McqMate
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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