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BOOLEAN ALGEBRA

(Contains CBSE Board Exam QP from 1998 to 2010)

[1998-Outside Delhi]

6(a)

a) State Absorption Law.Verify one of the absorption law using truth table. [2]

Ans: Absorption Law:

(i) A+AB=A

(ii)A(A+B)=A

|1 |2 |3 |4 |

|A |B |AB |A+AB |

|0 |0 |0 |0 |

|0 |1 |0 |0 |

|1 |0 |0 |1 |

|1 |1 |1 |1 |

Col. No.1 = Col.No.4

Therefore : A+AB=A

b) Prove X+Y’Z = (X+Y’+Z’) (X+Y’+Z) (X+Y+Z) algebraically. [2]

Ans: LHS.= X+Y’Z

=(X+Y’)(X+Z) (Distributive Law)

=(X+Y’+ZZ’)(X+Z+YY’) (ZZ’=0 and YY’=0)

= (X+Y’+Z)(X+Y’+Z’)(X+Y+Z)(X+Y’+Z) ( Distributive Law)

=R.H.S.

c) Write the dual of the Boolean Expression (UV+W) (V’+U). [1]

Ans: Dual is (U+V)W+V’U

d) Obtain a simplified form for a Boolean Expression : F(x,y,z,w) = Σ(0,1,3,4,5,6,7,9,10,11,13,15) using karnaugh map. [2]

Ans: K-Map:

| |Z’W’ |Z’W |ZW |ZW’ |

|X’Y |1 |1 |1 |0 |

|X’Y |1 |1 |1 |1 |

|XY |0 |1 |1 | 0 |

|XY’ |0 |1 |1 |1 |

There are 2 groups of four 1s and 2 groups of two 1s and 1 group of 8 ones. Therefore the expression will be as follows:

F(x,y,z,w)= W + X’Y + X’Z’+ XY’Z

e) Draw the logic circuit for a half-adder. [1]

Ans: Half Adder Logic Circuit Diagram: Using XOR and AND gate

A Sum=AB’+A’B

B

A Carry=AB

B

f) Represent the boolean expression X + Y.Z’ with the help of NAND gates only. [1]

Ans: X+YZ’ =((X+YZ’)’)’=(X’.(YZ’)’)’

X

Y

Z

g) Write the sum of product form of this function G(U,V,W), truth table representation of G is as

follows: [1]

|U |V |W |G |

|0 |0 |0 |1 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |0 |

Ans: G(U,V,W)=U’V’W’+U’V’W+UV’W’+UV’W

YEAR 1998-Delhi

6(a) State De Morgan’s Law. Verify one of the De Morgan’s Laws using truth tables.

(b) Prove X’Y+Z=(X’+Y’+Z)(X’+Y+Z)(X+Y+Z) algebraically.

(c) Write the dual of the Boolean expression (U+W)(V’U+W)

(d) Obtain the simplified form for a Boolean expression:

F(u,v,w,z)=∑(0,1,3,5,7,9,10,11,12,13,15)

(e) Draw the logic circuit for half adder.

(f) Represent the Boolean expression X+YZ’ with the help of NOR gates only.

(g) Write the Product of Sum(POS) of the function H(U,V,W), truth table representation of H is as follows:

|U |V |W |H |

|0 |0 |0 |1 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |0 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |1 |

Q 2. [1999]

a) State the commutative Law. Verify one of the commutative law using truth table. [2]

Ans: Commutative Law:

A+B=B+A

A.B=B.A

Proving Commutative Law using Truth Table:

|A |B |A.B |B.A |A+B |B+A |

|0 |0 |0 |0 |0 |0 |

|0 |1 |0 |0 |1 |1 |

|1 |0 |0 |0 |1 |1 |

|1 |1 |1 |1 |1 |1 |

b) Prove xy+yz+y’z=xy+z algebraically. [2]

Ans:

LHS

=xy+yz+y’z

=xy(z+z’)+yz(x+x’)+y’z(x+x’)

=xyz+xyz’+xyz+x’yz+xy’z+x’y’z

=xy(z+z’)+z(xy+x’y+xy’+x’y’)

=xy(1)+z(y(x+x’)+y’(x+x’))

=xy+z(y(1)+y’(1))

=xy+z(y+y’)

=xy+z(1)

=xy+z

=RHS.

c) Write the dual of the Boolean Expression xy+yz=1. [1]

Ans: Dual of xy+yz=1 is (x+y).(y+z)=0

d) Obtain a simplified form for a Boolean Expression : F(x,y,z) = Σ(1,3,5,7) using karnaugh map. [2]

Ans:

| yz-> |00 |01 |11 |10 |

|x |y’z’ |y’z |Yz |Yz’ |

|0 |0 |1 |1 | 0 |

|x’ | | | | |

|1 |0 |1 |1 | 0 |

|x | | | | |

From the above K-Map we get one group of four 1’s and in this group the only variable which does not change

Is z hence the Boolean Expression for F(x,y,z) = z

e) Draw the logic circuit (three input) which gives high output only when there is even number of low

or high inputs . [1]

Ans: The Truth Table :

|A |B |C |O/P |

| | | |F |

|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 |0 |

Logic Circuit Diagram for the above Truth Table is given below:

A A’ B B’ C C’

f) Represent the boolean expression x(y’+z) with the help of NOR gates only. [1]

Ans. X(Y’+Z) = ((X(Y’+Z))’)’=(X’+(Y’+Z)’)’ . Now the logic circuit diagram can be drawn using only NOR gates.

X’

X

Y’

Y

Z

g) Write the sum of product form of this function F(x,y,z), truth table representation of F is as follows: [1]

|x |y |z |F |

|0 |0 |0 |1 |

|0 |0 |1 |0 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |0 |

Ans.: F(x,y,z) = x’y’z’ + x’yz + xyz’

Q 3. [2000]

a) State DeMorgan’s Law. Verify them using truth tables. [2]

Ans.: De Morgan’s Laws:

i) (AB)’=A’+B’

ii) (A+B)’=A’B’

|1 |2 |3 |4 |5 |

|X’Y’ |1 |1 |0 |0 |

|X’Y |1 |1 |1 |0 |

|XY |1 |1 |1 |0 |

|XY’ |1 |1 |0 |0 |

There is 1 group of eight 1’s and 1 group of four 1’s. Therefore the expression will be as follows:

F(x,y,z,w)= z’+yw

b) Give the truth table for a full-adder. [1]

Ans.: Full Adder: Truth Table

|Inputs |Output |

|A |B |C |CARRY |SUM |

|0 |0 |0 |0 |0 |

|0 |0 |1 |0 |1 |

|0 |1 |0 |0 |1 |

|0 |1 |1 |1 |0 |

|1 |0 |0 |0 |1 |

|1 |0 |1 |1 |0 |

|1 |1 |0 |1 |0 |

|1 |1 |1 |1 |1 |

c) Draw the circuit diagram for the Boolean Function F(x,y,z)=(x’+y)(y’+z) using NOR gates only. [1]

Ans. (((x’+y)(y’+z))’)’= ((x’+y)’+(y’+z)’)’ Now we can draw the logic circuit diagram using only NOR gates.

X

Y

Z

d) Express in the POS form the Boolean function F(A,B,C), the truth table for which is given below. [1]

|A |B |C |F |

|0 |0 |0 |0 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |1 |

Ans.: F(A.B.C)= (A+B+C)(A+B’+C)(A’+B+C)(A’+B’+C)

Q 4 [2001]

a) State and verify duality principle. [2]

Ans.: Principal Of Duality states that every Boolean expression has its dual, which can be obtained by

i) Interchanging 1s and Zeros and

ii) Interchanging And (.) and Or (+) operators

b) Prove algebraically x’y’z’+x’y’z+x’yz+x’yz’+xy’z’+xy’z=x’+y’. [2]

Ans.:L.H.S.

= x’y’z’ + x’y’z + x’yz + x’yz’ + xy’z’+xy’z

= x’y’z’ + x’y’z + x’yz + x’yz’ + xy’z’+ xy’z+ x’y’z’ + x’y’z ( using A=A+A)

= x’(y’z’ + y’z + yz + yz’) + y’ (xz’ + xz + x’z’ + x’z) ( Using Distributive Law)

= x’(y’(z’+z)+y(z+z’) ) + y’ (x(z’+z)+x’(z’+z))

=x’(y’(1)+y(1)) + y’ (x(1)+x’(1)) (Using A’+A=1)

=x’(y’+y) + y’(x+x’)

=x’(1) + y’(1)

=x’ + y’

= R.H.S.

c) If F (a, b, c, d)=Σ (0,1,3,4,5,7,8,9,11,12,13,15), obtain the simplified form using k-map. [2]

Ans.: K-Map :

| |C’D’ |C’D |CD |CD’ |

|A’B’ |1 |1 |1 |0 |

|A’B |1 |1 |1 |0 |

|AB |1 |1 |1 |0 |

|AB’ |1 |1 |1 |0 |

There are 2 groups of eight 1’s. Therefore the expression will be as follows:

F (a, b, c, d) = c’ + d

d) Seven inverters are cascaded one after another. What is the output if the input is 1? [1]

Ans.: Since the number of inverters is odd, it will work as inverter. If the will be 0 (zero) if the input is 1.

e) Given the following circuit:

What is the output if : [1]

i)both inputs are false

ii)one is false and the other is true

Ans.: The above logic diagram will work like two input AND gate.

i) If both inputs are false the output will be false

ii) If one input is true and other is false then the output will be false.

f) Derive the expression for a full-adder. [2]

Ans.: The Expression for the Full- Adder can be derived from the truth table of the Full-Adder:

Truth Table of Full-Adder

|Inputs |Outputs |

|A |B |C |Carry |Sum |

|0 |0 |0 |0 |0 |

|0 |0 |1 |0 |1 |

|0 |1 |0 |0 |1 |

|0 |1 |1 |1 |0 |

|1 |0 |0 |0 |1 |

|1 |0 |1 |1 |0 |

|1 |1 |0 |1 |0 |

|1 |1 |1 |1 |1 |

The expression for sum and carry: (Carry can be derived from K-Map whereas Sum can be directly derived from the truth table)

K-Map for Carry:

| |B’C’ |B’C |BC |BC’ |

|A’ |0 |0 |1 |0 |

|A |0 |1 |1 |1 |

There will be 3 groups of two 1’s .

Carry= AB + BC + AC

Sum= A’B’C + A’BC’ + AB’C’ + ABC

Q 5. [2002]

a) State Distributive Law and verify the law using truth tables. [1]

Ans.: Distributive Law:

i) A(B+C)=AB+AC

ii) A+BC=(A+B).(A+C)

|1 |2 |3 |4 |5 |

|W’X’ |0 |0 |1 |1 |

|W’X |0 |0 |0 |1 |

|WX |0 |0 |0 |1 |

|WX’ |0 |0 |1 |1 |

There are 2 groups of four 1’s .

F (w, x, y, z)= yz’ + x’y

b) Draw the logic circuit of a full-adder. Give truth table of full-adder. [2]

Ans.:

Truth Table of Full-Adder

|Inputs |Outputs |

|A |B |C |Carry |Sum |

|0 |0 |0 |0 |0 |

|0 |0 |1 |0 |1 |

|0 |1 |0 |0 |1 |

|0 |1 |1 |1 |0 |

|1 |0 |0 |0 |1 |

|1 |0 |1 |1 |0 |

|1 |1 |0 |1 |0 |

|1 |1 |1 |1 |1 |

Carry= AB + BC + AC

Sum= A’B’C + A’BC’ + AB’C’ + ABC

Logic Circuit Diagram for Full Adder

A A’ B B’ C C’

Sum

Carry

c) Represent the Boolean expression (x + y)(y + z)(x + z) with help of NOR gate only. [1]

Ans.: The expression is : (x + y) (y + z) (x + z)

Applying Involution Law we get

(((x + y) (y + z) (x + z))’)’

= ((x + y)’ + (y + z)’ + (x + z)’)’

Now we can use NOR gates to implement this expression as follows:

X

Y

Y

Z

X

Z

d) Write the sum of product form of this function F(x, y, z), truth table representation of F is as follows: [2]

|x |y |z |F |

|0 |0 |0 |0 |

|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 |1 |

Ans.: F( x, y, z) = x’y’z + x’yz’ + xy’z’ + xyz

Q 6 [2003]

a) State DeMorgan’s Law. Verify one of them using truth tables. [2]

Ans.: Ans.: De Morgan’s Laws:

iii) (AB)’=A’+B’

iv) (A+B)’=A’B’

|1 |2 |3 |4 |5 |

|U’V’ |1 |0 |1 |1 |

|U’V |1 |0 |1 |0 |

|UV |0 |1 |1 |1 |

|UV’ |0 |1 |0 |1 |

There are 5 groups of two 1’s. Therefore the expression will be :

F(u, v, w, z) = u’w’z’ + u’wz + uw’z + uvw + v’wz’

c) Draw the logic circuit of a half-adder. [1]

Ans.: Ans: Half Adder Logic Circuit Diagram: Using XOR and AND gate

A Sum=AB’+A’B

B

A Carry=AB

B

d) Represent the boolean expression (x+y’).z with the help of NAND gates only. [2]

Ans.: The expression = (xz + y’z)

= (((xz + y’z))’)’

=((xy)’ . (y’z)’)’

X

Z

Y

Z

e) Write the sum of product form of this function H(u,v,w), truth table representation of H is as follows: [1]

|u |v |w |H |

|0 |0 |0 |0 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans. H(u,v,w) = u’vw’+uv’w’+uvw’+uvw

Q 7 CBSE 1

a) State Absorption Laws. Verify one of the absorption law using truth table. [2]

Ans.: : Absorption Law:

X+XY=X

L.H.S.= X+XY

= X.1+X.Y

= X(1+Y)

= X.1

= X

= R.H.S

X(X+Y)=X

L.H.S = X(X+Y)

= X.X+X.Y

= X+X.Y

=X(1+Y)

=X(1)

=X

= R.H.S

b) Verify XY’+Y’Z=XY’Z+XY’Z’+X’Y’Z algebraically. [1]

Ans.: R.H.S.=

= XY’Z + XY’Z’ + X’Y’Z

= XY’Z + XY’Z’ + X’Y’Z + XY’Z ( By A+A=A)

=XY’(Z+Z’) + Y’Z(X’+X)

=XY’(1)+Y’Z(1)

=XY’ + Y’Z

=L.H.S.

c) Write the dual of the Boolean Expression A+B’.C. [1]

Ans.: A.(B+C) is the dual of A+B’C

d) Obtain a simplified form for a Boolean Expression : F(u,v,w,z) = Σ(0,1,3,4,5,6,7,9,10,11,13,15) using karnaugh map. [2]

Ans.: K-Map:

| |W’Z’ |W’Z |WZ |WZ’ |

|U’V’ |1 |1 |1 |0 |

|U’V |1 |1 |1 |1 |

|UV |0 |1 |1 |0 |

|UV’ |0 |1 |1 |1 |

Ans.: F (U,V,W,Z) = Z+ U’W’+ U’V+UV’W

e) Draw the logic circuit for a half-adder. [1]

Ans: Half Adder Logic Circuit Diagram: Using XOR and AND gate

A Sum=AB’+A’B

B

A Carry=AB

B

f) Represent the boolean expression XY’+Z with the help of NOR gates only. [2]

Ans.

X

Y

Z

g) Write the product of sum form of the function H(u,v,w), truth table representation of H is as follows: [1]

|u |v |w |H |

|0 |0 |0 |1 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |0 |

Ans.: H(u ,v, w)= (u + v’ +w) . (u’ + v + w) . (u’ + v’ + w) . ( u’ + v’ + w’)

Q 8 [CBSE 2]

a) State DeMorgan’s Law. Verify one of them using truth tables. [2]

Ans.: De Morgan’s Laws:

v) (AB)’=A’+B’

vi) (A+B)’=A’B’

|1 |2 |3 |4 |5 |

|U’V’ |1 |0 |1 |1 |

|U’V |1 |0 |1 |0 |

|UV |0 |1 |1 |1 |

|UV’ |0 |1 |0 |1 |

There are 5 groups of two 1’s. Therefore the expression will be:

F(u, v, w, z) = u’w’z’ + u w’z + vwz + uwz’ + u’v’w

b) Draw the logic circuit for a half-adder. [1]

Ans: See Year-1998 1(e).

c) Represent the boolean expression (x+y’).z with the help of NAND gates only. [2]

Ans.: (x+y’).z=(xz+y’z)’)’=((xz)’.(y’z)’)’

x

z

y

z

d) Write the product of sum form of the function H(u,v,w), truth table representation of H is as follows: [1]

|u |v |w |H |

|0 |0 |0 |0 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans.: H=(u+v+w).(u+v+w’).(u+v’+w’).(u’+v+w’)

CBSE-I: 2004

(a) State and verify DeMorgan’s Laws

Ans.: DeMorgan’s Laws

(X+Y)’=X’.Y’

(X.Y)’=X’+Y’

|X |Y |X+Y |(X+Y)’ |

|0 |0 |0 |1 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans. F(U,V,W)=(U+V+W’).(U+V’+W’).(U’+V+W’)

(d) Reduce the following Boolean Expression using K-Map:

F(A,B,C,D)=Σ(0,1,2,4,5,8,10)

Answer:

| |C’D’ |C’D |CD |CD’ |

|A’B’ |1 |1 |0 |1 |

|A’B |1 |1 |0 |1 |

|AB |0 |0 |0 |0 |

|AB’ |1 |0 |0 |1 |

There are 3 groups of four 1’s therefore the expression will be as follows:

F(A,B,C,D)=A’C’+A’D’+B’D’

CBSE-II:2004

(a) State and algebraically verify Absorption Law: 2

Answer: Absorption Law:

X+XY=X

L.H.S.= X+XY

= X.1+X.Y

= X(1+Y)

= X.1

= X

= R.H.S

X(X+Y)=X

L.H.S = X(X+Y)

= X.X+X.Y

= X+X.Y

=X(1+Y)

=X(1)

=X

= R.H.S

(b) Write the equivalent Boolean Expression for the following Logic Circuit 2

U

V

F(U,V)=U’V+UV’

(c) Write the SOP from of a Boolean function G, which is represented in a truth table as follows:

|P |Q |R |G |

|0 |0 |0 |0 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans.: G(P,Q,R)=P’.Q.R’+P.Q’.R’+P.Q.R’+P.Q.R 1

(d) Reduce the following Boolean Expression using K-Map: 3

F(U,V,W,Z) = π ( 0,1,2,4,5,6,8,10)

Ans.:

| |W’Z’ |W’Z |WZ |WZ’ |

|U’V’ |0 |0 |1 |0 |

|U’V |0 |0 |1 |0 |

|UV |1 |1 |1 |1 |

|UV’ |0 |1 |1 |0 |

There are 3 groups of four 1’s. Therefore the expression will be as follows:

Answer: F(U,V,W,Z)=UV+WZ+UZ

Year: 2004:

6 (a) State and Prove the Absorption Law algebraically. (2)

Ans.: Absorption Law:

(i) A(A+B)=A

(ii) A+AB=A

Proof:

(i) A(A+B)=A

LHS=A(A+B)

=A.A+A.B

=A+A.B

=A(1+B)

=A.1

=A

=RHS

(ii) A+AB=A

LHS= A+AB

=A(1+B)

=A.1

=A

=RHS

(b) Given the following truth table, derive a SOP and a POS form of the following expression from it:

|A |B |C |F(A,B,C) |

|0 |0 |0 |1 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |1 |

|1 |1 |0 |1 |

|1 |1 |1 |0 |

Ans.:

SOP: F(A,B,C)= A’B’C’+ A’B’C+ A’BC+AB’C+ABC’

POS: F(A,B,C)= (A+B’+C).(A’+B+C).(A’+B’+C’)

(c) Obtain a simplified form for the following Boolean Expression using Karnaugh’s Map: (3)

F(a,b,c,d) = Σ (0,1,2,4,5,7,8,9,10,11,14)

Ans.:

K-Map: 4-Variables

| |C’D’ |C’D |CD |CD’ |

|A’B’ |1 |1 |0 |1 |

|A’B |1 |1 |1 |0 |

|AB |0 |0 |0 |1 |

|AB’ |1 |1 |1 |1 |

There are 3 groups with four 1’s and 2 groups with two 1’s. Therefore the expression will be as follows:

F(a,b,c,d)= ab’ + a’c’ + a’bd + acd’ + b’d’

(d) Draw the logic circuit for a Half Adder using NAND gates only. (2)

Ans.: For Half Adder:

Sum=A’B+AB’

Carry=AB

Carry

(AB)’

A Sum

B

(e) Interpret the following Logic Circuit as Boolean Expression:

W

X F

Y

Z

Ans.: F=(W+X’).(Y’+Z)

YEAR: 2005 ( DELHI)

6(a) a) State and verify Associative law in Boolean Algebra. 2

(a) (i) X+(Y+Z)=(X+Y)+Z

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

|X |Y |Z |Y+Z |X+Y |X+(Y+Z) |(X+Y)+Z |

|0 |0 |0 |0 |0 |0 |0 |

|0 |0 |1 |1 |0 |1 |1 |

|0 |1 |0 |1 |1 |1 |1 |

|0 |1 |1 |1 |1 |1 |1 |

|1 |0 |0 |0 |1 |1 |1 |

|1 |0 |1 |1 |1 |1 |1 |

|1 |1 |0 |1 |1 |1 |1 |

|1 |1 |1 |1 |1 |1 |1 |

(ii) X.(Y.Z)=(X.Y).Z

X Y Z Y.Z X.Y X.(Y.Z) (X.Y).Z

|X |Y |Z |Y.Z |X.Y |X.(Y.Z) |(X.Y).Z |

|0 |0 |0 |0 |0 |0 |0 |

|0 |0 |1 |0 |0 |0 |0 |

|0 |1 |0 |0 |0 |0 |0 |

|0 |1 |1 |1 |0 |0 |0 |

|1 |0 |0 |0 |0 |0 |0 |

|1 |0 |1 |0 |0 |0 |0 |

|1 |1 |0 |0 |1 |0 |0 |

|1 |1 |1 |1 |1 |1 |1 |

(1 Mark for stating any one of the law)

(1 Mark for verification of any one of the law)

(b) Write the equivalent Boolean expression for the following Logic Circuit : 2

[pic]

F=(X+Y').(X'+Y').(X'+Y')

[Full 2 marks for writing Correct Expression for Level III]

[Only 1½ mark, if Level II expressions are correctly written and Level

III is wrong]

[½

mark if only Level I expressions are correct]

(c) Write the SOP form of a Boolean Function F, which is represented by the following

truth table: 1

|A |B |C |F |

|0 |0 |0 |1 |

|0 |0 |1 |0 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans.:

F= A'.B'.C' + A'.B.C + A.B.C' + A.B.C

[1 mark for correct SOP expression]

[½ mark if ONLY one term of the expression is wrong]

(d) Reduce the following Boolean expression using K - Map: 3

F (A, B, C, D) =π(O, 1, 2, 3, 4, 5, 10, 11, 15)

Ans.

|AB\CD |C’D’=00 |C’D=01 |CD=11 |CD’=10 |

|A’B’ =00 |0 |0 |1 |0 |

|A’B =01 |0 |0 |1 |1 |

|AB =11 |1 |1 |0 |1 |

|AB’ =10 |1 |1 |0 |0 |

F(A, B, C, D) = (A+C) . (B + C') . (A'+C'+D')

[½ mark for Drawing The K-Map Correctly]

[½ mark for placing the 1s/0s at correct positions]

[½ Mark for grouping in the K-Map]

[½ Mark for each reduced term]

Note :

Deduct ½ Mark for extra redundant term(s)/grouping(s)

YEAR: 2005 (OUTSIDE DELHI)

6 (a) State and verify Absorption Law in Boolean Algebra? [2]

Ans.: Absorption Law:

X+X.Y=X

OR

X.(X+Y)=X

OR

X+X’.Y=X+Y

OR

X.(X’+Y)=X.Y

|X |Y |X.Y |X+X.Y |

|0 |0 |0 |0 |

|0 |1 |0 |0 |

|1 |0 |0 |1 |

|1 |1 |1 |1 |

OR

|X |Y |X+Y |X(X+Y) |

|0 |0 |0 |0 |

|0 |1 |1 |0 |

|1 |0 |1 |1 |

|1 |1 |1 |1 |

OR

|X |Y |X’ |X’+Y |X.(X’+Y) |XY |

|0 |0 |1 |1 |0 |0 |

|0 |1 |1 |1 |0 |0 |

|1 |0 |0 |0 |0 |0 |

|1 |1 |0 |1 |1 |1 |

OR

|X |Y |X’ |X’.Y |X+(X’.Y) |X+Y |

|0 |0 |1 |0 |0 |0 |

|0 |1 |1 |1 |1 |1 |

|1 |0 |0 |0 |1 |1 |

|1 |1 |0 |0 |1 |1 |

OR

Algebraic Verification:

X+X.Y =X

X.1+X.Y =X

X.(1+Y) =X

X.1 =X

X=X

OR

X.(X+Y) =X

XX+X.Y =X

X.1+X.Y =X

X.(1+Y) =X

X.1 =X

X=X

OR

X+X’. Y =X+Y

(X+X’)(X+Y) =X+Y

1.(X+Y) =X+Y

X+Y =X+Y

OR

X(X’+Y) =X.Y

XX’+X.Y =X.Y

0+X.Y =X.Y

X.Y =X.Y

(1 Mark for stating the absorption law)

(1 Mark for correctly verifying the law using truth table)

OR

(1 Mark for correct verification by using algebraic method)

(b) Write equivalent Boolean expression for the following Logic Circuit : [2]

[pic]

Ans:

F(U,V)= UV’ + U’V +U’V’

(½ Mark for each term - full marks if all the terms are correct)

(c) Write the POS form of a Boolean Function F, which is represented by the following truth table: [1]

|X |Y |Z |F |

|0 |0 |0 |1 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |1 |

|1 |0 |0 |0 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |0 |

Ans:

SOP: F(A,B,C)= X’Y’Z’ + AX’Y’Z + X’YZ + XY’Z

POS: F(A,B,C)= (X + Y’ +Z) .(X’ + Y + Z). ( X’ + Y’ + X) . (X’ + Y’ + Z’)

(1/2 Mark for each two terms)

(d) Reduce the following Boolean expression using K-Map: [3]

F(A,B,C,D)=Σ (0,1,2,3,4,5,10,11,15)

|AB\CD |C’D’=00 |C’D=01 |CD=11 |CD’=10 |

|A’B’ =00 |1 |1 |1 |1 |

|A’B =01 |1 |1 |0 |0 |

|AB =11 |0 |0 |1 |0 |

|AB’ =10 |0 |0 |1 |1 |

Ans: There are 3 groups of four 1’s and 1 group of two 1’s

F(A,B,C,D)= A’C’ + ACD + B’C

(1 Mark for drawing the K-Map with right place values)

(½ Mark for grouping in the K-Map)

(½ Mark for each reduced term)

Note: Deduct ½ Mark for extra redundant term/groupings

YEAR 2006 (OUTSIDE DELHI)

6. (a) State and verify Associative Law. [2]

(a) (i) X+(Y+Z)=(X+Y)+Z

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

|X |Y |Z |Y+Z |X+Y |X+(Y+Z) |(X+Y)+Z |

|0 |0 |0 |0 |0 |0 |0 |

|0 |0 |1 |1 |0 |1 |1 |

|0 |1 |0 |1 |1 |1 |1 |

|0 |1 |1 |1 |1 |1 |1 |

|1 |0 |0 |0 |1 |1 |1 |

|1 |0 |1 |1 |1 |1 |1 |

|1 |1 |0 |1 |1 |1 |1 |

|1 |1 |1 |1 |1 |1 |1 |

(ii) X.(Y.Z)=(X.Y).Z

X Y Z Y.Z X.Y X.(Y.Z) (X.Y).Z

|X |Y |Z |Y.Z |X.Y |X.(Y.Z) |(X.Y).Z |

|0 |0 |0 |0 |0 |0 |0 |

|0 |0 |1 |0 |0 |0 |0 |

|0 |1 |0 |0 |0 |0 |0 |

|0 |1 |1 |1 |0 |0 |0 |

|1 |0 |0 |0 |0 |0 |0 |

|1 |0 |1 |0 |0 |0 |0 |

|1 |1 |0 |0 |1 |0 |0 |

|1 |1 |1 |1 |1 |1 |1 |

(1 Mark for stating any one of the law)

(1 Mark for verification of any one of the law)

(b) Write the equivalent expression for the following logical circuit: [2]

[pic]

(b) (A+C)’.(A+B)’.(B+C)’

(1/2 Mark each for (A+C)’, (A+B)’, (B+C)’ and final expression)

OR

(Full 2 marks even if students have mentioned the reduced form of

the same expression)

(c) Express P+Q’R in POS form. [1]

(c) (P+Q’).(P+R)

OR

(P+Q’+R).(P+Q’+R’).(P+Q+R)

(1 Mark for any of the two POS forms of the given expression)

(d) Obtain a simplified form for the following Boolean Expression using Karnaugh’s Map: [3]

F(P,Q,R,S) = π(0,3,5,6,7,11,12,15)

(d) K-Map:

| |R’S’ |R’S |RS |RS’ |

|P’Q’ |0 |1 |0 |1 |

|P’Q |1 |0 |0 |0 |

|PQ |0 |1 |0 |1 |

|PQ’ |1 |1 |0 |1 |

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

(1 mark for representing correct places in K-Maps)

AND

(1/2 mark for grouping and writing minimum 2 Groups)

OR

(1 mark for grouping and writing minimum 3 Groups)

OR

(1 ½ mark for grouping and writing 5 Groups)

OR

(2 mark for grouping and writing final expression in reduced form)

Important: 1/2 Mark to be deducted for redundant group(s)

OR

F(P,Q,R,S)=P’QR’S’+PQ’S’+PRS’+PR’S+Q’RS’+Q’R’S

(1 mark for representing correct places in K-Maps)

AND

(1/2 mark for grouping and writing minimum 2 Groups)

OR

(1 mark for grouping and writing minimum 4 Groups)

OR

(1 ½ mark for grouping and writing 6 Groups)

OR

(2 mark for grouping and writing final expression in reduced form)

Important: 1/2 Mark to be deducted for redundant group(s)

YEAR 2006 (DELHI)

a) State and verify Distributive Law. 2

(a) Distributive Laws are :

(a) A (B + C) = A B + A C

(b) A + (B C) = (A + B) (A + C)

Verification of first distributive law using truth table:

|A |B |C |B+C |A(B+C) |AB |AC |AB+AC |

|0 |0 |0 |0 |0 |0 |0 |0 |

|0 |0 |1 |1 |0 |0 |0 |0 |

|0 |1 |0 |1 |0 |0 |0 |0 |

|0 |1 |1 |1 |0 |0 |0 |0 |

|1 |0 |0 |0 |0 |0 |0 |0 |

|1 |0 |1 |1 |1 |0 |1 |1 |

|1 |1 |0 |1 |1 |1 |0 |1 |

|1 |1 |1 |1 |1 |1 |1 |1 |

Comparing Column Numbers 5 and 8, Distributive law is verified.

(1 Mark for writing either of the distributive law)

(1 Mark for verification using truth table)

(b) Write the equivalent expression for the following Logical Circuit : 2

[pic]

(b) (AC)’+(AB)’+(BC)’

(2 Marks for correct expression)

OR

( ½ Mark each for (AC)’, (AB)’,( BC)’ and the final expression)

(c) Express P + Q’R in canonical SOP form. 1

(c) (P + Q’R)

= P ( Q+Q’)(R+R’) + Q’R(P+P’)

= (PQ +PQ’) (R + R’) + PQ’R + P’Q’R

= PQR + PQ’R + PQR’ + PQ’R’ + PQ’R + P’Q’R

= PQR + PQ’R + PQR’ + PQ’R’ + P’Q’R

(1 mark for the canonical expression)

(d) Reduce the following Boolean expression using K-Map.

F(P, Q, R, S) = ∑(0, 3, 5, 6, 7, 11, 12, 15)

(d) K-Map:

| |R’S’ |R’S |RS |RS’ |

|P’Q’ |1 |0 |1 |0 |

|P’Q |0 |1 |1 |1 |

|PQ |1 |0 |1 |0 |

|PQ’ |0 |0 |1 |0 |

F= RS + P’QS + P’QR + P’Q’R’S’ + PQR’S’

(1 Mark for correct placing of 1s)

(1 Mark for correct grouping)

(1 Mark for final reduced expression)

YEAR 2007 (DELHI)

6(a) State De Morgan’s Theorems and verify the same using truth table. 2

(a) If X,Y€B

(X+Y)’=X’.Y’

(X.Y)’=X’+Y’

|X |Y |X’ |Y’ |X+Y |(X+Y)’ |X’.Y’ |

|0 |0 |1 |1 |0 |1 |1 |

|0 |1 |1 |0 |1 |0 |0 |

|1 |0 |0 |1 |1 |0 |0 |

|1 |1 |0 |0 |1 |0 |0 |

|X |Y |X’ |Y’ |X.Y |(X.Y)’ |X’+Y’ |

|0 |0 |1 |1 |0 |1 |1 |

|0 |1 |1 |0 |0 |1 |1 |

|1 |0 |0 |1 |0 |1 |1 |

|1 |1 |0 |0 |1 |0 |0 |

(½ Mark for each form of DeMorgan’s Theorem)

(1 Mark for correct verification of any of the form of DeMorgan’s

Theorem using Truth Table)

(b) Write the equivalent Canonical Product of Sum Expression for the following

Sum of Product Expression

F(X, Y, Z) = Σ (0, 2, 4, 5) 2

(b) F(X, Y, Z) = π (1, 3, 6, 7)

OR

F=(X+Y+Z’)(X+Y’+Z’)(X’+Y’+Z)(X’+Y’+Z’)

(½ Mark for each term of POS)

(c) Write the equivalent Boolean expression for the following Logic Circuit. 2

[pic]

(c) F=A.B’+C’.D

(2 Marks for the correct expression)

OR

(½ Mark for each term of the expression)

(d) Reduce the following Boolean expression using K-Map : 2

F(A, B, C, D) = π(5,6,7,8,9,12,13,14,15)

(d)

| |C’D’ |C’D |CD |CD’ |

|A’B’ |1 |1 |1 |1 |

|A’B |1 |0 |0 |0 |

|AB |0 |0 |0 |0 |

|AB’ |0 |0 |1 |1 |

F(A,B,C,D)= A’.B’+B’.C’+A’.C’.D’

OR

F(A,B,C,D)=(A’+C).(B’+D’).(B’+C’)

(½ Mark for representing the terms in K-Map)

(½ Mark for obtaining each reduced term i.e. 1 ½ Marks for reducing

to the correct minimal form)

YEAR 2007 (OUTSIDE DELHI)

a) State Distributive law and verify the same using truth table. 2

|6(a) If X |Y |Z are Boolean Variables then |

X . ( Y + Z ) = X . Y + X . Z

OR

X + Y . Z = (X + Y) . (X + Z)

|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 |1 |0 |0 |0 |0 |

|0 |1 |0 |1 |0 |0 |0 |0 |

|0 |1 |1 |1 |0 |0 |0 |0 |

|1 |0 |0 |0 |0 |0 |0 |0 |

|1 |0 |1 |1 |1 |0 |1 |1 |

|1 |1 |0 |1 |1 |1 |0 |1 |

|1 |1 |1 |1 |1 |1 |1 |1 |

VERIFIED

(1 Mark for stating any one of the Distributive Law correctly)

(1 Mark for verification using Truth Table)

(b) Write the equivalent Canonical Sum of Product expression for the following

Product of Sum Expression 2

F(X,Y,Z) = π(1,3,6,7)

(b)

F(X,Y,Z) =∑(0,2,4,5)

= X’.Y’.Z’ + X’.Y.Z’ + X.Y’.Z’ + X.Y’.Z

(2 Marks for correct expression)

OR

(1 Mark, if a reduced non-canonical SOP expression is given)

(c) Write the equivalent Boolean Expression for the following Logic Circuit. 2

(c)

[pic]

F = W.X’ + Y’.Z

(2 Marks for correct expression)

OR

( ½ Mark each for W.X’ and Y’.Z)

(D)Reduce the following Boolean expression using K-Map 2

F(U,V,W,Z) = Σ (0, 1, 2, 3, 4, 10, 11)

(d)

| |W’Z’ |W’Z |WZ |WZ’ |

|U’V’ |1 |1 |1 |1 |

|U’V |1 |0 |0 |0 |

|UV |0 |0 |0 |0 |

|UV’ |0 |0 |1 |1 |

F = U'.V'+W.V'+U'.W'.Z'

(½ Mark for drawing correct K-Map )

(½ Mark for plotting 1’s correctly)

(½ Mark for correct grouping)

(½ Mark for correct Answer)

YEAR 2008 (DELHI)

6 a) State and verify De Morgan’s law in Boolean Algebra 2

Ans:

(X + Y )’ = X’ . Y’

OR

(X . Y)’ = X’ + Y’

Verification:

[pic]

or

(X+Y)’ = X’ . Y’

If (X+Y)’ . (X+Y) = (X’ . V’) . (X+Y)

If 0 - = X’ . Y’ . X + X’ . Y’ . Y

If 0 = 0 + 0

Hence Proved and Verified

(1 Mark for stating anyone of the De Morgan’s Law)

(1 Mark for verifying anyone of the De Morgan’s Law)

(b) Draw a Logical Circuit Diagram for the following Boolean Expression 1

X’ . (y’ + Z)

Ans:

[pic]

(1 Marks for Drawing Logic Circuit Diagram for all )

(c) Convert the following Boolean expression into its equivalent Canonical Sum

of Product Form«SOP)

(X’ + Y + Z’) . (X’ + Y + Z) . (X’ + Y’ + Z) . (X’ + Y’ + Z’) 2

Ans: F ( X , Y , Z) =π(4 , 5 , 6 , 7)

=∑(0 , 1 , 2 , 3)

= X’. Y’. Z’ + X’. Y’. Z + X’. Y. Z’ + X’. Y. Z

(2 Marks for writing correct Canonical SOP Expression)

OR

(1 Mark for only deriving Product Terms)

(d) Reduce the following Boolean expression using K - Map

F (A,B,C,D) = ∑(0,2,3,4,6,7,8,10,12)

Ans:

[pic]

F= C’.D’ + A’.C + B’.D’

(½ Mark for drawing correct K-Map)

(½ Mark for plotting 1‘s correctly)

(1 Mark for correct grouping)

(1 Mark for correct Answer)

YEAR 2008 (OUTSIDE DELHI)

6.

(a) State any verify Absorption law in Boolean- Algebra.

Ans:

X+X. Y=X

|X |Y |X.Y |X+XY |X |

|0 |0 |0 |0 |0 |

|0 |1 |0 |0 |0 |

|1 |0 |0 |1 |1 |

|1 |1 |1 |1 |1 |

OR

X. (X+Y) = X

|X |Y |X+Y |X.(X+Y) |X |

|0 |0 |0 |0 |0 |

|0 |1 |1 |0 |0 |

|1 |0 |1 |1 |1 |

|1 |1 |1 |1 |1 |

OR

X+X’ . Y=X+Y

|X |Y |X’ |X+X’.Y |X+Y |

|0 |0 |1 |0 |0 |

|0 |1 |1 |1 |1 |

|1 |0 |0 |1 |1 |

|1 |1 |0 |1 |1 |

OR

X. (X’ + Y) =X.Y

|X |Y |X’ |X’+Y |X.(X’+Y) |X.Y |

|0 |0 |1 |1 |0 |0 |

|0 |1 |1 |1 |0 |0 |

|1 |0 |0 |0 |0 |0 |

|1 |1 |0 |1 |1 |1 |

OR

X+X.Y=X

L.H.S=X+X.Y

=X.1+X. Y

=X. (l+Y)

=X.1

=X

=R.H.S

Verified

OR

X(X+Y)=X

L.H.S=X. (X+Y)

L.H.S=X.X+X.Y

=X+X.Y

=X.1+X.Y

=X. (l+Y)

=X.1

=X

=R.H.S

Verified

OR

X+X’.Y=X+Y

L.H.S=(X+X’).(X+Y)

=1.(X+Y)

=X+Y

=R.H.S

Verified

OR

X.(X’+Y)=X.Y

L.H.S=X.(X’ +Y)

=X.X’+X.Y

=0+X.Y

=X.Y

=R.H.S

Verified

(2 Marks for verification of anyone form of the Absorption Law)

OR

(1 Mark for stating anyone form of the Absorption Law)

(b) Draw a Logical Circuit Diagram for the following Boolean Expression :

A . (B + C’)

Ans:

[pic]

(1 mark for the correct circuit diagram)

(c) Convert the following Boolean expression into its equivalent Canonical

Product of Sum Form (POS) :

A. B’ . C + A’. B . C + A’. B . C’

Ans:

=π(0, 1, 4, 6, 7)

OR

= (A+B+C) . (A+B+C’ ) . (A’ +B+C) . (A’ +B’ +C) . (A’ +B’ +C’ )

(2 Marks for the correct expression)

OR

(1 Mark if only truth table is given for the expression without deriving

POS expression)

Note: No marks to be awarded for partial answers

(d) Reduce the following Boolean expression using K - Map:

F (A, B, C, D) =∑(0, l, 2, 4, 5, 8, 9, 10, 11)

Ans:

[pic]

F(A,B,C,D)= A’ .C’+A.B’ + B’ .D’

(1 Mark for drawing correct K-Map )

(½ Mark for plotting 1‘s correctly)

(½ Mark for each correct grouping)

Note: No marks should be deducted even if the solution is arrived with the

help of POS form.

CBSE Sample Paper Set 1 (2009)

6(a) State and verify Demorgan’s Laws. 2

Answer:

(X+Y)’ =X’.Y’

(X.Y)’ =X’+Y’

|X |Y |X+Y |(X+Y)’ |

|0 |0 |0 |1 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Answer:

F(U,V,W) = (U+V+W’).(U+V’+W’).(U’+V+W’)

(1 mark for correct POS representation)

c) Reduce the following Boolean Expression using K-Map: 3

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

Answer:

| |A’B’ |A’B |AB |AB’ |

|C’D’ |1 |1 | |1 |

| |0 |4 |12 |8 |

|C’D |1 |1 | | |

| |1 |5 |13 |9 |

|CD | | | | |

| |3 |7 |15 |11 |

|CD’ |1 |1 | |1 |

| |2 |6 |14 |10 |

F(A,B,C,D)=A’C’+A’D’+B’D’

(1 mark for correctly drawing K-Map with 1s represented on right places)

(1 mark for minimizing each Quad)

(1 mark for writing the complete Boolean Expression)

CBSE Sample Paper Set 1 (2009)

6(a) State and algebraically verify Absorbtion Laws. 2

Answer:

X+X.Y = X

L.H.S = X+X.Y

= X.1+X.Y

= X.(1+Y)

= X.1

= X

= R.H.S

X+X’.Y = X+Y

L.H.S. = X+X’.Y

= (X+X’).(X+Y)

= 1.(X+Y)

= X+Y

= R.H.S

(1 mark for stating the correct law)

(1 mark for the appropriate verification using algebraic method)

(b) Write the equivalent Boolean Expression for the following Logic Circuit 2

Answer:

F(U,V)=U’.V+U.V’

(Full 2 marks for obtaining the correct Boolean Expression for the Logic Circuit)

OR

(1 mark correctly interpreting Product terms)

d) Write the SOP form of a Boolean function G, which is represented in a truth table as follows:1

Answer:

|P |Q |R |G |

|0 |0 |0 |0 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

G(P,Q,R) = P’.Q.R’+P.Q’.R’+P.Q.R’+P.Q.R

(1 mark for correct SOP representation)

e) Reduce the following Boolean Expression using

K-Map: 3

F(U,V,W,Z)=((0,1,2,4,5,6,8,10)

Answer:

| |U’V’ |U’V |UV |UV’ |

|W’Z’ | | |1 | |

| |0 |4 |12 |8 |

|W’Z | | |1 |1 |

| |1 |5 |13 |9 |

|WZ |1 |1 |1 |1 |

| |3 |7 |15 |11 |

|WZ’ | | |1 | |

| |2 |6 |14 |10 |

F(U,V,W,Z)=UV+WZ+UZ

(1 mark for correctly drawing K-Map with 1s represented on right places)

(1 mark for minimizing each Quad)

(1 mark for writing the complete Boolean Expression)

Year 2009 (Delhi)

6. (a) Verify X’Y + X.Y’ + X’Y’ = (X’ + Y’) using truth table. 2

Ans

|X |Y |X’ |Y’ |

|0 |0 |0 |0 |

|0 |0 |1 |1 |

|0 |1 |0 |1 |

|0 |1 |1 |1 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |0 |

|1 |1 |1 |1 |

Ans (A+B+C).(A’+B+C’).(A’+B’+C)

OR

H(A,B,C) = n (0, 5, 6)

(1 Mark for the correct POS form)

Note: Deduct ½ mark if wrong variable names are used

(d) Reduce the following Boolean Expression using K-Map : 3

F(P, Q, R, S) = Σ(1, 2, 3, 5,6, 7, 9, 11, 12, 13, 15)

Ans

| |R’S’ |R’S |RS |RS’ |

|P’Q’ |0 |1 |1 |1 |

|P’Q |0 |1 |1 |1 |

|PQ |1 |1 |1 |0 |

|PQ’ |0 |1 |1 |0 |

F (P,Q,R,S = S+P’R+PQR’

( ½ Mark for placing aU1 s at correct positions in K-Map)

( ½ Mark for each grouping)

(1 Mark for writing final expression in reduced/minimal form)

Note: Deduct ½ mark if wrong variable names are used

Year 2009 (Outside Delhi)

6. (a) State and verify absorption law using truth table. 2

Ans Absorption Law : For every X , Y ∈ B

i) X + X . Y = X

X . (X + Y) = X (by Duality)

ii) X + X’ . Y = X + Y

(i)

|X |Y |XY |X+XY |

|0 |0 |0 |0 |

|0 |1 |0 |0 |

|1 |0 |0 |1 |

|1 |1 |1 |1 |

Verified as column 1= column 4

(ii)

|X |Y |X’ |X’Y |X+X’Y |XY’ |

|0 |0 |1 |0 |0 |0 |

|0 |1 |1 |1 |1 |1 |

|1 |0 |0 |0 |1 |1 |

|1 |1 |0 |0 |1 |1 |

Verified as column 5= column 6

(1 Mark for stating anyone of the Absorption Laws)

(1 Mark for verifying the law using Truth Table)

(b) Write the equivalent Boolean Expression for the following Logic Circuit: 2

P

Q

R

Ans P.Q’ + P’.R

(2 Marks for the final expression P.Q’+P’R)

OR

(1 Mark for anyone of the correct terms out of P.Q’ or P’.R)

(c) Write the POS form of a Boolean function G, which is represented in a truth

table as follows: 1

|U |V |W |G |

|0 |0 |0 |1 |

|0 |0 |1 |1 |

|0 |1 |0 |0 |

|0 |1 |1 |0 |

|1 |0 |0 |1 |

|1 |0 |1 |1 |

|1 |1 |0 |0 |

|1 |1 |1 |1 |

Ans (U+V’+W).(U+V’+W’) . (U’+V’+W)

OR

G(U,V,W) = Π (2, 3, 6)

(1 Mark for the correct POS form)

Note: Deduct ½ mark if wrong variable names are used

(d) Reduce the following Boolean Expression using K-Map : 3

H(U,V,W,Z) = Σ(0, 1, 4, 5, 6, 7, 11, 12, 13, 14, 15)

| |W’Z’ |W’Z |WZ |WZ’ |

|U’V’ |1 |1 |0 |0 |

|U’V |1 |1 |1 |1 |

|UV |1 |1 |1 |1 |

|UV’ |0 |0 |1 |0 |

Ans

H(U,V,W,Z) = V + U’W’+ UWZ

(½ Mark for placing all 1s at correct positions in K-Map)

(½ Mark for each grouping)

(1 Mark for writing Final expression in reduced/minimal form)

Note: Deduct ½ mark if wrong variable names are used

Year 2010 (Outside Delhi)

6. (a) Verrify the following algebraically: 2

X’.Y + X.Y’ =(X’+Y’).(X+Y)

Ans.:

R.H.S. = (X’+Y’).(X+Y)

= X’.(X+Y)+Y’.(X+Y) (DISTRIBUTIVE LAW)

=X’.X+X’.Y+Y’.X+Y’.Y

=0+X’Y+Y’.X+0

=X’.Y+X.Y’

=L.H.S.

(b) Write the equivalent Boolean Expression for the following Logic Circuit: 2

U

V

W

Ans.:

Boolean Expression: (U’+V).(V’+W)

(c) Write the SOP form of a Boolean function G, which is represented in a truth table as follows: 1

|P |Q |R |G |

|0 |0 |0 |0 |

|0 |0 |1 |0 |

|0 |1 |0 |1 |

|0 |1 |1 |1 |

|1 |0 |0 |1 |

|1 |0 |1 |0 |

|1 |1 |0 |1 |

|1 |1 |1 |1 |

Ans.:

SOP Expression:

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

(d).Reduce the following Boolean Expression using K-Map : 3

F(A,B,C,D)= ∑(3,4,5,6,7, 13,15)

C’D’ CD’ CD CD’

00 01 11 10

|0 |0 |1 |0 |

|1 |1 |1 |1 |

|0 |1 |1 |0 |

|0 |0 |0 |0 |

A’B’ 00

A’B 01

AB 11

AB’ 10

Ans.: Reduced Boolean Expression using K-Map:

F(A,B,C,D)= A’.B+ B.D+ A’.CD

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