Question 1 
If (1235)_{x}\:=\:(3033)_{y}, where x and y indicate the bases of the corresponding numbers, then
x\:=\:7 and y\:=\:5
 
x\:=\:8 and y\:=\:6
 
x\:=\:6 and y\:=\:4
 
x\:=\:9 and y\:=\:7

Question 1 Explanation:
x^{3}+2 x^{2}+3 x+5=3 y^{3}+3 y+3
Option (B) will satisfy the equation.
Option (B) will satisfy the equation.
Question 2 
P, Q, and R are the decimal integers corresponding to the 4bit binary number 1100
considered in signed magnitude, 1's complement, and 2's complement representations,
respectively. The 6bit 2's complement representation of (P+Q+R) is
110101  
110010  
111101  
111001 
Question 2 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 3 
The number of bytes required to represent the decimal number 1856357 in packed
BCD (Binary Coded Decimal) form is_______.
2  
3  
4  
5 
Question 3 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 4 
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 4 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 \\ PQ=P+(2 ' s& \text { complement of }Q) \\ &=00010011 \\ &\frac{+11100110}{11111001} \\ \text{2's complement of }\\ (PQ)&=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 \\ PQ=P+(2 ' s& \text { complement of }Q) \\ &=00010011 \\ &\frac{+11100110}{11111001} \\ \text{2's complement of }\\ (PQ)&=00000111 \end{aligned}
Question 5 
X = 01110 and Y =11001 are two 5bit 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 5 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 6 
A new Binary Coded Pentary (BCP) number system is proposed in which every
digit of a base5 number is represented by its corresponding 3bit binary code.
For example, the base5 number 24 will be represented by its BCP code 010100.
In this numbering system, the BCP code 10001001101 corresponds of the
following number is base5 system
423  
1324  
2201  
4321 
Question 6 Explanation:
100010011001 \rightarrow 4231
Question 7 
Decimal 43 in Hexadecimal and BCD number system is respectively
B2, 0100 0011  
2B, 0100 0011  
2B, 0011 0100  
B2, 0100 0100 
Question 7 Explanation:
(43)_{10}\rightarrow \begin{array}{ccc}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 8 
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 8 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 9 
The range of signed decimal numbers that can be represented by 6bits 1's complement number is
31 to +31  
63 to +63  
64 to +63  
32 to +31 
Question 9 Explanation:
\begin{aligned} \text { Range } &=\left(2^{n1}1\right) \text { to }+\left(2^{n1}1\right) \\ &=\left(2^{61}1\right) \text { to }+\left(2^{61}1\right) \\ &=31 \text { to }+31 \end{aligned}
Question 10 
4bit 2's complement representation of a decimal number is 1000. The number is
8  
0  
7  
8 
Question 10 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
There are 10 questions to complete.