1. |
## The 2s compliment form (Use 6 bit word) of the number 1010 is |

A. | 111100 |

B. | 110110 |

C. | 110111 |

D. | 1011 |

Answer» B. 110110 |

2. |
## AB+(A+B)’ is equivalent to |

A. | a ex-nor b |

B. | a ex or b |

C. | (a+b)a |

D. | (a+b)b |

Answer» A. a ex-nor b |

3. |
## The hexadecimal number equivalent to (1762.46)8 is |

A. | 3f2.89 |

B. | 3f2.98 |

C. | 2f3.89 |

D. | 2f3.98 |

Answer» B. 3f2.98 |

4. |
## A three input NOR gate gives logic high output only when |

A. | one input is high |

B. | one input is low |

C. | two input are low |

D. | all input are low |

Answer» D. all input are low |

5. |
## The absorption law in Boolean algebra say that |

A. | x + x = x |

B. | x . y = x |

C. | x + x . y = x |

D. | none of the above |

Answer» C. x + x . y = x |

6. |
## Logic X-OR operation of (4ACO)H & (B53F)H results |

A. | aacb |

B. | 0 |

C. | abcd |

D. | ffff |

Answer» D. ffff |

7. |
## What is decimal equivalent of (11011.1000)2 ? |

A. | 22 |

B. | 22.2 |

C. | 20.2 |

D. | 27.5 |

Answer» D. 27.5 |

8. |
## The negative numbers in the binary system can be represented by |

A. | sign magnitude |

B. | 2\s complement |

C. | 1\s complement |

D. | all of the above |

Answer» A. sign magnitude |

9. |
## Negative numbers cannot be represented in |

A. | signed magnitude form |

B. | 1’s complement form |

C. | 2’s complement form |

D. | none of the above |

Answer» D. none of the above |

10. |
## The answer of the operation (10111)2*(1110)2 in hex equivalence is |

A. | 150 |

B. | 241 |

C. | 142 |

D. | 1.01e+08 |

Answer» C. 142 |

11. |
## The Hexadecimal number equivalent of (4057.06)8 is |

A. | 82f.027 |

B. | 82f.014 |

C. | 82f.937 |

D. | 83f.014 |

Answer» B. 82f.014 |

12. |
## The NAND gate output will be low if the two inputs are |

A. | 0 |

B. | 1 |

C. | 10 |

D. | 11 |

Answer» D. 11 |

13. |
## A binary digit is called a |

A. | bit |

B. | character |

C. | number |

D. | byte |

Answer» A. bit |

14. |
## What is the binary equivalent of the decimal number 368 |

A. | 101110000 |

B. | 110110000 |

C. | 111010000 |

D. | 1.11e+08 |

Answer» A. 101110000 |

15. |
## What is the binary equivalent of the Octal number 367 |

A. | 11110111 |

B. | 11100111 |

C. | 110001101 |

D. | 10111011 1 |

Answer» A. 11110111 |

16. |
## What is the binary equivalent of the Hexadecimal number 368 |

A. | 1111101000 |

B. | 1101101000 |

C. | 1101111000 |

D. | 1.11e+09 |

Answer» B. 1101101000 |

17. |
## What is the binary equivalent of the decimal number 1011 |

A. | 1111110111 |

B. | 1111000111 |

C. | 1111110011 |

D. | 1.11e+09 |

Answer» C. 1111110011 |

18. |
## The gray code equivalent of (1011)2 is |

A. | 1101 |

B. | 1010 |

C. | 1111 |

D. | 1110 |

Answer» D. 1110 |

19. |
## The decimal equivalent of hex number 1A53 is |

A. | 6793 |

B. | 6739 |

C. | 6973 |

D. | 6379 |

Answer» B. 6739 |

20. |
## ( 734)8 =( )16 |

A. | c 1 d |

B. | d c 1 |

C. | 1 c d |

D. | 1 d c |

Answer» D. 1 d c |

21. |
## The simplification of the Boolean expression (A'BC')'+ (AB'C)' is |

A. | 0 |

B. | 1 |

C. | a |

D. | bc |

Answer» B. 1 |

22. |
## The hexadecimal number ‘A0’ has the decimal value equivalent to |

A. | 80 |

B. | 256 |

C. | 100 |

D. | 160 |

Answer» D. 160 |

23. |
## The Boolean expression A.B+ A.B+ A.B is equivalent to |

A. | a + b |

B. | a\.b |

C. | (a + b)\ |

D. | a.b |

Answer» A. a + b |

24. |
## The 2’s complement of the number 1101101 is |

A. | 101110 |

B. | 111110 |

C. | 110010 |

D. | 10011 |

Answer» D. 10011 |

25. |
## When simplified with Boolean Algebra (x + y)(x + z) simplifies to |

A. | x |

B. | x + x(y + z) |

C. | x(1 + yz) |

D. | x + yz |

Answer» D. x + yz |

Tags

Question and answers in Digital Electronics and Logic Design,
Digital Electronics and Logic Design multiple choice questions and answers,
Digital Electronics and Logic Design Important MCQs,
Solved MCQs for Digital Electronics and Logic Design,
Digital Electronics and Logic Design MCQs with answers PDF download