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The
16 bit
2` s complement representation of an integer is
1111 1111 1111 0101 ; its decimal representation is ______________.
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
Let X be the number of distinct 16bit integers in 2`s complement representation. Let Y be the number of distinct bit integers in sign magnitude representation Then X Y is______.
For N bits, Distinct values represented in 2's complement is 2^{n1} to 2^{n1} 1
Distinct values represented in Signed Magnitude is (2^{n1} 1) to 2^{n1} 1
Difference is 1.
4r+1=40+9
4r=48
r=12
Consider the number given by the decimal expression:
The number of 1’s in the unsigned binary representation of the number is ______
Hex representation of given no. is (9753)_{16}
Its binary representation is (1001 0111 0101 0011)_{2} The no. of 1's is 9
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. 1 * 2^{2} + 1 * 2^{1} + 0 * 2^{0} + 1 * 2^{1} + 0 * 2^{2} + 1 * 2^{3} = 6.625
b. 1118 mod 16 = 14, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4.
Writing the mods in the reverse order (in hex) gives (45E)_{H}
Both can be done using calculator also.
The number of 1's in the binary representation of (3*4096 + 15*256 + 5*16 + 3) are:
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.
Consider the following floatingpoint 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
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 12^(23) .
so range will be .5 to (12^(23))
2r^{2} + 2r + 4 = r^{2} + 6r+ 9
r^{2}  4r 5 = 0
r^{2} 5r +r  5 = 0
(r5)(r+1)=0
r can not be 1
so r = 5 is correct answer
The octal representation of an integer is (342)_{8}. If this were to be treated as an eightbit integer in an 8085 based computer, its decimal equivalent is
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
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?
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
The number 43 in 2's complement representation is
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.
Consider the values of A = 2.0 x 10^{30}, B = 2.0 x 10^{30} , C = 1.0, and the sequence
executed on a computer where floating point numbers are represented with 32 bits. The values for X and Y will be
Given 32 bits representation. So, the maximum precision can be 32 bits (In 32bit IEEE representation, maximum precision is 24 bits but we take best case here). This means approximately 10 digits.
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.
B choice.
The 2's complement representation of (539)_{10} in hexadecimal is
539 = 512 + 16 + 8 + 2 + 1 = 2^{9} + 2^{4} + 2^{3} + 2^{1} + 2^{0}
= (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
The 2's complement representation of the decimal value 15 is
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.
Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by 11111011?
MSB of 2's compliment number has a weight of  2^(n1).
(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.
If 73_{x} (in basex number system) is equal to 54_{y} (in base ynumber system), the possible values of x and y are
x * 7 + 3 = 5 * y + 4 => 7x = 5y + 1. Only option satisfying this is D.
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