ISRO2007-04

Question

When two numbers are added in excess-$3$ code and the sum is less than $9$, then in order to get the correct answer it is necessary to

• A. subtract $0011$ from the sum
• B. add $0011$ to the sum
• C. subtract $0110$ from the sum
• D. add $0110$ to the sum

Comments

$\\ 0 +3=3\\ 2 +3=5\\ 3+5=8(1000)<9\\ Subtract \ 0111 \ from\ 8(1000)\\ 1000-0011=0101(5)\\ \\ Actual\  sum\ of\ BCD=0+2=2\\ And \ 2 \ in \ excess-3: 2+3=5 (Which\ we\ got\ after\ subtracting\ 3\ from\ 8)$

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In Excess-3 addition:

If carry occurs => Sum is greater than 9 => Add 3. (Because we need to incorporate the carry in the result?)

If carry doesn't occur => Sum is less than or equal to 9 => Subtract 3.

Answer 1

Answer: A. Since addition is 9, means it does not produce the carry, hence 0011(3) will be subtracted from sum to get actual value of BCD.

Answer 2

The excess-3 binary code is an unweighted self-complementary BCD code.  It is an important BCD code, is a 4-bit code and used with BCD numbers. Self-Complementary property means that the 1’s complement of an excess-3 number is the excess-3 code of the 9’s complement of the corresponding decimal number. The excess-3 code for 3(0011) is 0110 and to find the excess-3 code of the complement of 3, we just need to find the 1’s complement of 0110 -> 1001, which is also the excess-3 code for the 9’s complement of 3 -> (9-3) = 6.

Excess-3 Addition Algo:

1. Convert BCD numbers to Excess-3 codes

2. Add the excess-3 codes

3.

If a carry is there

   Add 3 with the result

else

   Subtract 3 from the result

Ans A)

Answer 3

Ans.B

For more:http://www.electrical4u.com/excess-3-code-addition-subtraction/

Comments

5(8) + 3(6) = 8(11)

your answer is 14

so substract 3 will give excess 3 of th sum .

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it is asking for correct result which is 3+5=8

so we should need 8 as ans of the sum..not the excess 3 sum i.e 11

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@arjun sir look at this

Answer 4

Step 1 We have to convert the numbers (which are to be added) into excess 3 forms by adding 0011 with each of the four bit groups them or simply increasing them by 3.

Step 2 Now the two numbers are added using the basic laws of binary addition, there is no exception for this method.

Step 3 Now which of the four groups have produced a carry we have to add 0011 with them and subtract 0011 from the groups which have not produced a carry during the addition.

Step 4 The result which we have obtained after this operation is in Excess 3 form and this is our desired result