Question 1 |

P, Q, and R are the decimal integers corresponding to the 4-bit binary number 1100
considered in signed magnitude, 1's complement, and 2's complement representations,
respectively. The 6-bit 2's complement representation of (P+Q+R) is

110101 | |

110010 | |

111101 | |

111001 |

Question 1 Explanation:

Given, binary number 1100

1's complement of 1100 = -3

Sign magnitude of 1100 = -4

2's complement of 1100 = -4

P + Q + R = -4 - 3 - 4 = -11

The 6 digit 2's complement of (-11) = 110101

1's complement of 1100 = -3

Sign magnitude of 1100 = -4

2's complement of 1100 = -4

P + Q + R = -4 - 3 - 4 = -11

The 6 digit 2's complement of (-11) = 110101

Question 2 |

The number of bytes required to represent the decimal number 1856357 in packed
BCD (Binary Coded Decimal) form is_______.

2 | |

3 | |

4 | |

5 |

Question 2 Explanation:

To represent decimal number into BCD number each
decimal number is represented in 4 -bits while
converting in BCD numbers, as

\begin{array}{ll}1 \rightarrow 0001 & 6 \rightarrow 0110 \\ 8 \rightarrow 1000 & 3 \rightarrow 0011 \\ 5 \rightarrow 0101 & 5 \rightarrow 0101 \\ & 7 \rightarrow 0111\end{array}

So total 4 bytes are required.

\begin{array}{ll}1 \rightarrow 0001 & 6 \rightarrow 0110 \\ 8 \rightarrow 1000 & 3 \rightarrow 0011 \\ 5 \rightarrow 0101 & 5 \rightarrow 0101 \\ & 7 \rightarrow 0111\end{array}

So total 4 bytes are required.

Question 3 |

The two numbers represented in signed 2's complement form are P = 11101101
and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2's
complement is

1000001111 | |

00000111 | |

11111001 | |

111111001 |

Question 3 Explanation:

\because Signed 2 's complement of

\begin{aligned} P&=11101101\\ \therefore \text{No.} \quad P&=00010011 \end{aligned}

\because Signed 2 's complement of

\begin{aligned} Q &=11100110 \\ P-Q=P+(2 ' s& \text { complement of }Q) \\ &=00010011 \\ &\frac{+11100110}{11111001} \\ \text{2's complement of }\\ (P-Q)&=00000111 \end{aligned}

\begin{aligned} P&=11101101\\ \therefore \text{No.} \quad P&=00010011 \end{aligned}

\because Signed 2 's complement of

\begin{aligned} Q &=11100110 \\ P-Q=P+(2 ' s& \text { complement of }Q) \\ &=00010011 \\ &\frac{+11100110}{11111001} \\ \text{2's complement of }\\ (P-Q)&=00000111 \end{aligned}

Question 4 |

X = 01110 and Y =11001 are two 5-bit binary numbers represented in two's
complement format. The sum of X and Y represented in two's complement
format using 6 bits is

100111 | |

001000 | |

000111 | |

101001 |

Question 4 Explanation:

\begin{array}{cccccccc} x &=&&0&1&1&1&0 \\ Y &=&&1&1&0&0&1 \\ x+y &=&1&0&0&1&1&1 \end{array}

Carry is discarded in the addition of numbers represented in 2's complement form. X + Y in 6 bits is 000111.

Carry is discarded in the addition of numbers represented in 2's complement form. X + Y in 6 bits is 000111.

Question 5 |

A new Binary Coded Pentary (BCP) number system is proposed in which every
digit of a base-5 number is represented by its corresponding 3-bit binary code.
For example, the base-5 number 24 will be represented by its BCP code 010100.
In this numbering system, the BCP code 10001001101 corresponds of the
following number is base-5 system

423 | |

1324 | |

2201 | |

4321 |

Question 5 Explanation:

100010011001 \rightarrow 4231

Question 6 |

Decimal 43 in Hexadecimal and BCD number system is respectively

B2, 0100 0011 | |

2B, 0100 0011 | |

2B, 0011 0100 | |

B2, 0100 0100 |

Question 6 Explanation:

(43)_{10}\rightarrow \begin{array}{c|c|c}16&43&\\\hline16&2&B\\\hline&0&2\\\hline\end{array}

\therefore (2B)_{H} (43)_{10}=(01000011)_{BCD}

\therefore (2B)_{H} (43)_{10}=(01000011)_{BCD}

Question 7 |

11001, 1001, 111001 correspond to the 2's complement representation of which one of the following sets of number

25, 9, and 57 respectively | |

-6, -6, and -6 respectively | |

-7, -7 and -7 respectively | |

-25, -9 and -57 respectively |

Question 7 Explanation:

11001 \rightarrow 00111(+7)

1001 \rightarrow 0111(+7)

111001 \rightarrow 000111(+7)

\therefore Numbers given in question in 2 's complement correspond to -7

1001 \rightarrow 0111(+7)

111001 \rightarrow 000111(+7)

\therefore Numbers given in question in 2 's complement correspond to -7

Question 8 |

The range of signed decimal numbers that can be represented by 6-bits 1's complement number is

-31 to +31 | |

-63 to +63 | |

-64 to +63 | |

-32 to +31 |

Question 8 Explanation:

\begin{aligned} \text { Range } &=-\left(2^{n-1}-1\right) \text { to }+\left(2^{n-1}-1\right) \\ &=-\left(2^{6-1}-1\right) \text { to }+\left(2^{6-1}-1\right) \\ &=-31 \text { to }+31 \end{aligned}

Question 9 |

4-bit 2's complement representation of a decimal number is 1000. The number is

8 | |

0 | |

-7 | |

-8 |

Question 9 Explanation:

1000

MSB is 1 so, -ve number

Take 2's complement for magnitude.

+\begin{array}{cccc} 0&1&1&1\\ &&&1\\ \hline1&0&0&0 \end{array}=8

MSB is 1 so, -ve number

Take 2's complement for magnitude.

+\begin{array}{cccc} 0&1&1&1\\ &&&1\\ \hline1&0&0&0 \end{array}=8

Question 10 |

The 2's complement representation of -17 is

101110 | |

101111 | |

111110 | |

110001 |

Question 10 Explanation:

\begin{aligned} 17 &=010001 \\ -17 &=101111(2 \text { 's complement }) \end{aligned}

There are 10 questions to complete.