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

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:

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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:

1. X.(Y + Z) = X.Y + X.Z
2. 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:

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:

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:

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:

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:

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 2ⁿ 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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