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The quantity 837 in excess-3 BCD code would be represented as
- a)1001 0011 0111
- b)1000 0001 1001
- c)1011 0110 1010
- d)1000 0011 0111
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The quantity 837 in excess-3 BCD code would be represented asa)1001 00...
The quantity 837 is first converted to BCD representation then it is incremented by 3 bit by bit.
Most Upvoted Answer
The quantity 837 in excess-3 BCD code would be represented asa)1001 00...
Excess-3 BCD code is a self-complementary code, which means that the 10's complement of a number is obtained by flipping all the bits. It is also called XS-3 code or Stibitz code.
To represent the quantity 837 in excess-3 BCD code, we need to convert it into BCD first.
- Step 1: Convert 837 into binary
837 = 512 + 256 + 64 + 4 + 1 = 1101001101 (in binary)
- Step 2: Convert binary into BCD
1101 0011 0111 (in BCD)
- Step 3: Add 0011 (decimal 3 in excess-3 code) to each BCD digit
1101 0110 1010 (in excess-3 BCD)
Therefore, the quantity 837 in excess-3 BCD code would be represented as 1011 0110 1010.
To represent the quantity 837 in excess-3 BCD code, we need to convert it into BCD first.
- Step 1: Convert 837 into binary
837 = 512 + 256 + 64 + 4 + 1 = 1101001101 (in binary)
- Step 2: Convert binary into BCD
1101 0011 0111 (in BCD)
- Step 3: Add 0011 (decimal 3 in excess-3 code) to each BCD digit
1101 0110 1010 (in excess-3 BCD)
Therefore, the quantity 837 in excess-3 BCD code would be represented as 1011 0110 1010.
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Community Answer
The quantity 837 in excess-3 BCD code would be represented asa)1001 00...
We have to go from right to left , so first I check B having code 1011= 1×2^0 + 1×2^1 + 0×2^2 + 1×2^3 = 11 so, option C is correct.
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The quantity 837 in excess-3 BCD code would be represented asa)1001 0011 0111b)1000 0001 1001c)1011 0110 1010d)1000 0011 0111Correct answer is option 'C'. Can you explain this answer?
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