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Question

1+110+110×1+110+1101110+110×1110+110÷1+110+110+1110+110 simplifies to

A
100101
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B
90101
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C
20101
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D
101100
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Solution

The correct option is

B

20101



1+110+11021110+11021+110+110+1110+110 .....(a2b2)=(a+b)(ab)
1+110+110+1110+1101+110+1101110+1101+110+110+1110+110
1+110+1101110+110
⎜ ⎜ ⎜1+110110⎟ ⎟ ⎟⎜ ⎜ ⎜1110110⎟ ⎟ ⎟
(1+10101)(110101)
(101+10101)(10110101)=(111101)(91101)=20101

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