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’