# 600+ Digital Electronics and Logic Design Solved MCQs

1.

A. 111100
B. 110110
C. 110111
D. 1011
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.

A. 3f2.89
B. 3f2.98
C. 2f3.89
D. 2f3.98
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.

A. aacb
B. 0
C. abcd
D. ffff
7.

A. 22
B. 22.2
C. 20.2
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.

A. 150
B. 241
C. 142
D. 1.01e+08
11.

A. 82f.027
B. 82f.014
C. 82f.937
D. 83f.014
12.

A. 0
B. 1
C. 10
D. 11
13.

A. bit
B. character
C. number
D. byte
14.

A. 101110000
B. 110110000
C. 111010000
D. 1.11e+08
15.

A. 11110111
B. 11100111
C. 110001101
D. 10111011 1
16.

A. 1111101000
B. 1101101000
C. 1101111000
D. 1.11e+09
17.

A. 1111110111
B. 1111000111
C. 1111110011
D. 1.11e+09
18.

A. 1101
B. 1010
C. 1111
D. 1110
19.

A. 6793
B. 6739
C. 6973
D. 6379
20.

A. c 1 d
B. d c 1
C. 1 c d
D. 1 d c
Answer» D. 1 d c
21.

A. 0
B. 1
C. a
D. bc
22.

A. 80
B. 256
C. 100
D. 160
23.

A. a + b
B. a\.b
C. (a + b)\
D. a.b
Answer» A. a + b
24.

A. 101110
B. 111110
C. 110010
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.

A. binary code.
B. bcd.
C. excess – 3.
D. gray.
27.

A. 10001000
B. oooo1000
C. 10000000
D. 11000000
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.

A. 26
B. 36
C. 16
D. 23
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.

A. 0100 1001
B. 1011001
C. 1000 1001
D. 1001101
33.

A. 4
B. 5
C. 3
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.

A. 11
B. 1
C. 10
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.
39.

A. 12
B. 16
C. 18
D. 20
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.

A. 22
B. 23
C. 33
D. none of these
42.

A. 1100
B. 1001
C. 1011
D. 1010
43.

A. 111
B. 110
C. 10101
D. 11001
44.

A. dd
B. e0
C. f0
D. ef
45.

A. or
B. and
C. xor
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.

A. 10001
B. 10001
C. 10010
D. none
48.

A. 21
B. 31
C. 26
D. 28
49.

A. 11
B. 42
C. 32
D. 23
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.

A. 1.111e+11
B. 1.1111e+11
C. 1.111e+11
D. 1.11e+11
52.

A. (171)16
B. (271)16
C. (179)16
D. (181)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.

A. 1101
B. 110101
C. 10110
D. 10101
55.

A. c
B. bc
C. abc
D. a+bc
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.

A. 11001.1
B. 11011.101
C. 10101.11
D. 11001.010 1
59.

A. 1010.101
B. 1110.101
C. 1001.11
D. 1001.101
60.

A. 0.1011
B. 0.1111
C. 0.10111
D. 0.0101
61.

A. 1
B. 10
C. 11
D. 11110
62.

A. 1001
B. 1000
C. 1010
D. 110
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.

A. 1135
B. 1136
C. 1235
D. 1138
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.

A. 110110001
B. 110110000
C. 110110011
D. 11010000 1
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.

A. 8ae
B. 8be
C. 93c
D. fff
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.

A. 11
B. 111
C. 10
D. 10011
76.

A. 10
B. 111
C. 11
D. 10011
77.

B. b\
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.

A. (b26e)16
B. (a26e)16
C. (b26b)16
D. (b32e)16
80.

A. (261.2)8
B. (260.2)8
C. (361.2)8
D. (251.2)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.

A. 150
B. 241
C. 142
D. 10101111 0
85.

A. 2095.75
B. 2095.075
C. 2095.937
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.

A. 22
B. 22.2
C. 20.2
D. 21.2
88.

A. 101110000
B. 110110000
C. 111010000
D. 11110000 0
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.

A. 47.21
B. 12.74
C. 12.71
D. 17.21
91.

A. aacb
B. 0
C. ffff
D. abcd
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.

A. 1
B. 0
C. inverse
D. none
97.

A. 100
B. 111
C. 1
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
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.

A. 1234
B. abcd
C. 1001
D. defh