Answer: A

Solution:

Let $p$ be a pencil's cost and $s$ the cost of one pencil plus one eraser, in cents. Then $$13p + 3e = 3s + 10p = 100,$$ so $3s$ is a multiple of $10$ less than $100.$ Hence $s \in \{10, 20, 30\},$ with $p = 7, 4, 1$ respectively.

Since a pencil costs more than an eraser, $p \gt \tfrac{s}{2},$ which holds only for $s = 10$ (pencil $7,$ eraser $3$). So one pencil and one eraser cost $10$ cents.

Thus, the correct answer is A.