NumberSystems
Question 1 
If a variable can take only integral values from 0 to n, where n is an integer, then the variable can be represented as a bit field whose width is (the log in the Solutions are to the base 2, and [log n] means the floor of log n)
[log(n)] + 1 bits  
[log (n1)) + 1 bits  
[log (n+1)] + 1 bits  
None of the above 
Question 1 Explanation:
Question 2 
Given √(224)_{r} = 13_{r} the value of radix r is
10  
8  
6  
5 
Question 2 Explanation:
√(224)_{r} = 13_{r}
For f(x) to be maximum
f'(x) = 4x  2 = 0
⇒ x = 1/2
So at x = 1/2, f(x) is an extremum (either maximum or minimum).
f(2) = 2(2)^{2}  2(2) + 6 = 10
f(1/2) = 2 × (1/2)^{2}  2 × 1/2 + 6 = 5.5
f(0) = 6
So, the maximum value is at x=2 which is 10 as there are no other extremum for the given function.
For f(x) to be maximum
f'(x) = 4x  2 = 0
⇒ x = 1/2
So at x = 1/2, f(x) is an extremum (either maximum or minimum).
f(2) = 2(2)^{2}  2(2) + 6 = 10
f(1/2) = 2 × (1/2)^{2}  2 × 1/2 + 6 = 5.5
f(0) = 6
So, the maximum value is at x=2 which is 10 as there are no other extremum for the given function.
Question 3 
When two nbit binary numbers are added the sum will contain at the most
n bits  
(n+3) bits  
(n+2) bits  
(n+1) bits 
Question 3 Explanation:
→ When two nbit binary numbers are added the sum will contain at the most (n+1) bits
Example = 2 Decimal numbers are (7)_{10} and (7)_{10}
= Equivalent binary numbers are (111)_{2} + (111)_{2}
= Adding two binary numbers, the final result will be n+1 number (1110)_{2}
Example = 2 Decimal numbers are (7)_{10} and (7)_{10}
= Equivalent binary numbers are (111)_{2} + (111)_{2}
= Adding two binary numbers, the final result will be n+1 number (1110)_{2}
Question 4 
(1217)_{8} is equivalent to
(1217)_{16}  
(028F)_{16}  
(2297)_{1o}  
(0B17)_{16} 
Question 4 Explanation:
(1217)_{8}=(001 010 001 111)_{2}
=(0010 1000 1111)_{2}
=(2 8 F)_{16}
=(0010 1000 1111)_{2}
=(2 8 F)_{16}
Question 5 
The logic circuit given below converts a binary code y1,y2,y3 into
Excess3 code  
Gray code  
BCD code  
Hamming Code 
Question 5 Explanation:
X1= Y1
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
Question 6 
If 12A7C_{16} = X_{8}, then the value of X is
224174  
425174  
6173  
225174 
Question 6 Explanation:
Given, (12A7C)_{16} = (0001 0010 1010 0111 1100)_{2}
MAke blocks of 3 bits each from LSB to MSB.
(Note: In the last block append zeros (as MSBs) if number bits is not three)
(000 010 010 101 001 111 100)
Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.
= (0 2 2 5 1 7 4)_{8}
MAke blocks of 3 bits each from LSB to MSB.
(Note: In the last block append zeros (as MSBs) if number bits is not three)
(000 010 010 101 001 111 100)
Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.
= (0 2 2 5 1 7 4)_{8}
Question 7 
The Excess3 code is also called
Cyclic Redundancy Code  
Weighted Code  
SelfComplementing Code  
Algebraic Code 
Question 7 Explanation:
Excess3 code is also called SelfComplementing Code. Because 1’s complement of an excess3 number is equivalent to 9’s complement of the corresponding decimal digit.
→ In excess3 code, each of the 4bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.
Excess3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.
→ In excess3 code, each of the 4bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.
Excess3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.
Question 8 
Which of the following binary number is the same as its 2’s complement?
1010  
0101  
1000  
1001 
Question 8 Explanation:
Hint: Number of bits=4
(Decimal value of maximum 4bit number +1 )/2= (15+1)/2=8
(Decimal value of maximum 4bit number +1 )/2= (15+1)/2=8
Question 9 
The addition of 4bit, two’s complement, binary numbers 1101 and 0100 results in
0001 and an overflow
 
1001 and no overflow
 
0001 and no overflow  
1001 and an overflow

Question 9 Explanation:
Number one is 0100 (4Decimal value)
Another number is 1101(3 is decimal value)
Adding of 3 and 4, the result is 1 and there is no overflow
Another number is 1101(3 is decimal value)
Adding of 3 and 4, the result is 1 and there is no overflow
Question 10 
What is the decimal value of the floatingpoint number C1D00000 (hexadecimal notation)? (Assume 32bit, single precision floating point IEEE representation)
28  
15  
26  
28 
Question 10 Explanation:
Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32bit, single precision floating point IEEE representation, first MSB represents sign of mantissa: 1 is used to represent a negative mantissa and 0 for a positive value of mantissa, next 8 bits are for exponent value and then 23 bits represents mantissa.
Value of exponent = 131127 = 4
Mantissa = 1.1010000 0000 0000 0000 0000
Floating point number = 1.1010000 0000 0000 0000 0000
Converting the above one into decimal no (1*2^{0}+1*2^{1}*0*2^{2}+1*2^{2}+0* 2^{3} +.....)
= (1+½+⅛)=13/8
Decimal value =sign*Exponent*mantissa=1*4*13/8
= 26
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32bit, single precision floating point IEEE representation, first MSB represents sign of mantissa: 1 is used to represent a negative mantissa and 0 for a positive value of mantissa, next 8 bits are for exponent value and then 23 bits represents mantissa.
Value of exponent = 131127 = 4
Mantissa = 1.1010000 0000 0000 0000 0000
Floating point number = 1.1010000 0000 0000 0000 0000
Converting the above one into decimal no (1*2^{0}+1*2^{1}*0*2^{2}+1*2^{2}+0* 2^{3} +.....)
= (1+½+⅛)=13/8
Decimal value =sign*Exponent*mantissa=1*4*13/8
= 26
Question 11 
In the standard IEEE 754 single precision floating point representation, there is 1 bit for sign, 23 bits for fraction and 8 bits for exponent. What is the precision in terms of the number of decimal digits?
5  
6  
7  
8 
Question 11 Explanation:
A floatingpoint variable can represent a wider range of numbers than a fixedpoint variable of the same bit width at the cost of precision. A signed 32bit integer variable has a maximum value of 2^{31} − 1 = 2,147,483,647, whereas an IEEE 754 32bit base2 floatingpoint variable has a maximum value of (2 − 2^{−23}) × 2^{127} ≈ 3.402823 × 10^{38}.
In the IEEE 7542008 standard, the 32bit base2 format is officially referred to as binary32; it was called single in IEEE 7541985. IEEE 754 specifies additional floatingpoint types, such as 64bit base2 double precision and, more recently, base10 representations.
We can convert the binary into decimal representation by using the following steps
let the number of digits in decimal digits be ‘x’
2^{23} = 10^{x }
After taking log on both sides
log_{2}10^{x} =log2 2^{23}
x log_{2}10 = 23log_{2}2 (The value of log_{2}2=1)
3.322 x = 23 (The value of log_{2}10 = 3.321928)
x = 6.92
In the IEEE 7542008 standard, the 32bit base2 format is officially referred to as binary32; it was called single in IEEE 7541985. IEEE 754 specifies additional floatingpoint types, such as 64bit base2 double precision and, more recently, base10 representations.
We can convert the binary into decimal representation by using the following steps
let the number of digits in decimal digits be ‘x’
2^{23} = 10^{x }
After taking log on both sides
log_{2}10^{x} =log2 2^{23}
x log_{2}10 = 23log_{2}2 (The value of log_{2}2=1)
3.322 x = 23 (The value of log_{2}10 = 3.321928)
x = 6.92
Question 12 
How many different BCD numbers can be stored in 12 switches? (Assume two position or onoff switches)
2^{12}  
2^{12}1  
10^{12}  
10^{3} 
Question 12 Explanation:
Step1: A binarycoded decimal (BCD) is a class of binary encodings of decimal numbers where each decimal digit is represented by a fixed number of bits, usually four (or) eight.
Step2: Decimal number 0 can be represented 0000 and 9 can be represented by using 1001.
Step3: A switch can store maximum 1 bit data that may be either 0 (or) 1. In switch terminology, 0 means “off” and 1 means “on”. With 4 bit we can represent 10 BCD numbers.
Step4: A BCD digit can be from 0 to 9 (total 10 possibility).
Step5: Different possible BCD numbers in 12 switches are = 10*10*10
= 1000
= 10^{3}
Step2: Decimal number 0 can be represented 0000 and 9 can be represented by using 1001.
Step3: A switch can store maximum 1 bit data that may be either 0 (or) 1. In switch terminology, 0 means “off” and 1 means “on”. With 4 bit we can represent 10 BCD numbers.
Step4: A BCD digit can be from 0 to 9 (total 10 possibility).
Step5: Different possible BCD numbers in 12 switches are = 10*10*10
= 1000
= 10^{3}
Question 13 
The range of the numbers which can be stored in an eight bit register is
128 to +127  
128 to +128  
999999 + +999999  
none of these 
Question 13 Explanation:
There are 2 ^{8} (256) different possible values for 8 bits. When unsigned, it has possible values ranging from 0 to 255; when signed, it has 128 to 127.
Question 14 
The excess 3 code is also called
Cyclomatic redundancy code  
Weighted code  
Self complementing code  
algebraic code 
Question 14 Explanation:
Excess3 is a selfcomplementing code. This is because in Excess3 code we get the 9's complement of a number by just complementing each bit that means by replacing a '0' by '1' and '1' by '0'
Question 15 
In computers, subtraction is generally carried out by
1’s complement
 
10’s complement
 
2’s complement
 
9’s complement

Question 15 Explanation:
• In computers, subtraction is generally carried out by 2’s complement.
• In two'scomplement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value.
• In the subtraction there may possibility of negative number as a result.
• In two'scomplement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value.
• In the subtraction there may possibility of negative number as a result.
Question 16 
A decimal number has 30 digits. Approximately, how many digits would the binary representation have?
30  
60  
90  
120 
Question 16 Explanation:
Here, 30 digits numbers means 123....30.
10^{30} 1=10000000000000000000000000000001=999999999999999999999999999999
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Question 17 
The result of the subtraction FD_{16}  88_{16} is
75 _{16}  
65 _{16}  
5E _{16}  
10 _{16} 
Question 17 Explanation:
Step1: Convert Hexadecimal numbers into decimal numbers.
(FD) _{16</sub> = (253) 10 (88) 16</sub> = (136) 10 Step2: Perform subtraction 253136=(117) 10 Step3: Convert (117) 10 =(75) 16}
(FD) _{16</sub> = (253) 10 (88) 16</sub> = (136) 10 Step2: Perform subtraction 253136=(117) 10 Step3: Convert (117) 10 =(75) 16}
Question 18 
What will be the Excess3 code for 1001?
1001  
1010  
1011  
1100 
Question 18 Explanation:
Excess3 number starts with 3. Here, 1001 menas decimal number9. So, we have to add +3.
12 equivalent binary number is 1100.
12 equivalent binary number is 1100.
Question 19 
The Decimal equivalent of the Hexadecimal number (A09D)_{16} is
31845  
41117  
41052  
32546 
Question 19 Explanation:
Given Hexadecimal number is (A09D)_{16}
A decimal number is the sum of the digits multiplied with its power of 10.
(A09D)_{16} is equal to each digit multiplied with its corresponding power of 16:
Ax16 ^{3} +0x16 ^{2} +9x16 ^{1} +Dx16 ^{0} =(10x4096+144+13x1) [ Where A=10,D=13]
=40960+144+13=41117
A decimal number is the sum of the digits multiplied with its power of 10.
(A09D)_{16} is equal to each digit multiplied with its corresponding power of 16:
Ax16 ^{3} +0x16 ^{2} +9x16 ^{1} +Dx16 ^{0} =(10x4096+144+13x1) [ Where A=10,D=13]
=40960+144+13=41117
Question 20 
Which of the given number has its IEEE754 32 bit floating point representation as
2.5  
3.0  
3.5  
4.5 
Question 20 Explanation:
→ Sign bit S = 0 (It means positive number)
→ E=1000 0000B = 128D (in normalized form)
→ Fraction is 1.11B (with an implicit leading 1) = 1 + 1×2^{1} + 1×2^{2}
= 1.75D
→ The number is +1.75 × 2^{(128127) }
= +3.5D
→ E=1000 0000B = 128D (in normalized form)
→ Fraction is 1.11B (with an implicit leading 1) = 1 + 1×2^{1} + 1×2^{2}
= 1.75D
→ The number is +1.75 × 2^{(128127) }
= +3.5D
Question 21 
The range of integers that can be represented by an n bit 2’s complement number system is:
– 2^{n – 1} to 2^{n – 1} – 1  
– (2^{n – 1} – 1) to (2^{n – 1} – 1)  
– 2^{n – 1} to 2^{n – 1}  
– (2^{n – 1} + 1) to (2^{n – 1} – 1) 
Question 21 Explanation:
In 2’s complement numbers, the range of integers are from 2^{n1} to 2^{n1} – 1
Question 22 
The code which uses 7 bits to represent a character is
ASCII  
BCD  
EBCDIC  
Gray 
Question 22 Explanation:
→ ISO/IEC 646, like ASCII, is a 7bit character set. It does not make any additional codes available, so the same code points encoded different characters in different countries. Escape codes were defined to indicate which national variant applied to a piece of text, but they were rarely used, so it was often impossible to know what variant to work with and, therefore, which character a code represented, and in general, textprocessing systems could cope with only one variant anyway.
→ Extended Binary Coded Decimal Interchange Code (EBCDIC) is an 8bit binary code for numeric and alphanumeric characters.
→ BCD encoding uses 4 bits to represent each digit from the range 0 to 9 in its binary form.
→ In case of Gray codes, any number of bits can be used to represent a character, according to the requirement.
→ Extended Binary Coded Decimal Interchange Code (EBCDIC) is an 8bit binary code for numeric and alphanumeric characters.
→ BCD encoding uses 4 bits to represent each digit from the range 0 to 9 in its binary form.
→ In case of Gray codes, any number of bits can be used to represent a character, according to the requirement.
Question 23 
The number of 1’s in the binary representation of (3*4096 + 15*256 + 5*16 + 3) are:
8  
9  
10  
12 
Question 23 Explanation:
Binary expression of (3*4096 + 15*256 + 5*16 + 3)
=(12,288+3840+80+3)
=(16211)_{10}
=(0011111101010011)_{2}
Total number of 1’s in binary representation is 10.
=(12,288+3840+80+3)
=(16211)_{10}
=(0011111101010011)_{2}
Total number of 1’s in binary representation is 10.
Question 24 
The decimal number has 64 digits.The number of bits needed for its equivalent binary representation is?
200  
213  
246  
277 
Question 24 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{3}1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 >10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =3.322]
n=212.608
n=213
Then, Decimal number has 64 digits, so maximum number is 10^{64}1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n}1
So we can write 10^{64} –1 = 2^{n} – 1 >10^{64} = 2^{n}
After taking log_{2} on both sides
log_{2}2^{n}=log_{2}10^{64}
n log_{2}2=64 log_{ 2}10
n=64*(3.322) [ log_{2}2=1 & log_{2}10 =3.322]
n=212.608
n=213
Question 25 
Suppose x and y are floating point variables that have been assigned the values x = 8.8 and y = 3.5. What will be the value of the following arithmetic expression?
2 * x / 3 * y
20.33335
 
24.45453
 
16.35353
 
20.53333 
Question 25 Explanation:
x = 8.8 y=3.5
= 1 (equal priority
Associativity (left to Right)
(((2 * x)/3) * y)
((2 x *) / 3) * y)
(2x * 31) * y
2x * 31) y*
(2x * 31 y*
Put the 2 * 8.8 * 3 / 3.5 *
= 20.53333
= 1 (equal priority
Associativity (left to Right)
(((2 * x)/3) * y)
((2 x *) / 3) * y)
(2x * 31) * y
2x * 31) y*
(2x * 31 y*
Put the 2 * 8.8 * 3 / 3.5 *
= 20.53333
Question 26 
A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
50  
74  
40  
60  
None of these 
Question 26 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 >10 ^{25} = 2^{ n} After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2^{n} 1
So we can write 10^{ 25} –1 = 2^{ n} – 1 >10 ^{25} = 2^{ n} After taking log ^{2} on both sides
log _{2} 2^{ n} =log_{ 2} 10^{ 25}
n log _{2} 2=25 log_{ 2} 10
n=25*(3.322) [ log _{2} 2=1 & log_{ 2} 10 =3.322]
n=83
Question 27 
What is the base(radix) of the number system whose numbers 312, 20 and 13.1 satisfy the following equation?
312/20 = 13.18  
4  
5  
6 
Question 27 Explanation:
Let base of the number system is r.
(3r^{2} + r + 2) / 2r = (r + 3 + 1/r)
(3r^{2} + r + 2) / 2r = (r^{2} + 3r + 1) / r
(3r^{2} + r + 2) = (2r^{2} + 6r + 2)
r^{2}  5r = 0
Therefore, r = 5
(3r^{2} + r + 2) / 2r = (r + 3 + 1/r)
(3r^{2} + r + 2) / 2r = (r^{2} + 3r + 1) / r
(3r^{2} + r + 2) = (2r^{2} + 6r + 2)
r^{2}  5r = 0
Therefore, r = 5
Question 28 
What is the hexadecimal representation of the decimal number 8537?
(2059)_{16}
 
(2159)_{16}
 
(2195)_{16}
 
(2157)_{16}

Question 28 Explanation:
Step1: First convert decimal number into binary number
Step2: (8537)_{10} = (0010000101011001)_{2}
Step3: Divide binary number into 4 segments (0010 0001 0101 1001)_{2}
Step4: Write equivalent number of hexadecimal (2159)_{16}
Step2: (8537)_{10} = (0010000101011001)_{2}
Step3: Divide binary number into 4 segments (0010 0001 0101 1001)_{2}
Step4: Write equivalent number of hexadecimal (2159)_{16}
Question 29 
Shifting a register content to left by one bit is equivalent to____
Division by 2  
Addition by 2
 
Multiplication by 2  
Subtraction by 2 
Question 29 Explanation:
→ The leftshift operator (<<), which moves the bits of shiftexpression to the left.
→ The bit positions that have been vacated by the shift operation are zerofilled.
→ For example a=5 and equivalent binary value is 101 and shifting one bit left side means the result binary value is 1010 whose decimal value is 10
→ The bit positions that have been vacated by the shift operation are zerofilled.
→ For example a=5 and equivalent binary value is 101 and shifting one bit left side means the result binary value is 1010 whose decimal value is 10
Question 30 
What shall be the 2’s complement represented of 24 in a 16 bit computer?
1111 1111 1110 1011  
1111 1111 1110 1001  
1111 1111 1110 0111  
1111 1111 1110 1000 
Question 30 Explanation:
Negative numbers are represented in 2’s complement form.
The binary equivalent of 24 is 0000 0000 0001 1000
One’s complement is 1111 1111 1110 0111 (Flipping the bits 1 by 0 and 0 by 1)
Two’s complement is 1111 1111 1110 1000 (adding 1 to the LSB bit)
The binary equivalent of 24 is 0000 0000 0001 1000
One’s complement is 1111 1111 1110 0111 (Flipping the bits 1 by 0 and 0 by 1)
Two’s complement is 1111 1111 1110 1000 (adding 1 to the LSB bit)
Question 31 
A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
50  
74  
40  
None of the above 
Question 31 Explanation:
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10^{ 3} 1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Then, Decimal number has 25 digits, so maximum number is 10^{ 25} 1
Similarly, in the binary representation with “n” bits the maximum number is 2 ^{25} 1
So we can write 10 ^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}
After taking log _{2} on both sides
log _{2} 2 ^{n} =log _{ 2} 10 ^{25}
n log _{2} 2=25 log _{2} 10
n = 25 log _{2} 10
n = 25 x 3.3 [ log _{2} 2=1 & log _{2} 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Question 32 
Which of the following is minimum error code?
Octal code  
Binary Code  
Gray code  
Excess3 Code 
Question 32 Explanation:
→ "Gray code" as an alternative name is "reflected binary code". one of those also lists "minimum error code" and "cyclic permutation code" among the names.
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Question 33 
In singleprecision, doubleprecision and extendedprecision representation of floating point numbers, as defined by ANSI/IEEE standard 7541985, the no.of bits used are____ respectively.
32,64 and 80  
32,64 and 128  
16,32 and 64  
16,32 and 80 
Question 33 Explanation:
Extended precision, the third format, is usually an 80bit word, with 1 bit sign, 15 bit exponent and 64 bit significand, with leading bit of a normalized number not hidden
Question 34 
If the original size of data is 40 then after adding error detection redundancy bit the size of data length is
26  
36  
46  
56 
Question 34 Explanation:
→ Imagine that we want to design a code with m message bits and r check bits that will allow all single errors to be corrected.
→ Each of the 2 m legal messages has n illegal codewords at a distance of 11 from it.
→ These are formed by systematically inverting each of the n bits in the nbit codeword formed from it. Thus, each of the 2 ^{m} legal messages requires n+1 bit patterns dedicated to it.
→ Since the total number of bit patterns is 2 ^{n} , We must have (n+1)2 ^{m} ≤ 2^{ n} .
→ Using n=m+r, this requirement becomes
= (m+r+1) ≤ 2 ^{r}
= 40+6+1 ≤ 2^{6}
= 47 ≤ 2
r=6
Message size will be 6+40=46
→ Each of the 2 m legal messages has n illegal codewords at a distance of 11 from it.
→ These are formed by systematically inverting each of the n bits in the nbit codeword formed from it. Thus, each of the 2 ^{m} legal messages requires n+1 bit patterns dedicated to it.
→ Since the total number of bit patterns is 2 ^{n} , We must have (n+1)2 ^{m} ≤ 2^{ n} .
→ Using n=m+r, this requirement becomes
= (m+r+1) ≤ 2 ^{r}
= 40+6+1 ≤ 2^{6}
= 47 ≤ 2
r=6
Message size will be 6+40=46
Question 35 
Which of the following is a recursive algorithm to convert a positive decimal integers into equivalent binary integers?
Question 35 Explanation:
Algorithm:
Take decimal number is 12.
Step1 → 12%2 which is equal to 0+10*(⌊12/2⌋)%2
Step2 → 6%2 which is equal to 0+10*(⌊6/2⌋)%2
Step3 → 3%2 which is equal to 1+10*(⌊3/2⌋)%2
Step4 → 1%2 which is equal to 1+10*(⌊1/2⌋)%2
Take decimal number is 12.
Step1 → 12%2 which is equal to 0+10*(⌊12/2⌋)%2
Step2 → 6%2 which is equal to 0+10*(⌊6/2⌋)%2
Step3 → 3%2 which is equal to 1+10*(⌊3/2⌋)%2
Step4 → 1%2 which is equal to 1+10*(⌊1/2⌋)%2
Question 36 
Consider an arbitrary number system with the independent digits as 0,1 and X. What is the radix of this number system?
1  
2  
3  
4 
Question 36 Explanation:
→ The radix of a number system is the number of unique digits, including, zero, that are used to represent larger numbers. In the decimal system that would be 0 to 9.
→ In the question, the unique digits are 0,1 and X(possible x value is 2) then number system is 3;
→ In the question, the unique digits are 0,1 and X(possible x value is 2) then number system is 3;
Question 37 
The hamming(7,4) code for 0000 using even parity is
0000000  
1111111  
2222222  
12121212 
Question 37 Explanation:
● Hamming(7,4) is a linear errorcorrecting code that encodes four bits of data into seven bits by adding three parity bits.
● The data is 0000 and Hamming(7,4) transmitted is 0000000
● The data is 0000 and Hamming(7,4) transmitted is 0000000
Question 38 
The number of bits required to represent decimal number 4096 in binary form is___
16  
10  
12  
13 
Question 38 Explanation:
(4098)_{ 10} =(1000000000010)_{ 2}
So, total 13 bits required to represent 4096 decimal number.
So, total 13 bits required to represent 4096 decimal number.
Question 39 
The signed 2’s complement representation of 33 is:
11011111
 
00100001  
01011111
 
10100001

Question 39 Explanation:
Step1: 33 corresponding positive number 00100001
Step2: Perform 2’s complement.
Step2: Perform 2’s complement.
Question 40 
Using signed 2’s complement subtraction the result of 1111101011110011 is:
10000111  
00000111  
10001101  
00001101 
Question 40 Explanation:
11111010 → 250(Decimal)
11110011 → 243(Decimal)
Step1: convert signed number into 2’s complement
11110011
00001100 → (1’s complement)
+1

00001101 → (2’s complement)
Step 2: Add 11111010 into 2’s complement number
11111010
00001101

00000111 → solution
11110011 → 243(Decimal)
Step1: convert signed number into 2’s complement
11110011
00001100 → (1’s complement)
+1

00001101 → (2’s complement)
Step 2: Add 11111010 into 2’s complement number
11111010
00001101

00000111 → solution
Question 41 
How many bits are used in the exponent part of IEEE single precision for the representation of floating point numbers?
32 bits  
8 bits  
16 bits  
24 bits 
Question 41 Explanation:
The IEEE 754 standard specifies a binary floating point format(binary32) as having:
→ Sign bit: 1 bit
→ Exponent width: 8 bits
→ Significand precision: 24 bits (23 explicitly stored)
→ Sign bit: 1 bit
→ Exponent width: 8 bits
→ Significand precision: 24 bits (23 explicitly stored)
Question 42 
Convert the following octal number into its decimal equivalent:
2 3 7 4 in octal
(10208)_{10}  
(1276)_{10}  
(2374)_{10}  
(1272)_{10} 
Question 42 Explanation:
(2374)_{8}=2x8^{3}+3x8^{2}+7x8^{1}+4x8^{0}=1024+192+56+4=1276
Question 43 
Represent the decimal number 3.248*10^{4} into a single precision floating point binary number(using standard format).
01000110111111011100000000000000  
01100110111111011100000000000000  
11100110111111011100000000000000  
01000111011111011100000000000000 
Question 43 Explanation:
Given number is in base 10. Convert it to base10.
3.248x10^{4} =32480= 1111110111
= 1.111110111 x 2^{14}
Mantissa = 11111011100...00
Biased exponent = 14 +127= 141 = 10001101
3.248x10^{4} =32480= 1111110111
= 1.111110111 x 2^{14}
Mantissa = 11111011100...00
Biased exponent = 14 +127= 141 = 10001101
Question 44 
What is the octal equivalent of the hexadecimal number 132A?
46252  
11450  
11452  
45250 
Question 44 Explanation:
Step1: Convert hexadecimal number in binary format. It is nothing but representing 4 binary values of each character.
Step2: (132A) 16 =(0001 0011 0010 1010) _{2} Step3: Divide 4 binary numbers in to 3 binary numbers from right to left.
Step4: (0 001 001 100 101 010) _{2}
(0 1 1 4 5 2) _{8}
Step2: (132A) 16 =(0001 0011 0010 1010) _{2} Step3: Divide 4 binary numbers in to 3 binary numbers from right to left.
Step4: (0 001 001 100 101 010) _{2}
(0 1 1 4 5 2) _{8}
Question 45 
The Octal equivalent of the binary number 1011101011 is :
7353  
1353  
5651  
5657 
Question 45 Explanation:
We have to divide binary number into 3 bit pairs from LSB.
1 011 101 011
1 3 5 3
(1011101011)_{ 2} = (1353)_{ 8}
1 011 101 011
1 3 5 3
(1011101011)_{ 2} = (1353)_{ 8}
Question 46 
Which is the hexadecimal number equivalent to the octal number 46250
4AC8  
4CA8  
CCA8  
4CA4 
Question 46 Explanation:
Given octal number is 46250 and the corresponding binary number we will get by writing each decimal digit into three binary digits form which is 100110010101000
Now group the four digits from LSB and write corresponding equivalent of Hexadecimal digit of binary digits.
The Hexadecimal number of 100(4) 1100(C) 1010(A) 1000(8) which is nothing but 4CA8
Now group the four digits from LSB and write corresponding equivalent of Hexadecimal digit of binary digits.
The Hexadecimal number of 100(4) 1100(C) 1010(A) 1000(8) which is nothing but 4CA8
Question 47 
Let m=(313)_{ 4} and n=(322)_{ 4} . Find the base 4 expansion of m+n.
(635)_{ 4}  
(32312) _{4}  
(21323) _{4}  
(1301) _{4} 
Question 47 Explanation:
In this problem, they are asking to find (m+n) _{4}
→ We are using addition for decimal number system. But in this problem m and n values are base 4. So, we can’t add directly.
Step1: We have to perform base 4 into decimal values of m and n.
m = 3*42 + 1*41 + 3*40
m = 48 + 4 + 3
m = 55
n= 3*42 + 2*41 +2*40
n= 48 + 8 +2
n= 58
Step2: The resultant decimal values should perform addition.
m+n = 55 + 58
m+n = 113
Step3: Finally we have to convert decimal value into base 4 value.
(1301)_{4}
→ We are using addition for decimal number system. But in this problem m and n values are base 4. So, we can’t add directly.
Step1: We have to perform base 4 into decimal values of m and n.
m = 3*42 + 1*41 + 3*40
m = 48 + 4 + 3
m = 55
n= 3*42 + 2*41 +2*40
n= 48 + 8 +2
n= 58
Step2: The resultant decimal values should perform addition.
m+n = 55 + 58
m+n = 113
Step3: Finally we have to convert decimal value into base 4 value.
(1301)_{4}
Question 48 
Convert the octal number 0.4051 into its equivalent decimal number.
0.5100098  
0.2096  
0.52  
0.4192 
Question 48 Explanation:
Question 49 
The hexadecimal equivalent of the octal number 2357 is :
2EE  
2FF  
4EF  
4FE 
Question 49 Explanation:
Step1: Convert octal number into binary number
(2357)_{ 8} = (010 011 101 111)_{ 2}
Step2: Divide 4 bits from LSB then will get hexadecimal number
0100 1110 1111
2 E F
(2EF) _{16} = (2357)_{ 8}
(2357)_{ 8} = (010 011 101 111)_{ 2}
Step2: Divide 4 bits from LSB then will get hexadecimal number
0100 1110 1111
2 E F
(2EF) _{16} = (2357)_{ 8}
Question 50 
If X is a binary number which is power of 2, then the value of X & (X – 1) is :
11....11  
00.....00  
100.....0  
000......1 
Question 50 Explanation:
Given data,
→ X is binary number which is power of 2. It means, we have to take powers of 2 numbers only.
Ex: 1,2,4,8,16,32,..,
Let X=4
X=4 equivalent binary number is 100
X1=3 equivalent binary number is 011
100
011

000 (AND operation)

Ex2:
X=8 and X1=7
8 binary value is
1000
7 binary number is 0111

0000(AND operation)

So, Option B is correct answer.
→ X is binary number which is power of 2. It means, we have to take powers of 2 numbers only.
Ex: 1,2,4,8,16,32,..,
Let X=4
X=4 equivalent binary number is 100
X1=3 equivalent binary number is 011
100
011

000 (AND operation)

Ex2:
X=8 and X1=7
8 binary value is
1000
7 binary number is 0111

0000(AND operation)

So, Option B is correct answer.
Question 51 
The octal number 326.4 is equivalent to
(214.2) _{10} and (D6.8))_{ 16}  
(212.5) _{10} and (D6.8)) _{ 16}  
(214.5) _{10} and (D6.8)) _{ 16}  
(214.5) _{10} and (D6.4)) _{ 16} 
Question 51 Explanation:
(326.4) _{8} = ( ?)
Step1: First convert given octal no. to binary number because it will be easier to solve this way.
Step 2: Now convert above binary no. into decimal .
(011010110.100) _{2}
= ( 1 * 2 ^{7} ) + ( 1 × 2^{ 6} ) + ( 1 × 2^{ 4} ) + ( 1 × 2 ^{2} ) + ( 1 × 2^{ 1} ) · [ 1 × ( 1/2) ]
= (214.5) _{10}
(326.4) _{8} = (?)_{ 16} Step 1: Convert given octal no. into binary no.
Step 2: Now convert above binary no. into decimal .
(011010110.100) _{2}
= ( 1 * 2 ^{7} ) + ( 1 × 2^{ 6} ) + ( 1 × 2^{ 4} ) + ( 1 × 2^{ 2} ) + ( 1 × 2^{ 1} ) · [ 1 × ( 1/2) ]
= (214.5) _{10}
(326.4)_{ 8} = (?)_{ 16}
Step1: First convert given octal no. to binary number because it will be easier to solve this way.
Step 2: Now convert above binary no. into decimal .
(011010110.100) _{2}
= ( 1 * 2 ^{7} ) + ( 1 × 2^{ 6} ) + ( 1 × 2^{ 4} ) + ( 1 × 2 ^{2} ) + ( 1 × 2^{ 1} ) · [ 1 × ( 1/2) ]
= (214.5) _{10}
(326.4) _{8} = (?)_{ 16} Step 1: Convert given octal no. into binary no.
Step 2: Now convert above binary no. into decimal .
(011010110.100) _{2}
= ( 1 * 2 ^{7} ) + ( 1 × 2^{ 6} ) + ( 1 × 2^{ 4} ) + ( 1 × 2^{ 2} ) + ( 1 × 2^{ 1} ) · [ 1 × ( 1/2) ]
= (214.5) _{10}
(326.4)_{ 8} = (?)_{ 16}
Question 52 
Which of the following is the most efficient to perform arithmetic operations on the numbers?
Signmagnitude  
1’s complement  
2’s complement  
9’s complement 
Question 52 Explanation:
2’s complement has single representation for zero , but Signmagnitude, 1’s complement and 9’s complement have two representations for 0 (i.e., both positive zero and negative zero).
While doing arithmetic operations like addition or subtraction using 1's complement(or 9's complement), we have to add an extra carry bit, i.e 1 to the result to get the correct answer. 2's complement doesn't require such extra calculation.
Question 53 
An example of a binary number which is equal to its 2 ’s complement is :
1100  
1001  
1000  
1111 
Question 53 Explanation:
OptionA: 1100 convert into 2’s complement is
1100
1’s complement: 0011
2’s complement: 1

0100
OptionB: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement:
1

0111

OptionC: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement:
1

1000

OptionD: 1111 convert into 2’s complement is
1111
1’s complement: 0000
2’s complement:
1

0001

So, OptionC is correct answer.
1100
1’s complement: 0011
2’s complement: 1

0100
OptionB: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement:
1

0111

OptionC: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement:
1

1000

OptionD: 1111 convert into 2’s complement is
1111
1’s complement: 0000
2’s complement:
1

0001

So, OptionC is correct answer.
Question 54 
Which of the following is divisible by 4 ?
100101100  
1110001110001  
11110011  
10101010101010 
Question 54 Explanation:
OptionA: (100101100) _{2} = (300) _{10}
300 is divisible by 4
OptionB: (1110001110001) _{2} = (7281)_{ 10}
7281 is not divisible by 4
OptionC: (11110011) _{2} = (243)_{ 10}
243 is not divisible by 4
OptionD: (10101010101010) _{2} = (10,922)_{ 10}
10,922 is not divisible by 4.
300 is divisible by 4
OptionB: (1110001110001) _{2} = (7281)_{ 10}
7281 is not divisible by 4
OptionC: (11110011) _{2} = (243)_{ 10}
243 is not divisible by 4
OptionD: (10101010101010) _{2} = (10,922)_{ 10}
10,922 is not divisible by 4.
Question 55 
(101011) _{2} =(53) _{b} , then ‘b’ is equal to :
4  
8  
10  
16 
Question 55 Explanation:
We are dividing binary number into 3 then
101 011
5 3
(101011) _{2} = (53) _{8}
Suppose, we are dividing binary digits into 4 then
0010 1011
2 B
So, it is not hexadecimal number.
101 011
5 3
(101011) _{2} = (53) _{8}
Suppose, we are dividing binary digits into 4 then
0010 1011
2 B
So, it is not hexadecimal number.
Question 56 
Which of the following binary number is the same as its 2’s complement :
1010  
0101  
1000  
1001 
Question 56 Explanation:
OptionA: 1010 convert into 2’s complement is
1010
1’s complement: 0101
2’s complement: 1

0110
 OptionB: 0101 convert into 2’s complement is
0101
1’s complement: 1010
2’s complement: 1

1011

OptionC: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement: 1

1000

OptionD: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement: 1

0111

So, OptionC is correct answer.
1010
1’s complement: 0101
2’s complement: 1

0110
 OptionB: 0101 convert into 2’s complement is
0101
1’s complement: 1010
2’s complement: 1

1011

OptionC: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement: 1

1000

OptionD: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement: 1

0111

So, OptionC is correct answer.
Question 57 
The hexadecimal equivalent of (10111)_{ 2} ×(1110)_{ 2} is :
150  
241  
142  
101011110 
Question 57 Explanation:
Step1: First convert binary number into decimal number
(10111) _{2} = (23)_{ 10}
(1110) _{2} = (14) _{10}
Step2: Perform multiplication 23*14=(322) _{10}
Step3: Convert (322) _{10} into hexadecimal number (322) _{10} = (142)_{ 16}
(10111) _{2} = (23)_{ 10}
(1110) _{2} = (14) _{10}
Step2: Perform multiplication 23*14=(322) _{10}
Step3: Convert (322) _{10} into hexadecimal number (322) _{10} = (142)_{ 16}
Question 58 
An example of a self complementing code is :
8421 code  
Gray code  
Excess3 code  
7421 code 
Question 58 Explanation:
→ Excess3 code is also called SelfComplementing Code. Because 1’s complement of excess3 number is equivalent to 9’s complement of corresponding decimal digit.
→ In excess3 code, each of the 4bit number represents decimal digit which is 3 less than actual decimal digit. So the bits have no fixed weight.
→ Excess3 code is neither CRC nor Algebraic Code which are used for error detection and/or correction.
→ In excess3 code, each of the 4bit number represents decimal digit which is 3 less than actual decimal digit. So the bits have no fixed weight.
→ Excess3 code is neither CRC nor Algebraic Code which are used for error detection and/or correction.
Question 59 
The number of 1 ’s present in the binary representation of (3*512 + 7*64 +5*8 +3) _{10} is :
8  
9  
10  
11 
Question 59 Explanation:
(3*512 + 7*64 +5*8 +3) =2027
(2027) _{10} = (111 1110 1011)_{ 2}
Here, total number of 1’s are 9.
(2027) _{10} = (111 1110 1011)_{ 2}
Here, total number of 1’s are 9.
Question 60 
In a weighted code with weight 6, 4, 2, 3 the decimal 5 is represented by :
0101  
0111  
1011  
1000 
Question 60 Explanation:
The decimal value 5 is represented by 1011.
= 6*1 + 4*0 + 2*1 + 3*1
= 6+23
=5
= 6*1 + 4*0 + 2*1 + 3*1
= 6+23
=5
Question 61 
The BCD adder to add two decimal digits needs minimum of
6 full adders and 2 half adders  
5 full adders and 3 half adders  
4 full adders and 3 half adders  
5 full adders and 2 half adders 
Question 61 Explanation:
→ Each digit is represented by a 4bit BCD code.
→ To add two 4bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).
→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.
→ This can be done with 1 Half adder and 2 Full Adder
(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)
→ Here, Half adder is used to add next significant bits.
Total 5 Full Adders and 2 Half Adders are needed
→ To add two 4bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).
→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.
→ This can be done with 1 Half adder and 2 Full Adder
(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)
→ Here, Half adder is used to add next significant bits.
Total 5 Full Adders and 2 Half Adders are needed
Question 62 
The Excess3 decimal code is a selfcomplementing code because
The binary sum of a code and its 9’s complement is equal to 9.  
It is a weighted code  
Complement can be generated by inverting each bit pattern  
The binary sum of a code and its 10’s complement is equal to 9 
Question 62 Explanation:
The Excess3 decimal code is a selfcomplementing code because complement can be generated by inverting each bit pattern.
Question 63 
The range of representable normalized numbers in the floating point binary fractional representation in a 32bit word with 1bit sign, 8bit excess 128 biased exponent and 23bit mantissa is
2 ^{128} to (1 – 2^{ –23} ) * 2^{ 127}  
(1 – 2 ^{ –23} ) * 2 ^{ 127} to 2 ^{128}  
(1 – 2 ^{–23} ) * 2^{ –127} to 2^{ 23>}  
2 ^{–129} to (1 – 2^{ –23} ) * 2^{ 127} 
Question 63 Explanation:
The range of representable normalized numbers in the floating point binary fractional representation in a 32bit word with 1bit sign, 8bit excess 128 biased exponent and 23bit mantissa is 2 ^{–129} to (1 – 2^{ –23} ) * 2^{ 127}
Question 64 
The decimal floating point number 40.1 represented using IEEE754 32bit representation and written in hexadecimal form is
0xC2206000  
0xC2006666  
0xC2006000  
0xC2206666 
Question 64 Explanation:
1. Fraction part can be converted into binary form by multiplying it with 2 and taking non fractional part of the product. Take fractional part and multiply again as explained above.
0.1 x 2= 0.2 → 0
0.2 x 2= 0.4 → 0
0.4 x 2= 0.8 → 0
0.8 x 2= 1.6 → 1
0.6 x 2= 1.2 → 1
(0.1) _{10} = (0.00011)_{ 2}
(40) _{10} = (101000)_{ 2}
101000.00011
Normalize the number
1.0100000011 x 2 ^{5}
Biased exponent= 5+127= 132=(1000 0100)_{ 2}
Mantissa= 01000000110000000000000
Sign= 1
0.1 x 2= 0.2 → 0
0.2 x 2= 0.4 → 0
0.4 x 2= 0.8 → 0
0.8 x 2= 1.6 → 1
0.6 x 2= 1.2 → 1
(0.1) _{10} = (0.00011)_{ 2}
(40) _{10} = (101000)_{ 2}
101000.00011
Normalize the number
1.0100000011 x 2 ^{5}
Biased exponent= 5+127= 132=(1000 0100)_{ 2}
Mantissa= 01000000110000000000000
Sign= 1
Question 65 
The IEEE singleprecision and doubleprecision format to represent floatingpoint numbers, has a length of ______ and ______ respectively.
8 bits and 16 bits  
16 bits and 32 bits  
32 bits and 64 bits
 
64 bits and 128 bits 
Question 65 Explanation:
The IEEE singleprecision and doubleprecision format to represent floatingpoint numbers, has a length of 32 bits and 64 bits respectively.
Question 66 
Given that (292)_{10} = (1204)_{x} in some number system x. The base x of that number system is
2  
8  
10  
None of the above 
Question 66 Explanation:
OptionA: It is false because it is equivalent to (292)_{10} =(100100100)_{2}
OptionB: It is false because it is equivalent to (292)_{10} =(444)_{8}
OptionB: It is false because it is equivalent to (292)_{10} =(292)_{10}
So, optionD is the correct answer. The actual x value is 6.
Question 67 
If an integer needs two bytes of storage, then the maximum value of a signed integer is
2^{16} – 1  
2^{15} – 1  
2^{16}  
2^{15} 
Question 67 Explanation:
= If an integer needs two bytes of storage, then the maximum value of a signed integer is 2^{n1} 1
= 2^{15} 1
= 2^{15} 1
Question 68 
If an integer needs two bytes of storage, then the maximum value of unsigned integer is
2^{16} – 1  
2^{15} – 1  
2^{16}  
2^{15} 
Question 68 Explanation:
= If an integer needs two bytes of storage, then the maximum value of a signed integer is 2^{n1} 1
= 2^{15} 1
= 2^{15} 1
Question 69 
Negative numbers cannot be represented in
signed magnitude form  
1’s complement form  
2’s complement form  
none of the above 
Question 69 Explanation:
Negative numbers can be represented in
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
Question 70 
Two’s complement of a binary number 1010 is
0101  
0101  
0110  
1001 
Question 71 
If integer needs two by storage, then maximum value of an unsigned integer is
2 ^{16} − 1  
2 ^{15} − 1  
2 ^{16}  
2 ^{15} 
Question 72 
The hexadecimal number equivalent to (1762.46)_{8} is
3F2.89  
3F2.98  
2F3.89  
2F3.98 
Question 72 Explanation:
(1762.46)_{8}
→ For making the conversion easy, first convert number into binary
(001 111 110 010. 100 110)_{2}
→ Now convert above binary no. into hexadecimal, we need 4bits to represent a no. into hexadecimal
(3F2.98)_{16}
→ For making the conversion easy, first convert number into binary
(001 111 110 010. 100 110)_{2}
→ Now convert above binary no. into hexadecimal, we need 4bits to represent a no. into hexadecimal
(3F2.98)_{16}
Question 73 
8bit 1’s complement form of –77.25 is
01001101.0100  
01001101.0010  
10110010.1011  
10110010.1101 
Question 73 Explanation:
(77.25)_{10}
Step 1: Convert (77)_{10}into binary form.
Question 74 
In which of the following codes the successive numbers differ in only one bit position ?
ASCII  
Gray Code  
Excess3 Code  
BCD 
Question 75 
The binary addition of 1+1+1 is
111  
10  
110  
11 
Question 76 
Which one of the following is decimal value of a signed binary number 1101010, if it is in 2’s complement form ?
42  
22  
21  
106 
Question 76 Explanation:
Step1: 2's complement number are weighted number. So,
1101010= 1*(0.26) + 1*25 + 1*23 + 1*21
= 64 + 32 +8 +2
= 22
Hence (1101010)=(22) in 2's complement form
Question 77 
Which of the following weights makes the complement operation easier in BCD form ?
8421  
Excess3  
2421  
3210 
Question 78 
Let a _{n} a _{n−1} ...a_{ 1} a_{ 0} be the binary representation of an integer b . The integer b is divisible by 3 if
the number of one’s is divisible by 3  
the number of one’s is divisible by 3, but not by 9  
the number of zeroes is divisible by 3  
the difference of alternate sum, i.e., ( a _{0} + a_{ 2} + . ..) − ( a _{1} + a_{ 2} + . ..) is divisible by 3 
Question 79 
The decimal number equivalent of (4057.06)_{8} is
2095.75  
2095.075  
2095.937  
2095.0937 
Question 79 Explanation:
Question 80 
12bit 2’s complement of –73.75 is
01001001.1100  
11001001.1100  
10110110.0100  
10110110.1100 
Question 80 Explanation:
Question 81 
Encoding of data bits 0011 into 7bit even Parity Hamming Code is
0011110  
0101110  
0010110  
0011100 
Question 81 Explanation:
(m) Data = 0011
No. of parity bits needed is decided using
2^{p}≥ m+p+1
p=3
Bit pattern:
P_{1}(check even parity at 1, 3, 5, 7 bit) = 0
P_{2}(check even parity at 2, 3, 6, 7 bit) = 1
P_{4}(check even parity at 4, 5, 6, 7 bit) = 1
So encoded data = 0011110
No. of parity bits needed is decided using
2^{p}≥ m+p+1
p=3
Bit pattern:
P_{1}(check even parity at 1, 3, 5, 7 bit) = 0
P_{2}(check even parity at 2, 3, 6, 7 bit) = 1
P_{4}(check even parity at 4, 5, 6, 7 bit) = 1
So encoded data = 0011110
Question 82 
What is decimal equivalent of BCD 11011.1100 ?
22.0  
22.2  
20.2  
21.2  
None of the above 
Question 82 Explanation:
The question is not properly framed.
In the question it is mentioned that 11011.1100 is BCD.
We know that BCD code is different from Binary number system.
Each of the decimal digit has a 4bit binary code or in other words every block of 4 bits has a corresponding decimal digit.
0000 0
0001 1
:
:
1001 9
1010  1111 are not valid codes.
If the given code 11011.1100 is BCD then divide it into blocks of 4 bits.
0001 1011. 1100 (Note: Appending zeros on the left side will not change value.)
But 1011 is not a valid BCD code.
In the question it is mentioned that 11011.1100 is BCD.
We know that BCD code is different from Binary number system.
Each of the decimal digit has a 4bit binary code or in other words every block of 4 bits has a corresponding decimal digit.
0000 0
0001 1
:
:
1001 9
1010  1111 are not valid codes.
If the given code 11011.1100 is BCD then divide it into blocks of 4 bits.
0001 1011. 1100 (Note: Appending zeros on the left side will not change value.)
But 1011 is not a valid BCD code.
Question 83 
In order that a code is ‘t’ error correcting, the minimum Hamming distance should be :
t  
2t1  
2t  
2t+1 
Question 83 Explanation:
Question 84 
The octal equivalent of hexadecimal (A.B)_{16} is:
47.21  
12.74  
12.71  
17.21  
None of the above 
Question 84 Explanation:
Question 85 
The answer of the operation (10111)_{2}* (1110)_{2} in hex equivalence is
150  
241  
142  
101011110 
Question 85 Explanation:
Question 86 
How many 1’s are present in the binary representation of
3 × 512 + 7 × 64 + 5 × 8 + 3
8  
9  
10  
11 
Question 86 Explanation:
Given expression is 3 × 512 + 7 × 64 + 5 × 8 + 3
We can write above statement based on precedence is (3 * 512) + (7 * 64) + (5 * 8) + 3
Step1: 1536+448+40+3
Step2: (2027)_{10}
Step3: Equivalent binary number is (011111101011)_{2}
Note: Total number of 1’s is 9.
We can write above statement based on precedence is (3 * 512) + (7 * 64) + (5 * 8) + 3
Step1: 1536+448+40+3
Step2: (2027)_{10}
Step3: Equivalent binary number is (011111101011)_{2}
Note: Total number of 1’s is 9.
Question 87 
If a code is ‘t’ error detecting, the minimum hamming distance should be equal to :
t1  
t  
t+1  
2t+1 
Question 87 Explanation:
Question 88 
If a code is terror correcting, the minimum Hamming distance is equal to :
2t+1  
2t  
2t1  
t1 
Question 88 Explanation:
Question 89 
The octal equivalent of the hexadecimal number FF is :
100  
150  
377  
737 
Question 89 Explanation:
Step1: Convert FF into binary number
(FF)_{16} = (1111 1111)_{2}
Step2: Divide binary number into 3 segments from LSB(Least significant bit).
(FF)_{16} = (1111 1111)_{2}
Step2: Divide binary number into 3 segments from LSB(Least significant bit).
Question 90 
2’s complement of 100 is :
00011100  
10011101  
10011100  
11100100 
Question 90 Explanation:
Question 91 
How many 1’s are present in the binary representation of 15*256 + 5*16 + 3 :
8  
9  
10  
11 
Question 91 Explanation:
Step1: The precedence to given decimal number is (15*256) + (5*16) + 3
Step2: 3840+80+3=(3923)^{10}
Step3: Convert decimal number into binary
(3923)^{10} = (111101010011 )_{2}
Step4: Total number of 1’s are 8.
Step2: 3840+80+3=(3923)^{10}
Step3: Convert decimal number into binary
(3923)^{10} = (111101010011 )_{2}
Step4: Total number of 1’s are 8.
Question 92 
Let C be a binary linear code with minimum distance 2t + 1 then it can correct upto _____ bits of error.
t + 1  
t  
t  2  
t / 2 
Question 92 Explanation:
A binary linear code with minimum distance 2t + 1 then it can correct up to ‘T’ bits of error.
There are 92 questions to complete.