Number Representation MCQ Quiz in मल्याळम - Objective Question with Answer for Number Representation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is option 1

Concept:

When we multiply a number by 2 this means a binary representation of this number is shifted left

shifting depends upon how many time the number is multiplied by 2

Explanation:

Given a hexadecimal number in 2's complement and required in 2's complement

So, nothing want to change it

P = (F87B)16 = (1111 1000 0111 1011)₂

8P =2³*P

23*P this means that the binary number that is represented by P is shifted 3 times left

So, it becomes

(1111 1000 0111 1011)2 = (1100 0011 1101 1000)₂ = (C3D8)₁₆

P = (F87B)16 = (1111 1000 0111 1011)2

Most significant bit is 1 here its mean number is negative

To get the value 2's complement of this is (0000 0111 1000 0101)₂

(0000 0111 1000 0101)2 =(-1925)₁₀

8P =8 × -1925 =-15400

now need to find 2's complement of -15400

(15400)₁₀= (0011 1100 0010 1000)2

take 2's complement of (0011 1100 0010 1000)2

(1100 0011 1101 1000)2 = (C3D8)16

Since 8 bit number is in 2's complement, weight of the last bit will be negative

(11111011)2 = - 1 × 27 + 1 × 26+ 1 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 = -5

(00001010 )2 = - 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20 =  10

(11111011)2 × (00001010 )2 = -5 × 10 = -50

Option 2:

11001110 = - 1 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 0 × 20 

11001110 = -128 + 64 + 8 + 4 + 2= -50

Therefore option 2 is correct.

Concept:

In 2's complement form, for n bits

Minimum number = –2n-1

Maximum number = 2n-1 – 1

Range is –(2n-1) to  (2n-1 – 1)

Calculation:

n = 8 bits

Minimum number = –(2n-1) = –(28-1 ) = – 128

Maximum number = 2n-1 – 1 = 28-1 – 1 = 127

The weighted codes are those that obey the position weighting principle. It states that the position of each number represents a specific weight. In these codes, each decimal digit is represented by a group of four bits.

The BCD is a weighted code and the weights used in the binary coded decimal code are 8, 4, 2, 1

8421 is weighted code and the weights are 8, 4, 2, 1

6421 is weighted code and the weights are 6, 4, 2, 1

Given Number is 10100

The Right shift of the content in register is same as the content divided by 2 

Apply Right Shift ⇒ 11010              

Operation right shift is equivalent to divided by 2

Addition of integers in One’s complement-

One’s complement of 011000 → 100111

One’s complement of 011000 → 100111

Now, addition-

 100111 + 100111 = 1001110

Therefore, in this 1001110 the digit 1-001110 is an overflow.

Addition of integers in Two’s complement-

Two’s complement of 011000 → 101000

Two’s complement of 011000 → 101000

Now, addition-

101000 + 101000 = 1010000

Therefore, in this 1010000 the digit 1-010000 is an overflow.

Addition of integers in sign and magnitude format

If start with 0 that means +ve.

If start with 1 that means –ve.

So, given numbers 011000 and 011000 both are positive

Now, addition

11000 + 11000 = 110000

Therefore, in this 110000 the digit 1-10000 is an overflow.

(C28000000)₁₆ = (1100 0010 1000 0000 …. 0000)₂

M = 0000 … 0000 = 0

E = (1000 0101)₂ = (133)₁₀

S = 1

(-1)^s × 1.M × 2^E-127

(-1)¹ × 1.00 × 2^133-127

Data:

number of bits = n = 8 

Formula:

Range of 2's compliment number with n bits is (–2^n-1 ) to +(2^n-1 – 1)

Smallest integer = –2n-1

largest integer = 2n-1 – 1

Calculation:

Smallest integer = –28-1 = –128 

The correct answer is Gray code.

Key Points

• The Gray code, also referred to as the reflected binary code, is a binary numeral system in which two consecutive numbers differ in only one bit. The unique property of Gray code is that each transition from one value to the next value involves changing only one bit.
• This system was invented by Frank Gray at Bell Labs to prevent spurious output from electromechanical switches. For instance, while switching from one position to another in standard binary code, there is a risk that switches will change at different times leading to invalid numbers, but with Gray code, since only one bit changes at a time, such misinterpretations are avoided.
• So, Gray code is also known as reflected binary code because the sequence of binary values reflects it about its midpoint. For example:
• Binary: 000, 001, 010, 011, 100, 101, 110, 111
• If we reflect this sequence, i.e., reverse it, we get:111, 110, 101, 100, 011, 010, 001, 000

• Now let's take the first half of the original sequence and the first half of the reflected sequence.

• By inverting the bits in the second half, we get the sequence of Gray codes:  000, 001, 011, 010, 110, 111, 101, 100

So, every binary number has a unique Gray code, and vice versa, attributing to why the term "reflected binary code" is used to describe the Gray code.

The correct answer is option 2.

Concept:

Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.

The given data,

Given two four-bit registers R1 and R2.

Option 1: R1 = 1011 and R2 = 1110

False,  

R1      =  1 0 1 1 = -(0101)= -5

+ R2   =  1 1 1 0 = -(0010)= -2

-----------------------------------------------

               1 0 0 1  =           = -7        

Here No overflow occurred, because sign bit is same for (R1 + R2 ).

Option 2: R1 = 1100 and R2 = 1010

True,

R1      =  1 1 0 0 = -(0100)= -4

+ R2   =  1 0 1 0 = -(0110)= -6

  --------------------------------------------

              0 1 1 0 =            = -10       

Here Overflow occurred because the sign bit is different for (R1 + R2 ).

Option 3: R1 = 0011 and R2 = 0100

False,

R1      =   0 0 1 1 = +(0011)= +3

+ R2   =   0 1 0 0 = +(0100)= +4

  --------------------------------------------

                0 1 1 1                 =   +7       

Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).

Option 4: R1 = 1001 and R2 = 1111

False, 

R1      =   1 0 0 1 = -(0111)  = -7

+ R2   =   1 1 1 1 = -(0001) = -1

  --------------------------------------------

                1 0 0 0 =               = -8

Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).

Hence the correct answer is R1 = 1100 and R2 = 1010.