NumberSystems
Question 1 
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 2 
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 3 
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 4 
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 5 
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 6 
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 7 
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 8 
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 9 
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 10 
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 11 
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 13 
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 14 
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 15 
xy+x'y'  
x⊕y'  
x'⊕y  
x'⊕y' 
x’ ⊕ y’ = xy’ + x’y = x⊕y. Hence option D is correct.
Question 16 
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 17 
(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 18 
(1217)_{16}
 
(028F)_{16}
 
(2297)_{10}
 
(0B17)_{16} 
Divide the bits into groups, each containing 4 bits.
=(0010 1000 1111)_{2}
=(28F)_{16}
Question 19 
the normalized value 2^{127}  
the normalized value 2^{126}
 
the normalized value +0
 
the special value +0

Question 20 
decimal 10
 
decimal 11
 
decimal 10 and 11  
any value >2

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 21 
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 22 
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 23 
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 24 
(135103.412)_{O}  
(564411.412)_{O}  
(564411.205)_{O}  
(135103.205)_{O} 
= 1100000000010010.00100101  0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)_{O}
Question 25 
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 26 
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 27 
2Y and Y  
2Y and 2Y  
2Y and 0  
0 and Y 
⇒ 2Y and 0
Question 28 
 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 29 
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 30 
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 31 
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 32 
(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 33 
8, 16
 
10, 12  
9, 13
 
8, 11

7x+3 = 5y+4
7x5y = 1
Only option (D) satisfies above equation.
Question 34 
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 35 
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 36 
(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 37 
I and III  
I, II and III  
II and IV  
I, II, III and IV

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 38 
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 39 
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 40 
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 41 
1111  
11111  
111111  
10001 
15 = 11111
1's complement = 10000
2's complement = 10001
Question 42 
floating point multiplication  
signed 16 bit integer addition  
arithmetic left shift  
converting a signed integer from one size to another 
Question 43 
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 45 
01010101  
11010101  
00101011  
10101011 
Question 46 
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 47 
0 – 100 + 1000  
0 – 100 + 100 1  
0 – 1 + 100 – 10 + 1  
00 – 10 + 100  1 
Question 48 
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 49 
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 50 
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 52 
0 to 1  
0.5 to 1  
2^{23} to 0.5  
0.5 to (12^{23}) 
Question 54 
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 55 
(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 56 
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 57 
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 58 
9 
(9753)_{16}
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
Question 59 
c_{1} = b_{1}a_{0} ⊕ a_{1}b_{0} 
⇒ c_{1} = b_{1}a_{0} ⊕ a_{1}b_{0}
Question 60 
The dynamic range is large.  
The precision is high.  
The smallest number is represented by all zeros.  
Overflow is avoided. 