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} $



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