| 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 | |