Digital Electronics Set 4
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This set of Digital Electronics Multiple Choice Questions & Answers (MCQs) focuses on Digital Electronics Set 4
Q1 | Convert 59.7210 to BCD.
- 111011
- 01011001.01110010
- 1110.11
- 0101100101110010
Q2 | Convert 8B3F16 to binary.
- 35647
- 011010
- 1011001111100011
- 1000101100111111
Q3 | Which is typically the longest: bit, byte, nibble, word?
- bit
- byte
- nibble
- word
Q4 | Assign the proper odd parity bit to the code 111001.
- 1111011
- 1111001
- 0111111
- 0011111
Q5 | Convert decimal 64 to binary.
- 01010010
- 01000000
- 00110110
- 01001000
Q6 | Convert hexadecimal value C1 to binary.
- 11000001
- 1000111
- 111000100
- 111000001
Q7 | The given hexadecimal number (1E.53)16 is equivalent to
- (35.684)8
- (36.246)8
- (34.340)8
- (35.599)8
Q8 | The octal number (651.124)8 is equivalent to
- 16
- (1b0.10)16
- (1a8.a3)16
- (1b0.b0)16
Q9 | The octal equivalent of the decimal number (417)10 is
- (641)8
- (619)8
- (640)8
- (598)8
Q10 | Convert the hexadecimal number (1E2)16 to decimal:
- 480
- 483
- 482
- 484
Q11 | (170)10 is equivalent to
- (fd)16
- (df)16
- (aa)16
- (af)16
Q12 | Convert the binary number (01011.1011)2 into decimal:
- (11.6875)10
- (11.5874)10
- (10.9876)10
- (10.7893)10
Q13 | 1011)2 = (11.6875)10
- (111101)2
- (010100)2
- (111100)2
- (101010)2
Q14 | On addition of +38 and -20 using 2’s complement, we get
- 11110001
- 100001110
- 010010
- 110101011
Q15 | On addition of -46 and +28 using 2’s complement, we get
- 00101110
- 0101110
- 00101111
- 1001111
Q16 | On subtracting +28 from +29 using 2’s complement, we get
- 11111010
- 111111001
- 100001
- 1
Q17 | The decimal number 10 is represented in its BCD form as
- 10100000
- 01010111
- 00010000
- 00101011
Q18 | When numbers, letters or words are represented by a special group of symbols, this process is called
- decoding
- encoding
- digitizing
- inverting
Q19 | Carry out BCD subtraction for (68) – (61) using 10’s complement method.
- 00000111
- 01110000
- 100000111
- 011111000
Q20 | How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes?
- 12
- 14
- 16
- 18
Q21 | The decimal equivalent of the excess-3 number 110010100011.01110101 is
- 970.42
- 1253.75
- 861.75
- 1132.87
Q22 | In boolean algebra, the OR operation is performed by which properties?
- associative properties
- commutative properties
- distributive properties
- all of the mentioned
Q23 | The expression for Absorption law is given by
- a + ab = a
- a + ab = b
- ab + aa’ = a
- a + b = b + a
Q24 | According to boolean law: A + 1 = ?
- 1
- a
- a’
Q25 | The involution of A is equal to
- a
- a’
- 1