100% FREE ACCESS12th Standard CBSE Syllabus & Materials
12th Standard CBSE
CBSE 12th Economics Government Budget and the Economy Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Computer Science Interface Python with MySQL - New Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Computer Science Database Concept - New Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Computer Science Data Communication - New Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Computer Science Data Structures - New Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Computer Science Functions - New Previous year Question Papers Study Material - QB365 Set A
Published on: 31/07/2019
Boolean Algebra
Download CBSE Class 12th Standard CBSE Computer Science question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Computer Science
Questions + Answers key
Take MCQ Computer Science Test
1.
Prove that (a’+b’)(a’+b)(a+b’)=a’b’
2.
Write the equivalent Boolean Expression for the following Logic Circuit
3.
Draw a Logical Circuit Diagram for the following Boolean Expression: A.(B+C’)
4.
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’)
5.
Convert the following boolean expression into the equivalent canonical Product of Sum(POS) form \(A.\bar { B } .C+\bar { A } .B.C+\bar { A } .B.\bar { C } \)
6.
Verify \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \) using truth table.
7.
Name the law shown below and verify it using a truth table
A+B.C=(A+B).(A+C)
8.
Express \(P+\bar { Q } .R\) in canonical SOP form
9.
Why are NAND and NOR gates called Universal gates?
10.
Find the complement of the following boolean function \({ F }_{ 1 }=A.B+\bar { C } .\bar { D } \) :
11.
Write the dual of the boolean expression \((A+0).(A.1.\bar { A } )\)
12.
Write the Product of Sum(POS) form of the function F(X, Y, Z) for the following truth table representation of F:
| 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 | 1 |
13.
Give the logic function F=AB+A’B’+A’B. Assuming the complements are available. Simplify the function using De Morgan’s Theorem. Synthesize F by using NOR gates only.
14.
Draw the truth table for a full adder and then simplify the Sum and Carry by using K-map. After that draw its circuit diagram
15.
Reduce the following Boolean Expression with the help of Karnnaugh Map F(U,V,W,Z)=S(0,1,2,3,12,13,14,15)
16.
Solve the following F(a,b,c,d)=∑(0,1,3,4,5,7,8,9,11,12,13,15). Obtain the simplified form using K-Map.
17.
If F(a,b,c,d)=(0,2,4,5,7,8,10,12,13,15), obtain the simplified form using K-Map
18.
Reduce the following Boolan Expression using K-Map:
F(U,V,W,Z)=\(\pi\) (0,1,2,3,4,5,6,810,1)
19.
Reduce the following Boolean expression using
K – Map F (A, B, C, D) =Σ (0,2,3,4,6,7,8,10,12)
20.
Reduce the following Boolean expression using K-Map:
F(A,B,C,D)=(0,1,2,4,5,8,9,10,11)
1.
LHS=(a’+b’)(a’+b)(a+b’)
=(a’a’+a’b+a’b’+b’b)(a+b’)
=(a’+a’b+a’b’+0)(a+b’)
=aa’+a’b’+aa’b+a’bb’+a’ab’+a’b’b’
=0+a’b’+0+)+0+0+a’b’=a’b’=RHS
2.
F(P,Q)=(P'+Q).(P+Q')
3.
4.
F( X , Y , Z ) = ∏ (4 , 5 , 6 , 7)
= Σ (0 , 1 , 2 , 3)
= X’. Y’. Z’ + X’. Y’. Z + X’. Y. Z’ + X’. Y. Z
5.
F= \(A.\bar { B } .C+\bar { A } .B.C+\bar { A } .B.\bar { C } \)
i.e., (101)+(011)+(010)=m5+m3+m2=\(\sum (2,3,5)\)
POS expression will be :
\(=\prod (0,1,4,6,7)\)
= M0.M1.M4.M6.M7
=(000).(001).(100).(110).(111)
=\((A+B+C).(A+B+\bar { C } ).(\bar { A } +B+C).(\bar { A } +\bar { B } +C).(\bar { A } +\bar { B } +\bar { C } )\)
6.
\(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \)
| X | Y | \(\bar { X } \) | \(\bar { Y } \) | \(\bar { X } .Y\) | \(X.\bar { Y } \) | \(\bar { X } .\bar { Y } \) | \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } \) | \(\bar { X } +\bar { Y } \) |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
Both columns \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } \) and \(\bar { X } +\bar { Y } \) are identical.
So, \(\bar { X } .Y+X.\bar { Y } +\bar { X } .\bar { Y } =\bar { X } +\bar { Y } \) Hence proved.
7.
A+B.C=(A+B).(A+C)
The above stated law is called distributive law.
| A | B | C | B.C | A+B.C | A+B | A+C | (A+B).(A+C) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Since, column five and column eight are identical, hence, the given law A+B.C=(A+B).(A+C) is verified
8.
( )
\(\Longrightarrow P+\bar { Q } .R\)
\(\Longrightarrow P.(Q+\bar { Q } ).(R+\bar { R } )+(P+\bar { P } ).\bar { Q } .R\)
\(\Longrightarrow (P.Q+P.\bar { Q } ).(R+\bar { R } )+P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
\(\Longrightarrow R.(P.Q+P.\bar { Q } ).\bar { R } .(P.Q+P.\bar { Q } )+P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
\(\Longrightarrow P.Q.R+P.\bar { Q } .R+P.Q.\bar { R } +P.\bar { Q } .\bar { R } +P.\bar { Q } .R+\bar { P } .\bar { Q } .R\)
Now, remove duplicate term,
\(\Longrightarrow P.Q.R+P.\bar { Q } .R+P.Q.\bar { R } +P.\bar { Q } .\bar { R } +\bar { P } .\bar { Q } .R\)
9.
( )
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.
10.
( )
\(\overset { \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ }{ (A.\bar { B } +\bar { C } .\bar { D } ) } \) \(=\overset { \_ \_ \_ \_ \_ \_ \_ \_ \_ \quad \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ }{ (A.\bar { B } ).(\bar { C } +\bar { D } ) } \) [DeMorgan's law]
\(=(\bar { A } +\overset { \_ \_ \_ \_ }{ \bar { B } } ).(\overset { \_ \_ \_ }{ \bar { C } } +\overset { \_ \_ \_ \_ }{ \bar { D } } )\) [Demorgan's law]
\(=(\bar { A } +B).(C+D)\)
11.
( )
Using duality principle, changing (+) to (.) and vice-versa and by replacing 0's with 1's and 1's with 0's, the dual for the given expression is as follows:\((A.1)+(A+0+\bar { A } )\)
12.
( )
| X | Y | Z | F | Maxterms |
| 0 | 0 | 0 | 1 | X+Y+Z |
| 0 | 0 | 1 | 0 | \(X+Y+\bar { Z } \) |
| 0 | 1 | 0 | 0 | \(X+\bar {Y}+ { Z } \) |
| 0 | 1 | 1 | 1 | \(X+ \bar {Y}+\bar { Z } \) |
| 1 | 0 | 0 | 0 | \(\bar {X}+ Y+Z \) |
| 1 | 0 | 1 | 0 | \(\bar {X}+ Y+\bar { Z } \) |
| 1 | 1 | 0 | 1 | \(\bar {X}+ \bar {Y}+Z \) |
| 1 | 1 | 1 | 1 | \(\bar {X}+ \bar {Y}+\bar { Z } \) |
POS form of function F(X, Y, Z) is: \((X+Y+\bar { Z } ).(X+\bar { Y } +Z).(\bar { X } +Y+Z).(\bar { X } +Y+\bar { Z } )\)
13.
14.
15.
16.
17.
18.
F(U,V,W,Z)= (0,1,2,4,5,6,8,10)
=\(\Sigma\)(3,7,9,11,12,13,14,15)
19.
F= C’.D’ + A’.C + B’.D’
20.
F(A,B,C,D)= A’.C’+A.B’ + B’.D’
12th Standard CBSE Syllabus & Materials
12th Standard CBSE
CBSE 12th Computer Science Python Revision Tour I - New Previous year Question Papers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Business Studies Planning Important Questions And Answers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Business Studies Business Environment Important Questions And Answers Study Material - QB365 Set A
NEW12th Standard CBSE
CBSE 12th Business Studies Principles of Management Important Questions And Answers Study Material - QB365 Set A
CBSE 12th Standard CBSE Subjects
CBSE Standards
