity Law: A + 0 = A, A · 1 = A
- Null Law: A + 1 = 1, A · 0 = 0
- Complement Law: A + A’ = 1, A · A’ = 0
- Idempotent Law, Associative, Commutative, Distributive, and De Morgan’s Theorems.

Simplification of Boolean Expressions

Use Boolean identities to reduce expressions to their simplest form.
Simplified expressions help design efficient logic circuits.

Canonical Forms

Sum of Products (SOP) and Product of Sums (POS) forms.
SOP: OR of multiple AND terms.
POS: AND of multiple OR terms.

Karnaugh Maps (K-Maps)

A visual method to simplify Boolean expressions.
Used for minimizing logical expressions in 2, 3, or 4 variables.
Reduces complexity in circuit design.

Logic Gates and Circuits

Basic gates: AND, OR, NOT
Derived gates: NAND, NOR, XOR, XNOR
Constructing and interpreting logic circuits using gates.
Conversion between Boolean expressions and logic circuits.

Download Chapter 8 Solutions PDF – Boolean Algebra

Our downloadable PDF includes:

Complete NCERT textbook question-answers
Topic-wise theory with diagrams and solved examples
Truth tables, Boolean expressions, and circuit diagrams
Step-by-step simplification using Boolean laws and K-maps
Exam-ready revision notes and high-scoring tips

Highlights of Our NCERT Solutions

Easy-to-follow explanations of Boolean laws and operations
Visual aids: truth tables, logic gate diagrams, and Karnaugh maps
Solved examples with real-world logic circuit applications
Board exam-focused solutions with structured format
Practice exercises for SOP/POS and expression minimization

Recommended Preparation Tips

Learn and memorize key Boolean identities and law
s.
Practice constructing and simplifying expressions using truth tables and K-maps.
Solve SOP and POS problems to understand canonical forms.
Master the basic and derived logic gates with circuit diagrams.
Understand how to draw logic circuits from Boolean expressions and vice versa.

Additional Study Resources

Flashcards: Boolean laws and logic symbols
Worksheets on SOP/POS conversions and truth tables
K-map templates and simplification practice sheets
CBSE sample papers and PYQs for Boolean Algebra
Interactive quizzes on logic gate combinations and simplification

Mastering Chapter 8 – Boolean Algebra

Boolean Algebra plays a vital role in understanding how digital computers and logic circuits operate. Mastery of this chapter gives students a strong foundation in logical reasoning and circuit design — key skills for computer science, electronics, and engineering careers.

Strong conceptual knowledge of Boolean expressions, simplification techniques, and logic gate design not only boosts board exam scores but also empowers students to excel in technical domains like programming, AI, embedded systems, and competitive exams.

Class 12 Computer Science (Python) – Chapter 8 Boolen Algebra

NCERT Textbook Questions Solved – Class 12 Computer Science (Python)

Question 1: Express the OR operator in terms of AND and NOT operator.

Express the OR operator in terms of AND and NOT operator.

Answer: A + B = (A′ . B′)′ (using De Morgan’s Law)

Question 2: State the following Boolean laws:

State the following Boolean laws:

x + x′ = 1 — Complementary Law
xy + x = x — Absorption Law

Question 3: Which gates are known as universal gates? Why?

Which gates are known as universal gates? Why?

Answer: NAND and NOR gates are universal gates because any Boolean function can be implemented using only NAND or only NOR gates.

Question 4: State and verify Associative Law using Truth Table:

State and verify Associative Law using Truth Table:

A	B	C	(A + B) + C	A + (B + C)	(A . B) . C	A . (B . C)
0	0	0	0	0	0	0
0	0	1	1	1	0	0
0	1	0	1	1	0	0
0	1	1	1	1	0	0
1	0	0	1	1	0	0
1	0	1	1	1	0	0
1	1	0	1	1	1	1
1	1	1	1	1	1	1

Question 5: Draw the equivalent logic circuit for the following Boolean expression:

Draw the equivalent logic circuit for the following Boolean expression:

Expression: (A + B) . (A + C)

Answer: Use two OR gates for A + B and A + C, and one AND gate for the final output.

   A ─┬─────┐
      │     │
      │     └────► OR ─────┐
      │                   │
   B ─┘                   │
                          ▼
                        AND ───► Output
                          ▲
   A ─────┐               │
          │               │
          └────► OR ──────┘
                ▲
              C ─

Short Answer Type Questions

Question 1: State and define principle of duality. Why is it so important in Boolean Algebra?

Answer:

Principle of Duality: The principle of duality in Boolean algebra states that for every Boolean expression, another valid Boolean expression (called its dual) can be obtained by:

Replacing every OR (+) operator with an AND (·) operator
Replacing every AND (·) operator with an OR (+) operator

Example:
Original expression: A + A′B
Dual expression: A · (A′ + B)

Importance: The duality principle is useful because once a Boolean identity is proven true, its dual is also automatically true. This helps reduce the effort required when proving Boolean theorems.

Question 2: Write the equivalent Boolean expression for the following logic circuit:

Answer:

The equivalent Boolean expression for the given circuit is:

((X′ · Y′)′ + (X′ · Y′)′)′

This expression represents a combination of NAND and OR gates applied to the complements of variables X and Y.

Question 3: Write the equivalent Boolean expression for the following logic circuit:

Answer:

Expression: X · (Y + Z)

This expression shows that variable X is ANDed with the result of Y OR Z, using one OR gate and one AND gate.

Question 4: Write the equivalent Boolean expression for the following logic circuit:

Answer:

Expression: (X + Y) · (Y + Z)

This circuit uses two OR gates and one AND gate to produce the final output.

Question 5: Name the law shown below and verify it using a truth table:

Answer:

This is the Absorption Law (also called Redundancy Law).

Law: A + AB = A

Truth Table:

A	B	AB	A + AB
0	0	0	0
0	1	0	0
1	0	0	1
1	1	1	1

Since the output of A + AB is always equal to A, the law is verified.

Class 12 Computer Science (Python) – Chapter 8: Boolean Algebra

Short Answer Type Questions (Continued)

Question 6: Obtain the Boolean expression for the logic circuit shown below:

Answer: (A + B) · (C + D)

Question 7: Verify the following Boolean identities using truth tables:

Identities:

X + 0 = X
X + X′ = 1

(i) Truth Table for X + 0 = X

X	X + 0
0	0
1	1

(ii) Truth Table for X + X′ = 1

X	X′	X + X′
0	1	1
1	0	1

Question 8: Obtain the Boolean expression for the following logic circuit:

Answer: A · B + A′ · C

Question 9: State De Morgan’s Laws:

De Morgan’s Laws:

(A · B)′ = A′ + B′
(A + B)′ = A′ · B′

Verification of: (A · B)′ = A′ + B′ using truth table

A	B	A · B	(A · B)′	A′	B′	A′ + B′
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Conclusion: The column for (A·B)′ matches with A′ + B′, hence verified.

Question 10: Draw the logic circuit for the expression: (A + B′) · C

Answer:

Use a NOT gate on B to get B′
Use an OR gate to combine A and B′
Use an AND gate to combine output of OR gate with C

Question 11: Obtain the Boolean expression for the following logic circuit:

Answer: A′ · B + C

Question 12: Verify the Boolean identity: X + Z = X + X′Z + YZ

Verification using Boolean Laws:

Given RHS: X + X′Z + YZ

Step 1: Apply Distributive Law on X′Z + YZ
= Z(X′ + Y)

LHS becomes:
X + Z(X′ + Y)

Step 2: Apply Distributive again:
= X + Z · X′ + Z · Y

Step 3: Apply Absorption:
= X + Z

Conclusion: LHS = RHS. Verified.

Question 6: Obtain the Boolean expression for the logic circuit shown below:

Answer: (A + B) · (C + D)

Question 7: Verify the following Boolean identities using truth tables:

Identities:

X + 0 = X
X + X′ = 1

(i) Truth Table for X + 0 = X

X	X + 0
0	0
1	1

(ii) Truth Table for X + X′ = 1

X	X′	X + X′
0	1	1
1	0	1

Question 8: Obtain the Boolean expression for the following logic circuit:

Answer: A · B + A′ · C

Question 9: State De Morgan’s Laws:

De Morgan’s Laws:

(A · B)′ = A′ + B′
(A + B)′ = A′ · B′

Verification of: (A · B)′ = A′ + B′ using truth table

A	B	A · B	(A · B)′	A′	B′	A′ + B′
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Conclusion: The column for (A·B)′ matches with A′ + B′, hence verified.

Question 10: Draw the logic circuit for the expression: (A + B′) · C

Answer:

Use a NOT gate on B to get B′
Use an OR gate to combine A and B′
Use an AND gate to combine output of OR gate with C

Question 11: Obtain the Boolean expression for the following logic circuit:

Answer: A′ · B + C

Question 12: Verify the Boolean identity: X + Z = X + X′Z + YZ

Verification using Boolean Laws:

Given RHS: X + X′Z + YZ

Step 1: Apply Distributive Law on X′Z + YZ
= Z(X′ + Y)

LHS becomes:
X + Z(X′ + Y)

Step 2: Apply Distributive again:
= X + Z · X′ + Z · Y

Step 3: Apply Absorption:
= X + Z

Conclusion: LHS = RHS. Verified.

Question 13: Verify the Boolean identity A + C = A + A′C + BC

Verification using Boolean Laws:

RHS: A + A′C + BC

Step 1: Apply Distributive Law: A + C(A′ + B)

Step 2: Apply Absorption Law: A + C

Conclusion: LHS = RHS. Verified.

Question 14: Obtain the Boolean expression for the following logic circuit

Answer: (A · B) + (C · D)

Question 15: Verify the following Boolean identities using truth tables

(i) X · X′ = 0

X	X′	X · X′
0	1	0
1	0	0

(ii) X + 1 = 1

X	1	X + 1
0	1	1
1	1	1

Question 16: Write the Boolean expression for the given logic circuit

Answer: (A · B) + (C · D)

Question 17: Distribution Law

(i) x(y + z) = xy + xz
(ii) x + yz = (x + y)(x + z)

Verification of (i) x(y + z) = xy + xz using truth table

x	y	z	y + z	x(y + z)	xy	xz
0	0	0	0	0	0	0
0	0	1	1	0	0	0
0	1	0	1	0	0	0
0	1	1	1	0	0	0
1	0	0	0	0	0	0
1	0	1	1	1	0	1
1	1	0	1	1	1	0
1	1	1	1	1	1	1

Verified: LHS = RHS

Question 18: Verify the Distributive Law A + BC = (A + B)(A + C)

A	B	C	BC	A + BC	A + B	A + C
0	0	0	0	0	0	0
0	0	1	0	0	0	0
0	1	0	0	0	1	0
0	1	1	1	1	1	1
1	0	0	0	1	1	1
1	0	1	0	1	1	1
1	1	0	0	1	1	1
1	1	1	1	1	1	1

Verified: A + BC = (A + B)(A + C)

Question 19: Write the POS (Product of Sums) form from given truth table

Answer: F(P, Q, R) = (P + Q + R)(P′ + Q + R)(P′ + Q′ + R)

Question 20: Write the Boolean expression for the given logic circuit

Answer:

I = [(A + B) · (A′ + C′)] · (B + C)

Question 21: Canonical POS (Product of Sums) form

Answer: From the truth table, take the rows where output = 0 and write sum terms for each.

Example:

G(P, Q, R) = (P + Q′ + R)(P′ + Q + R)(P + Q + R′)

Question 22: Identify and correct Boolean expression mistakes

Incorrect: (A + A′) = A → Correct: A + A′ = 1 (Complement Law)
Incorrect: A · A = 0 → Correct: A · A = A (Idempotent Law)

Question 23: Canonical SOP (Sum of Products) form

Answer: From the truth table, take the rows where output = 1 and write product terms for each.

Example: G(P, Q, R) = P′Q′R + PQ′R + PQR′

Question 24: Derive a Canonical POS expression from the given truth table

A	B	C
0	0	0
0	1	0
0	1	1
1	0	1

Canonical POS expression:
F = (A + B + C)(A + B′ + C)(A + B′ + C′)(A′ + B + C′)

Question 25: Write the Boolean Expression for the given logic circuit

Answer: C = P.Q′ + P.R + Q.R′

Question 26: Verify the identity P′ + Q′.R = P′.Q′.R′ + P′.Q.R′ + P′.Q.R + P′.Q′.R + P.Q′.R

Verification:


RHS = P′(Q′R′ + QR′ + QR + Q′R) + P.Q′.R
    = P′(Q′ + Q)(R′ + R) + P.Q′.R
    = P′.1.1 + P.Q′.R
    = P′ + P.Q′.R = LHS

Question 27: Verify the identity x′ + y′z = x′y′z′ + x′y′z + x′yz + xy′z

Verification:


RHS = x′y′z′ + x′y′z + x′yz + xy′z
     = x′(y′z′ + y′z + yz) + xy′z
     = x′(y′ + y) + xy′z
     = x′.1 + xy′z = x′ + xy′z = LHS

Question 28: State and prove De Morgan’s laws in Boolean algebra

De Morgan’s Laws:

(X + Y)’ = X’.Y’
(X · Y)’ = X’ + Y’

Verification of (X + Y)’ = X’.Y’ using truth table:

X	Y	X + Y	(X + Y)′	X′	Y′	X′ · Y′
0	0	0	1	1	1	1
0	1	1	0	1	0	0
1	0	1	0	0	1	0
1	1	1	0	0	0	0

Verification of (X · Y)’ = X’ + Y’ using truth table:

X	Y	X · Y	(X · Y)′	X′	Y′	X′ + Y′
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

Conclusion: Both of De Morgan’s laws are verified using truth tables.

Question 29: Write the equivalent Boolean expression for the following logic circuit

Answer: F = (A + B) · (A + C)

Question 30: Verify A + B.C = (A + B)(A + C) using truth table

A	B	C	B·C	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

Question 31: Write the equivalent Boolean expression for the following logic circuit

Answer: F = (X + Y′) · (X + Z′)

Question 32: Verify X + Y.Z = (X + Y)(X + Z) using truth table

X	Y	Z	Y·Z	X + Y·Z	X + Y	X + Z	(X + Y)(X + Z)
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

Question 33: Write the equivalent Boolean expression for the following logic circuit

Answer: F = P·Q′ + P·R′ = P(Q′ + R′)

Question 34: State and verify De Morgan’s laws algebraically

Answer:

(X + Y)′ = X′ · Y′
(X · Y)′ = X′ + Y′

Verification: Apply truth tables and distributive laws as shown in earlier questions.

Question 35: Write the equivalent Boolean expression for the following logic circuit

Answer: F(P, Q) = (P′ + Q) · (P + Q′)

Question 36: State and algebraically verify the Absorption Law

Answer:

A + A′B = A + B
A · (A + B) = A

Verification: Apply the distributive and complement laws as shown in earlier questions.

Question 37: Write the equivalent Boolean expression for the following logic circuit

Answer: F(U, V) = (U′ + V) · (U + V′)

Question 38: Represent the expression X′ · Y + Y′ · Z + Z′ · X using NAND gates

Answer:

X′ · Y = NAND(NAND(X, X), Y)
Y′ · Z = NAND(NAND(Y, Y), Z)
Z′ · X = NAND(NAND(Z, Z), X)
Combine using NAND-implemented OR: OR = NAND(NAND(A, A), NAND(B, B))

Question 39: Write the equivalent expression for the given logic circuit

Answer: F = A′ · B′ + B′ · C′

Question 40: Verify algebraically (A′ + B′)(A + B) = A′B + AB′

Verification:


(A′ + B′)(A + B) = A′A + A′B + B′A + B′B
                = 0 + A′B + AB′ + 0
                = A′B + AB′

Hence verified.

Question 41: Write the equivalent boolean expression for the following logic circuit

Answer: Y = (P + Q’) . (Q + R’)

Question 42: Verify algebraically – X’ . Y + X . Y’ = (X’ + Y’) . (X + Y)

Answer:
Apply distributive law:
(X’ + Y’) . (X + Y) = X’.X + X’.Y + Y’.X + Y’.Y
Since X’.X = 0 and Y’.Y = 0:
The expression becomes: X’.Y + X.Y’ which matches LHS.
Verified.

Question 43: Write the equivalent boolean expression for the following logic circuit

Answer: (U’ + V) . (V’ + W)

Topic-2: Karnaugh Map Minimization and Applications of Boolean Algebra

Question 1: Convert to Canonical SOP

Answer: (U + V + W)(U + V + W’)(U’ + V + W)(U’ + V’ + W’)

Question 2 to Question 7

Answer: Refer to the respective truth tables for detailed SOP/POS expressions.

Question 8: SOP form of F(X, Y, Z)

Answer: F = X’.Y’.Z’ + X’.Y’.Z’ + X.Y’.Z’ + X.Y.Z

Question 9: POS form of G(A, B, C)

Answer: G = (A + B + C)(A + B’ + C’)(A’ + B + C)(A’ + B + C’)

Question 10: SOP form of F(X, Y, Z)

Answer: F = X’Y’Z’ + X’Y’Z + X’YZ + XY’Z’ + XYZ

Question 11: POS form of F(A, B, C)

Answer: F = (A + B + C)(A + B’ + C’)(A’ + B + C’)(A’ + B’ + C)

Question 12: SOP form of F(U, V, W)

Answer: F = U’.V’.W’ + U’.V.W + U.V.W’ + U.V.W

Question 13: POS form of F(U, V, W)

Answer: F = (U + V + W’)(U + V’ + W’)(U’ + V + W’)

Question 14: SOP form of G(P, Q, R)

Answer: G = P’.Q.R’ + P.Q’.R’ + P.Q.R’ + P.Q.R

Question 15: POS form of H(x, y, z)

Answer: H = (x + y + z’)(x’ + y + z’)(x’ + y’ + z)

Question 16:

Write the SOP form of a Boolean function G, which is represented in a truth table as follows:

Answer: G(P, Q, R) = P’QR’ + P’QR + PQ’R’ + PQR’ + PQR

Question 17:

Write the POS form of a Boolean function G, which is represented in a truth table as follows:

Answer: G(U, V, W) = (U + V + W) · (U’ + V + W’) · (U’ + V’ + W)

Question 18:

Write the POS form of a Boolean function F, which is represented in a truth table as follows:

Answer: F(U, V, W) = (U + V + W’) · (U + V’ + W’) · (U’ + V + W’)

Question 19:

Reduce the following Boolean Expression to its simplest form using K-map:

Answer: Refer to the K-map steps for simplification of the Boolean expression.

Question 20:

Write the SOP form of a Boolean function G, which is represented in a truth table as follows:

Answer: G(P, Q, R) = P’Q’R’ + PQ’R’ + PQR’ + PQR

Question 21:

Write the POS equivalent of ABC + AB’C’ + ABC’ + A’B’C:

Answer: (A + B + C) · (A’ + B + C) · (A + B’ + C’) · (A’ + B’ + C)

Truth Tables and Explanation

Question 16 Truth Table:

P	Q	R	G(P, Q, R)
0	0	0	0
0	0	1	1
0	1	0	1
0	1	1	1
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	1

Question 17 Truth Table:

U	V	W	G(U, V, W)
0	0	0	1
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	1
1	0	1	0
1	1	0	0
1	1	1	0

Question 18 Truth Table:

U	V	W	F(U, V, W)
0	0	0	1
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	0
1	1	1	1

Question 21 Truth Table:

A	B	C	F(A, B, C)
0	0	0	1
0	0	1	1
0	1	0	1
0	1	1	1
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	0