# Number Systems

 Question 1
If $(1235)_{x}\:=\:(3033)_{y}$, where x and y indicate the bases of the corresponding numbers, then
 A $x\:=\:7$ and $y\:=\:5$ B $x\:=\:8$ and $y\:=\:6$ C $x\:=\:6$ and $y\:=\:4$ D $x\:=\:9$ and $y\:=\:7$
GATE EC 2021   Digital Circuits
Question 1 Explanation:
$x^{3}+2 x^{2}+3 x+5=3 y^{3}+3 y+3$
Option (B) will satisfy the equation.
 Question 2
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
 A 110101 B 110010 C 111101 D 111001
GATE EC 2020   Digital Circuits
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
 Question 3
The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is_______.
 A 2 B 3 C 4 D 5
GATE EC 2014-SET-2   Digital Circuits
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.
 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
 A 1000001111 B $00000111$ C 11111001 D 111111001
GATE EC 2008   Digital Circuits
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 \\ P-Q=P+(2 ' s& \text { complement of }Q) \\ &=00010011 \\ &\frac{+11100110}{11111001} \\ \text{2's complement of }\\ (P-Q)&=00000111 \end{aligned}
 Question 5
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
 A 100111 B 001000 C 000111 D 101001
GATE EC 2007   Digital Circuits
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.
 Question 6
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
 A 423 B 1324 C 2201 D 4321
GATE EC 2006   Digital Circuits
Question 6 Explanation:
$100010011001 \rightarrow 4231$
 Question 7
Decimal 43 in Hexadecimal and BCD number system is respectively
 A B2, 0100 0011 B 2B, 0100 0011 C 2B, 0011 0100 D B2, 0100 0100
GATE EC 2005   Digital Circuits
Question 7 Explanation:
$(43)_{10}\rightarrow \begin{array}{c|c|c}16&43&\\\hline16&2&B\\\hline&0&2\\\hline\end{array}$
$\therefore (2B)_{H} (43)_{10}=(01000011)_{BCD}$
 Question 8
11001, 1001, 111001 correspond to the 2's complement representation of which one of the following sets of number
 A 25, 9, and 57 respectively B -6, -6, and -6 respectively C -7, -7 and -7 respectively D -25, -9 and -57 respectively
GATE EC 2004   Digital Circuits
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
 Question 9
The range of signed decimal numbers that can be represented by 6-bits 1's complement number is
 A -31 to +31 B -63 to +63 C -64 to +63 D -32 to +31
GATE EC 2004   Digital Circuits
Question 9 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 10
4-bit 2's complement representation of a decimal number is 1000. The number is
 A 8 B 0 C -7 D -8
GATE EC 2002   Digital Circuits
Question 10 Explanation:
1000
MSB is 1 so, -ve number
Take 2's complement for magnitude.
$+\begin{array}{cccc} 0&1&1&1\\ &&&1\\ \hline1&0&0&0 \end{array}=8$
There are 10 questions to complete.