Digital Electronics - Digital Arithmetic Operations and Circuits - Discussion
Discussion Forum : Digital Arithmetic Operations and Circuits - General Questions (Q.No. 34)
34.
Discussion:
12 comments Page 1 of 2.
Mani said:
8 years ago
It's so simple if you follow the BCD rule i.e., the no. which is greater than 9 (1001) add 6 (0110) to it for getting a valid one.
so, here
01101000 + 00110110 = 10011110
Here you can see last 4-bit group is invalid so add 6 to it.
10011110
+ 0110
---------------
10100100
Now the first 4 bit group becomes invalid so add 6 to it +0110 ---------------> 100000100.
so the answer (c) is correct.
so, here
01101000 + 00110110 = 10011110
Here you can see last 4-bit group is invalid so add 6 to it.
10011110
+ 0110
---------------
10100100
Now the first 4 bit group becomes invalid so add 6 to it +0110 ---------------> 100000100.
so the answer (c) is correct.
(1)
Tamil said:
8 years ago
My method is 68 for the first BCD.
36 for 2nd bcd.
ADD 68 ,36 --104.
104 changed to bcd 000100000100.
36 for 2nd bcd.
ADD 68 ,36 --104.
104 changed to bcd 000100000100.
Priya said:
8 years ago
Way of solving is right @Anjana.
Anjana said:
9 years ago
In bcd no 4 bit is greater than 1001 so after adding all we get,
1001 1110
Here initially 1110 >1001.
So, 1110 + 0110 = 1 0100.
This carry 1 is added to next 4 bit 1001.
Which results in 1010 which is again>1001.
So again 1010+0110 = 10000.
So the answer is 1 0000 0100.
Or 0001 0000 0100.
1001 1110
Here initially 1110 >1001.
So, 1110 + 0110 = 1 0100.
This carry 1 is added to next 4 bit 1001.
Which results in 1010 which is again>1001.
So again 1010+0110 = 10000.
So the answer is 1 0000 0100.
Or 0001 0000 0100.
Vinay said:
9 years ago
Given:
0110 (6) 1000 (8).
+ 0011 (3) 0110 (6).
Sum = 1001 (9) 1110 (13) > 9.
As we see that 1110 (13) is greater than 9 so add 6 to this no as 1110.
+ 0110.
Sum = 10100.
Since in number 10100 has a carry MSB digit so add this carry to the 1st number 1001 we get.
1001,
+ 1,
Sum = 1010.
So, after addition the BCD no. Is = 1010 0100.
So, this is the right answer.
0110 (6) 1000 (8).
+ 0011 (3) 0110 (6).
Sum = 1001 (9) 1110 (13) > 9.
As we see that 1110 (13) is greater than 9 so add 6 to this no as 1110.
+ 0110.
Sum = 10100.
Since in number 10100 has a carry MSB digit so add this carry to the 1st number 1001 we get.
1001,
+ 1,
Sum = 1010.
So, after addition the BCD no. Is = 1010 0100.
So, this is the right answer.
Shashi said:
9 years ago
In BCD format 01101000 = 6 8
=> 00110110 = 36.
Add to numbers we will get the answer is 104, in Bcd format 1 0 4 = 0001 0000 0100.
So answer is C.
=> 00110110 = 36.
Add to numbers we will get the answer is 104, in Bcd format 1 0 4 = 0001 0000 0100.
So answer is C.
CHIRAG said:
10 years ago
Please give correct answer and explanation.
Laxman said:
1 decade ago
I think as per the rules given in @Anand Kumar, we need to add 0110 only in the invalid term and add carry to its ahead term.
So,
After addition we get,
1001 1110, add 0110 in second term only.
+0000 0110.
-------------
1010 0100.
Sol : 1010 0100 is right answer.
So,
After addition we get,
1001 1110, add 0110 in second term only.
+0000 0110.
-------------
1010 0100.
Sol : 1010 0100 is right answer.
Pooja said:
1 decade ago
After addition we have to add 0110 (6) to both the term i.e. after addition we get 1001 1110 and now we add 0110 to 1001 and also 0110 to 1110.
i.e 0110 0110 will be added to 1001 1110 because no term would be greater than or equal to 9.
After addition final result will be 100000100. For making it a three 4 bit number we add extra 3 zero.
i.e 0110 0110 will be added to 1001 1110 because no term would be greater than or equal to 9.
After addition final result will be 100000100. For making it a three 4 bit number we add extra 3 zero.
Ganesh said:
1 decade ago
Why you have added 6 as carry and again the result is not clear?
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