NumberSystems
Question 1 
bias of 127.
S : 1 E : 10000001 F : 11110000000000000000000
Here S, E and F denote the sign, exponent and fraction components of the floating point representation.
The decimal value corresponding to the above representation (rounded to 2 decimal places) is _______
7.75 
Sign bit S= 1. The given number is a negative number.
Biased Exponent E = 2^{7} + 1= 129
Actual Exponent e = E127
= 129 127
= 2
The decimal value= (1)^{s} x 1.M x 2^{e}
= (1) 1 x 1.1111 x 2^{2}
=  (111.11)
=  (7 + 0.75)
= 7.7
Question 2 
21  
528
 
D2  
15 
On converting (210)3 in decimal, we will get:=>
2*3^{2}+1*3=2*9+3=21_{10 }
=>(15)_{16}
Question 3 
Which one of the following choices gives the correct values of x and y?
x is 1 and y is 1  
x is 0 and y is 1  
x is 1 and y is 0  
x is 0 and y is 0 
C2 checks the bits d1, d3, d4, d6, d7.
C2=1, d1= 1, d3= 1, d4= 0, d6= 0, d7= 1.
The number of 1s is even. So, even parity is used in this problem.
C1 checks the bits d1, d2, d4, d5, d7.
C1=0, d1= 1, d2= 0, d4= 0, d5= x, d7= 1.
As the parity used is even parity, the value of d5 should be 0.
x=d5=0
C8 checks the bitsa d5, d6, d7, d8.
C8=y, d5= x=0, d6= 0, d7= 1, d8= 1.
As the parity used is even parity, the value of C8 should be 0.
C8=y=0.
x=y=0.
Question 4 
Consider nbit (including sign bit) 2’s complement representation of integer number. The range of integer values, N, that can be represented is _________ ≤ N ≤ _________
2^{n1} to 2^{n1}  1 
Question 5 
Write a program in 8085 Assembly language to Add two 16bit unsigned BCD(8421 Binary Coded Decimal) number. Assume the two input operands are in BC and DE Register pairs. The result should be placed in the register pair BC. (Higher order register in the register pair contains higher order digits of operand)
Theory Explanation. 
Question 6 
The number of 1’s in the binary representation of
(3*4096 + 15*256 + 5*16 + 3) are:
8  
8  
10  
12 
= (11000000000000)_{2}
15 × 256 = 15 × 2^{8}
= (111100000000)_{2}
5 × 16 = 5 × 2^{4}
= (1010000)_{2}
3 = (11)_{2}
Hence, all binary numbers,
∴ 101's
Question 7 
(a) An asynchronous serial communication controller that uses a start stop scheme for controlling the serial I/O of a system is programmed for a string of length seven bits, one parity bit (odd parity) and one step bit. The
transmission rate is 1200 bits/second.
(i) What is the complete bit stream that is transmitted for the string ‘0110101’?
(ii) How many such strings can be transmitted per second?
(b) Consider a CRT display that has a text mode display format of 80 × 25 characters with a 9 × 12 character cell. What is the size of the video buffer RAM for the display to be used in monochrome (1 bit per pixel) graphics mode?
Theory Explanation. 
Question 8 
Consider three registers R1, R2 and R3 that store numbers in IEEE754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.
If R3 = R1/R2, what is the value stored in R3?
0x40800000  
0x83400000  
0xC8500000  
0xC0800000 
R1 = 1.0100..0 X 2^{132127}
= 1.0100..0 X 2^{5}
= 101.0 X 2^{3}
= 5 X 8
= 40
R2 = (1) x 1.0100..0 X 2^{130127}
= (1) x 1.0100..0 X 2^{3}
= (1) x 101.0 X 2^{1}
= (1) x5 X 2
= 10
R3 = R1/R2
= 4
= (1)x 1.0 x 2^{2}
Sign = 1
Mantissa = 000..0
Exponent = 2+127 = 129
R3 = 1100 0000 1000 000..0
= 0x C 0 8 0 0 0 0 0
Question 9 
The decimal value 0.5 in IEEE single precision floating point representation has
fraction bits of 000…000 and exponent value of 0  
fraction bits of 000…000 and exponent value of −1  
fraction bits of 100…000 and exponent value of 0  
no exact representation 
So, value of the exponent = 1
and
fraction is 000…000 (Implicit representation)
Question 10 
Consider the following floating point number representation
The exponent is in 2's complement representation and mantissa is in the sign magnitude representation. The range of the magnitude of the normalized numbers in this representation is
0 to 1  
0.5 to 1  
2^{23} to 0.5  
0.5 to (12^{23}) 
Question 11 
Given √224)_{r} = 13)_{r}.
The value of the radix r is:
10  
8  
5  
6 
Convert r base to decimal.
√2r^{2} + 25 + 4 = r + 3
Take square both sides,
2r^{2} + 2r + 4 = r^{2} + 6r + 9
r^{2}  4r  5 = 0
r^{2}  5r + r  5 = 0
r(r  5) + (r  5) = 0
r = 1, 5
r cannot be 1,
So r = 5 is correct answer.
Question 12 
The octal representation of an integer is (342)_{8}. If this were to be treated as an eightbit integer is an 8085 based computer, its decimal equivalent is
226  
98  
76  
30 
If this can be treated as 8 bit integer, then the first becomes sign bit i.e., '1' then the number is negative.
8085 uses 2's complement then
⇒ 30
Question 13 
Suppose the domain set of an attribute consists of signed four digit numbers. What is the percentage of reduction in storage space of this attribute if it is stored as an integer rather than in character form?
80%  
20%  
60%  
40% 
We have four digits. So to represent signed 4 digit numbers we need 5 bytes, 4 bytes for four digits and 1 for the sign.
So required memory = 5 bytes.
Now, if we use integer, the largest no. needed to represent is 9999 and this requires 2 bytes of memory for signed representation.
9999 in binary requires 14 bits. So, 2 bits remaining and 1 we can use for sign bit.
So, memory savings,
= 5  2/5 × 100
= 60%
Question 14 
Booth’s coding in 8 bits for the decimal number –57 is
0 – 100 + 1000  
0 – 100 + 100  1  
0 – 1 + 100 – 10 + 1  
00 – 10 + 100  1 
Question 15 
Zero has two representations in
Sign magnitude  
1’s complement  
2’s complement  
None of the above  
Both A and B 
+0 = 0000
0 = 1000
1's complement:
+0 = 0000
0 = 1111
Question 16 
The number 43 in 2’s complement representation is
01010101  
11010101  
00101011  
10101011 
Question 17 
Consider the values A = 2.0 x 10^{30}, B = 2.0 x 10^{30}, C = 1.0, and the sequence
X: = A + B Y: = A + C X: = X + C Y: = Y + B
executed on a computer where floatingpoint numbers are represented with 32 bits. The values for X and Y will be
X = 1.0, Y = 1.0  
X = 1.0, Y = 0.0  
X = 0.0, Y = 1.0  
X = 0.0, Y = 0.0 
A = 2.0 * 10^{30}, C = 1.0
So, A + C should make the 31^{st} digit to 1, which is surely outside the precision level of A (it is 31^{st} digit and not 31^{st} bit). So, this addition will just return the value of A which will be assigned to Y.
So, Y + B will return 0.0 while X + C will return 1.0.
Question 18 
The 2’s complement representation of (539)_{10} in hexadecimal is
ABE  
DBC  
DE5  
9E7 
For (539)_{10} = (1101 1110 0100)_{2}
1's complement = (1101 1110 0100)_{2}
2's complement = (1101 1110 0101)_{2}
= (DE5)_{16}
Question 19 
Consider the circuit shown below. The output of a 2:1 Mux is given by the function (ac' + bc).
Which of the following is true?
f = x1' + x2  
f = x1'x2 + x1x2'  
f = x1x2 + x1'x2'  
f = x1 + x2' 
g = (1 and x1’) or (0 and x1)
g = x1’
f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2
Question 20 
The decimal value 0.25
is equivalent to the binary value 0.1  
is equivalent to the binary value 0.01  
is equivalent to the binary value 0.00111…  
cannot be represented precisely in binary 
Multiply 0.25 by 2.
0.25×2 = 0.50 (product)
Fractional part = 0.50
Carry = 0
2^{nd} Multiplication iteration:
Multiply 0.50 by 2.
0.50×2 = 1.00 (product)
Fractional part = 0.00
Carry = 1
The fractional part in the 2^{nd} iteration becomes zero and so we stop the multiplication iteration.
Carry from 1^{st} multiplication iteration becomes MSB and carry from 2^{nd} iteration becomes LSB. So the result is 0.01.
Question 21 
The 2’s complement representation of the decimal value 15 is
1111  
11111  
111111  
10001 
15 = 11111
1's complement = 10000
2's complement = 10001
Question 22 
Sign extension is a step in
floating point multiplication  
signed 16 bit integer addition  
arithmetic left shift  
converting a signed integer from one size to another 
Question 23 
In 2’s complement addition, overflow
is flagged whenever there is carry from sign bit addition  
cannot occur when a positive value is added to a negative value  
is flagged when the carries from sign bit and previous bit match  
None of the above 
Question 24 
Consider the following 32bit floatingpoint representation scheme as shown in the formal below. A value is specified by 3 fields, a one bit sign field (with 0 for positive and 1 for negative values), a 24 bit fraction field (with the binary point being at the left end of the fraction bits), and a 7 bit exponent field (in excess64 signed integer representation, with 16 being the base of exponentiation). The sign bit is the most significant bit.
(a) It is required to represent the decimal value –7.5 as a normalized floating point number in the given format. Derive the values of the various fields. Express your final answer in the hexadecimal.
(b) What is the largest values that can be represented using this format? Express your answer as the nearest power of 10.
Theory of Explanation is given below. 
Question 25 
Assuming all numbers are in 2's complement representation, which of the following numbers is divisible by 11111011?
11100111  
11100100  
11010111  
11011011 
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = 5
(A) 11100111  (25)_{10}
(B) 11100100  (28)_{10 (C) 11010111  (41)10 (D) 11011011  (37)10 Answer: Option A (25 is divisible by 5)}
Question 26 
The following is a scheme for floating point number representation using 16 bits.
Let s, e, and m be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:
What is the maximum difference between two successive real numbers representable in this system?
2^{40}  
2^{9}  
2^{22}  
2^{31} 
The largest number is 1.111111111× 2^{6231} = (2−2^{−9})×2^{31}
Second largest number is 1.111111110×2^{6231} = (2−2^{8})×2^{31}
Difference = (2−2^{−9})×2^{31}  (2−2^{8})×2^{31}
= (2^{8}−2^{−9}) ×2^{31}
= 2^{−9}×2^{31}
= 2^{22}
Question 27 
Convert the following numbers in the given bases into their equivalents in the desired bases.
(a) 110.101)_{2} = x)_{10}
(b) 1118)_{10} = y)_{H}
(a) 6.625, (b) (45E)_{H} 
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)_{H}.
Question 28 
Consider the number given by the decimal expression:
16^{3} * 9 + 16^{2} * 7 + 16 * 5 + 3
The number of 1’s in the unsigned binary representation of the number is ________.
9 
(9753)_{16}
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
Question 29 
When two 4bit binary number A = a_{3}a_{2}a_{1}a_{0} and B = b_{3}b_{2}b_{1}b_{0} are multiplied, the digit c_{1} of the product C is given by _________
c_{1} = b_{1}a_{0} ⊕ a_{1}b_{0} 
⇒ c_{1} = b_{1}a_{0} ⊕ a_{1}b_{0}
Question 30 
exponent = 00000001 and mantissa = 00000000000000000000001  
exponent = 00000001 and mantissa = 00000000000000000000000  
exponent = 00000000 and mantissa = 00000000000000000000000  
exponent = 00000000 and mantissa = 00000000000000000000001 
The smallest biased exponent for the normalized number is E= 1.
The smallest Mantissa M = 000...0
The smallest positive normalized number = 1.M x 2 ^{E127}
= 1.0 x 2 ^{126 .}
= 2 ^{126}
Question 31 
0x4243  
0x6665  
0x0001  
0x0100 
It is given that the unsigned integer is 2bytes long. It needs 4 hexadecimal digits.
We know the bigendian and littleendian computers differ in how the data is stored in memory.
In little endian machines, the last byte of binary representation of the multibyte datatype is stored first. On the other hand, in big endian machines, the first byte of binary representation of the multibyte datatype is stored first.
In the hexadecimal representation of the 2byte number on the little endian machine, the first two hexadecimal digits are for one byte and the last two hexadecimal digits are for the second byte. On the big endian machine it is the other way round.
It is given that the value in little endian representation is 255 more than the value in the big endian machine.
From the given options
A). In little endian = 0x4243 in binary (0100 0010 0100 0011) which has decimal value = 16963
On big endian = 0x4342 in binary (0100 0011 0100 0010) which has decimal value = 17218
Here the big endian value is higher than the little endian value. So this is not the correct option.
B). In little endian 0x6665 in binary (0110 0110 0110 0101) which has the decimal value = 26213
In big endian 0x6566 in binary (0110 0101 0110 0110) which has the decimal value = 25958
The difference = 26213  25958 = 255. So this is also the correct option.
C). In little endian 0x0001 in binary (0000 0000 0000 0001) which has the decimal value = 1.
In big endian 0x0100 in binary (0000 0001 0000 0000) has decimal value = 256.
But here also the big endian value is higher than the little endian. So this is not the correct option.
D). On the little endian machine for hexadecimal number 0x0100 in binary (0000 0001 0000 0000) which has decimal value = 256. On a big endian machine it is 0x 0001 in binary (0000 0000 0000 0001) which has decimal value = 1
The difference in value of little endian to big endian is 2561 = 255.
Hence 2, 4 are the correct options.
Question 32 
3 
x=1 and y=2
x+y= 1+2=3
Question 33 
The number (123456)_{8} is equivalent to
(A72E)_{16} and (22130232)_{4}  
(A72E)_{16} and (22131122)_{4}  
(A73E)_{16} and (22130232)_{4}  
(A62E)_{16} and (22120232)_{4}

= (00 1010 0111 0010 1110)_{2}
= (A72E)_{16}
Also,
(001 010 011 100 101 110)_{2}
= (00 10 10 01 11 00 10 11 10)_{2}
= (22130232)_{4}
Question 34 
Consider a parity check code with three data bits and four parity check bits. Three of the code words are 0101011, 1001101 and 1110001. Which of the following are also code words?
1. 0010111 2. 0110110 3. 1011010 4. 0111010
1 and 3  
1, 2 and 3  
2 and 4  
1, 2, 3 and 4 
Given transmitted codewords are
By inspection we can find the rule for generating each of the parity bits,
Now from above we can see that (I) and (III) are only codewords.
Question 35 
(34.4)_{8} × (23.4)_{8} evaluates to
(1053.6)_{8}  
(1053.2)_{8}  
(1024.2)_{8}  
None of these 
(34.4)_{8} = 3×8^{1} + 4×8^{0} + 4×8^{1}
= 24 + 4 + 0.5
= (28.5)_{10}
(23.4)_{8} = 2×8^{1} + 3×8^{0} + 4×8^{1}
= 16 + 3 + 0.5
= (19.5)_{10}
Now,
(28.5)_{10} × (19.5)_{01} = (555.75)_{10}
Now,
(555.75)_{10} = ( ? )_{8}
To convert the integer part,
We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,
∴ (555.75)_{10} = (1053.6)_{8}
Question 36 
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 
2's complement of 1100 = 1100
Add = 1111
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
Question 37 
When multiplicand Y is multiplied by multiplier X = x_{n1}x_{n2} ...x_{0} using bitpair recoding in Booth's algorithm, partial products are generated according to the following table.
The partial products for rows 5 and 8 are
2Y and Y  
2Y and 2Y  
2Y and 0  
0 and Y 
⇒ 2Y and 0
Question 38 
(C012.25)_{H} – (10111001110.101)_{B} =
(135103.412)_{O}  
(564411.412)_{O}  
(564411.205)_{O}  
(135103.205)_{O} 
= 1100000000010010.00100101  0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)_{O}
Question 39 
The following bit pattern represents a floating point number in IEEE 754 single precision format
110000011101000000000000000000000The value of the number in decimal form is
10  
13  
26  
None of these 
Exponent bits  10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011  127
= 131  127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 2^{4} = (11010)_{2} = 26
Question 40 
A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2's complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be:
1, 1, 0  
1, 0, 0  
0, 1, 0  
1, 0, 1 
Carry flag = 1
Overflow flag = 0
Sign bit = 0 (MSB bit is 0)
Overflow flag:
In computer processors, the overflow flag is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation.
Question 41 
The two numbers given below are multiplied using the Booth's algorithm.
Multiplicand : 0101 1010 1110 1110 Multiplier: 0111 0111 1011 1101How many additions/Subtractions are required for the multiplication of the above two numbers?
6  
8  
10  
12 
Now we have some values defined for pair of bits in Booth’s Algorithm,
00 → 0
11 → 0
01 → 1
10 → 1
Now after adding 0 to the LSB of the multiplier, start traversing from left to right and accordingly put the values defined above.
Hence, total 8 additions / subtractions required.
Question 42 
R1 = 1011 and R2 = 1110  
R1 = 1100 and R2 = 1010  
R1 = 0011 and R2 = 0100  
R1 = 1001 and R2 = 1111 
Question 43 
Which one of the following is FALSE?
A + C = 0  
C = A + B  
B = 3 C  
( B  C ) > 0 
A= 12, B= +36 and C= +12
A+C= 0
B=3C
(BC)>0
C≠A+B
Question 44 
The exponent of a floatingpoint number is represented in excessN code so that:
The dynamic range is large.  
The precision is high.  
The smallest number is represented by all zeros.  
Overflow is avoided. 
Question 45 
If 73_{x} (in basex number system) is equal to 54_{y} (in basey number system), the possible values of x and y are
8, 16  
10, 12  
9, 13  
8, 11 
7x+3 = 5y+4
7x5y = 1
Only option (D) satisfies above equation.
Question 46 
What is the result of evaluating the following two expressions using threedigit floating point arithmetic with rounding?
(113. + 111.) + 7.51 113. + (111. + 7.51)
9.51 and 10.0 respectively
 
10.0 and 9.51 respectively
 
9.51 and 9.51 respectively  
10.0 and 10.0 respectively 
= (2) + 7.51
= 9.51 (✔️)
113. + (111. + 7.51)
= 113. + (103.51)
= 113. + 103
= 10 (✔️)
Question 47 
Let A = 1111 1010 and B = 0000 1010 be two 8bit 2's complement numbers. Their product in 2's complement is
1100 0100
 
1001 1100
 
1010 0101
 
1101 0101 
B = 0000 1010 = 10_{10} [2's complement number]
A×B = 6×10 =  60_{10}
⇒ 60_{10} = 10111100_{2}
= 11000011_{2} (1's complement)
= 11000100_{2} (2's complement)
Question 48 
Consider the following minterm expression for F:
F(P,Q,R,S) = Σ0,2,5,7,8,10,13,15
The minterms 2, 7, 8 and 13 are 'do not care' terms. The minimal sumofproducts form for F is:
Question 49 
Let ⊕ denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:
F(P,Q) = ((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0))
The equivalent expression for F is
P+Q  
P⨁Q  
⊕ is associative i.e P ⊕ (Q ⊕ R) = (P⊕Q) ⊕ R.
P ⊕ P = 0, 1 ⊕ P = P’ and 0 ⊕ Q = Q
(1 ⊕ P) ⊕ ((P ⊕ Q) ⊕ (P ⊕ Q)) ⊕ (Q ⊕ 0)
= P’⊕ (0) ⊕ Q
= P’ ⊕ Q
= (P ⊕ Q)’
Question 50 
Let X be the number of distinct 16bit integers in 2’s complement representation. Let Y be the number of distinct 16bit integers in sign magnitude representation.
Then XY is _________.
1  
2  
3  
4 
Since range is  2^{15} to 2^{15}  1
Y = 2^{16}  1
Here, +0 and 0 are represented separately.
X  Y = 2^{16}  (2^{16}  1)
= 1
Question 51 
110 
Question 52 
0x404C2EF4  
0x405C2EF4  
0xC15C2EF4  
0xC14C2EF4 
Question 53 
The 16bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is __________.
11  
12  
13  
14 
It is a negative number because MSB is 1.
Magnitude of 1111 1111 1111 0101 is 2’s complement of 1111 1111 1111 0101.
1111 1111 1111 0101
0000 0000 0000 1010 : 1’s Complement
0000 0000 0000 1011 : 2’s complement
= (11)_{10}
Hence, 1111 1111 1111 0101 = 11
Question 54 
The representation of the value of a 16bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is
136251  
736251  
571247  
136252 
Each hexadecimal digit is equal to a 4bit binary number. So convert
X = (BCA9)_{16} to binary
Divide the binary data into groups 3 bits each because each octal digit is represented by 3bit binary number.
X = (001 011 110 010 101 001)_{2}
Note: Two zeroes added at host significant position to make number bits of a multiple of 3 (16 + 2 = 18)
X = (136251)_{8}
Question 55 
Given the following binary number in 32bit (single precision) IEEE754 format:
The decimal value closest to this floatingpoint number is
1.45 × 10^{1}  
1.45 × 10^{1}  
2.27 × 10^{1}  
2.27 × 10^{1} 
For singleprecision floatingpoint representation decimal value is equal to (1)^{5} × 1.M × 2^{(E127)}
S = 0
E = (01111100)_{2} = (124).
So E – 127 =  3
1.M = 1.11011010…0
= 2^{0} + 2^{(1)} + 2^{(1)} + 2^{(4)} + 2^{(5)} + 2^{(7)}
= 1+0.5+0.25+0.06+0.03+0.007
≈ 1.847
(1)^{5} × 1.M × 2^{(E127)}
= 1^{0} × 1.847 × 2^{3}
≈ 0.231
≈ 2.3 × 10^{1}
Question 56 
Consider a quadratic equation x^{2}  13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b=________.
8  
9  
10  
11 
Generally if a, b are roots.
(x  a)(x  b) = 0
x^{2}  (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)_{b} + (6)_{b} = (13)_{b}
Convert them into decimal value
5_{b} = 5_{10}
6_{10} = 6_{10}
13_{b} = b+3
11 = b+3
b = 8
Now check with ab = 36
5_{b} × 6_{b} = 36_{b}
Convert them into decimals
5_{b} × 6_{b} = (b×3) + 6_{10}
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8