Test: Number System- 2 - Computer Science Engineering (CSE) MCQ

Given:

(110)_x = (132)₄,

The decimal equivalent of this number will be:

1 x x² + 1 x x¹ + 0 x x⁰ = 1 x 4² + 3 x 4 + 2 x 4⁰

x² + x + 0 = 16 + 12 + 2

x² + x - 30 = 0 

On solving this quadratic equation we'll get:

x = 5, -6

Base can't be a negative so;

x = 5

A = 11111010
Since MSB is 1 hence number is negative i.e. - (00000110)
= -(00000110)
A = - 6
B = 00001010,
There is number 2’s complement representation for positive number B = +10 
A x B = 10 x (- 6)
= - 60
Binary representation of - 60 = 00111100 
Now take 2’s complement = 11000100

Since in 1 ’s complete every bits get complemented so, no such number exist whose complement is same while in 2’s complement 8 and 0 are exist.

Decimal number 3.248 x 10⁴
So sign bit should be 0 since number is positive. 
3.248 x 10⁴ = 32480
= 111111011100000₂ 
= 1.11111011100000 x 2¹⁴
In normalized representation:
Biased = 14 + 127 = 14
= 10001101₂
So, number is

a = 0.052 and b = 28 E + 1
So, b + a - b will b
a = 0.52 exponent
= -1
b = 28 E + 11 mantissa
= 0,28 exponent
= 13
b + a = 28000000000000 E - 1 + 0.52 E-1 
= 280000000000.52 E - 1
With is equal to 28 E + 11 
So, b - b = 0

Sign extension is the operation, in computer arithmetic, of increasing the number of bits of a binary number while preserving the number’s sign (positive/negative) and value.

The decimal number is 0.239 x 2¹³
We have to find hexadecimal representation without normalization.
Biased exponent = 13 + 64 = 77
Representing 77 in binary (77)¹⁰ = (1001101)₂
Representing mantissa in binary
senting mantissa in binary (0.239)₁₀ = 0.00111101000101 

(x 567)₈ 
(2yx5)₈ 
(71yx)8
x will be 4 and y will be 3.

Smallest number stored in 8 bit 2’s complement arithm etic is -2^n-1
=-2^{8- 1} 
= -128

F will represents x₀ x₁ x₂ .... x_n-1 x_n.