Blog: CS302 Assignment Solution 2

A	B	C	(A XOR B)	X =NOT(A XOR B)	Y =B XOR C	X AND Y
0  0 0 0 1 1 1 1	0 0 1 1 0 0 1 1	0 1 0 1 0 1 0 1	0  0 1 1 1 1 0 0	1 1 0 0 0 0 1 1	0 1 1 0 0 1 1 0	0 1 0 0 0 0 1 0

BCD conversion codes representing the decimal digits 0 to 9:-
Decimal  BCD  Decimal  BCD
0            0000     5            0101
1            0001     6            0110
2            0010     7            0111
3            0011     8            1000
4            0100     9            1001

478  Decimal to Binary conversion: 478/2             239             Remainder  0

239/2             119             Remainder  1

119/2              59              Remainder  1

59/2               29               Remainder  1

29/2                14              Remainder  1

14/2               7                 Remainder  0

7/2                3                  Remainder  1

3/2                1                  Remainder  1

1/ 2               0                  Remainder  1      478   = 111011110

1. a.      BCD  conversion

478   = 111011110     
= 0001 1101   1110
129   = 10000001
= 1000 0001

BCD

129
0001 0010 1001
478
0100 0111 1000
 Question 2                                                                                                                                          [5Marks]

   b.  BCD addition of “129”  and “478”
            
                   0001 0010 1001
                  0100 0111 1000
                  0101 1010  0001 
                    1001          9
                   1000           8
                   0001          17
                   0110         adding 6              
                  0111


               0010         2
               0111         7       
              1010        10
              0110         adding 6               
         1   0000
  Carry  bit is added to next  most significant digit             
          0001
         0100
        0110                                       
  The answer is=   0110   0000  0111