Digital Principles And System Design Set 14

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This set of Digital Principles and System Design Multiple Choice Questions & Answers (MCQs) focuses on Digital Principles And System Design Set 14

Q1 | The parameter through which 16 distinct values can be represented is known as                  
  • bit
  • byte
  • word
  • nibble
Q2 | If the decimal number is a fraction then its binary equivalent is obtained by                  the number continuously by 2.
  • dividing
  • multiplying
  • adding
  • subtracting
Q3 | The representation of octal number (532.2)8 in decimal is                  
  • (346.25)10
  • (532.864)10
  • (340.67)10
  • (531.668)10
Q4 | The decimal equivalent of the binary number (1011.011)2 is                  
  • (11.375)10
  • (10.123)10
  • (11.175)10
  • (9.23)10
Q5 | An important drawback of binary system is
  • it requires very large string of 1’s and 0’s to represent a decimal number
  • it requires sparingly small string of 1’s and 0’s to represent a decimal number
  • it requires large string of 1’s and small string of 0’s to represent a decimal number
  • it requires small string of 1’s and large string of 0’s to represent a decimal number
Q6 | The decimal equivalent of the octal number (645)8 is              
  • (450)10
  • (451)10
  • (421)10
  • (501)10
Q7 | The largest two digit hexadecimal number is
  • (fe)16
  • (fd)16
  • (ff)16
  • (ef)16
Q8 | What is the addition of the binary numbers 11011011010 and 010100101?
  • 0111001000
  • 1100110110
  • 11101111111
  • 10011010011
Q9 | Representation of hexadecimal number (6DE)H in decimal:
  • 6 * 162 + 13 * 161 + 14 * 160
  • 6 * 162 + 12 * 161 + 13 * 160
  • 6 * 162 + 11 * 161 + 14 * 160
  • 6 * 162 + 14 * 161 + 15 * 160
Q10 | 100101 × 0110 = ?
  • 1011001111
  • 0100110011
  • 101111110
  • 0110100101
Q11 | Perform multiplication of the binary numbers: 01001 × 01011 = ?
  • 001100011
  • 110011100
  • 010100110
  • 101010111
Q12 | Divide the binary numbers: 111101 ÷ 1001 and find the remainder
  • 0010
  • 1010
  • 1100
  • 0011
Q13 | Binary coded decimal is a combination of
  • two binary digits
  • three binary digits
  • four binary digits
  • five binary digits
Q14 | The decimal number 10 is represented in its BCD form as                      
  • 10100000
  • 01010111
  • 00010000
  • 00101011
Q15 | Carry out BCD subtraction for (68) – (61) using 10’s complement method.
  • 00000111
  • 01110000
  • 100000111
  • 011111000
Q16 | When numbers, letters or words are represented by a special group of symbols, this process is called
  • decoding
  • encoding
  • digitizing
  • inverting
Q17 | A three digit decimal number requires                  for representation in the conventional BCD format.
  • 3 bits
  • 6 bits
  • 12 bits
  • 24 bits
Q18 | How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes?
  • 12
  • 14
  • 16
  • 18
Q19 | The excess-3 code for 597 is given by                      
  • 100011001010
  • 100010100111
  • 010110010111
  • 010110101101
Q20 | The decimal equivalent of the excess-3 number 110010100011.01110101 is                            
  • 970.42
  • 1253.75
  • 861.75
  • 1132.87
Q21 | In boolean algebra, the OR operation is performed by which properties?
  • associative properties
  • commutative properties
  • distributive properties
  • all of the mentioned
Q22 | The expression for Absorption law is given by
  • a + ab = a
  • a + ab = b
  • ab + aa’ = a
  • a + b = b + a
Q23 | DeMorgan’s theorem states that                    
  • (ab)’ = a’ + b’
  • (a + b)’ = a’ * b
  • a’ + b’ = a’b’
  • (ab)’ = a’ + b
Q24 | (A + B)(A’ * B’) = ?
  • 1
  • 0
  • ab
  • ab’
Q25 | According to boolean law: A + 1 = ?                    
  • 1
  • a
  • a’