NumberSystems
Question 1 
In 16bit 2's complement representation, the decimal number 28 is:
1111 1111 1110 0100  
1111 1111 0001 1100  
0000 0000 1110 0100  
1000 0000 1110 0100 
1’s complement = 1111 1111 1110 0011
2’s complement = 1’s complement + 1
2’s complement = 1111 1111 1110 0100 = (28)
Question 2 
Two numbers are chosen independently and uniformly at random from the set {1, 2, ..., 13}. The probability (rounded off to 3 decimal places) that their 4bit (unsigned) binary representations have the same most significant bit is ______.
0.502  
0.461  
0.402  
0.561 
1  0001
2  0010
3  0011
4  0100
5  0101
6  0110
7  0111
8  1000
9  1001
10  1010
11  1011
12  1100
13  1101
The probability that their 4bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
Question 3 
Consider Z = X  Y, where X, Y and Z are all in signmagnitude form. X and Y are each represented in n bits. To avoid overflow, the representation of Z would require a minimum of:
n bits  
n + 2 bits  
n  1 bits  
n + 1 bits 
To store overflow/carry bit there should be extra space to accommodate it.
Hence, Z should be n+1 bits.
Question 4 
Consider the unsigned 8bit fixed point binary number representation below,

b_{7}b_{6}b_{5}b_{4}b_{3} ⋅ b_{2}b_{1}b_{0}
where the position of the binary point is between b_{3} and b_{2} . Assume b_{7} is the most significant bit. Some of the decimal numbers listed below cannot be represented exactly in the above representation:

(i) 31.500 (ii) 0.875 (iii) 12.100 (iv) 3.001
Which one of the following statements is true?
None of (i), (ii), (iii), (iv) can be exactly represented
 
Only (ii) cannot be exactly represented  
Only (iii) and (iv) cannot be exactly represented  
Only (i) and (ii) cannot be exactly represented 
= 16 + 8 + 4 + 2 + 1 + 0.5
= (31.5)_{10}
(ii) (0.875)_{10} = (00000.111)_{2}
= 2^{1} + 2^{2} + 2^{3}
= 0.5 + 0.25 + 0.125
= (0.875)_{10}
(iii) (12.100)_{10}
It is not possible to represent (12.100)_{10}
(iv) (3.001)_{10} It is not possible to represent (3.001)_{10}
Question 5 
The nbit fixedpoint representation of an unsigned real number X uses f bits for the fraction part. Let i = nf. The range of decimal values for X in this representation is
2^{f} to 2^{i}  
2^{f} to (2^{i}  2^{f})  
0 to 2^{i}  
0 to (2^{i}  2^{f }) 
Number of bits in fraction part → fbits
Number of bits in integer part → (n – f) bits
Minimum value:
000…0.000…0 = 0
Maximum value:
= (2^{ nf }  1) + (1  2 ^{f}
= (2^{nf}  2 ^{f})
= (2^{i}  2 ^{ f })
Question 6 
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 7 
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 8 
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
Question 9 
Consider a binary code that consists of only four valid code words as given below:
Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are
p=3 and q=1  
p=3 and q=2  
p=4 and q=1  
p=4 and q=2 
Minimum Distance = p = 3
Error bits that can be corrected = (p1)/2 = (31)/2 = 1
∴ p=3 and q=1
Question 10 
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 11 
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 12 
The base (or radix) of the number system such that the following equation holds is_________.
312/20 = 13.1
5  
6  
7  
8 
(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
Therefor r = 5
Question 13 
Consider the equation (123)_{5} = (x8)_{y} with x and y as unknown. The number of possible solutions is __________.
3  
5  
6  
7 
(123)_{5} = (x8)_{y}
In R.H.S. since y is base so y should be greater than x and 8, i.e.,
y > x
y > 8
Now, to solve let's change all the above bases number into base 10 number,
5^{2} × 1 +2 × 5 + 3 = y × x + 8
38 = xy + 8
xy = 30
⇒ yx = 30
So the possible combinations are
(1,30), (2,15), (3,10), (5,6)
But we will reject (5,6) because it violates the condition (y > 8).
So, total solutions possible is 3.
Question 14 
The value of a float type variable is represented using the singleprecision 32bit floating point format of IEEE754 standard that uses 1 bit for sign, 8 bits for biased exponent and 23 bits for mantissa. A float type variable X is assigned the decimal value of −14.25. The representation of X in hexadecimal notation is
C1640000H  
416C0000H  
41640000H  
C16C0000H 
(14.25)_{10} = 1110.01000
= 1.11001000 x 2^{3}
23 bit Mantissa = 11001000000000000000000
Biased Exponent = exponent + bias
= 3 + 127 = 130 = 1000 0010
(14.25) in 32bit IEEE754 floating point representation is
1 10000010 11001000000000000000000
= 1100 0001 0110 0100 0000 0000 000 0000
= (C 1 6 4 0 0 0 0)_{16}
Question 15 
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 16 
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 17 
The smallest integer that can be represented by an 8bit number in 2’s complement form is
256  
128  
127  
0 
The smallest 8bit 2’s complement number is 1000 0000.
MSB is 1. So it is a negative number.
To know the magnitude again take 2’s complement of 1000 0000.
1000 0000
0111 1111 ← 1’s complement
1000 0000 ← 2’s complement (1’s complement +1)
= 128
128 is 1000 0000 in 2’s complement representation.
Question 18 
Which one of the following expressions does NOT represent exclusive NOR of x and y?
xy+x'y'  
x⊕y'  
x'⊕y  
x'⊕y' 
x’ ⊕ y’ = xy’ + x’y = x⊕y. Hence option D is correct.
Question 19 
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 20 
P is a 16bit signed integer. The 2's complement representation of P is (F87B)_{16}. The 2's complement representation of 8*P is
(C3D8)_{16}  
(187B)_{16}  
(F878)_{16}  
(987B)_{16} 
(F87B)_{16}=(1111 1000 0111 1011)_{2}. (It is a negative number which is in 2's complement form)
P = 1111 1000 0111 1011 (2's complement form)
8 * P = 2^{3}* P = 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2^{k})
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)_{16}.
Question 21 
(1217)_{8} is equivalent to
(1217)_{16}
 
(028F)_{16}  
(2297)_{10}  
(0B17)_{16} 
Divide the bits into groups, each containing 4 bits.
= (0010 1000 1111)_{2}
= (28F)_{16}
Question 22 
In the IEEE floating point representation, the hexadecimal value 0×00000000 corresponds to
the normalized value 2^{  127}  
the normalized value 2^{  126}  
the normalized value + 0  
the special value + 0 
Question 23 
Let r denote number system radix. The only value(s) of r that satisfy the equation is/are
decimal 10  
decimal 11  
decimal 10 and 11  
any value > 2 
(r^{2} + 2r + 1)^{1/2} = r + 1
(r + 1)^{2} * 1/2 = r + 1
r + 1 = r + 1 Any value of r will satisfy the above equation. But the radix should be greater than 2 because the 121 has 2. So r > 2 is correct.
Question 24 
Consider numbers represented in 4bit gray code. Let h_{3}h_{2}h_{1}h_{0} be the gray code representation of a number n and let g_{3}g_{2}g_{1}g_{0} be the gray code of (n+1)(modulo 16) value of the number. Which one of the following functions is correct?
g_{0}(h_{3}h_{2}h_{1}h_{0}) = Σ(1,2,3,6,10,13,14,15)  
g_{1}(h_{3}h_{2}h_{1}h_{0}) = Σ(4,9,10,11,12,13,14,15)  
g_{2}(h_{3}h_{2}h_{1}h_{0}) = Σ(2,4,5,6,7,12,13,15)  
g_{3}(h_{3}h_{2}h_{1}h_{0}) = Σ(0,1,6,7,10,11,12,13)

g_{2}(h_{3}h_{2}h_{1}h_{0}) = Σ(2,4,5,6,7,12,13,15)
Question 25 
The range of integers that can be represented by an n bit 2's complement number system is:
 2^{n1} to (2^{n1}  1)  
 (2^{n1}  1) to (2^{n1}  1)  
 2^{n1} to 2^{n1}
 
 (2^{n1} + 1) to (2^{n1}  1) 
The smallest (negative) n bit number is 100..0 (i.e., 1 followed by n1 zeros) which is equal to  2^{n1}.
1000...00
0111...11 < 1’s complement
1000..00 < 2’s complement
=  2^{n1}
Question 26 
The hexadecimal representation of 657_{8} is
1AF  
D78  
D71  
32F 
Make 3 zeros on the left side so that the number of bits is multiple of 4.
= (0001 1010 1111)_{2}
= (1 A F)_{16}
Question 27 
Consider the following floating point format.
Mantissa is a pure fraction in signmagnitude form.
The decimal number 0.239 × 2^{13} has the following hexadecimal representation (without normalization and rounding off:
0D 24  
0D 4D  
4D 0D  
4D 3D 
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)_{2}
Bias = 64. So biased exponent is 13+64 = 77= (1001101)_{2}
0.239 × 2^{13} = 0 1001101 00111101
= 0100 1101 0011 1101
= 4 D 3 D
Question 28 
Consider the following floating point format.
Mantissa is a pure fraction in signmagnitude form.
The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0's are padded in while shifting a field. The normalized representation of the above number (0.239 × 2^{13}) is
0A 20  
11 34  
4D D0
 
4A E8 
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)_{2}
0.239 × 2^{13} = 1.11101000 x 2^{10} < Normalized Mantissa
Bias = 64. So biased exponent is 10+64 = 74 = (1001010)_{2}
0.239 × 2^{13} = 0 1001010 11101000
= 0100 1010 1110 1000
= (4 A E 8)_{16}
Question 29 
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 30 
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 31 
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 32 
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 33 
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 34 
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 35 
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 36 
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 37 
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 38 
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 39 
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 40 
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 41 
The number 43 in 2’s complement representation is
01010101  
11010101  
00101011  
10101011 
Question 42 
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 43 
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 44 
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 45 
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 46 
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 47 
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 48 
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 49 
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 50 
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 51 
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 52 
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 53 
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 54 
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 55 
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 56 
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 57 
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 58 
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 59 
(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 60 
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 61 
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 62 
(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 63 
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 64 
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 65 
3+n ternary digits  
2n/3 ternary digits  
n(log_{2}3) ternary digits  
n(log_{3}2 ) ternary digits 
→ But in question they are given ternary numbers, it means 3x1.
→ Both will take different no. of bits to represent the same number.
3^{x} 1 = 2^{n} 1
3^{x} = 2^{n}
Apply log on both side
x= log_{3}( 2^{n})
x=n*log_{3}2 .
Question 66 
subtract 0011 from the sum  
add 0011 to the sum
 
subtract 0110 from the sum
 
add 0110 to the sum

Example:
x+3
y+3

(x+y+6)
Here, sum is excess6. Hence, subtract 0011 to make it excess3.
Question 67 
8  
9  
10  
12 
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.
Question 68 
0.60  
0.52  
0.54  
0.50 
= (0.6)_{8}
Option (A) is correct.
Question 69 
1 × 10^{128} and 2^{15}× 10^{15}
 
1 × 10^{256} and 2^{15}× 10^{255}  
1 × 10^{128} and 2^{15}× 10^{127}  
1 × 10^{128}and 2^{15}– 1 × 10^{127} 
According to question 16 bit mantissa and 8 bit Exponent.
Since the mantissa is always 1.xxxxxxxxx in the normalised form, no need to represent the leading 1.
Single Precision: mantissa ===> 1 bit + 15 bits
The largest mantissa value value is 2^{15}1 (one bit meant for sign)
The largest exponent value is 2^{7}1=127
The smallest mantissa value is 0000 0000 0000 0000(one bit is always 1) =1
The Smallest (largest negative) exponent value is 1111 1111 (which is 2’s complement form) 2^{8}=128
Question 70 
CD73E  
ABD3F  
7CDE3  
FA4CD 
7 C D E 3
(7CDE3)16
Question 71 
4  
5  
6  
7 
Question 72 
[log(n)] + 1 bits  
[log (n1)) + 1 bits  
[log (n+1)] + 1 bits  
None of the above 
Question 73 
10  
8  
6  
5 
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 74 
n bits  
(n+3) bits  
(n+2) bits  
(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}
Question 75 
(1217)_{16}  
(028F)_{16}  
(2297)_{1o}  
(0B17)_{16} 
=(0010 1000 1111)_{2}
=(2 8 F)_{16}
Question 76 
Excess3 code  
Gray code  
BCD code  
Hamming Code 
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
Question 77 
224174  
425174  
6173  
225174 
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 78 
Cyclic Redundancy Code  
Weighted Code  
SelfComplementing Code  
Algebraic Code 
→ 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 79 
1010  
0101  
1000  
1001 
(Decimal value of maximum 4bit number +1 )/2= (15+1)/2=8
Question 80 
0001 and an overflow
 
1001 and no overflow
 
0001 and no overflow  
1001 and an overflow

Another number is 1101(3 is decimal value)
Adding of 3 and 4, the result is 1 and there is no overflow
Question 81 
28  
15  
26  
28 
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 82 
5  
6  
7  
8 
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 83 
2^{12}  
2^{12}1  
10^{12}  
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 84 
To guarantee correction of upto t errors, the minimum Hamming distance dmin in a block code must be
t+1  
t−2
 
2t−1  
2t+1 
Question 85 
The hexadecimal equivalent of the binary integer number 110101101 is :
D 2 4
 
1 B D
 
1 A E  
1 A D 
2^{4} = 16
So, 4bits in binary will represent one integer in Hexadecimal.
So,
Question 86 
128 to +127  
128 to +128  
999999 + +999999  
none of these 
Question 87 
Cyclomatic redundancy code  
Weighted code  
Self complementing code  
algebraic code 
Question 88 
In computers, subtraction is generally carried out by
1’s complement
 
10’s complement
 
2’s complement
 
9’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.
Question 89 
30  
60  
90  
120 
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Question 90 
75 _{16}  
65 _{16}  
5E _{16}  
10 _{16} 
(FD) _{16</sub> = (253) 10 (88) 16</sub> = (136) 10 Step2: Perform subtraction 253136=(117) 10 Step3: Convert (117) 10 =(75) 16}
Question 91 
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 92 
1001  
1010  
1011  
1100 
12 equivalent binary number is 1100.
Question 93 
31845  
41117  
41052  
32546 
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 94 
2.5  
3.0  
3.5  
4.5 
→ 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 95 
– 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 96 
ASCII  
BCD  
EBCDIC  
Gray 
→ 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 97 
8  
9  
10  
12 
=(12,288+3840+80+3)
=(16211)_{10}
=(0011111101010011)_{2}
Total number of 1’s in binary representation is 10.
Question 98 
200  
213  
246  
277 
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 99 
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 
= 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 100 
50  
74  
40  
60  
None of these 
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 101 
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 
(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 102 
What is the hexadecimal representation of the decimal number 8537?
(2059)_{16}
 
(2159)_{16}
 
(2195)_{16}
 
(2157)_{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 103 
Division by 2  
Addition by 2
 
Multiplication by 2  
Subtraction by 2 
→ 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 104 
1111 1111 1110 1011  
1111 1111 1110 1001  
1111 1111 1110 0111  
1111 1111 1110 1000 
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 105 
50  
74  
40  
None of the above 
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 106 
Octal code  
Binary Code  
Gray code  
Excess3 Code 
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Question 107 
32,64 and 80  
32,64 and 128  
16,32 and 64  
16,32 and 80 
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 108 
26  
36  
46  
56 
→ 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 109 
Which of the following is a recursive algorithm to convert a positive decimal integers into equivalent binary integers?
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 110 
1  
2  
3  
4 
→ In the question, the unique digits are 0,1 and X(possible x value is 2) then number system is 3;
Question 111 
0000000  
1111111  
2222222  
12121212 
● The data is 0000 and Hamming(7,4) transmitted is 0000000
Question 112 
16  
10  
12  
13 
So, total 13 bits required to represent 4096 decimal number.
Question 113 
The signed 2’s complement representation of 33 is:
11011111
 
00100001  
01011111
 
10100001

Step2: Perform 2’s complement.
Question 114 
Using signed 2’s complement subtraction the result of 1111101011110011 is:
10000111  
00000111  
10001101  
00001101 
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 115 
32 bits  
8 bits  
16 bits  
24 bits 
→ Sign bit: 1 bit
→ Exponent width: 8 bits
→ Significand precision: 24 bits (23 explicitly stored)
Question 116 
(10208)_{10}  
(1276)_{10}  
(2374)_{10}  
(1272)_{10} 
Question 117 
01000110111111011100000000000000  
01100110111111011100000000000000  
11100110111111011100000000000000  
01000111011111011100000000000000 
3.248x10^{4} =32480= 1111110111
= 1.111110111 x 2^{14}
Mantissa = 11111011100...00
Biased exponent = 14 +127= 141 = 10001101
Question 118 
46252  
11450  
11452  
45250 
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 119 
7353  
1353  
5651  
5657 
1 011 101 011
1 3 5 3
(1011101011)_{ 2} = (1353)_{ 8}
Question 120 
4AC8  
4CA8  
CCA8  
4CA4 
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 121 
(635)_{ 4}  
(32312) _{4}  
(21323) _{4}  
(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 122 
0.5100098  
0.2096  
0.52  
0.4192 
Question 123 
2EE  
2FF  
4EF  
4FE 
(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 124 
11....11  
00.....00  
100.....0  
000......1 
→ 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 125 
(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} 
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 126 
Signmagnitude  
1’s complement  
2’s complement  
9’s complement 
Question 127 
16  
32  
48  
64 
→ In the IEEE 7542008 standard, the 64bit base2 format is officially referred to as binary64 called double in IEEE 7541985.
→ IEEE 754 specifies additional floatingpoint formats, including 32bit base2 single precision and, more recently, base10 representations.
Question 128 
1100  
1001  
1000  
1111 
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 129 
100101100  
1110001110001  
11110011  
10101010101010 
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 130 
4  
8  
10  
16 
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 131 
1010  
0101  
1000  
1001 
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 132 
150  
241  
142  
101011110 
(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 133 
8421 code  
Gray code  
Excess3 code  
7421 code 
→ 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 134 
8  
9  
10  
11 
(2027) _{10} = (111 1110 1011)_{ 2}
Here, total number of 1’s are 9.
Question 135 
0101  
0111  
1011  
1000 
= 6*1 + 4*0 + 2*1 + 3*1
= 6+23
=5
Question 136 
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 
→ 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 137 
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 138 
0xC2206000  
0xC2006666  
0xC2006000  
0xC2206666 
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 139 
8 bits and 16 bits  
16 bits and 32 bits  
32 bits and 64 bits
 
64 bits and 128 bits 
Question 140 
2  
8  
10  
None of the above 
Question 141 
2^{16} – 1  
2^{15} – 1  
2^{16}  
2^{15} 
= 2^{15} 1
Question 142 
2^{16} – 1  
2^{15} – 1  
2^{16}  
2^{15} 
= 2^{15} 1
Question 143 
signed magnitude form  
1’s complement form  
2’s complement form  
none of the above 
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
Question 144 
0101  
0101  
0110  
1001 
Question 145 
2 ^{16} − 1  
2 ^{15} − 1  
2 ^{16}  
2 ^{15} 
Question 146 
3F2.89  
3F2.98  
2F3.89  
2F3.98 
→ 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 147 
01001101.0100  
01001101.0010  
10110010.1011  
10110010.1101 
Question 148 
ASCII  
Gray Code  
Excess3 Code  
BCD 
Question 149 
111  
10  
110  
11 
Question 150 
42  
22  
21  
106 
Question 151 
8421  
Excess3  
2421  
3210 
Question 152 
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 153 
2095.75  
2095.075  
2095.937  
2095.0937 
Question 154 
01001001.1100  
11001001.1100  
10110110.0100  
10110110.1100 
Question 155 
0011110  
0101110  
0010110  
0011100 
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 156 
22.0  
22.2  
20.2  
21.2  
None of the above 
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 157 
t  
2t1  
2t  
2t+1 
Question 158 
47.21  
12.74  
12.71  
17.21  
None of the above 
Question 159 
150  
241  
142  
101011110 
Question 160 
8  
9  
10  
11 
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 161 
t1  
t  
t+1  
2t+1 
Question 162 
2t+1  
2t  
2t1  
t1 
Question 163 
100  
150  
377  
737 
(FF)_{16} = (1111 1111)_{2}
Step2: Divide binary number into 3 segments from LSB(Least significant bit).
Question 164 
00011100  
10011101  
10011100  
11100100 
Question 165 
8  
9  
10  
11 
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 166 
t + 1  
t  
t  2  
t / 2 
Question 167 
8  
7  
10  
16 
Substitute 7 in b,
(146)_{7}+(317)_{72}=(246)_{8}
(146)_{7}+(317)_{5}=(246)_{8}
(146)_{7}=1*7^{2}+4*7^{1}+6*7^{0}
=49+28+6
=83
(317)_{5}=3*5^{2}+1*5^{1}+7*5^{0}
=75+5+7
= 87
(246)_{8}=2*8^{2}+4*8^{1}+6*8^{0}
=128+32+6
= 166
∴ (146)_{7}+(317)_{5}=(246)_{8}
=83+87
=166
166=166
LHS = RHS equal only if b is 7.
Question 168 
320  
480  
640  
768 
→ Number of bit strings of length 10 that end with 00: 2^{8 }= 256
→ Number of bit strings of length 10 that start with 1 and end with 00: 2^{7 }= 128
→ Applying the subtraction rule, the number is 512+256128 = 640
Question 169 
10011001  
00011001  
00111000  
11000110 
So data bits are:
00011001
Question 170 
FADED  
AEOBE  
ADOBE  
ACABE 
Question 171 
2^{ r}  1  
2 ^{r}  r  1  
2 ^{r}  r + 1  
2 ^{r} + r  1 
where p is the no. of parity bits.
m = no. of msg. digits
No. of check bits in a msg = p
In question no. of checkbits are given as ‘r’
So r = p
Put it in equation (1)
2 ^{r} = m + r + 1
m = 2^{ r} − r − 1
Question 172 
(a) 8421 code
(b) 2421 code
(c) excess3 code
(d) excess3 gray code
Choose the correct option:
(a) and (b)  
(b) and (c)  
(c) and (d)  
(d) and (a) 
Note: 8421 is not self complementing. Self complementing is nothing but reverse order from starting and last number.
Ex: 0 value is 0000
9 value is 1111
So, it is self complementing.
Note: Excess3 and Gray code is different. So, it is not relevant.
Question 173 
Let the population of chromosomes in genetic algorithm is represented in terms of binary number. The strength of fitness of a chromosome in decimal form, x, is given by
The population is given by P where:
P = {(01101, (11000), (01000), (10011)}
The strength of fitness of chromosome (11000) is ___________
24  
576  
14.4  
49.2 
Question 174 
The following program is stored in the memory unit of the basic computer. Give the content of accumulator register in hexadecimal after the execution of the program.
A1B4  
81B4  
A184  
8184 
Question 175 
Given following equation:
(142)_{b} + (112)_{b2} = (75)_{8}. Find base b.
3  
6  
7  
5 
OptionB:
Step1: Converting (142)_{5 } to base 10.
1*5^{2}=25
4*5^{1}=20
2*5^{0}=2
Adding all to get (47)_{10}
Step2: Converting (112)_{3} to base 10.
1*3^{2}=9
1*3^{1}=3
2*3^{0}=2
Adding all to get (14)_{10}
Step3: Convert (61)_{10} to (?)_{8} // 47+14=61. Both are in decimal so we can add directly.
8_61
8_7 → 5
8_7 → 7
Ans:(75)_{8}
No need to check OptionC & D.
Question 176 
12; 9  
7; 12  
9; 12  
12; 7 
Total 9 digits of binary number required to convert decimal number
Question 177 
What will be the minimum Hamming distance for the following coding scheme?
1  
2  
3  
4 
The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. Note that the Hamming distance is a value greater than zero
The minimum hamming distance is between the words 10 and 11
10101 exor 11100 = 01001 which is “2”.
Question 178 
Biquinary code  
BCD code  
2421 code  
Excess3 code 
The term biquinary indicates that the code comprises both a twostate (bi) and a fivestate (quinary) component.
Question 179 
Address Data
0 X 104 78
0 X 103 56
0 X 102 34
0 X 101 12
0 X 104 12 0 X 103 56 0 X 102 34 0 X 101 12  
0 X 104 12 0 X 103 34 0 X 102 56 0 X 101 78  
0 X 104 56 0 X 103 78 0 X 102 12 0 X 101 34  
0 X 104 56 0 X 103 12 0 X 102 78 0 X 101 34 
In the given question each integer is two bytes. 0x101, 0x102 are part of one word.
In the big endian 0x101 has 12 and 0x102 has 34. In the little endian 0x101 will have 34 and 0x102 will have 12.
Similarly in the big endian 0x103 has 56 and 0x104 has 78...in the little endian 0x103 will have 78 and 0x104 will have 56.
Question 180 
Gates required for parity generation differ from those required for parity checking  
ExclusiveOR gates  
RSflipflop  
Counters 
Question 181 
2  
7  
3  
5 
Question 182 
Logical shift right  
Logical shift left  
Arithmetic shift right  
Arithmetic shift left 
Question 183 
The gray code for decimal number 6 is equivalent
1100  
1001  
0101  
0110 
The binary value b3, b2, b1, b0 and find the gray code g3, g2, g1, g0 based on the below concept, g3 = b3 = 0
g2 = b3 XOR b2 = 0 xor 1 = 1
g1 = b2 XOR b1 = 1 xor 1 = 0
g0 = b1 XOR b0 = 1 xor 0 = 1
Question 184 
Excess3 code of decimal number 2 is:
0101  
1010
 
1100
 
0011 
2+3 = 5
Binary form of 5 is 0101.
Question 185 
MSB stands for
Middle significant bit
 
Measured Significant bit  
Maximum significant bit
 
Most significant bit

LSB stands for Least Significant Bit. It is the right most bit of any binary string.
Question 186 
Which is the correct method to obtain 10's complement of a decimal number?
The minuend is added to the 10's complement of the subtrahend and carry is dropped  
The 10's complement of a decimal number is equal to 9's complement of given number minus one
 
The subtrahend is added to the 10's complement of the minuend and carry is dropped  
The 10's complement of a decimal number is equal to 9's complement of given number plus one 
Question 187 
The multiplicand register and multiplier register of hardware circuit implementation of booth's algorithm have (11101) and (1100). the result shall be:
(812)_{10}  
(812)_{10}
 
(12)_{10}
 
(12)_{10}

Hence the decimal value of multiplicand,
11101 = 101 = 3
The decimal value of multiplier,
1100 = 100 = 4
Hence the required result is,
3 × 4 = 12
Question 188 
The number 105 in decimal system is equal to
(140)_{5}  
(410)_{5}
 
(042)_{5}
 
(321)_{5} 
∴ Answer is (410)_{5}
Question 189 
Detect d bit errors and correct (d – 1) bit errors.  
Detect (d – 1) bit errors only with no error correction.  
Detect (d – 1) bit errors and correct (d – 1)/2 bit errors.  
Detect and correct all d bit errors. 
So using above solution most appropriate answer is option C.
Question 190 
1110 0011 01  
1110 1100 10  
1110 1010 10  
1110 0011 00 
13 × 6 = 78
Let’s first find binary value of 78,
Now to get 78 let’s take 2’s complement of above binary no.,
10110010
which can be also written as,
1110110010
Question 191 
0.1  
0.2  
0.4  
0.5  
All the above 
Question 192 
the number of possible Gray codes is even  
the number of possible Gray codes is odd  
in any Gray code, if two strings are separated by k other strings in the ordering, then they must diﬀer in exactly k +1 bits  
in any Gray code, if two strings are separated by k other strings in the ordering, then they must diﬀer in exactly k bits  
none of the above 
Question 193 
3.5  
14  
7/8  
2 
Question 194 
00000000  
10101110  
01000000  
10000000  
11000000 
Question 195 
{n ∈ N : n ≤ 182 and n has exactly four ones in its binary representation}
91  
70  
54  
35  
27 