Ques 1: Consider the equation (43)x = (y3)8 where x and y are unknown. The number of possible solutions is ________.
GATE 2015 Set 3
Ans: 5
Solution: 4x+3=8y+3
4x=8y
x=2y
Also 4<x i.e x>4
y<8
So possible combinations are:
x=6, y=3
x=8,y=4
x=10,y=5
x=12,y=6
x=14,y=7
So total 5 possibilities
Ques 2: The 16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is
GATE 2016 Set 1
Ans: -11
Solution. Since MSB is 1
So number is negative
And its numerical value is 2’s complement of given number i.e 1’s compliment +1=
=0000 0000 0000 1010 +1
=0000 0000 0000 1011
=-11
Ques 3: Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is _____ .
GATE 2014 Set 2
Ans: 3
Solution: 1x5x5 + 2×5 + 3 = xy + 8
25+10+3=xy+8
38=xy+8
xy=30
Now, y>x
Possible Values are
x=1, y=30
x=2, y=15
x=3, y=10
So total 3 possibilities
Ques 4: The smallest integer that can be represented by 8-bit number in 2’s complement form is
GATE 2013
Ans: -128
Solution: Total 8 bits
1 bit is used for sign
remaining 7 bits
So -128 to 127 can be represented
Ques 5: The representation of the value of a 16-bit unsigned integer X in a hexadecimal number system is BCA9. The representation of the value of X in octal number system is:
GATE 2017 Set 2
(A) 571244
(B) 736251
(C) 571247
(D) 136251
Ans: (D) 136251
Solution : (BCA9)16
=(1011 1100 1010 1001 )2
=(1 011 110 010 101 001 )2
=(136251)8
Ques 6: 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 _________
GATE 2016 SET 2
Ans: 1
Solution: For 16- bit representation,
2’s Complement range X : (-215) to (215-1) = -32768 to 32767 = 65536
signed Magnitude range Y : -(215-1) to (215-1) = -32767 to 32767 = 65535
X – Y = 65536 – 65535 = 1
Ques 7: The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n – f. The range of decimal values for X in this representation is
GATE 2017 Set 1
(A) 2-f to 2i
(B) 2-f to (2i-2-f )
(C) 0 to 2i
(D) 0 to (2i-2-f )
Ans: (D) 0 to (2i-2-f )
Solution: Suppose n=5
f=2
So, i=5-2=3
Min value : 000.00 = 0
Max Value: 111.11
= 7.75
= 8 – 0.25
=23 – 2-2