MCQ
Sum of the two binary numbers ${(1000010)_2}$ and ${(11011)_2}$ is
- ✓${(111101)_2}$
- B${(111111)_2}$
- C${(101111)_2}$
- D${(111001)_2}$
✓
Answer
Correct option: A.
${(111101)_2}$
a
(a)${(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$$ = 32 + 0 + 0 + 0 + 2 + 0 = {(34)_{10}}$
and ${(11011)_2} = {2^4} \times 1 + {2^3} \times 1 + {2^2} \times 0 + {2^1} \times 1 + {2^0} \times 1$
$ = 16 + 8 + 0 + 2 + 1 = {(27)_{10}}$
Sum ${(100010)_2} + {(11011)_2} = {(34)_{10}} + {(27)_{10}} = {(61)_{10}}$
Now
Required sum (in binary system)
${(100010)_2} + {(11011)_2} = {(111101)_2}$
(a)${(100010)_2} = {2^5} \times 1 + {2^4} \times 0 + {2^3} \times 0 + {2^2} \times 0 + $
${2^1} \times 1 + {2^0} \times 0$$ = 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
Required sum (in binary system)
${(100010)_2} + {(11011)_2} = {(111101)_2}$
| $2$ | $61$ | Remainder |
| $2$ | $30$ | $1$ $LSD$ |
| $2$ | $15$ | $0$ |
| $2$ | $7$ | $1$ |
| $2$ | $3$ | $1$ |
| $2$ | $1$ | $1$ |
| $0$ | $1$ $MSD$ |
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