SOLUTION: It takes 1 hour longer to fly to St. Paulat 200mi/h than it does to return at 250 mi/h. How far away is St. Paul?

Make a DRT-chart:

             DISTANCE    RATE    TIME
Going                                
Returning                                

------------------------------------

Let the time required to return be t.
So fill in t for the "Returning time".

             DISTANCE    RATE    TIME
Going                                  
Returning                           t

------------------------------------
We read:
>>...1 hour longer to fly to St. Paul...<<

So we fill in the "Going time" box with t+1:

             DISTANCE    RATE    TIME
Going                             t+1
Returning                           t

------------------------------------
We read:
>>...fly to St. Paul at 200mi/h...<<

So we fill 200 in the "Going rate" box
in the chart:

             DISTANCE    RATE    TIME
Going                     200     t+1
Returning                           t

------------------------------------
We read:
>>...to return at 250 mi/h...<<

So we fill 250 in the "Returning rate"
box in the chart:

             DISTANCE    RATE    TIME
Going                     200     t+1
Returning                 250       t

------------------------------------

Now we use the formula 

DISTANCE = RATE × TIME

to fill in the "Distance Going" and the 
"Distance Returning" boxes.

             DISTANCE    RATE    TIME
Going        200(t+1)     200     t+1
Returning     250t        250       t

------------------------------------

Now we form the equation by realizing that
the "Distance Going" equals the "Distance
Returning". So we set what's in the "Distance
Going" box in the chart above, namely 200(t+1),
equal to what's in the "DISTANCE returning" box,
just below it, namely, 250t:

     200(t+1) = 250t

Then we solve that equation and get t=4

But the question is: 

>>...How far away is St. Paul?...<<

It doesn't ask for the time.  It asks 
for the distance. So we have to substitute
t=4 into either the expression for the 
"distance Going" or the "distance Returning".  

Substituting t=4 into the expression
for the "distance Returning":

    250t
    250(4)
    1000

So the "distance Returning" is 1000
miles.  That's the answer.

As a check, we substitute t=4 into 
the expression for the "distance Going":

    200(t+1)
    200(4+1)
    200(5)
    1000

So the distance Going is 1000
miles also. So that checks, since the
distance going should be the same as
the distance returning.

Edwin