Person $A$ starts to walk from A to B at 4 miles per hour, and on the way is overtaken by person $B$; if $B$ had started an hour later, $A$ could have walked 8 miles more before being overtaken. Find $B$'s speed.
I have no idea how to solve this.
What I do know is that when A is overtaken by B, the distances they will have traveled will be equal. Since $d=rt$, A's rate multiplied by how long he runs for until he is overtaken by B (the distance for which he walks) will be equal to B's rate times the time for which B walks. Letting x equal the time for which A walks until he is overtaken, then 4x is the distance A has covered until B catches up to him.
Applying the same to B by letting his rate equal y and the time for which he has walked until catching up to B, B's distance is yz. So 4x=yz. However, I am certain my reasoning is probably flawed and this is all wrong. But, if not, I have no clue as to how to use the other information given in the problem (the information about B starting later and A being able to walk 8 more miles until being overtaken).
Consider the point where B was overtaken the first time as reference.
Next time he was overtaken $8$ miles further, thus $2$ hours later.
But next time A reaches the point starting one hour later,
so he has covered $8$ miles in the hour he "missed"