McqMate

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These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) , Electronics and Telecommunication Engineering [ENTC] .

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» D. all of the above |

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 |

26. |
## The code where all successive numbers differ from their preceding number by single bit is |

A. | binary code. |

B. | bcd. |

C. | excess – 3. |

D. | gray. |

Answer» D. gray. |

27. |
## -8 is equal to signed binary number |

A. | 10001000 |

B. | oooo1000 |

C. | 10000000 |

D. | 11000000 |

Answer» A. 10001000 |

28. |
## DeMorgan’s first theorem shows the equivalence of |

A. | or gate and exclusive or gate. |

B. | nor gate and bubbled and gate. |

C. | nor gate and nand gate. |

D. | nand gate and not gate |

Answer» B. nor gate and bubbled and gate. |

29. |
## When signed numbers are used in binary arithmetic, then which one of the following notations would have unique representation for zero. |

A. | sign- magnitude. |

B. | 1’s complement. |

C. | 2’s complement. |

D. | 9’s compleme nt. |

Answer» A. sign- magnitude. |

30. |
## The decimal equivalent of Binary number 11010 is |

A. | 26 |

B. | 36 |

C. | 16 |

D. | 23 |

Answer» A. 26 |

31. |
## 1’s complement representation of decimal number of -17 by using 8 bit |

A. | 1110 1110 |

B. | 1101 1101 |

C. | 1100 1100 |

D. | 0001 0001 |

Answer» A. 1110 1110 |

32. |
## The excess 3 code of decimal number 26 is |

A. | 0100 1001 |

B. | 1011001 |

C. | 1000 1001 |

D. | 1001101 |

Answer» B. 1011001 |

33. |
## How many AND gates are required to realize Y = CD+EF+G |

A. | 4 |

B. | 5 |

C. | 3 |

D. | 2 |

Answer» D. 2 |

34. |
## The hexadecimal number for (95.5)10 is |

A. | (5f.8) 16 |

B. | (9a.b) 16 |

C. | ( 2e.f) 16 |

D. | ( 5a.4) 16 |

Answer» A. (5f.8) 16 |

35. |
## The octal equivalent of (247) 10 is |

A. | ( 252) 8 |

B. | (350) 8 |

C. | ( 367) 8 |

D. | ( 400) 8 |

Answer» C. ( 367) 8 |

36. |
## The number 140 in octal is equivalent to |

A. | (96)10 . |

B. | ( 86) 10 |

C. | (90) 10 . |

D. | none of these. |

Answer» A. (96)10 . |

37. |
## The NOR gate output will be low if the two inputs are |

A. | 11 |

B. | 1 |

C. | 10 |

D. | all |

Answer» D. all |

38. |
## Convert decimal 153 to octal. Equivalent in octal will be |

A. | (231)8 |

B. | ( 331) 8 |

C. | ( 431) 8 . |

D. | none of these. |

Answer» A. (231)8 |

39. |
## The decimal equivalent of ( 1100)2 is |

A. | 12 |

B. | 16 |

C. | 18 |

D. | 20 |

Answer» A. 12 |

40. |
## The binary equivalent of (FA)16 is |

A. | 1010 1111 |

B. | 1111 1010 |

C. | 10110011 |

D. | none of these |

Answer» B. 1111 1010 |

41. |
## How many two-input AND and OR gates are required to realize Y=CD+EF+G |

A. | 22 |

B. | 23 |

C. | 33 |

D. | none of these |

Answer» A. 22 |

42. |
## The excess-3 code of decimal 7 is represented by |

A. | 1100 |

B. | 1001 |

C. | 1011 |

D. | 1010 |

Answer» D. 1010 |

43. |
## When an input signal A=11001 is applied to a NOT gate serially, its output signal is |

A. | 111 |

B. | 110 |

C. | 10101 |

D. | 11001 |

Answer» B. 110 |

44. |
## The result of adding hexadecimal number A6 to 3A is |

A. | dd |

B. | e0 |

C. | f0 |

D. | ef |

Answer» B. e0 |

45. |
## A universal logic gate is one, which can be used to generate any logic function. Which of the following is a universal logic gate? |

A. | or |

B. | and |

C. | xor |

D. | nand |

Answer» D. nand |

46. |
## Karnaugh map is used for the purpose of |

A. | reducing the electronic circuits used. |

B. | to map the given boolean logic function. |

C. | to minimize the terms in a boolean expression. |

D. | to maximize the terms of a given a boolean expression . |

Answer» C. to minimize the terms in a boolean expression. |

47. |
## The 2’s complement of the number 1101110 is |

A. | 10001 |

B. | 10001 |

C. | 10010 |

D. | none |

Answer» C. 10010 |

48. |
## The decimal equivalent of Binary number 10101 is |

A. | 21 |

B. | 31 |

C. | 26 |

D. | 28 |

Answer» A. 21 |

49. |
## How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB |

A. | 11 |

B. | 42 |

C. | 32 |

D. | 23 |

Answer» C. 32 |

50. |
## Which of following are known as universal gates |

A. | nand & nor |

B. | and & or. |

C. | xor & or. |

D. | none. |

Answer» A. nand & nor |

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