How many vertices does this tree have? 
Suppose that $T$ is a tree. It has $e$ edges and $n$ vertices, and
  $\overline{T}$ has $10e$ edges. What is n?

I think $n = 1$ is a solution, because $T$ can have no edges then, so $0=10*0$.
A tree with $n$ vertices has $n-1$ edges. So it's complimentary has $n(n-1) - (n-1) = (n-1)^2$ edges.
Therefore, I think, solutions of
$$10(n-1)=(n-1)^2$$
are the $n$'s that fulfill the requirements. So $n=1$ comes this way as well. And there is another possible solution, $n=11$. Is this the right solution? If so, is there a better way to come to the same conclusion?
 A: Apart from the a miscalculation of the number of edges in the complete graph you had done well. Just wanted to make a couple of things clear. 
Well first of all there are few better ways to come to a solution to the  problem. I would actually venture to say there are non. As you said we have that $e = n-1$ and $10 e = |E(\overline T)| = |E(K_n) - E(T)| = \dfrac{n(n - 1)}{2} -(n - 1)$ which gives us the equation;
$$ 20(n - 1) = n(n - 1) - 2(n - 1) \implies (n - 1)( 20 - n + 2 ) = 0 \implies (n - 1)(22 - n)  = 0$$
Therefore the solutions are $n = 1 $ and $n = 22$. Both mus be considered. Since the problem does not state that $e \gt 0$. Teherefore it is very possible that $e = 0$ and hence the singleton graph satisfies our conditions. Why? Well if $T$ is the singleton graph then it has $e = 0$ edges. And its complement too has $0 = 10e$ edges . This is perfectly valid. 
A: Since $T$ is a tree, we have $e = n-1$. Furthermore, $T$ and $\bar{T}$ together contain all $\binom{n}{2}$ edges of the complete graph $K_n$. This means that $\binom{n}{2} = e + 10 e = 11e = 11(n-1)$ or $\frac{n(n-1)}{2} = 11(n-1)$, thus $n=22$ (or $n=1$ if we consider the empty graph to be a tree as well).
