Your Answer 3 is wrong because here the addition is greater than 9 and there the final carry is 1 (not 0 ok, if you have 0 in your final carry, you can add 0110 , )
Last H.W. ques is a bit tricky. Here is its solution: (83)+(34) 10000011+ 00110100= 10110111. So divide it in 4 bits then we get 1011 and 0111. Here 0111 i.e.79 . So the problem here is when we add 0110 i.e. 6 in the answer we will get wrong answer because 10110111+ 00000110 Here we are adding 0110 i.e.6 in the side of seven i.e.0111 and in sir's problem we were adding 0110 in the side of 13>9. So we only have to take number 11 out and write 1011+ 0110= 10001 now just merge this result and seven we get 100010111 when we write this result in 4 bits we get 000100010111 so its decimal form is 117.
Another example could be when you also have to watch out for when the carry (after adding 6) from a digit in units place makes the digit in the tens place (and/or a carry from a digit in tens place makes the digit in the hundreds place) exceed 9, and so on. You need to add 6 more than once. eg 57+46=103, and 657+346=1003, etc. eg 1 57+46 0101 0111 0100 0110+ ------------------ 1001 1101 (initial sum) 0110+ (add 6) ----------------- 1010 0011 0110+ (add 6) ------------------ 0000 0011 Final Carry=1 0001 0000 0011= 103
@@hungryabhinav1022 7 + 9 = 16 Therefore doing it in BCD addition goes as following: 0111 is the same as 7.... plus 1001 is the same as 9............. Therefore, 0111 plus 1001 = 1, 0000.............. but because you carried the final 1, you have to follow the RULES for whenever you carrying a 1 at the end of the equation.................. We have to follow the rule of ADDING (SIX) to the solution because we had a final carry of 1............. 1, 0000 + 0110 ________ =1, 0110 THEN, you must go ahead and separate the five bit answer of 1, 0110 into eight bits by ADDING three more zero in front of the first 1............... and you will get TWO SETS OF (four bits) of 0001 and 0110................... The first four bits of 0001 is equal to 1 The second four bits of 0110 is equal to 6 CONCLUSION is 16 PROOF: 7+9 = 16
@@bhogichandini2998 83 + 34 = 117 Therefore doing it in BCD addition goes as following: 1000 , 0011 is the same as 8 and 3.... plus 0011 , 0100 is the same as 3 and 4............. Therefore, 1000 , 0011 PLUS 0011 , 0100 = 1011, 0111.............. we DID NOT carry anything over at the end of the equation.............We DO NOT add (SIX) to the solution. 1000 , 0011 + 0011 , 0100 _______________ = 1011, 0111 The first four bits of 1011 is equal to 8+2+1 = 11 The second four bits of 0111 is equal to 4+2+1 = 7 CONCLUSION is 117 PROOF: 83+34 = 117
In the last homework question we don't need to add 6 at all as 7 which is less than 9 with final carry 0 it's correct then 11 which is indeed greater than 9 but final carry is 1 which is also correct as mentioned in the lecture. So 83 + 34 which is 117... We get the same result by making pairs of 4.
Good work .im watching some of your lectures every day and i m using a book for taking notes just like in the class. I prefer it as good formal academic program.
(i)0111 (ii)10110 (iii)10101 (iv)100010111 Use a small trick :- add the number in decimal [a+b] = (c) then represent c in BCD {with the help of table } Thanks me later
just do simple addition and then write bcd equivalents of each digit of the answer you obtained by doing simple addition 1) 3+4 = 7 ....BCD equivalent of 7 is 0111 , so that is our answer 2) 7+9 = 16 BCD equivalent of 1 is 0001 BCD equivalent of 6 is 0110 now simply write both BCD equivalents together so our answer is 00010110 3) 7+8 = 15 ..... our answer is 00010101 83+34 = 117 ..... our answer is 000100010111
Then this is not called BCD addition....You just did Decimal Addition and coded it to binary....This is not BCD addition...Humans can understand this , not machines ✌️🙂..
After error correction (i.e add 0110) if msb of the number generates a carry it can be considered as a bcd bit, which is happening in the home work problem 4.
divide your final answer to 4-bit blocks and consider each 4-bit result with the 1,2,3 conditions. for example 68 + 79 = 0110 1000 + 0111 1001 = 1110 0001 . For last 4 bit (sum 9 F.C:0) add 0110 so the correct answer will be (wrong answer) + (0110 each of the 4 bits), so 1110 0001 + 0110 0110 = 1 0100 0111 is also equals 0001 0100 0111 = 147 in decimal --> Correct answer. Good luck.
For the last example, the initial sum is (10111100) which is incorrect because if we group the digits by 4 I get 11 and 12 which are both bigger than 9. I added (0110) and got (11000010) which is 12 and 2 on groups by 4 which gives 122 and it is the correct answer, BUT if you try to convert 122 base 10 to BCD it is not (11000010)... so then I added (0110) to (11000010) but to the msb (left side) and then got the correct BCD equivalent to 122 which is (000100100010). I think I did this correctly. Thank you for your amazing videos!
+Roberto Garelli i think you did incorrectly somehow. Because 1. Initial sum should be = 1000 0011 (83) + 0011 0100 (34) = 1011 0111 2. After adding 6 to left half part it will be 1 0001 0111 = (000)1 0001 0111 (117) 3. Check my result : 83+34 = 117
knowledge unlimited Depends. first, is it 83+39, or 83+34? If it is the first case, answer is 122. 1000 0011 0011 1001+ __________ 1011 1100 0110 0110+ ____________ 0001 0010 1+ C=1 ____________ 0010 0010 C=1 = 0001 0010 0010 (BCD) Another example could be when you also have to watch out for when the carry from a digit in units place makes the digit in the tens place exceed 9, and so on. eg 57+46=103, or 657+346=1003
I have some prob sir, u added 8+9 ans would be 17 why are separating sum and final carry,I think two conditions are sufficient if sum >9 then add 6 if not then not add 6
any 4 bit number greater than 9 is actually A (1010b) to F (1111b) in hexadecimal. Hence there cannot be any carry. But when the sum exceeds F, the sum will be 0, and there will be a carry. So your case does not exist.
83 + 34 = 117 Therefore doing it in BCD addition goes as following: 1000 , 0011 is the same as 8 and 3.... plus 0011 , 0100 is the same as 3 and 4............. Therefore, 1000 , 0011 PLUS 0011 , 0100 = 1011, 0111.............. we DID NOT carry anything over at the end of the equation.............We DO NOT add (SIX) to the solution. 1000 , 0011 + 0011 , 0100 _____________ = 1011, 0111 The first four bits of 1011 is equal to 8+2+1 = 11 The second four bits of 0111 is equal to 4+2+1 = 7 CONCLUSION is 117 PROOF: 83+34 = 117
+Alfred Prah Sir told us in one video that the answers to all these homework problems is in the website of Neso Academy. You can go and check out there
Nope... in 15 +15 carry is not one 🙂..carry is 0 & it follows case 3.. U have to add 0110 to the result obtained after addition...it becomes 0011 0000 which is actually 30 ...hope it helps 😁👍
In this world of paid lectures of substandard quality,
Neso is keeping the standards of education better while free🤗
YEAH totally companies like whitehat are just looting and scamming parents for stuff that are freely available. respect for neso
@Keaton Heath wow did u get anything from it
@@mandyhiman436 yes, Virus
answers for the HW questions--
a) 0111
b) 00010110
c) 00010101
d)000100010111
..
Loved it for sure....
thank You Neso Academy
mere 3 ans wrong aaraha hai
3 wrong hai iska
@@divyarajsinhpadhiyar9739 no
@@divyarajsinhpadhiyar9739 Why? 7+8 = 15 (0001 0101 in BCD)
Your Answer 3 is wrong because here the addition is greater than 9 and there the final carry is 1 (not 0 ok, if you have 0 in your final carry, you can add 0110 , )
Last H.W. ques is a bit tricky. Here is its solution:
(83)+(34)
10000011+
00110100=
10110111. So divide it in 4 bits then we get 1011 and 0111.
Here 0111 i.e.79 .
So the problem here is when we add 0110 i.e. 6 in the answer we will get wrong answer because
10110111+
00000110 Here we are adding 0110 i.e.6 in the side of seven i.e.0111 and in sir's problem we were adding 0110 in the side of 13>9.
So we only have to take number 11 out and write
1011+
0110=
10001
now just merge this result and seven
we get 100010111
when we write this result in 4 bits we get 000100010111
so its decimal form is 117.
Thanks
Bro,ur ans is correct but I have some doubt,when u add 6 to 11 side then result violates the 2nd rule of sir please check it
@@techandnontech1574 yeah thats the point if u got answer can u tell me.
Thnk u bro
Superb 👌
Another example could be when you also have to watch out for when the carry (after adding 6) from a digit in units place makes the digit in the tens place (and/or a carry from a digit in tens place makes the digit in the hundreds place) exceed 9, and so on. You need to add 6 more than once.
eg 57+46=103, and 657+346=1003, etc.
eg 1
57+46
0101 0111
0100 0110+
------------------
1001 1101 (initial sum)
0110+ (add 6)
-----------------
1010 0011
0110+ (add 6)
------------------
0000 0011
Final Carry=1
0001 0000 0011= 103
good stuff
what happens when we find both the initial sum to be greater than 9 for example we got the initial sum 1011 1101
should we add 6 to both??
There was a dilemma whether should I listen your channel which is one from my two selected channels but now I found that your's the best channel..
My exam is tomorrow and you cleared things on this subject! :D Thanks man! you're the best :)
Lmao
us
How did it go?
Graduated 7 years ago xD@@aspect6688
was able to pass it easily cuz of it
It's very helpful and your Besic electronics lactures concept is cool
what a timing! Tomorrow i have a test about this stuff!
Same dude :)
I have after an hour
U r from
@@obstutors lol, I am writing exam now
Thanks neso academy for giving such unpaid content better than paid ❤️❤️
To all those who are watching this lecture, please don't miss to try the last HW question.
YEAH what is the problem ?
i couldn't do the 2nd no. hw question.
Sir I am getting wrong answer for last question
@@hungryabhinav1022 7 + 9 = 16 Therefore doing it in BCD addition goes as following: 0111 is the same as 7.... plus 1001 is the same as 9............. Therefore, 0111 plus 1001 = 1, 0000.............. but because you carried the final 1, you have to follow the RULES for whenever you carrying a 1 at the end of the equation.................. We have to follow the rule of ADDING (SIX) to the solution because we had a final carry of 1.............
1, 0000
+ 0110
________
=1, 0110
THEN, you must go ahead and separate the five bit answer of 1, 0110 into eight bits by ADDING three more zero in front of the first 1............... and you will get TWO SETS OF (four bits) of 0001 and 0110...................
The first four bits of 0001 is equal to 1
The second four bits of 0110 is equal to 6
CONCLUSION is 16
PROOF: 7+9 = 16
@@bhogichandini2998 83 + 34 = 117
Therefore doing it in BCD addition goes as following: 1000 , 0011 is the same as 8 and 3.... plus 0011 , 0100 is the same as 3 and 4............. Therefore, 1000 , 0011 PLUS 0011 , 0100 = 1011, 0111.............. we DID NOT carry anything over at the end of the equation.............We DO NOT add (SIX) to the solution.
1000 , 0011
+ 0011 , 0100
_______________
= 1011, 0111
The first four bits of 1011 is equal to 8+2+1 = 11
The second four bits of 0111 is equal to 4+2+1 = 7
CONCLUSION is 117
PROOF: 83+34 = 117
Very organized , very neat !!! kudos neso brother
I've been looking for an explanation like this for hours. Thank you so much!!!
Thank you!!!! Happy teachers day !!!
In the last homework question we don't need to add 6 at all as 7 which is less than 9 with final carry 0 it's correct then 11 which is indeed greater than 9 but final carry is 1 which is also correct as mentioned in the lecture. So 83 + 34 which is 117... We get the same result by making pairs of 4.
yeah, but then you are representing two digits (11) with a single bcd code (1011) and it isn't even a valid bcd code.That makes no sense
Thanks for the explanation ..
U saved me from a presentation tomorrow
Thank you so much. I was really confused about it.
Thank you so much sir ❤️
tomorrow I have a test thank you for this explaining.
Thank you sir
Thanks mate you saved my day...
This is too good, thanks
nice lectures sir thanks
Good work .im watching some of your lectures every day and i m using a book for taking notes just like in the class. I prefer it as good formal academic program.
I understand more from u videos thanks a lot,ur way of presentation is perfect ♥️
Thank you
Great tutorial.
Thank you for explaining this so well!
thankyou sir God bless u
For those wondering, you have to only add 6 once and then not again if the answer again in bigger than 9, just group them into bits of 4.
Thankyou so much sir for teaching us so easily.
Thank uh so much!!!
(i)0111
(ii)10110
(iii)10101
(iv)100010111
Use a small trick :- add the number in decimal [a+b] = (c) then represent c in BCD {with the help of table }
Thanks me later
lol
4th is 100100010
Tq bro
1st - 0111
2nd - 00010110
3rd - 00010101
4th - 000100010111
Thank you...
A very good explanation sir
Love this channel
Thank you 😊
Very clearly explained Sir..ur way of teaching and explaining is awsm...thnks
Great Video!
helpful video thanks so much!
Can we represent a decimal number containing farctional part also into BCD representaion ??? For example 183.556 can we convert it into BCD ?
Yes...We can represent....
183.556 in BCD =
000110000011.010101010110
just do simple addition and then write bcd equivalents of each digit of the answer you obtained by doing simple addition
1) 3+4 = 7 ....BCD equivalent of 7 is 0111 , so that is our answer
2) 7+9 = 16
BCD equivalent of 1 is 0001
BCD equivalent of 6 is 0110
now simply write both BCD equivalents together
so our answer is 00010110
3) 7+8 = 15 ..... our answer is 00010101
83+34 = 117 ..... our answer is 000100010111
Then this is not called BCD addition....You just did Decimal Addition and coded it to binary....This is not BCD addition...Humans can understand this , not machines ✌️🙂..
saved me straight from hell. Thank you
Thank you so much sir
Exellent Explained❤
Thank you so much for this wonderful explanation
Well done!!
Excellent teaching thankyou
all thing are well explained and it is easy way to lean
After error correction (i.e add 0110) if msb of the number generates a carry it can be considered as a bcd bit, which is happening in the home work problem 4.
God bless you sir....
I just love this channel......🥰
Great man thanks
thanks a lot!
Shaandaar. Very nice
Simply super thank you sir
Thanks sir for valuable lectures..
Thanks a lot sir
What an actual savior
Useful lecture
tnx sir .. it helps me much ...
divide your final answer to 4-bit blocks and consider each 4-bit result with the 1,2,3 conditions. for example 68 + 79 = 0110 1000 + 0111 1001 = 1110 0001 . For last 4 bit (sum 9 F.C:0) add 0110 so the correct answer will be (wrong answer) + (0110 each of the 4 bits), so 1110 0001 + 0110 0110 = 1 0100 0111 is also equals 0001 0100 0111 = 147 in decimal --> Correct answer. Good luck.
you are great sir ....thnkuuuu so much ......this is the osm presentatin
For the last example, the initial sum is (10111100) which is incorrect because if we group the digits by 4 I get 11 and 12 which are both bigger than 9. I added (0110) and got (11000010) which is 12 and 2 on groups by 4 which gives 122 and it is the correct answer, BUT if you try to convert 122 base 10 to BCD it is not (11000010)... so then I added (0110) to (11000010) but to the msb (left side) and then got the correct BCD equivalent to 122 which is (000100100010). I think I did this correctly. Thank you for your amazing videos!
+Roberto Garelli i think you did incorrectly somehow. Because
1. Initial sum should be = 1000 0011 (83) + 0011 0100 (34) = 1011 0111
2. After adding 6 to left half part it will be 1 0001 0111 = (000)1 0001 0111 (117)
3. Check my result : 83+34 = 117
in step 2 adding 6 to left half gives rise to case 2 where sum
Roberto Garelli it is 117 ....not 122
knowledge unlimited Depends. first, is it 83+39, or 83+34? If it is the first case, answer is 122.
1000 0011
0011 1001+
__________
1011 1100
0110 0110+
____________
0001 0010
1+
C=1
____________
0010 0010
C=1
= 0001 0010 0010 (BCD)
Another example could be when you also have to watch out for when the carry from a digit in units place makes the digit in the tens place exceed 9, and so on.
eg 57+46=103, or 657+346=1003
Thank you. Very Informative. (Y)
1)111
2)10110
3)10101
4)000100010111
Thanks for the content
what is the "final carry" in BCD addition. Didn't quite understand that part.
last one was little bit tricky:83+34=0001 0001 0111(in bcd)
Thank you so much now cleared about this ❤️❤️
I have some prob sir, u added 8+9 ans would be 17 why are separating sum and final carry,I think two conditions are sufficient if sum >9 then add 6 if not then not add 6
Wonderful
You r great sir❤️
thnx for ur efforts
what happens when we find both the initial sum to be greater than 9 for example we got the initial sum 1011 1101
should we add 6 to both??
Try it using an example
Thanks Sir🙏🙏
1. 0111=7
2. 0001 0110=16
3. 0001 0101=15
4. 0001 0001 0111=117
Radio Mafia 122
Radio Mafia 117 is correct
@@anirudhcodes wrong it will be 117
i love neso academy
Thank so much sir. I learned a lot 😊
Third home work problem i am unable to solve pls clarify this
good job .
very helpful thanks a lot
Thanks a lot, you're awesome!
you are the best of the best
Please make a video on BCD subtraction using 9's complement
This is so helpful thanks a lot man :)
sir, what would happen when final carry is 1 and no is greater the 9?
Danish Bhatia
u ll get a big 0 (ZERO)
no such case exist, adding two single digit bcd no. and getting sum greater then 9 and carry 1
any 4 bit number greater than 9 is actually A (1010b) to F (1111b) in hexadecimal. Hence there cannot be any carry. But when the sum exceeds F, the sum will be 0, and there will be a carry. So your case does not exist.
sir what if carry is 1 and sum is > 9?
sohil=>actually 83+34=117 . if u r convert ubr ans into decimal 120.
83 + 34 = 117
Therefore doing it in BCD addition goes as following: 1000 , 0011 is the same as 8 and 3.... plus 0011 , 0100 is the same as 3 and 4............. Therefore, 1000 , 0011 PLUS 0011 , 0100 = 1011, 0111.............. we DID NOT carry anything over at the end of the equation.............We DO NOT add (SIX) to the solution.
1000 , 0011
+ 0011 , 0100
_____________
= 1011, 0111
The first four bits of 1011 is equal to 8+2+1 = 11
The second four bits of 0111 is equal to 4+2+1 = 7
CONCLUSION is 117
PROOF: 83+34 = 117
Why the final carry is 0 in first example ?
What is meant by the final carry ?
It is the last digit of the addition of any two numbers .
There is some technical glitch at your website,sir . We are facing problem in accessing homework problems of all pages.
Great videos but I wish you'll comment on answers posted, especially for complex problems
+Alfred Prah Sir told us in one video that the answers to all these homework problems is in the website of Neso Academy. You can go and check out there
In the last question, the left pair's sum was greater than 9 ..? Obviously, the answer is 117 but what bout this case ?
But what about if the sum is > 9 and carry is 1? For example 15 + 15. Do you add six twice?
Nope... in 15 +15 carry is not one 🙂..carry is 0 & it follows case 3..
U have to add 0110 to the result obtained after addition...it becomes 0011 0000 which is actually 30 ...hope it helps 😁👍
when doing 8 + 9, we get binary sequence of 10001. My question is, why are we excluding the carry over 1 when taking the sum of 10001.?
Same my questions, I think its unintentional mistakes
in the second sum,the sum is binary cuz from 9 to less than 9 the binary and bcd is the same,right?
why do we add 6
In 2) homework the addition is 10 which is greater than 9 and carry is 1 then we have to add 6 or not??
yes
Tq so much
Can't we add first in decimal form and then convert in BCD?
Ashish Ranjan This is for a computer though, which can only add in binary
@@vibaj16 yeah i know now..earlier i was in first year ..didn't study digital 😂
i like the way that you explain this subject but sorry can i ask what is this app that you use it for explain ?