# Electronics - Number Systems and Codes - Discussion

### Discussion :: Number Systems and Codes - General Questions (Q.No.11)

11.

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

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

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.