add 648 and 487 in bcd codeans in bcd code

Question

Add 648 and 487 in BCD code.

Ans. In BCD Code, addition of 648 and 487:

6 4 8  =  0 1 1 0   0 1 0 0      1 0 0 0
4 8 7  =  0 1 0 0   1 0 0 0      0 1 1 1
--------------------------------------------------
             1 0 1 0   1 1 0 0     1 1 1 1

                10           12          15
--------------------------------------------------
In the given problem every three groups are invalid, since the four bit sum is more than 9. In these cases, add +6(that is 0110) to the 4 bit sum to skip the 6 invalid states. Whether a carry is generated while adding 6, so add the carry to the next four bit group that is,

6 4 8  =  0 1 1 0   0 1 0 0  1 0 0 0
4 8 7  =  0 1 0 0   1 0 0 0   0 1 1 1
--------------------------------------------------
             1 0 1 0     1 1 0 0    1 1 1 1
           0 1 1 0     0 1 1 0    0 1 1 0
      1   1 1    1     1       1    1 1
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0001   0 0 0 1      0 0 1 1    0 1 0 1

1            1              3              5
-------------------------------------------------------

Hence addition of 648 and 487 is 1135 in BCD Code.