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.