NCERT Solutions Class 12 Computer Science Boolean Algebra

Question 5: State and verify Associative law using Truth Table.
Аnswer: Associative Law: This law states that:
(A + B) + C = A + (B + C)
(A.B).C = A. (B.C)

Short Answer Type Questions-I 

Question 1: Correct the following Boolean statements :
1. X+1 = X
2. (A’)’ = A’
3. A+A’ = 0
4. (A+B)’ = A.B
Аnswer:
1. X+l=l or X+0=X
2. ((A’)’) = A
3. A + A’ = 1 or A. A’ = 0
4. (A 4- B)’ = A’.B1

Question 2: Write the POS form of a Boolean Function F, which is represented in a truth table as follows

Аnswer:
(P+Q+R).(P’+Q+R).(P’+Q’+R)

Short Answer Type Questions-II 

Laws and Theorems

Question 1: State and Verify Absorption law algebraically.
Аnswer: Absorption law states that :
A + AB = A and A. (A + B) = A
Algebraic method :
Taking LHS
A + AB = (A.l) + (A.B) by Identity
= A. (1 + B) by Distribution
= A.l by Null Element
= A

Question 2: State and define principle of duality. Why is it so important in Boolean Algebra ?
Аnswer:
Principle of duality : Duality principle states that from every Boolean relation another Boolean relation can be derived by :
(i) Changing each OR sign (+) to an AND sign (-).
(ii) Changing each AND sign (-) to an OR sign (+)
ex : Dual of A + A’B = A. (A’ + B)
Importance in Boolean Algebra : The principle of duality is an important concept in Boolean algebra, particularly in proving various theorems. The principle of duality is used extensively in proving Boolean algebra theorem. Once we prove that an expression is valid, by the principle of duality, its dual is also valid. Hence, our effort in proving various theorems is reduced to half.

Question 3: Name the law shown below & verify it using a . truth table.

Question 6:
Draw the Logic Circuit for the following Boolean Expression :
(U + V). w + z

Question 7: Verify the following using Boolean Laws :
LT + V = LTV + LP.V + U.V
Аnswer:
L.H.S.
= U’ + V
= U’ . (V + V) + V (LP + U)
= U’ . V + LP . V + U . V + U. V
= U’. V + LP. V + U. V
= R.H.S.
OR
R.H.S.
= U’V’ + U’. V + U. V
= LP . (V + V) + U. V
= U’ 1 + U.V
= U’ + U.V
= U’ +V
= L.H.S.

Question 8: Draw the Logic Circuit for the following Boolean Expression :
(X’ + Y). Z + W’
Аnswer:

Question 9: Write the equivalent Boolean expression for the following logic circuit.

Question 17: Verify the following using Boolean Laws X + Z = X + X’. Z + Y. Z
Аnswer:
Taking RHS
X + X’Z + YZ
= (X + X’). (X + Z) + YZ (Distribution Law)
= 1. (X + Z) + YZ (A + A’ = 1)
= X + Z + YZ
= X + Z (1 + Y)
= X + Z (1 + A = 1; 1. A = A)
= Hence verified

Question 18: Verify the following using Boolean Laws : A + C = A + A. C + B.C
Аnswer:
A + C = A + A’.C + BC
Solve RHS
A + A’C + BC
(A + A). (A + C) + BC [Using distributive law]
1. (A + C) + BC
= A + C + BC
= A + C(1 + B)
= A + C.1
= A + C
= LHS (Hence, verified)

Question 19: Obtain the Boolean Expression for the logic circuit shown below :

Question 24: Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table :

Аnswer:
F(A, B, C) = A’B’C + A’BC + AB’C + ABC
OR
F(A,B,C) =Σ(0, 3,4,7)

Question 25: Derive a Canonical POS expression for a Boolean function F, represented by the following truth table :

Аnswer:
F(RQ,R) = (P+Q+R’)(P+Q,+R)(P’+Q,+R’) (P’+Q’+R)
OR
F(RQ,R)=II(1,2,5,6)

Question 26: Obtain a simplified form for a Boolean expression :
F(U, V, W, Z) = II (0,1,3,5, 6, 7,15)
Аnswer:

Question 27: Reduce the following Boolean Expression to its simplest form using K-Map :
F (X, X Z, W) = X (0,1, 6, 8, 9,10,11,12,15)
Аnswer:

Question 29: Verify the following using Boolean Laws.
X + Y’ = X. Y + X. Y + X’. Y
Аnswer:
L. H. S.
= X + Y’
= X. (Y+Y’) + (X + X’). Y’
= X. Y + X. Y’ + X. Y’ + X’. Y’
= X. Y + X. Y’ + X’. Y’
= R. H. S
OR
= X. Y + X. Y’ + X’. Y’
= X. (Y + Y’) + X’. Y’
= X. 1 + X’. Y’
= X + X’. Y’
= X + Y
= L. H. S

Question 30: State Distributive law and verify it using truth table.
Аnswer:
Distributive law : This law states that
(i) x(y + z) = xy + x.z.

Question 31: Reduce the following Boolean Expression using K Map :
F(A, B, C, D) = Σ{0,1,3, 5, 6, 7,9,11,13,14,15}
Аnswer:

TOPIC-2 Karnaugh Map Minimization and Applications of Boolean Algebra

Very Short Answer Type Questions 

Question 1: Write Product of Sum expression of the function F (a, b, c, d) from the given truth table

Аnswer:

F (a, b, c, d) =
(a + b + c + d).(a + b + c + d’). (a + b’ + c + d) . (a + b’ + c’ + d’). (a’ + b + c + d).
(a’ + b + c + d’). (a’ + b’ + c + d). (a’ + b’ + c + d’) . (a’ + b’ + c’ + d)

Question 2: Convert the following Boolean expression inti! its equivalent Canonical Sum of Products form (SOP) :
(U + V + W) (U + V + W’) (U’ + V + W) (U’ + V’ + W’)
Аnswer:
π (0,1, 4, 7)
Σ(2, 3, 5, 6)
010 011 101 110
= U’VW’ + U’VW + UV’W + UVW’

Question 3: Write the Sum of Product form of the function F(R Q, R) for the following truth table representation of F :

Question 5:Write the Product of Sum form of the function G(U, V W) for the following truth table representation of G :

Question 6: Write the Product of Sum form of function G(U, V, W) for the following truth table representation of G :

Question 8: Write the SOP form of a Boolean function F, which is represented in a truth table as follows:

Аnswer:
F(X, Y, Z) = X’.Y’. Z’ + X’. Y. Z’ + X. Y’. Z’+ X.Y.Z

Question 9: Write the POS form of Boolean function G, which is represented in a truth table as follows :

Аnswer:
G (A, B, C) = (A + B + C). (A + B’ + C’). (A’ + B + C). (A’ + B + C’)

Short Answer Type Questions-II

Question 1: Obtain the minimal SOP form for the following Boolean expression using K-Map.
F(A,B,C,D) = Σ (0,2,3,5,7,8,10,1143,15)
Аnswer:

Quad 1 = m0 + m2 + m8 + m10 = B’D’
Quad 2 = m3 + m7 + m15 + m11 = CD
Quad 3 = m5 + m7 + m15 + m13 = BD
Minimal SOP = B’D’ + CD + BD

Question 2: Reduce the following Boolean expression using
K-Map :
F(A,B,C,D) = 7r (0,1,2,4,5,6,8,10)
Аnswer:

F(A, B, C, D) = π(0,1,2,4, 5, 6,8,10) F = (A + C).(A + D).(B + D)

Question 3: Reduce the following using K-Map :
F (A, B,C,D) = Σ(1,3,4,5,6,7,12,13)
Аnswer:

Question 4: Reduce the following Boolean expression using K-map.
F(EQ,R,S) = 2(0,2,4,5,6,7,8,10,13,15).
Аnswer:

Question 5: Reduce the following Boolean Expression using K-Map :
F(P, Q, R, S) = Σ(1,2, 3,4,5, 6, 7, 8,10)
Аnswer:

Question 6: Reduce the following Boolean Expression using
K-Map :
F (A, B, C, D) = Σ(2, 3,4,5, 6, 7,8,10,11)

Long Answer Type Questions 

Question 1: Verify the following using Boolean Laws :
A ‘ + B’ . C=A’ . B ‘ . C ‘ + A’ . B . C ‘ + A’ .B.C + A’ .B’ .C+ A.B’ .C
Аnswer:
A’+ B’C = A’B’C’ + A,BCI + A’BC’ + A’BC + A’B’C + ABC
=A’C'(B’+B)+A’C (Grouping)
(B+B’)+AB’C
=> A’C’+ A’C + AB’C
(x+x’y=x+y)
=> A’ (C+C’ ) +AB’ C
=> A’+AB’C
(x+x’=1)
=> A’+B’C
X=A’ y=B’C
= LHS
Hence Proved.

Question 2: Write the Boolean Expression for the result of the Logic Circuit as shown below :

Аnswer:
F = (u+v’).(u+w).(v+w’)

Question 3: Derive a Canonical POS expression for a Boolean function F, represented by the following truth table :

Аnswer:
F = Σ(0, 3,4,5)
= (P + Q + R) (P + Q’ + R’) (P’ + Q + R) (P’ + Q + R’)

Question 4: Reduce the following Boolean Expression to its simplest form using K-Map :
F (X, Y, Z, W)
Σ(2,6,7,8,9,10,11,13,14,15)
Аnswer:
Σ(2,6,7,8,9,10,11,13,14,15)

Question 7: Verify the following using Boolean Laws :
X’+ Y’Z = X’ .Y’ .Z’+X’ .Y.Z’+X’ .Y.Z+X’ .Y’ .Z + X.Y’.Z.
Аnswer:
X ‘+Y’ Z=X ‘ Y ‘ Z ‘ +X ‘ YZ ‘ +X ‘ YZ+X ‘ Y ‘ Z+XY ‘ Z
Taking RHS
Grouping terms
=> x’Z’ (Y’+Y)+X’ Z(Y+Y’)+XY’Z
=> X’Z’+X’Z+XY’Z
(Y+Y’=1)
=> X'(Z’+Z)+XY’Z
(Grouping)
=> X’+XY’Z (Z+Z’=l)
=> X’+Y’Z (Substitute X=X’ Y=Y’Z X+X’ Y = X+Y)
= LHS

Question 8: Reduce the following Boolean Expression to its simplest form using K-Map :
F(P,Q,R,S) = Σ(0,4,5,8,9,10,11,12,13,15)
Аnswer: F(P,Q,R,S) = Σ(0,4,5,8,9,10,11,12,13,15)