## Number-Systems

 Question 1
Consider the following representation of a number in IEEE 754 single-precision floating point format with a
bias of 127.
S : 1      E : 10000001      F : 11110000000000000000000
Here S, E and F denote the sign, exponent and fraction components of the floating point representation.
The decimal value corresponding to the above representation (rounded to 2 decimal places) is _______
 A -7.75
Question 1 Explanation:

Sign bit S= 1. The given number is a negative number.

Biased Exponent E = 27 + 1= 129

Actual Exponent e = E-127

= 129- 127

= 2

The decimal value= (-1)s x 1.M x 2e

= (-1) 1 x 1.1111 x 22

= - (111.11)

= - (7 + 0.75)

= -7.7

 Question 2
Let the representation of a number in base 3 be 210. What is the hexadecimal representation of the number?
 A 21 B 528 C D2 D 15
Question 2 Explanation:

On converting (210)3 in decimal, we will get:=>

2*32+1*3=2*9+3=2110

=>(15)16

 Question 3

Which one of the following choices gives the correct values of x and y?
 A x is 1 and y is 1 B x is 0 and y is 1 C x is 1 and y is 0 D x is 0 and y is 0
Question 3 Explanation:

C2 checks the bits d1, d3, d4, d6, d7.

C2=1, d1= 1, d3= 1, d4= 0, d6= 0, d7= 1.

The number of 1s is even. So, even parity is used in this problem.

C1 checks the bits d1, d2, d4, d5, d7.

C1=0, d1= 1, d2= 0, d4= 0, d5= x, d7= 1.

As the parity used is even parity, the value of d5 should be 0.

x=d5=0

C8 checks the bitsa d5, d6, d7, d8.

C8=y, d5= x=0, d6= 0, d7= 1, d8= 1.

As the parity used is even parity, the value of C8 should be 0.

C8=y=0.

x=y=0.

 Question 4

Consider n-bit (including sign bit) 2’s complement representation of integer number. The range of integer values, N, that can be represented is _________ ≤ N ≤ _________

 A -2n-1 to 2n-1 - 1
 Question 5

Write a program in 8085 Assembly language to Add two 16-bit unsigned BCD(8-4-2-1 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)

 A Theory Explanation.
 Question 6

The number of 1’s in the binary representation of
(3*4096 + 15*256 + 5*16 + 3) are:

 A 8 B 8 C 10 D 12
Question 6 Explanation:
3 × 4096 = 3 × 212
= (11000000000000)2
15 × 256 = 15 × 28
= (111100000000)2
5 × 16 = 5 × 24
= (1010000)2
3 = (11)2
Hence, all binary numbers,

∴ 101's
 Question 7

(a) An asynchronous serial communication controller that uses a start stop scheme for controlling the serial I/O of a system is programmed for a string of length seven bits, one parity bit (odd parity) and one step bit. The transmission rate is 1200 bits/second.
(i) What is the complete bit stream that is transmitted for the string ‘0110101’?
(ii) How many such strings can be transmitted per second?

(b) Consider a CRT display that has a text mode display format of 80 × 25 characters with a 9 × 12 character cell. What is the size of the video buffer RAM for the display to be used in monochrome (1 bit per pixel) graphics mode?

 A Theory Explanation.
 Question 8

Consider three registers R1, R2 and R3 that store numbers in IEEE-754 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?

 A 0x40800000 B 0x83400000 C 0xC8500000 D 0xC0800000
Question 8 Explanation:
Given numbers are 0x42200000 and 0xC1200000 which are stored in the registers R1 and R2, respectively.

R1 = 1.0100..0 X 2132-127
= 1.0100..0 X 25
= 101.0 X 23
= 5 X 8
= 40

R2 = (-1) x 1.0100..0 X 2130-127
= (-1) x 1.0100..0 X 23
= (-1) x 101.0 X 21
= (-1) x5 X 2
= -10
R3 = R1/R2
= -4
= (-1)x 1.0 x 22
Sign = 1
Mantissa = 000..0
Exponent = 2+127 = 129

R3 = 1100 0000 1000 000..0
= 0x C 0 8 0 0 0 0 0
 Question 9

The decimal value 0.5 in IEEE single precision floating point representation has

 A fraction bits of 000…000 and exponent value of 0 B fraction bits of 000…000 and exponent value of −1 C fraction bits of 100…000 and exponent value of 0 D no exact representation
Question 9 Explanation:
(0.5)10 = (1.0)2 × 2–1
So, value of the exponent = -1
and
fraction is 000…000 (Implicit representation)
 Question 10

Consider the following floating point number representation

The exponent is in 2's complement representation and mantissa is in the sign magnitude representation. The range of the magnitude of the normalized numbers in this representation is

 A 0 to 1 B 0.5 to 1 C 2-23 to 0.5 D 0.5 to (1-2-23)
Question 10 Explanation:
Maximum value of mantissa will be 23, is where a decimal point is assumed before first 1. So the value is 1 - 2-23.
 Question 11

Given √224)r = 13)r.
The value of the radix r is:

 A 10 B 8 C 5 D 6
Question 11 Explanation:
(√224)r = (13)r
Convert r base to decimal.
√2r2 + 25 + 4 = r + 3
Take square both sides,
2r2 + 2r + 4 = r2 + 6r + 9
r2 - 4r - 5 = 0
r2 - 5r + r - 5 = 0
r(r - 5) + (r - 5) = 0
r = -1, 5
r cannot be -1,
So r = 5 is correct answer.
 Question 12

The octal representation of an integer is (342)8. If this were to be treated as an eight-bit integer is an 8085 based computer, its decimal equivalent is

 A 226 B -98 C 76 D -30
Question 12 Explanation:
(342)8 = (011 100 010)2 = (1110 0010)2
If this can be treated as 8 bit integer, then the first becomes sign bit i.e., '1' then the number is negative.
8085 uses 2's complement then

⇒ -30
 Question 13

Suppose the domain set of an attribute consists of signed four digit numbers. What is the percentage of reduction in storage space of this attribute if  it is stored as an integer rather than in character form?

 A 80% B 20% C 60% D 40%
Question 13 Explanation:
We assume byte addressable memory - nothing smaller than a byte can be used.
We have four digits. So to represent signed 4 digit numbers we need 5 bytes, 4 bytes for four digits and 1 for the sign.
So required memory = 5 bytes.
Now, if we use integer, the largest no. needed to represent is 9999 and this requires 2 bytes of memory for signed representation.
9999 in binary requires 14 bits. So, 2 bits remaining and 1 we can use for sign bit.
So, memory savings,
= 5 - 2/5 × 100
= 60%
 Question 14

Booth’s coding in 8 bits for the decimal number –57 is

 A 0 – 100 + 1000 B 0 – 100 + 100 - 1 C 0 – 1 + 100 – 10 + 1 D 00 – 10 + 100 - 1
Question 14 Explanation:
 Question 15

Zero has two representations in

 A Sign magnitude B 1’s complement C 2’s complement D None of the above E Both A and B
Question 15 Explanation:
Sign magnitude:
+0 = 0000
-0 = 1000
1's complement:
+0 = 0000
-0 = 1111
 Question 16

The number 43 in 2’s complement representation is

 A 01010101 B 11010101 C 00101011 D 10101011
Question 16 Explanation:
Positive integers are represented in its normal binary form while negative numbers are represented in its 2′s complement form. Binary representation of 43 is 00101011.
 Question 17

Consider the values A = 2.0 x 1030, B = -2.0 x 1030, C = 1.0, and the sequence

```             X: = A + B    Y: = A + C
X: = X + C    Y: = Y + B  ```

executed on a computer where floating-point numbers are represented with 32 bits. The values for X and Y will be

 A X = 1.0, Y = 1.0 B X = 1.0, Y = 0.0 C X = 0.0, Y = 1.0 D X = 0.0, Y = 0.0
Question 17 Explanation:
Given: 32 bits representation. So, the maximum precision can be 32 bits (In 32-bit IEEE representation, maximum precision is 24 bits but we take best case here). This means approximately 10 digits.
A = 2.0 * 1030, C = 1.0
So, A + C should make the 31st digit to 1, which is surely outside the precision level of A (it is 31st digit and not 31st bit). So, this addition will just return the value of A which will be assigned to Y.
So, Y + B will return 0.0 while X + C will return 1.0.
 Question 18

The 2’s complement representation of (-539)10 in hexadecimal is

 A ABE B DBC C DE5 D 9E7
Question 18 Explanation:
(539)10 = (0010 0001 1011)2
For (-539)10 = (1101 1110 0100)2
1's complement = (1101 1110 0100)2
2's complement = (1101 1110 0101)2
= (DE5)16
 Question 19

Consider the circuit shown below. The output of a 2:1 Mux is given by the function (ac' + bc).

Which of the following is true?

 A f = x1' + x2 B f = x1'x2 + x1x2' C f = x1x2 + x1'x2' D f = x1 + x2'
Question 19 Explanation:
g = (a and x1′) or (b and x1)
g = (1 and x1’) or (0 and x1)
g = x1’
f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2
 Question 20

The decimal value 0.25

 A is equivalent to the binary value 0.1 B is equivalent to the binary value 0.01 C is equivalent to the binary value 0.00111… D cannot be represented precisely in binary
Question 20 Explanation:
1st Multiplication iteration:
Multiply 0.25 by 2.
0.25×2 = 0.50 (product)
Fractional part = 0.50
Carry = 0
2nd Multiplication iteration:
Multiply 0.50 by 2.
0.50×2 = 1.00 (product)
Fractional part = 0.00
Carry = 1
The fractional part in the 2nd iteration becomes zero and so we stop the multiplication iteration.
Carry from 1st multiplication iteration becomes MSB and carry from 2nd iteration becomes LSB. So the result is 0.01.
 Question 21

The 2’s complement representation of the decimal value -15 is

 A 1111 B 11111 C 111111 D 10001
Question 21 Explanation:
15 = 1111
-15 = 11111
1's complement = 10000
2's complement = 10001
 Question 22

Sign extension is a step in

 A floating point multiplication B signed 16 bit integer addition C arithmetic left shift D converting a signed integer from one size to another
Question 22 Explanation:
Sign extension is a step in converting a signed integer from on size to another.
 Question 23

 A is flagged whenever there is carry from sign bit addition B cannot occur when a positive value is added to a negative value C is flagged when the carries from sign bit and previous bit match D None of the above
Question 23 Explanation:
The left most bit of positive value is zero. And left most bit for negative value is one. The value of 0+1 becomes 1. Then overflow never occurs.
 Question 24

Consider the following 32-bit floating-point 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 excess-64 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.

 A Theory of Explanation is given below.
 Question 25

Assuming all numbers are in 2's complement representation, which of the following numbers is divisible by 11111011?

 A 11100111 B 11100100 C 11010111 D 11011011
Question 25 Explanation:
Given: Binary numbers = 11111011
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = -5
(A) 11100111 - (-25)10
(B) 11100100 - (-28)10
(C) 11010111 - (-41)10
(D) 11011011 - (-37)10
Answer: Option A (-25 is divisible by -5)
 Question 26

The following is a scheme for floating point number representation using 16 bits.

Let s, e, and m be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:

What is the maximum difference between two successive real numbers representable in this system?

 A 2-40 B 2-9 C 222 D 231
Question 26 Explanation:
Largest gap will be in between two most largest numbers.
The largest number is 1.111111111× 262-31 = (2−2−9)×231
Second largest number is 1.111111110×262-31 = (2−2-8)×231
Difference = (2−2−9)×231 - (2−2-8)×231
= (2-8−2−9) ×231
= 2−9×231
= 222
 Question 27

Convert the following numbers in the given bases into their equivalents in the desired bases.
(a) 110.101)2 = x)10
(b) 1118)10 = y)H

 A (a) 6.625, (b) (45E)H
Question 27 Explanation:
(a) 1*22 + 1*21 + 0*20 + 1*2-1 + 0*2-2 + 1*2-3
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)H.
 Question 28

Consider the number given by the decimal expression:

`  163 * 9 + 162 * 7 + 16 * 5 + 3`

The number of 1’s in the unsigned binary representation of the number is ________.

 A 9
Question 28 Explanation:
Hexadecimal representation of a given no. is,
(9753)16
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
 Question 29

When two 4-bit binary number A = a3a2a1a0 and B = b3b2b1b0 are multiplied, the digit c1 of the product C is given by _________

 A c1 = b1a0 ⊕ a1b0
Question 29 Explanation:

⇒ c1 = b1a0 ⊕ a1b0
 Question 30
The format of the single-precision floating-point representation of a real number as per the IEEE 754 standard is as follows:   Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard?
 A exponent = 00000001 and mantissa = 00000000000000000000001 B exponent = 00000001 and mantissa = 00000000000000000000000 C exponent = 00000000 and mantissa = 00000000000000000000000 D exponent = 00000000 and mantissa = 00000000000000000000001
Question 30 Explanation:

The smallest biased exponent for the normalized number is E= 1.

The smallest Mantissa M = 000...0

The smallest positive normalized number = 1.M x 2 E-127

= 1.0 x 2 -126 .

= 2 -126

 Question 31
If the numerical value of a 2-byte unsigned integer on a little endian computer is 255 more than that on a big endian computer, which of the following choices represent(s) the unsigned integer on a little endian computer?
 A 0x4243 B 0x6665 C 0x0001 D 0x0100
Question 31 Explanation:

It is given that the unsigned integer is 2-bytes long. It needs 4 hexadecimal digits.

We know the big-endian and little-endian computers differ in how the data is stored in memory.

In little endian machines, the last byte of binary representation of the multibyte data-type is stored first. On the other hand, in big endian machines, the first byte of binary representation of the multibyte data-type is stored first.

In the hexadecimal representation of the 2-byte number on the little endian machine, the first two hexadecimal digits are for one byte and the last two hexadecimal digits are for the second byte. On the big endian machine it is the other way round.

It is given that the value in little endian representation is 255 more than the value in the big endian machine.

From the given options

A). In little endian = 0x4243 in binary (0100 0010 0100 0011) which has decimal value = 16963

On big endian = 0x4342 in binary (0100 0011 0100 0010) which has decimal value = 17218

Here the big endian value is higher than the little endian value. So this is not the correct option.

B). In little endian 0x6665 in binary (0110 0110 0110 0101) which has the decimal value = 26213

In big endian 0x6566 in binary (0110 0101 0110 0110) which has the decimal value = 25958

The difference = 26213 - 25958 = 255. So this is also the correct option.

C). In little endian 0x0001 in binary (0000 0000 0000 0001) which has the decimal value = 1.

In big endian 0x0100 in binary (0000 0001 0000 0000) has decimal value = 256.

But here also the big endian value is higher than the little endian. So this is not the correct option.

D). On the little endian machine for hexadecimal number 0x0100 in binary (0000 0001 0000 0000) which has decimal value = 256. On a big endian machine it is 0x 0001 in binary  (0000 0000 0000 0001) which has decimal value = 1

The difference in value of little endian to big endian is 256-1 = 255.

Hence 2, 4 are the correct options.

 Question 32
If x and y are two decimal digits and (0.1101)2=(0.8xy5)10, the decimal value of x+y is _______.
 A 3
Question 32 Explanation:
(0.1101)2 = 1/2 + 1/4 + 1/16= 0.5+0.25+ 0.0625 = (0.8125)10 .
x=1 and y=2
x+y= 1+2=3
 Question 33

The number (123456)8 is equivalent to

 A (A72E)16 and (22130232)4 B (A72E)16 and (22131122)4 C (A73E)16 and (22130232)4 D (A62E)16 and (22120232)4
Question 33 Explanation:
(123456)8 = (001 010 011 100 101 110)2
= (00 1010 0111 0010 1110)2
= (A72E)16
Also,
(001 010 011 100 101 110)2
= (00 10 10 01 11 00 10 11 10)2
= (22130232)4
 Question 34

Consider a parity check code with three data bits and four parity check bits. Three of the code words are 0101011, 1001101 and 1110001. Which of the following are also code words?

```1. 0010111
2. 0110110
3. 1011010
4. 0111010 ```
 A 1 and 3 B 1, 2 and 3 C 2 and 4 D 1, 2, 3 and 4
Question 34 Explanation:
Let x1, x2, x3 are data bits, and c1, c2, c3 and c4 are parity check bits.
Given transmitted codewords are

By inspection we can find the rule for generating each of the parity bits,

Now from above we can see that (I) and (III) are only codewords.
 Question 35

(34.4)8 × (23.4)8 evaluates to

 A (1053.6)8 B (1053.2)8 C (1024.2)8 D None of these
Question 35 Explanation:
First convert (34.4)8 and (23.4)8 to decimal.
(34.4)8 = 3×81 + 4×80 + 4×8-1
= 24 + 4 + 0.5
= (28.5)10
(23.4)8 = 2×81 + 3×80 + 4×8-1
= 16 + 3 + 0.5
= (19.5)10
Now,
(28.5)10 × (19.5)01 = (555.75)10
Now,
(555.75)10 = ( ? )8
To convert the integer part,

We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,

∴ (555.75)10 = (1053.6)8
 Question 36

The addition of 4-bit, two’s complement, binary numbers 1101 and 0100 results in

 A 0001 and an overflow B 1001 and no overflow C 0001 and no overflow D 1001 and an overflow
Question 36 Explanation:
2's complement of 1101 = 0011
2's complement of 1100 = 1100
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
 Question 37

When multiplicand Y is multiplied by multiplier X = xn-1xn-2 ...x0 using bit-pair recoding in Booth's algorithm, partial products are generated according to the following table.

The partial products for rows 5 and 8 are

 A 2Y and Y B -2Y and 2Y C -2Y and 0 D 0 and Y
Question 37 Explanation:

⇒ -2Y and 0
 Question 38

(C012.25)H – (10111001110.101)B =

 A (135103.412)O B (564411.412)O C (564411.205)O D (135103.205)O
Question 38 Explanation:
(C012.25)H – (10111001110.101)B
= 1100000000010010.00100101 - 0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)O
 Question 39

The following bit pattern represents a floating point number in IEEE 754 single precision format

` 110000011101000000000000000000000`
The value of the number in decimal form is
 A -10 B -13 C -26 D None of these
Question 39 Explanation:
Sign bit is 1 then given number is negative.
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 × 24 = -(11010)2 = -26
 Question 40

A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2's complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be:

 A 1, 1, 0 B 1, 0, 0 C 0, 1, 0 D 1, 0, 1
Question 40 Explanation:

Carry flag = 1
Overflow flag = 0
Sign bit = 0 (MSB bit is 0)
Overflow flag:
In computer processors, the overflow flag is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation.
 Question 41

The two numbers given below are multiplied using the Booth's algorithm.

```Multiplicand : 0101 1010 1110 1110
Multiplier: 0111 0111 1011 1101```
How many additions/Subtractions are required for the multiplication of the above two numbers?
 A 6 B 8 C 10 D 12
Question 41 Explanation:
Take the multiples and add 0 to the LSB.
Now we have some values defined for pair of bits in Booth’s Algorithm,
00 → 0
11 → 0
01 → -1
10 → 1
Now after adding 0 to the LSB of the multiplier, start traversing from left to right and accordingly put the values defined above.

Hence, total 8 additions / subtractions required.
 Question 42
Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1+R2, which one of the following values of R1 and R2 gives an arithmetic overflow?
 A R1 = 1011 and R2 = 1110 B R1 = 1100 and R2 = 1010 C R1 = 0011 and R2 = 0100 D R1 = 1001 and R2 = 1111
Question 42 Explanation:
 Question 43
Consider three floating point numbers A , B and C stored in registers R A , R B and R C , respectively as per IEEE-754 single precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows.

Which one of the following is FALSE?
 A A + C = 0 B C = A + B C B = 3 C D ( B - C ) > 0
Question 43 Explanation:

A= -12, B= +36 and C= +12
A+C= 0
B=3C
(B-C)>0
C≠A+B
 Question 44

The exponent of a floating-point number is represented in excess-N code so that:

 A The dynamic range is large. B The precision is high. C The smallest number is represented by all zeros. D Overflow is avoided.
Question 44 Explanation:
To avoid extra work, excess-N code is used so that all exponent can be represented in positive numbers, starting with 0.
 Question 45

If 73x (in base-x number system) is equal to 54y (in base-y number system), the possible values of x and y are

 A 8, 16 B 10, 12 C 9, 13 D 8, 11
Question 45 Explanation:
(73)x = (54)y
7x+3 = 5y+4
7x-5y = 1
Only option (D) satisfies above equation.
 Question 46

What is the result of evaluating the following two expressions using three-digit floating point arithmetic with rounding?

```   (113. + -111.) + 7.51
113. + (-111. + 7.51) ```
 A 9.51 and 10.0 respectively B 10.0 and 9.51 respectively C 9.51 and 9.51 respectively D 10.0 and 10.0 respectively
Question 46 Explanation:
(113. + -111.) + 7.51
= (2) + 7.51
= 9.51 (✔️)
113. + (-111. + 7.51)
= 113. + (-103.51)
= 113. + -103
= 10 (✔️)
 Question 47

Let A = 1111 1010 and B = 0000 1010 be two 8-bit 2's complement numbers. Their product in 2's complement is

 A 1100 0100 B 1001 1100 C 1010 0101 D 1101 0101
Question 47 Explanation:
A = 1111 1010 = -610 [2's complement number]
B = 0000 1010 = 1010 [2's complement number]
A×B = -6×10 = - 6010
⇒ -6010 = 101111002
= 110000112 (1's complement)
= 110001002 (2's complement)
 Question 48

Consider the following minterm expression for F:

`  F(P,Q,R,S) = Σ0,2,5,7,8,10,13,15  `

The minterms 2, 7, 8 and 13 are 'do not care' terms. The minimal sum-of-products form for F is:

 A B C D
Question 48 Explanation:
 Question 49

Let ⊕ denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:

`     F(P,Q) = ((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0)) `

The equivalent expression for F is

 A P+Q B C P⨁Q D
Question 49 Explanation:
((1 ⊕ P) ⊕ (P ⊕ Q)) ⊕ ((P ⊕ Q) ⊕ (Q ⊕ 0))
⊕ is associative i.e P ⊕ (Q ⊕ R) = (P⊕Q) ⊕ R.
P ⊕ P = 0, 1 ⊕ P = P’ and 0 ⊕ Q = Q
(1 ⊕ P) ⊕ ((P ⊕ Q) ⊕ (P ⊕ Q)) ⊕ (Q ⊕ 0)
= P’⊕ (0) ⊕ Q
= P’ ⊕ Q
= (P ⊕ Q)’
 Question 50

Let X be the number of distinct 16-bit integers in 2’s complement representation. Let Y be the number of distinct 16-bit integers in sign magnitude representation.

Then X-Y is _________.

 A 1 B 2 C 3 D 4
Question 50 Explanation:
X = 216
Since range is - 215 to 215 - 1
Y = 216 - 1
Here, +0 and -0 are represented separately.
X - Y = 216 - (216 - 1)
= 1
 Question 51
A particular number is written as 132 in radix-4 representation. The same number in radix-5 representation is ______?
 A 110
 Question 52
Consider the IEEE-754 single precision floating point numbers P=0xC1800000 and Q=0x3F5C2EF4. Which one of the following corresponds to the product of these numbers (i.e., P × Q), represented in the IEEE-754 single precision format?
 A 0x404C2EF4 B 0x405C2EF4 C 0xC15C2EF4 D 0xC14C2EF4
 Question 53

The 16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is __________.

 A -11 B -12 C -13 D -14
Question 53 Explanation:
Given number is 1111 1111 1111 0101.
It is a negative number because MSB is 1.
Magnitude of 1111 1111 1111 0101 is 2’s complement of 1111 1111 1111 0101.
1111 1111 1111 0101
0000 0000 0000 1010 : 1’s Complement
0000 0000 0000 1011 : 2’s complement
= (11)10
Hence, 1111 1111 1111 0101 = -11
 Question 54

The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is

 A 136251 B 736251 C 571247 D 136252
Question 54 Explanation:
X = (BCA9)16
Each hexadecimal digit is equal to a 4-bit 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 3-bit binary number.
X = (001 011 110 010 101 001)2
Note: Two zeroes added at host significant position to make number bits of a multiple of 3 (16 + 2 = 18)
X = (136251)8
 Question 55

Given the following binary number in 32-bit (single precision) IEEE-754 format:

00111110011011010000000000000000

The decimal value closest to this floating-point number is

 A 1.45 × 101 B 1.45 × 10-1 C 2.27 × 10-1 D 2.27 × 101
Question 55 Explanation:

For single-precision floating-point representation decimal value is equal to (-1)5 × 1.M × 2(E-127)
S = 0
E = (01111100)2 = (124).
So E – 127 = - 3
1.M = 1.11011010…0
= 20 + 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(E-127)
= -10 × 1.847 × 2-3
≈ 0.231
≈ 2.3 × 10-1
 Question 56

Consider a quadratic equation x2 - 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=________.

 A 8 B 9 C 10 D 11
Question 56 Explanation:
x2 - 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x - a)(x - b) = 0
x2 - (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
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8
There are 56 questions to complete.

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