Midpoint of a midpoint, then what?

This does 't have an unique solution. E could be at C, so that EB = 2ED, since CB = 15, CD = 30. The other "correct answer" would be 39.

Everything is on the same (infinite) line, isn't it ? So...

Let us create algebraic landmark. A(0,0) is the origin ; B(60;0) (on the right, in alphabetic order, not very original i am...)

By construction C(30;0). (Every notice that $\overline{AC} = \overline{CB} = 30$ )

By construction $\overline{CD}=15=\overline{DB}$

Instead of your "E is twice as far from B as it is from D such that it is between DB." i would have proposed "E is on segment [DB] so that EB=2ED." Reader has the work to interprete " $\overrightarrow{EB}=-\overrightarrow{ED}$ "

Till on quick scheme (i do not scan, but imagine everybody draw the same scheme or sketch, a "plan coté" = "Bemassung" = "quotatura") we note E on a segment with $\overline{DE}=5 and \overline{EB}=10$ .

"[Still on the same segment [AC],] F is four times as far from C as it is from A" (it would be lighter and simpler in algebraic notation : $\overline{FC}=-4\overline{FA}$ ). We conclude $\overline{AF}=6$ and $\overline{FC}=24$ . (If we are rigorous, it can be proven by Chasles relation : $\overline{FC}=-4\overline{FA} => \overline{FA}+\overline{AC}=-4\overline{FA} => \overline{AC}=-5\overline{FA}$ . So $\overline{AF}$ is one fifth of $\overline{AC}$ : +30/5=6 ; F is on the right of A, at 6 units from A, 24 from C.)

This exercise is not tricky : we read FC + DE = $\overline{FC}+\overline{DE}$ (all these one-dimension vectors disguised in algebraic measures are turned to the right) = 24+5 = 29. The reading of the scheme, the sketch *replaces * the one-D vectors - like on a good building plan.

It would have been more difficult if there was a trap (philosophical or geometrical...)