Electronics - Number Systems and Codes - Discussion

11. 

What is the resultant binary of the decimal problem 49 + 1 = ?

[A]. 01010101
[B]. 00110101
[C]. 00110010
[D]. 00110001

Answer: Option C

Explanation:

No answer description available for this question.

Pankaj said: (Oct 23, 2010)  
How?

Vidhya Utl said: (Jun 3, 2011)  
How it's possible ?

Narendra said: (Jul 16, 2011)  
49=110001
1=1

49+1= 110001+1
110010

Rupa said: (Aug 4, 2011)  
Ya Narendra said is correct.

Mani said: (Apr 13, 2012)  
Please explain me.

Pachai said: (May 11, 2012)  
2| 49
2| 24 1
2| 12 0
2| 6 0
2| 3 0
1 1
(49)10=(110001)2
(1)10=(1)2

49 00110001
1 00000001
00110010
(or)

49+1=50
(50)10=(00110010)2

B.Malathi said: (May 29, 2012)  
49=110001
1=1
110001
1(+)
110010

Komali said: (Dec 12, 2013)  
49+1=50. Decimal 50 is divided by binary 2. Result was 110010. Since a BCD can store a 4 bit. So we are adding 00 to MSB. Result is 00110010.

Bhaskar said: (Aug 26, 2015)  
'49' binary equivalent is 110001.

'1' binary equivalent is 000001.

After binary addition = 110010.

Jamuna said: (Dec 30, 2015)  
Why you can't take 49+1 =50 directly?

Hailey said: (Oct 25, 2016)  
Why are there two additional zeros in the beginning of 110010?

Sankar said: (Jun 7, 2018)  
49 = 00110001,
1 = 00000001,
49 + 1 = 00110001 + 00000001 = 00110010.

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