Electronic Devices Question 284
Question: Sum of the two binary numbers $ {(1000010)}_2 $ and $ {(11011)}_2 $ is
[DCE 2004]
Options:
A) $ {{(111101)} _{2}} $
B) $ {{(111111)} _{2}} $
C) $ {{(101111)} _{2}} $
D) $ {{(111001)} _{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(100010)} _{2}}=2^{5}\times 1+2^{4}\times 0+2^{3}\times 0+2^{2}\times 0+ $
$ 2^{1}\times 1+2^{0}\times 0 $
$ =32+0+0+0+2+0={{(34)} _{10}} $ and
$ {{(11011)} _{2}}=2^{4}\times 1+2^{3}\times 1+2^{2}\times 0+2^{1}\times 1+2^{0}\times 1 $
$ =16+8+0+2+1={{(27)} _{10}} $
Sum
$ {{(100010)}_2}+{{(11011)}_2}$=
${34 _{10}}+{27 _{10}}={61 _{10}} $
Now
| 2 | 61 | Remainder |
| 2 | 30 | 1 LSD |
| 2 | 15 | 0 |
| 2 | 7 | 1 |
| 2 | 3 | 1 |
| 2 | 1 | 1 |
| 0 | 1 | MSD |
Required sum (in binary system)
$ {{(100010)}_2}+{{(11011)}_2}={{(111101)}_2} $
BETA
