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 
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 
(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 51 
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 52 
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 53 
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 54 
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 55 
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.