McqMate
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 |
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 |
51. |
Convert the octal number 7401 to Binary. |
A. | 1.111e+11 |
B. | 1.1111e+11 |
C. | 1.111e+11 |
D. | 1.11e+11 |
Answer» A. 1.111e+11 |
52. |
Find the hex sum of (93)16 + (DE)16 . |
A. | (171)16 |
B. | (271)16 |
C. | (179)16 |
D. | (181)16 |
Answer» A. (171)16 |
53. |
Perform 2’s complement subtraction of (7)10 − (11)10 . |
A. | 1100 (or -4) |
B. | 1101 (or -5) |
C. | 1011 (or -3) |
D. | 1110 (or -6) |
Answer» A. 1100 (or -4) |
54. |
What is the Gray equivalent of (25)10 |
A. | 1101 |
B. | 110101 |
C. | 10110 |
D. | 10101 |
Answer» D. 10101 |
55. |
Simplify the Boolean expression F = C(B + C)(A + B + C). |
A. | c |
B. | bc |
C. | abc |
D. | a+bc |
Answer» A. c |
56. |
Simplify the following expression into sum of products using Karnaugh map F(A,B,C,D) = (1,3,4,5,6,7,9,12,13) |
A. | a\b+c\ d+ a\d+bc\ |
B. | a\b\+c\ d\+ a\d\+b\c\ |
C. | a\b\+c\ d+ a\d+bc\ |
D. | a\b+c\ d\+ a\d\+bc\ |
Answer» A. a\b+c\ d+ a\d+bc\ |
57. |
Simplify F = (ABC)'+( AB)'C+ A'BC'+ A(BC)'+ AB'C. |
A. | ( a\ + b\ +c\ ) |
B. | ( a\ + b +c ) |
C. | ( a + b +c ) |
D. | ( a + b\ +c\ ) |
Answer» A. ( a\ + b\ +c\ ) |
58. |
Determine the binary numbers represented by 25.5 |
A. | 11001.1 |
B. | 11011.101 |
C. | 10101.11 |
D. | 11001.010 1 |
Answer» A. 11001.1 |
59. |
Conversion of decimal number 10.625 into binary number: |
A. | 1010.101 |
B. | 1110.101 |
C. | 1001.11 |
D. | 1001.101 |
Answer» A. 1010.101 |
60. |
Conversion of fractional number 0.6875 into its equivalent binary number: |
A. | 0.1011 |
B. | 0.1111 |
C. | 0.10111 |
D. | 0.0101 |
Answer» A. 0.1011 |
61. |
Perform the following subtractions using 2’s complement method. 01000 – 01001 |
A. | 1 |
B. | 10 |
C. | 11 |
D. | 11110 |
Answer» A. 1 |
62. |
Subtraction of 01100-00011 using 2’s complement method. : |
A. | 1001 |
B. | 1000 |
C. | 1010 |
D. | 110 |
Answer» A. 1001 |
63. |
Minimize the logic functionY(A,B,C,D) = IZm(0,1,2,3,5,7,8,9,11,14) . Using Karnaugh map. |
A. | abc d\ + a\ b\ + b\ c\ + b\ d+ a\d |
B. | abc d + a b + b\ c\ + b\ d+ a\d |
C. | a\ b\ + b\ c\ + b\ d+ a\d |
D. | abc d\ + a\ b\ + b\ c\ + b\ d |
Answer» A. abc d\ + a\ b\ + b\ c\ + b\ d+ a\d |
64. |
Simplify the given expression to its Sum of Products (SOP) form Y = (A + B)(A + (AB)')C + A'(B+C')+ A'B+ ABC |
A. | ac+ bc+ a\b + a\ c\ |
B. | ac+ bc+ a\b |
C. | bc+ a\b + a\ c\ |
D. | ac+ a\b + a\ c\ |
Answer» A. ac+ bc+ a\b + a\ c\ |
65. |
Convert the decimal number 82.67 to its binary, hexadecimal and octal equivalents |
A. | (1010010.1010 1011)2; (52.ab)16 ; |
B. | (1010010.10 101011)2; (52.ab)16 ; |
C. | (1010010.10 101011)2; (52.ab)16 ; |
D. | (1010010. |
Answer» A. (1010010.1010 1011)2; (52.ab)16 ; |
66. |
Add 20 and (-15) using 2’s complement. |
A. | (100100 )2 or (+4)10 |
B. | (000100 )2 or (-4)10 |
C. | both (a) and (b) |
D. | none of the above |
Answer» A. (100100 )2 or (+4)10 |
67. |
Add 648 and 487 in BCD code. |
A. | 1135 |
B. | 1136 |
C. | 1235 |
D. | 1138 |
Answer» A. 1135 |
68. |
(23.6)10 = (X)2 FIND X |
A. | (10111.100110 0)2 |
B. | (10101.1001 100)2 |
C. | (10001.1001 100)2 |
D. | (10111.10 00011)2 |
Answer» A. (10111.100110 0)2 |
69. |
(65.535)10 =(X)16 FIND X |
A. | (41.88f5c28)16 . |
B. | (42.88f5c28 )16. |
C. | (41.88f5c)16. |
D. | (42.88f5c )16. |
Answer» A. (41.88f5c28)16 . |
70. |
Convert the decimal number 430 to Excess-3 code: |
A. | 110110001 |
B. | 110110000 |
C. | 110110011 |
D. | 11010000 1 |
Answer» A. 110110001 |
71. |
Minimize the following logic function using K-maps F(A,B,C,D) = m(1,3,5,8,9,11,15) + d(2,13) |
A. | a b\ c\ + c\ d + b\d + ad |
B. | c\ d + b\d + ad |
C. | a b\ c\ + b\d + ad |
D. | a b\ c\ + c\ d + b\d |
Answer» A. a b\ c\ + c\ d + b\d + ad |
72. |
Convert (2222)10 in Hexadecimal number. |
A. | 8ae |
B. | 8be |
C. | 93c |
D. | fff |
Answer» A. 8ae |
73. |
Divide ( 101110) 2 by ( 101)2. |
A. | quotient -1001 remainder - 001 |
B. | quotient - 1000 remainder - 001 |
C. | quotient - 1001 remainder - 011 |
D. | quotient - 1001 remainder -000 |
Answer» A. quotient -1001 remainder - 001 |
74. |
Minimise the logic function (POS Form) F A,B,C,D) = PI M (1, 2, 3, 8, 9, 10, 11,14)× d (7, 15) |
A. | f=[(b+d’)+(b+ c’)’(a’+c’)+(a’ +b)]’ |
B. | f=[(b+d’)+(b +c’)’(‘a’+c’)+ (a’+b)]’ |
C. | f=[(b+d’)+(b +c’)’(‘a’+c’)+ (a’+b)]’ |
D. | f=[(b+d’)+ (b+c’)’(‘a’ +c’)+(a’+b )]’ |
Answer» A. f=[(b+d’)+(b+ c’)’(a’+c’)+(a’ +b)]’ |
75. |
Perform following subtraction (i) 11001-10110 using 1’s complement |
A. | 11 |
B. | 111 |
C. | 10 |
D. | 10011 |
Answer» A. 11 |
76. |
Perform following subtraction(ii) 11011-11001 using 2’s complement |
A. | 10 |
B. | 111 |
C. | 11 |
D. | 10011 |
Answer» A. 10 |
77. |
Reduce the following equation using k-map Y = (ABC)'+ A(CD)'+ AB'+ ABCD'+ (AB)'C |
A. | b\+(ad)\ |
B. | b\ |
C. | (ad)\ |
D. | b\+ad |
Answer» A. b\+(ad)\ |
78. |
Write the expression for Boolean function F (A, B, C) = m (1,4,5,6,7) in standard POS form. |
A. | = (a+b+c)(a+b\ +c)(a+b\ +c\ ) |
B. | = (a+b\ +c)(a+b\ +c\ ) |
C. | = (a+b+c)(a+b \ +c) |
D. | = (a+b+c)(a +b\ +c\ ) |
Answer» A. = (a+b+c)(a+b\ +c)(a+b\ +c\ ) |
79. |
Convert the decimal number 45678 to its hexadecimal equivalent number. |
A. | (b26e)16 |
B. | (a26e)16 |
C. | (b26b)16 |
D. | (b32e)16 |
Answer» A. (b26e)16 |
80. |
Convert (177.25)10 to octal. |
A. | (261.2)8 |
B. | (260.2)8 |
C. | (361.2)8 |
D. | (251.2)8 |
Answer» A. (261.2)8 |
81. |
Reduce the following equation using k-map Y = B C' D'+ A' B C' D+ A B C' D+ A' B C D+ A B C D |
A. | bc’ + bd |
B. | bc’ + bd+a |
C. | bc’ + bd + ac |
D. | bc’ + bd + ad |
Answer» A. bc’ + bd |
82. |
8-bit 1’s complement form of –77.25 is |
A. | 1001101.01 |
B. | 10110010.10 11 |
C. | 01001101.00 10 |
D. | 10110010 |
Answer» B. 10110010.10 11 |
83. |
In computers, subtraction is generally carried out by |
A. | 9’s complement |
B. | 10’s complement |
C. | 1’s complement |
D. | 2’s compleme nt |
Answer» D. 2’s compleme nt |
84. |
The answer of the operation (10111)2*(1110)2 in hex equivalence is |
A. | 150 |
B. | 241 |
C. | 142 |
D. | 10101111 0 |
Answer» C. 142 |
85. |
The decimal number equivalent of (4057.06)8 is |
A. | 2095.75 |
B. | 2095.075 |
C. | 2095.937 |
D. | 2095.094 |
Answer» D. 2095.094 |
86. |
12-bit 2’s complement of –73.75 is |
A. | 01001001.110 0 |
B. | 11001001.11 00 |
C. | 10110110.01 00 |
D. | 10110110. 1100 |
Answer» C. 10110110.01 00 |
87. |
What is decimal equivalent of BCD 11011.1100 ? |
A. | 22 |
B. | 22.2 |
C. | 20.2 |
D. | 21.2 |
Answer» B. 22.2 |
88. |
What is the binary equivalent of the decimal number 368 |
A. | 101110000 |
B. | 110110000 |
C. | 111010000 |
D. | 11110000 0 |
Answer» A. 101110000 |
89. |
(2FAOC)16 is equivalent to |
A. | (195 084)10 |
B. | (001011111 010 0000 1100)2 |
C. | both (a) and (b) |
D. | none of these |
Answer» B. (001011111 010 0000 1100)2 |
90. |
The octal equivalent of hexadecimal (A.B)16 is |
A. | 47.21 |
B. | 12.74 |
C. | 12.71 |
D. | 17.21 |
Answer» B. 12.74 |
91. |
Logic X-OR operation of (4ACO)H & (B53F)H results |
A. | aacb |
B. | 0 |
C. | ffff |
D. | abcd |
Answer» C. ffff |
92. |
The simplified form of the Boolean expression (X+Y+XY)(X+Z) is |
A. | x + y + zx + y |
B. | xy – yz |
C. | x + yz |
D. | xz + y |
Answer» C. x + yz |
93. |
The output of a logic gate is 1 when all its inputs are at logic 0. the gate is either |
A. | nand or an xo |
B. | or or an xn |
C. | o an and or xor |
D. | a nor or an xnor |
Answer» D. a nor or an xnor |
94. |
2's complement of any binary number can be calculated by |
A. | adding 1\s complement twice |
B. | adding 1 to 1\s complement |
C. | subtracting 1 from 1\s complement. |
D. | calculating 1\s compleme nt and inverting most significant bit |
Answer» B. adding 1 to 1\s complement |
95. |
Sum-of-Weights method is used |
A. | to convert from one number system to other |
B. | to encode data |
C. | to decode data |
D. | to convert from serial to parralel data |
Answer» A. to convert from one number system to other |
96. |
The complement of a variable is always |
A. | 1 |
B. | 0 |
C. | inverse |
D. | none |
Answer» C. inverse |
97. |
The difference of 111 - 001 equals |
A. | 100 |
B. | 111 |
C. | 1 |
D. | 110 |
Answer» D. 110 |
98. |
The Unsigned Binary representation can only represent positive binary numbers |
A. | true |
B. | false |
C. | both (a) and (b) |
D. | none of above |
Answer» A. true |
99. |
which of the following rules states that if one input of an AND gate is always 1, the output is equal to the other input? |
A. | a +1 =1 |
B. | a +a =a |
C. | a.a = a |
D. | a.1= a |
Answer» C. a.a = a |
100. |
Which of the number is not a representative of hexadecimal system |
A. | 1234 |
B. | abcd |
C. | 1001 |
D. | defh |
Answer» D. defh |
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