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.
There are 5 questions to complete.