Addition of 1101011012, 1110111012, 0001111112, 1001011012, 1111101112 is A. 10110101110102 B. 11110110000102 C. 10001010001012 D. 010010000000102 E. None of the above

The correct answer is $\boxed{\text{A. }10110101110102}$.

To add binary numbers, you line them up by their place values, starting with the ones place. Then, you add each digit in the same place value column, carrying over any extra digits to the next column.

In this

$$\begin{array}{cccccccc}
\text{Place value} & 1 & 1 & 0 & 1 & 0 & 1 & 2 \
\cline{2-8}
\text{1101011012} & 1 & 1 & 0 & 1 & 0 & 1 & 2 \
\cline{2-8}
\text{1110111012} & 1 & 1 & 1 & 0 & 1 & 1 & 2 \
\cline{2-8}
\text{0001111112} & 0 & 0 & 0 & 1 & 1 & 1 & 1 \
\cline{2-8}
\text{1001011012} & 1 & 0 & 0 & 1 & 0 & 1 & 0 \
\cline{2-8}
\text{1111101112} & 1 & 1 & 1 & 1 & 1 & 0 & 1 \
\cline{2-8}
\text{10110101110102} &

As you can see, we have a carryover of 1 in the eights place. We add this to the ones place of the next number, giving us 10 in the eights place.

We continue this process until we reach the ones place. In this case, we have a carryover of 1, which we add to the zero in the tens place to give us 1.

The final answer is $\boxed{\text{A. }10110101110102}$.

The other options are incorrect because they do not represent the correct sum of the binary numbers.