Sam drives from City A to City B, and he takes 8 hours for the entire trip.
    Richard drives from City B to City A, and he takes 10 hours for the trip, 
    assuming they are driving on the same road. Sam leaves City A and 
    Richard leaves City B at the same time.

    When they meet on a highway, they revealed that the meeting point was 40 km away from the halfway. 
    How many hours will did they drive to get this point ?
    Express your answer as a mixed number.

I actually see nothing wrong with or unprofessional about the problem as
stated.

Ikleyn assumes all students who post here are from English-speaking
countries.  It doesn't occur to her that often problems are the
student's own translations from other languages.  So if a problem is not
stated in the best English, she will complain.  Don't take it personally.

But, as I said above, I think the wording is OK.

What she didn't notice was that it's easy to tell which driver is 
faster and thus which one will thus drive 40 km farther than halfway,
and which one is slower and will drive 40 km less than halfway. 

Let x = distance from A to B (or B to A). 

Make this chart, using rate = distance/time

        |  distance  | rate  | time  |
--------------------------------------
Sam     |      x     |  x/8  |   8   |        
Richard |      x     | x/10  |  10   |

Sam's rate is x/8 while Richard's rate is x/10.

Let t = the time at which they meet.

Sam is faster, so he will travel 40 km farther than halfway, which is
1/2x+40. 

Richard is slower, so he will travel 40 km less than halfway, which is
1/2x-40.  Add those two lines onto the chart:

        |  distance  | rate  | time  |
--------------------------------------
Sam     |      x     |  x/8  |   8   |        
Richard |      x     | x/10  |  10   |
Sam     | (1/2)x+40  |  x/8  |   t   |
Richard | (1/2)x-40  | x/10  |   t   |

Use distance = rate × time to form two equations:



Multiply the first equation through by 8 and the second one
through by 10:



Since xt = xt, we have the equation:

4x + 320 = 5x - 400 
     720 = x         

So the distance from A to B is 720 km.

Substitute 720 for x in:















That's the answer as a mixed number.

Edwin

Oh well, someone beat me to this one!!

Finding TIME, IMMEDIATELY
Since Sam takes 8 hours to complete the trip, while Richard takes 10 hours, “needless to say,” Sam’s speed is greater 
than Richard’s, which means that Sam will get to the 40-km mark, on Richard’s side, after passing the ½-way mark

Let time taken by both to get to 40 kms beyond the ½-way point, from A, be T, and let distance from A to B, or from B to A, be 2D
Then the ½-way point is 2D/2 = D, and Sam will travel a distance of D + 40, while Richard will travel a distance of D - 40

Since Sam takes 8 hours to complete the trip, then Sam’s speed = 
Since Richard takes 10 hours to complete the trip, his speed, from B to A, is 

We then get the following SPEED equations for:
      Sam                       and                  Richard 
                             
    ------ eq (i)                            ------ eq (ii)

With , and , we can say that: 
200(T - 4) = 160(5 - T) ------ Cross-multiplying
40(5)(T - 4) = 40(4)(5 - T)
5(T - 4) = 4(5 - T)
5T - 20 = 20 - 4T
5T + 4T = 20 + 20
9T = 40
T, or time taken by both to get to 40 kms beyond the ½-way point, from A is:  

OR

Finding Sam’s SPEED, in order to get the answer: TIME
Since Sam takes 8 hours to complete the trip, while Richard takes 10 hours, “needless to say,” Sam’s speed is greater
than Richard’s, which means that Sam will get to the 40-km mark, on Richard’s side, after passing the ½-way mark

Let Sam’s speed be S
Since Sam takes 8 hours to complete the trip, then distance, from A to B, or from B to A, is 8S
Since Richard takes 10 hours to complete the trip, his speed, from B to A, is 

As the distance is 8S, the ½-way point is: , and with Sam’s speed being greater than Richard’s, Sam will travel to the 40-km mark, on “Richard’s side,” after passing the ½-way point. 

As such, distance Sam will have traveled to get to 40 kms past the half-way point: 4S + 40, AND
         distance Richard will have traveled to get to “Sam’s point”: 8S - (4S + 40) = 4S - 40

Time Sam takes to travel to 40 kms past the half-way point (on Richard’s side): 
Time Richard takes to travel to “Sam’s point”: 

Since they left at the same time, then their times will be the same, to get to "Sam's point." Therefore, we get: 

4S + 40 = 5(S - 10) ------ Denominators are equal and so are the numerators
4S + 40 = 5S - 50
4S - 5S = - 50  -  40
- S = - 90
Sam’s speed, or 

Travelling 40 kms past the halfway mark (on “Richard’s side) - a distance of 4S + 40, or 4(90) + 40 = 400 kms, at S (90) km/h, or 400 km @ 90 km/h - time Sam takes is: 
OR
Travelling a distance of 4S - 40, or 4(90) - 40 = 320 kms, at , or 320 kms @ 72 km/h - time Richard takes is: