Number Systems and Codes Questions and Answers

Q: What is the single error-correcting code for the information code 10111 for odd parity ?

The code word is 100111110.Five data bits need four Hamming check bits (\(2^{4} \ge 5 + 4 + 1\)), so the code has 9 bits. Number the positions 9 down to 1 from left to right; check bits sit at positions 1, 2, 4 and 8 and the data 1 0 1 1 1 fills positions 9, 7, 6, 5 and 3.For odd parity, each check bit makes the count of 1s in its group odd:\[ \begin{aligned} P_{1} &: \text{bits } 3, 5, 7, 9 = 1, 1, 0, 1 \Rightarrow P_{1} = 0 \\ P_{2} &: \text{bits } 3, 6, 7 = 1, 1, 0 \Rightarrow P_{2} = 1 \\ P_{4} &: \text{bits } 5, 6, 7 = 1, 1, 0 \Rightarrow P_{4} = 1 \\ P_{8} &: \text{bit } 9 = 1 \Rightarrow P_{8} = 0 \end{aligned} \]Reading positions 9 to 1 gives 1 0 0 1 1 1 1 1 0. Options (C) and (D) have only 8 bits, too few for single-error correction of 5 data bits, and 111000001 does not contain the data bits at the data positions.

Q: Hamming distance of the following code words A and B is ................ .A = 001001001001 andB = 110100100110

The Hamming distance is 10.The Hamming distance between two code words of equal length is the number of positions in which they differ, that is, the number of 1s in their bitwise XOR.A0010 0100 1001B1101 0010 0110A XOR B1111 0110 1111The XOR has 4 + 2 + 4 = 10 ones, so the words differ in 10 of their 12 positions. They agree only in positions 5 and 8. Counting only part of the word, or counting the agreeing positions, produces the smaller distractors.

Q: The decimal equivalent of 8 DH in sign magnitude form is :

The decimal equivalent is -13.8D in hexadecimal is the 8-bit pattern 1000 1101. In sign-magnitude form the most significant bit is the sign and the remaining seven bits are the magnitude:Sign bit 1: the number is negative.Magnitude 000 1101 = 8 + 4 + 1 = 13.So the value is -13. Reading the pattern as unsigned gives 141, and dropping the sign gives 13, the first distractor. Reading it in two's complement would give -115, so the -11 and 11 options correspond to no standard reading of 8D.

Q: If an even parity Hamming code 0110011 is transmitted and that 0100011 is received. The receiver is unaware of what was transmitted ? Determine the bit location where error has occurred using received code.

The error is in the 5th location.In a 7-bit Hamming code the bit positions are numbered 7 down to 1 from left to right: D7 D6 D5 P4 D3 P2 P1, with parity bits at positions 1, 2 and 4. For the received word 0100011, b7 = 0, b6 = 1, b5 = 0, b4 = 0, b3 = 0, b2 = 1, b1 = 1. With even parity, each check over its group should be 0:C1 = b1 &oplus; b3 &oplus; b5 &oplus; b7 = 1 &oplus; 0 &oplus; 0 &oplus; 0 = 1C2 = b2 &oplus; b3 &oplus; b6 &oplus; b7 = 1 &oplus; 0 &oplus; 1 &oplus; 0 = 0C4 = b4 &oplus; b5 &oplus; b6 &oplus; b7 = 0 &oplus; 0 &oplus; 1 &oplus; 0 = 1The syndrome C4 C2 C1 = 101 = 5, so bit 5 is wrong. It is the third bit counted from the left, which is the trap in the '3rd location' option.

Q: The ASCII code for decimal digit 9 with an even parity, placing the parity bit in the most significant position is :

The code is 00111001.The ASCII code for the digit 9 is hexadecimal 39, which in 7 bits is 0111001. It contains four 1s.With even parity the total number of 1s, parity bit included, must be even. Four is already even, so the parity bit is 0, and placing it in the most significant position gives 0 0111001 = 00111001.10111001 has five 1s, which is odd parity. 00100101 and 11000110 are not built on the code for 9 at all (the second is the bitwise complement of 00111001).

Q: The 15's complement of A9BH is :

The 15's complement of A9BH is 564H.The 15's complement in hexadecimal is the analogue of the 1's complement in binary: subtract every digit from F (15).\[ \begin{aligned} F - A &= 15 - 10 = 5 \\ F - 9 &= 15 - 9 = 6 \\ F - B &= 15 - 11 = 4 \end{aligned} \]So the result is 564H. As a check, A9B + 564 = FFF. Adding 1 to the 15's complement would give the 16's complement, 565H, which is not offered; the other options do not add to FFF with A9B.

Q: "100" is a number in Decimal Number System. What will be its representation in Octal Number System ?

Decimal 100 is 144 in octal.Divide repeatedly by 8 and read the remainders from the bottom up:\[ \begin{aligned} 100 \div 8 &= 12 \text{ remainder } 4 \\ 12 \div 8 &= 1 \text{ remainder } 4 \\ 1 \div 8 &= 0 \text{ remainder } 1 \end{aligned} \]Check: \(1 \times 64 + 4 \times 8 + 4 = 100\). The digits 414 and 441 are rearrangements of the right digits, and 100 would be decimal 64.

Q: "100" is a number in Octal Number System. What will be its representation in Hexadecimal Number System ?

Octal 100 is 40 in hexadecimal.Convert through decimal or through binary:\[ \begin{aligned} 100_{8} &= 1 \times 8^{2} = 64_{10} \\ 64_{10} &= 4 \times 16 + 0 = 40_{16} \end{aligned} \]In binary, octal 1 0 0 is 001 000 000, regrouped in fours as 0100 0000, which is hexadecimal 40.64 is the decimal value, not the hexadecimal form. 100 assumes the digits carry over unchanged between bases, and 256 is the decimal value of hexadecimal 100.

Q: The American Standard Code for Information Interchange (ASCII) uses how many individual pulses for any given character ?

ASCII uses 7 pulses (bits) for each character.The American Standard Code for Information Interchange is a 7-bit code, giving 27 = 128 characters: letters, digits, punctuation and control codes. Each character is therefore sent as 7 binary pulses.8 bits are used by extended ASCII and when a parity bit is added, which is the usual trap. 1 or 2 bits could encode only 2 or 4 characters.

What is the single error-correcting code for the information code 10111 for odd parity ?

A.100111110
B.111000001
C.10101011
D.10101100

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The code word is 100111110.

Five data bits need four Hamming check bits (\(2^{4} \ge 5 + 4 + 1\)), so the code has 9 bits. Number the positions 9 down to 1 from left to right; check bits sit at positions 1, 2, 4 and 8 and the data 1 0 1 1 1 fills positions 9, 7, 6, 5 and 3.

For odd parity, each check bit makes the count of 1s in its group odd:

\[ \begin{aligned} P_{1} &: \text{bits } 3, 5, 7, 9 = 1, 1, 0, 1 \Rightarrow P_{1} = 0 \\ P_{2} &: \text{bits } 3, 6, 7 = 1, 1, 0 \Rightarrow P_{2} = 1 \\ P_{4} &: \text{bits } 5, 6, 7 = 1, 1, 0 \Rightarrow P_{4} = 1 \\ P_{8} &: \text{bit } 9 = 1 \Rightarrow P_{8} = 0 \end{aligned} \]

Reading positions 9 to 1 gives 1 0 0 1 1 1 1 1 0. Options (C) and (D) have only 8 bits, too few for single-error correction of 5 data bits, and 111000001 does not contain the data bits at the data positions.

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Hamming distance of the following code words A and B is ................ .

A = 001001001001 and
B = 110100100110

A.4
B.6
C.10
D.12

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The Hamming distance is 10.

The Hamming distance between two code words of equal length is the number of positions in which they differ, that is, the number of 1s in their bitwise XOR.

A	0010 0100 1001
B	1101 0010 0110
A XOR B	1111 0110 1111

The XOR has 4 + 2 + 4 = 10 ones, so the words differ in 10 of their 12 positions. They agree only in positions 5 and 8. Counting only part of the word, or counting the agreeing positions, produces the smaller distractors.

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The decimal equivalent of 8 D_H in sign magnitude form is :

A.13
B.-13
C.-11
D.11

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The decimal equivalent is -13.

8D in hexadecimal is the 8-bit pattern 1000 1101. In sign-magnitude form the most significant bit is the sign and the remaining seven bits are the magnitude:

Sign bit 1: the number is negative.
Magnitude 000 1101 = 8 + 4 + 1 = 13.

So the value is -13. Reading the pattern as unsigned gives 141, and dropping the sign gives 13, the first distractor. Reading it in two's complement would give -115, so the -11 and 11 options correspond to no standard reading of 8D.

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If an even parity Hamming code 0110011 is transmitted and that 0100011 is received. The receiver is unaware of what was transmitted ? Determine the bit location where error has occurred using received code.

A.Bit in 3rd location is wrong
B.Bit in 4th location is wrong
C.Bit in 5th location is wrong
D.Bit in 1st location is wrong

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The error is in the 5th location.

In a 7-bit Hamming code the bit positions are numbered 7 down to 1 from left to right: D7 D6 D5 P4 D3 P2 P1, with parity bits at positions 1, 2 and 4. For the received word 0100011, b7 = 0, b6 = 1, b5 = 0, b4 = 0, b3 = 0, b2 = 1, b1 = 1. With even parity, each check over its group should be 0:

C1 = b1 ⊕ b3 ⊕ b5 ⊕ b7 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
C2 = b2 ⊕ b3 ⊕ b6 ⊕ b7 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
C4 = b4 ⊕ b5 ⊕ b6 ⊕ b7 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1

The syndrome C4 C2 C1 = 101 = 5, so bit 5 is wrong. It is the third bit counted from the left, which is the trap in the '3rd location' option.

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Assertion (A) : 2's complement arithmetic is preferred in digital computers.
Reason (R) : The hardware required to obtain 2's complement of a number is simple.

A.Both (A) and (R) are true and (R) is the correct explanation of (A)
B.Both (A) and (R) are true but (R) is not the correct explanation of (A)
C.(A) is true but (R) is false
D.(A) is false and (R) is true

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

Both (A) and (R) are true and (R) is the correct explanation of (A).

Computers use 2's complement because the same adder then handles both addition and subtraction: A - B is computed as A + (2's complement of B). There is only one representation of zero, and no end-around carry is needed as in 1's complement.

The 2's complement itself is obtained with very simple hardware: invert every bit (an XOR or NOT gate per bit) and add 1 through the carry input of the adder. That simplicity is the reason the scheme is preferred, so the reason explains the assertion.

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The ASCII code for decimal digit 9 with an even parity, placing the parity bit in the most significant position is :

A.00111001
B.10111001
C.00100101
D.11000110

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

The code is 00111001.

The ASCII code for the digit 9 is hexadecimal 39, which in 7 bits is 0111001. It contains four 1s.

With even parity the total number of 1s, parity bit included, must be even. Four is already even, so the parity bit is 0, and placing it in the most significant position gives 0 0111001 = 00111001.

10111001 has five 1s, which is odd parity. 00100101 and 11000110 are not built on the code for 9 at all (the second is the bitwise complement of 00111001).

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The 15's complement of A9B_H is :

A.\(\mathrm{B02_{H}}\)
B.\(\mathrm{173_{H}}\)
C.\(\mathrm{CD2_{H}}\)
D.\(\mathrm{564_{H}}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The 15's complement of A9B_H is 564_H.

The 15's complement in hexadecimal is the analogue of the 1's complement in binary: subtract every digit from F (15).

\[ \begin{aligned} F - A &= 15 - 10 = 5 \\ F - 9 &= 15 - 9 = 6 \\ F - B &= 15 - 11 = 4 \end{aligned} \]

So the result is 564_H. As a check, A9B + 564 = FFF. Adding 1 to the 15's complement would give the 16's complement, 565_H, which is not offered; the other options do not add to FFF with A9B.

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"100" is a number in Decimal Number System. What will be its representation in Octal Number System ?

A.100
B.144
C.414
D.441

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Decimal 100 is 144 in octal.

Divide repeatedly by 8 and read the remainders from the bottom up:

\[ \begin{aligned} 100 \div 8 &= 12 \text{ remainder } 4 \\ 12 \div 8 &= 1 \text{ remainder } 4 \\ 1 \div 8 &= 0 \text{ remainder } 1 \end{aligned} \]

Check: \(1 \times 64 + 4 \times 8 + 4 = 100\). The digits 414 and 441 are rearrangements of the right digits, and 100 would be decimal 64.

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"100" is a number in Octal Number System. What will be its representation in Hexadecimal Number System ?

A.40
B.64
C.100
D.256

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

Octal 100 is 40 in hexadecimal.

Convert through decimal or through binary:

\[ \begin{aligned} 100_{8} &= 1 \times 8^{2} = 64_{10} \\ 64_{10} &= 4 \times 16 + 0 = 40_{16} \end{aligned} \]

In binary, octal 1 0 0 is 001 000 000, regrouped in fours as 0100 0000, which is hexadecimal 40.

64 is the decimal value, not the hexadecimal form. 100 assumes the digits carry over unchanged between bases, and 256 is the decimal value of hexadecimal 100.

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10.

The American Standard Code for Information Interchange (ASCII) uses how many individual pulses for any given character ?

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: C

ASCII uses 7 pulses (bits) for each character.

The American Standard Code for Information Interchange is a 7-bit code, giving 2⁷ = 128 characters: letters, digits, punctuation and control codes. Each character is therefore sent as 7 binary pulses.

8 bits are used by extended ASCII and when a parity bit is added, which is the usual trap. 1 or 2 bits could encode only 2 or 4 characters.

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