Contents

**Chapterwise Question Bank CBSE Class 12 Computer Science (C++) – Boolean Algebra**

### Topic – 1

Boolean Operations and Logic Gates

**Exam Practice**

**Short Answer Type Questions [2 Mark]**

**Question 1:
**Write the Sum of Product (SOP) form of the function F(P, Q, R)for the following truth table representation of F:

**Delhi 2014**

**Question 2:
**Write the Product of Sum (POS) form of the function F(X, Y, Z)for the following truth table representation of F:

**All India 2014**

**Аnswer:**

–

**Question 3:
**Write the Product of Sum (POS) form of the function G (U,V,W) for the following truth table representation ofG:

**Delhi 2013**

**Question 4:
**Write the Sum of Product (SOP) form of the function F(A, B, C) for the following truth table representation of F:

**All India 2013**

**Question 5:
**Write the dual of the boolean expression

(A+0).(A.1.\(\overline { A } \)).

**Аnswer:**

Using duality principle, changing (+) to (.) and vice-versa and by replacing 0’s with l’s and l’s with 0’s, the dual for the given expression is as follows:

(A.1) + (A + 0 + \(\overline { A } \))

**Question 6:
**Give the dual of the boolean function \(\overline { X } \).(Y + Z) + \(\overline { X } \) .(\(\overline { Y } \) + \(\overline { Z } \))

**Аnswer:**

Using duality principle, changing (+) to (.) and vice-versa and by keeping the form of variables as such the dual of the expression is:

(\(\overline { X } \) + Y.Z) .(\(\overline { X } \) + \(\overline { Y } \).\(\overline { Z } \))

**Question 7:
**Seven inverters are cascaded one after another. What is the output, if the input is 1?

**Аnswer:**

The logic circuit is given:

–

**Question 8:
**Represent F(a, b) = a.b using only NOT and OR gates.

**Аnswer:**

The expression F(a, b) = a.b using only NOT and OR gates is:

**Question 9:
**Find the complement of the following boolean function F1 = A.B + \(\overline { C } \).\(\overline { D } \);

**Аnswer:**

**Question 10:
**State the distributive laws of boolean algebra. How do they differ from the distributive laws of ordinary algebra?

**Аnswer:**

Distributive laws of boolean algebra state that:

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

Law X.(Y + Z) = X.Y + X.Z satisfies ordinary algebra for all values, whereas X + Y.Z = (X + Y).(X +‘Z) satisfies only for two values (0, 1) of X, Y and Z.

**Question 11:
**How many input combinations can be there in the truth table of a logic system having (N) input binary variables?

**Аnswer:**

2

^{N}input combinations can be there in the truth table of a logic system.

**Question 12:
**Why are NAND and NOR gates called Universal gates?

**Or**

Which gates are called Universal gates and why?

**Аnswer:**

NAND and NOR gates are easier to design and basic functions like AND, OR and NOT, etc., can be easily implemented using NAND/NOR gates. So, these gates are called Universal gates.

**Question 13:
**Draw a logic circuit diagram for the boolean expression A.(B + \(\overline { C } \)).

**All India2008**

**Аnswer:**

Logic circuit diagram for the boolean expression:

A.(B + \(\overline { C } \))

**Question 14:
**Draw a logic circuit diagram for the boolean expression \(\overline { A } \).(B + C).

**Аnswer:**

Logic circuit diagram for the boolean expression: \(\overline { A } \).(B + C)

**Question 15:
**Express P + \(\overline { Q } \).R in canonical SOP form.

**Аnswer:**

**Question 16:
**Draw a logic circuit diagram for the following

boolean expression \(\overline { X } \). (\(\overline { Y } \)+ Z)

**Delhi 2008**

**Аnswer:**

A logic diagram for the boolean expression \(\overline { X } \).(\(\overline { Y } \) +Z)

**Short Answer Type Questions [2 Marks] **

**Question 17:
**Write the equivalent boolean expression for the following logic circuit:

**Question 18:**

Interpret the following logic circuit as boolean expression:

**Question 19:
**Verify the following using boolean laws:

A + C = A + \(\overline { A } \).C + B.C

**Delhi 2013**

**Аnswer:**

A+C = A + \(\overline { A } \).C + B.C

R.H.S =A + \(\overline { A } \).C + B.C

=(A + \(\overline { A } \)).(A + C)+B.C [By using Distributive law]

= 1. (A + C)+B.C [By using A+\(\overline { A } \)=1]

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

= A + C.( 1 ) = A+C [∵ 1 + B = 1]

= L.H.S

**Question 20:
**Obtain the boolean expression for the logic circuit shown below:

**Question 21:
**Represent the boolean expression X.\(\overline { Y } \) + Y.\(\overline { Z } \) using only NAND gate.

**Аnswer:**

**Question 22:
**Represent the boolean expression (X + \(\overline { Y } \))Z with the help of NAND gate only.

**Аnswer:**

The boolean expression (X + \(\overline { Y } \)). Z can be implemented as:

**Question 23:
**Write the equivalent boolean expression for the following logic circuit:

**Question 24:
**Verify the following boolean expression using truth table:

(i) X.X = 0

(ii)X+ 1=1

**All India 2012**

Аnswer:

Аnswer:

**Question 25:
**Name the law shown below and verify it using a truth table.

**Delhi 2014**

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

**Аnswer:**

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

The above stated law is called distributive law.

Since, column five and column eight are identical, hence, the given law A + B.C = (A + B).(A + C) is verified.

**Question 26:
**Name the law shown below and verify it using a tru th table: All India 2014

X + \(\overline { X } \).Y = X+Y

**Аnswer:**

Name of the given law is redundant literal law.

Since, last two columns of the above tmth table are identical, hence, X+\(\overline { X } \). Y=X+Y is verified.

**Question 27:
**Draw the logic circuit for a half adder using NOR gate only.

**Аnswer:**

The boolean expression for the half adder,

Sum (S) = A.\(\overline { B } \) + \(\overline { A } \).B

Carry (C) = A.B

**Question 28:
**Draw the logic circuit for a half adder using NAND gate only.

**Аnswer:**

The boolean expression for the half adder,

Sum (S) = A.\(\overline { B } \) + \(\overline { A } \).B

Carry (C ) = A.B

**Question 29:
**State and prove DeMorgan’s laws in boolean algebra.

**Delhi 2011C, 2009C, 2008**

**Or**

State DeMorgan’s laws. Verify one of the DeMorgan’s laws using a truth table.

**Delhi 2013C**

**Аnswer:**

Refer to text on page no. 303-304.

**Question 30:**

Draw a logic circuit diagram for full adder. Give the truth table for full adder.

**Аnswer:**

Boolean function for full adder

**Question 31:
**Verify the following algebraically:

\(\overline { X } \).Y + X.\(\overline { Y } \) = (\(\overline { X } \) + \(\overline { Y } \)).(X + Y)

**All India 2010**

Аnswer:

Аnswer:

**Question 32:
**Given the following circuit:

**Question 33:
**Write the equivalent boolean expression for the following logic circuit:

**All India 2010**

**Question 34:
**Verify \(\overline { X } \).Y + X.\(\overline { Y } \) + \(\overline { X } \).\(\overline { Y } \) = \(\overline { X } \) + \(\overline { Y } \) using truth table.

**Delhi 2009**

Аnswer:

Аnswer:

**Question 35:
**State and verify absorption law using truth table.

**All India 2009,2008**

**Аnswer:**

Absorption Law It states that for boolean variables X and Y,

(i) X + X.Y = X

(ii) X.(X + Y) = X

**Question 36:
**Represent (P + \(\overline { Q } \) R) in POS form.

**Delhi 2009C**

To get the POS form, we need to product maxterms for all those input combinations that produce output as 0.

Thus, P + \(\overline { Q } \).R = (P +Q + R).(P + \(\overline { Q } \) + R).(P +\(\overline { Q } \)+\(\overline { R } \))

**Question 37:
**Convert the following boolean expression into the equivalent canonical Product of Sum (POS) form

A.\(\overline { B } \).C + \(\overline { A } \).B.C + \(\overline { A } \).B.\(\overline { C } \)

**All India 2008**

Аnswer:

Аnswer:

**Question 38:
**State and verify distributive law in boolean algebra.

**Delhi 2008C**

**Аnswer:**

Distributive law state that:

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

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

**Verification:**

(i) Truth table of: X.(Y + Z) = X.Y + X.Z

Both columns identical. X.(Y + Z) and X.Y + X.Z are identical

∴ X.(Y + Z) = X.Y + X.Z

Hence Proved.

(ii)Truth table for X + Y.Z = (X + Y).(X + Z)

Both columns X + Y.Z and (X + Y).(X + Z) are identical.

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

Hence Proved.

**Question 39:
**Convert the following boolean expression into its equivalent canonical Sum of Product (SOP) form.

**Delhi 2008C**

(\(\overline { U } \) + v + \(\overline { W } \)).(u + \(\overline { V } \) + \(\overline { W } \)).(u + v + w)

**Аnswer:**

**Question 40:
**State and verify associative law.

**Аnswer:**

The associative law states:

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

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

**Verification:**

(i) Truth table of X+ {Y + Z) = (X+ Y) +Z

Both columns X + (Y + Z) and (X + Y) + Z are identical.

∴ X +(Y + Z) = (X + Y) + Z Hence Proved.

(ii) Truth table of X.(Y.Z) = (X.Y).Z

Both columns X.(Y.Z) and (X.Y).Z are identical.

X .(Y .Z) = (X ,Y).Z Hence Proved.

**Question 41:
**Given the following truth table, derive a Sum of Product (SOP) and Product of Sum (POS) form of boolean expression from it:

**Question 42:**

**Question 43:**

**Question 44:
**Design a circuit (3 inputs) which gives a high output only when there is even number of low or high inputs.

**Аnswer:**

High output when even number of low inputs or even number of high inputs, i.e.

**Question 45:
**A boolean function F defined on three input variables X, Y and Z is, if and only if number of 1 input is odd. Draw the truth table for the above function and express it in canonical SOP form.

**Аnswer:**

Truth table

### Topic – 2

Karnaugh Map (K-Map)

**Exam Practice**

**Short Answer Type Questions [3 Mark]**

**Question 1:**

Obtain the minimal form for the following boolean expression using Karnaugh’s map: **All India 2014**

F(A, B, C, D) = ∑(1,3, 4,5,6,7,12,13)

**Аnswer:**

F (A, B,C, D) = ∑(1,3, 4,5,6,7,12,13)

**Question 2:
**Obtain the minimal form for the following boolean expression using Karnaugh’s map:

H (P, Q, R, S)=∑(0, 1,2, 3, 5, 7, 8, 9, 10, 14, 15)

**Delhi 2013**

Аnswer:

Аnswer:

**Question 3:
**What is the simplified boolean expression for the following K-map:

**Question 4:**

Simplify the expression:

**Question 5:
**Reduce the following boolean expression using K-map: F (A, B, C,D) = ∑(2,3, 4,5,6,7,8,10,11 )

**All India 2012**

**Аnswer:**

F (A, B,C, D) = ∑(2,3,4,5,6,7,8,10,11)

**Question 6:
**Reduce the following boolean expression using K-map: F(U, V, W, Z) = ∑(0, 1, 2, 4, 5, 6, 8, 10)

**Аnswer:**

F(U,V,W,Z) = ∑(0,1,2,4,5,6,8,10)

**Question 7:
**Reduce the following boolean expression using K-map:

F(A,B, C, D) = ∑(3, 4,5,6,7,13,15)

**All India 2010**

**Аnswer:**

F (A, B,C,D) = ∑(3,4,5,6,7,13,15)

**Question 8:
**Reduce the following boolean expression using K-map:

F(A,B,C,D) = ∑(0,1,2,3,4,5,10,11,15)

**Аnswer:**

**Question 9:
**Reduce the following boolean expression using K-map :

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

**Delhi 2009**

**Аnswer:**

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

**Question 10:
**Simplify using Karnaugh’s map for:

**Question 11:
**In the K-map shown below, which of the groupings shown in the K-map represents legal grouping?

**Аnswer:**

Groups must contain 2,4, 8 or in general 2^{n} cells either in horizontally or vertically. So, grouping A and E are legal groupings.

**Question 12:
**Draw the K-map and give the reduced SOP expression for the following truth table:

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