Ans.13 Given : (10 w1z)2 X (15)10 = (y 01011001)2
(22 + w22 + 21 +z) X15= y28 +26+ 24 +23+ 1
(16+ 4w+ 2 +z) X15 =256y+ 6y+ 16+ 8+ 1
(60w+15z+270)=265y+89
Putting value of w, y and z from option (C) its satisfied.
Hence, the correct option is (C).
Ans.14 Given : A = 11101001 (in 2’s complement)
B=11100101 (in 2’s complement)
Again, 2’s complement of B will be B=00011011
Now, adding we get
= A+B
= 11101001
00011011
________
1 00000100
Hence, the correct option is (B)
Ans.15 Given : (235)R1= (565)10= (865)R2
(235)R1 = (565)10
2R12 +3R1+ 5=(565)10
Similarly,
(565)10 =(865)R2
(565)10 = 8R22 + 6R2 +5
By checking option (B) is satisfy.
Hence, the correct option is (B).
Ans.16 Given
(54)b = (13)b X(4)b
5b+ 4= (b+ 3)4
5b+ 4 =4b+ 12
b=8
Ans.17 Given expression :
(34)x+ (14)x =(103)x
Converting to decimal, we get 3x+4+x +4=x2+ 3
4x+8= x2+ 3
x2-4x-5=0
x2-5x+x-5=0
x( x- 5)+ 1( x-5)= 0
x=-1,5
Ans.18 Given : 2's complement representation of 16-bit number = FFFF
= 2 (1111 1111 1111 1111)2
Actual number = 2's complement of given number
1's complement of number
(1000 0000 0000 0000)2
2's Complement of number =
1000000000000000
+
1
_________________
1000000000000001
Sign representation of 1000000000000001=-1.
Thus, magnitude will be 1.
Hence, the correct option is (B).
Ans.19 Given : N =(45)10 - (45)16
Conversion from Hexadecimal to decimal :
(45)16= 4 X16 + 5 X160= 69
From equation (i),
N= (45)10- (69)16
N =(-24)10
2’s complement representation of – 24
+24 → 011000
-24= 2’s complement of 011000
= 101000
Hence, the correct option is (C).
Ans.20 Given (73)x =(54)y
Where X and Y base number,
7X+3X0=5Y+4Y0
7X+3=5Y+4
7X-5Y=1
By checking options
X=8, Y=11
7 X 8-5 X 11=56-55=1
Ans.21 Given :
2's complement representation of an integer
(FA)16 = (1 1 1 1 1 0 1 0)2
Actual Number = 2's complement of given number
1’s complement of number (1 0 0 0 0 1 0 1)2
2’s complement of number
10000101
+1
10000110
Sign representation of 10000110=-6