Test: Number Representations- 1 - Computer Science Engineering (CSE) MCQ

for 2's complement  1 111 1111 1111 0101 →2′scomplement→2′scomplement  1 000 0000 0000 1011
1st bit is same not involved in 2's complement same with 1's complement. since msb bit for sign.
Take one's complement and add 1 we get 11, and as it is negative number we get answer as -11

For N bits, Distinct values represented in 2's complement is -2^n-1 to 2^n-1 -1
Distinct values represented in Signed Magnitude is -(2^n-1 -1) to 2^n-1 -1
Difference is 1.

4r+1=40+9
4r=48
r=12

Hex representation of given no. is (9753)₁₆
Its binary representation is (1001 0111 0101 0011)₂ The no. of 1's is 9

a. 1 * 2² + 1 * 2¹ + 0 * 2⁰ + 1 * 2^-1 + 0 * 2^-2 + 1 * 2^-3 = 6.625

I suggest following approach , here we can clearly see that numbers are getting multiplied by powers of 16. So this is nothing but Hexadecimal number in disguise.

Here, we are asked "magnitude" - so we just need to consider the mantissa bits.
Also, we are told "normalized representation"- so most significant bit of mantissa is always 1 (this is different from IEEE 754 normalized representation where this 1 is omitted in representation, but here it seems to be added on the right of decimal point as seen from options).
So, the maximum value of mantissa will be 23 1's where a decimal point is assumed before first 1. So, this value will be 1- 2^-23.
Due to the 1 in normalized representation, the smallest positive number will be 1 followed by 23 0's which will be 2^-1 = 0.5

Given that mantissa is sign magnitude representation so 1 bit for sign and remaining 23 bits for mantissa and  we have to find the range of mantissa in normalized form ....
so smallest will be = .100.....0(normalize form) which is 2^(-1) = .5
and for largest will be .111....1 which is 1-2^(-23) .
so range will be .5 to (1-2^(-23))

2r² + 2r + 4 = r² + 6r+ 9
r² - 4r -5 = 0
r² -5r +r - 5 = 0
(r-5)(r+1)=0
r can not be -1
so r = 5 is correct answer

If we treat this as an 8 bit integer, the first bit becomes sign bit and since it is "1", number is negative. Computer uses 2's complement representation for negative numbers and hence the decimal equivalent will be 

I 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 for four digits and 1 for the sign (like -7354). So, required memory = 5 bytes
Now, if we use integer, the largest number needed to represent is 9999 and this requires 2 bytes of memory for signed representation.

So, memory savings while using integer is 

Sign Magnitude
+0 = 0000
-0 = 1000
1's complement
+0 = 0000
-0 = 1111

2's complement representation is not same as 2's complement of a number. In 2's complement representation positive integers are represented in its normal binary form while negative numbers are represented in its 2's complement form. So (c) is correct here.

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 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. 
B choice. 

539 = 512 + 16 + 8 + 2 + 1 = 2⁹ + 2⁴ + 2³ + 2¹ + 2⁰
 = (1000011011)₂
Now all answers have 12 bits, so we add two 0's at beginning = (001000011011)₂
To convert to 2's complement invert all bits till the rightmost 1, which will be (110111100101)₂
= (1101 1110 0101)₂
= (DE5)₁₆

1st Multiplication Iteration Multiply 0.25 by 2
0.25 x 2 = 0.50(Product)         Fractional part=0.50  
Carry=0    (MSB)
2nd Multiplication Iteration
Multiply 0.50 by 2
0.50 x 2 = 1.00(Product)         Fractional part = 1.00
Carry = 1(LSB)   
The fractional part in the 2nd iteration becomes zero and hence we stop the multiplication iteration.
Carry from the 1st multiplication iteration becomes MSB and  carry from 2nd iteration becomes LSB
So the Result is 0.01

is the correct ans.In 2's complement representation, positive numbers are represented in simple binary form and negative numbers are represented in its 2's complement form. So, for -15, we have to complement its binary value -01111 and add a 1 to it, which gives 10001. Option D.

is the answer. Sign extension (filling the upper bits using the sign bit) is needed while increasing the number of bits for representing a number. For positive numbers, 0 is extended and for negative numbers 1 is extended.

is the answer. When a positive value and negative value are added overflow never happens. 

MSB of 2's compliment number has a weight of - 2^(n-1).
(Trick: (from reversing sign extension) just skip all leading 1's from MSB expect but 1, and then calculate the value as normal signed binary rep. ) so by calculating, we get the given number is -5 in decimal. and options are 
a. -25
b. -28
c. -41
d. -37
Therefore it is clear that - 25 is divisible by - 5. so we can say that (a.) is correct.

x * 7 + 3 = 5 * y + 4 => 7x = 5y + 1. Only option satisfying this is D.