NCERT Solutions for Class 12 Computer Science (C++) – Boolean Algebra

Computer ncert solution of class 12 Boolean Algebra, we have covered all the chapter questions. This solution will prove to be beneficial for the board examination as well as for students who are preparing for graduate(B.Tech) in information technology and preparing for Post Graduate(M.C.A).

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NCERT Solutions for Class 12 Computer Science (C++) having 14th Chapter whose Chapter wise Solution given below.

                          

                      Very Short Answer Type Questions [1 mark each]

                          

Question 1:
Which gates are known as universal gates ? Why?
Аnswer:
Universal gates are the ones which can be used for implementing any gate like AND, OR and NOT or any combination of these basic gates. NAND and NOR gates are universal gates.

Check the boolean algebra calculator, Look at How easily to solve Boolean Expressions.

Question 2:
Draw the equivalent logic circuit for the following Boolean expression :
Аnswer:

Question 3:
Express the OR operator in terms of AND and NOT operator.
Аnswer:
(A . B)’ =  + 
( + )’ = A + B

Question 4:
Specify which axioms/theorems are being used in the following Boolean reductions :
(a) (be)’ + be = 1
(b) xyz + zx = xz
Аnswer:
(a) x + x’ = 1 & Complementary law
(b) y + x = x & Absorption law.

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)
Proof:

∴ From above truth table,
(A + B) + C = A + (B + C)
Similarly, we can prove,
A. (B.C) = (A. B).C

Short Answer Type Questions-I    [2 mark each]

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  [3 mark each]

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.
X+ .Y = X + Y.
Аnswer:
This law is called “Absorption Law” also referred as redundance law.

Question 4:
Draw a logic circuit for the following Boolean expression : ab + c.d’.
Аnswer:

Question 5:
Write the SOP form of a Boolean function F, which is represented in a truth table as follows :

Аnswer:
A’B’C + A’BC + AB’C + AB’C

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

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.

Аnswer:
((X’.Y)’ + (X.Y’)’)’

Question 10:
Write the equivalent Boolean Expression for the. following Logic Circuit :

Аnswer:
Z = (A+B)(B’ +C)
= A.B’ + AC + B.B’ + BC
= A.B’ + AC + BC

Question 11:
Obtain the Boolean Expression for the logic circuit shown below :

Аnswer:
F = 
= 

Question 12:
Name the law shown below & verify it using a truth table.
A + B . C = (A + B). (A + C).
Аnswer:
This law is called “Distributive Law”.
Prove by Truth table

Question 13:
Obtain the Boolean Expression for the logic circuit shown below :

Аnswer:
F = ( X. ) + (  + W).
F =  + Y +  + W.

Question 14:
State Demorgan’s law. Verify one of them using truth table.
Аnswer:
There are two Demorgan’s law :
(i)   =  + 
(ii)   =  . 
Proof :
(i)   =  + 

Question 15:
Draw a logic Circuit for the boolean expression:
A .  + (C + ). 
Аnswer:

Question 16:
Obtain the Boolean Expression for the logic circuit shown below :

F = P’ Q + (Q + R’)
= Q. (P’ + R’)

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 :

Expression at F :
(  . Y) + (Y +  )
(  + 1) Y +     [Distributive law]
Y +     [ ∴ 1 + Z = 1]

Question 20:
Verify the following using truth table :
(i) X . X’ = 0
(ii) X + 1 = 1
Аnswer:

Question 21:
Write the equivalent boolean expression for the following logic circuit :

Аnswer:
Y = U  +  

Question 22:
Write the equivalent boolean expression for the following logic circuit :

Аnswer:
Y = (U +  ). (U +  )

Question 23:
Verify the following using truth table :
(i) X + 0 = X
(ii) X + X’ = 1
Аnswer:

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) =∑(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:

(u + v + w).(u+z’).(v’+w’).(u’+w’+z)

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:


Simplified Expression : XY + Y’Z’ + XZ’W’ + XZW + X’YZW’

Question 28:
Reduce the following Boolean Expression to its simplest form using K-Map :
F(X, Y, Z, W) = X (0,1,4, 5,6, 7,8, 9,11,15)
Аnswer:


Simplified Expression : Y’Z’ + XY + XZW

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.
(ii) x + yz = (x + y)(x + z)

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

A’B’C’+D+BC

TOPIC-2
Karnaugh Map Minimization and Applications of Boolean Algebra

Very Short Answer Type Questions [1 mark each]

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)
∑(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 4:
Write the Product of Sum form of the function F(X, Y, Z) 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 7:Write the Sum of Product form of the function F(A, B, C) for the following truth table reprsentation of

SOP = A’BC’ + A’BC + AB’C’ + ABC

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) = ∑ (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) = ∑(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:
 Q +  + QS.

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

F(P, Q, R, S) = P’Q + P’S + P’R’S’ + PQ’S’

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

F (A, B, C, D) = A’B + A’C + B’C + ABD’

Long Answer Type Questions [4 marks each]

Question 1:
Verify the following using Boolean Laws :
[Delhi, 2016]
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 Boolen function F, represented by the following truth table :

Аnswer:
F = ∑(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)
∑(2,6,7,8,9,10,11,13,14,15)
Аnswer:
∑(2,6,7,8,9,10,11,13,14,15)

F = XY’ + ZW’+ XW + YZ

Question 5:
Write the Boolean Expression for the result of the Logic Circuit as shown below :

Аnswer:
G = PQ’ + PR + QR’

Question 6:
Derive a Canonical SOP expression for a Boolean function G, represented by the following truth table :

Аnswer:
G = ∑{(0,2,6,7)
= A’B’C’ + A’BC’ + ABC’ + ABC

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) = ∑(0,4,5,8,9,10,11,12,13,15)
Аnswer:
F(P,Q,R,S) = ∑(0,4,5,8,9,10,11,12,13,15)

F = R’ S ‘ + PQ ‘ + QR’ + PS

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NCERT Solutions for Class 12 Computer Science