Question 1158
Sol: 
       <------------*--------------------->
    A(Bus Stop)     P                     B(Home)
                                          
      Assume that her mother left home M minutes before 4 pm and the driving speed of the
      mother is V mile/min. Also, as the diagram above, let P be the point of picking up
      Tasha last Friday.
      Since mother's car arrives the bus stop at 4 pm (usually),so the distance(AB) between
      the bus stop and home is VM miles.
      
      With constant speed, the time(M min) of driving from B to A is the same as the time
      of driving from A to B. Thus,their usual arrival time is M min after 4pm. 
      But on Friday, the time of their arriving home was M - 10 min after 4pm.
      Note, on that day, the total driving time of mother (from B to P and then back to B) 
      was M + M -10 = 2M -10 (min). And so the driving time  from B to P was M - 5(half of
      2M -10). Recall that her other left B at M min before 4 pm. 
      Hence, she picked up Tasha at 4 pm - M + M -5 = 4 pm - 5 min,that is,5 min before 4 pm.
      
      The distance from A to P was the distance that Tasha walked. We see that AP = AB - BP
      = VM - V(M-5) = 5 V. And we know that the distance AB between home and the bus stop 
      is VM.Therefore, to determine the distance AP we have to know the speed V of the are.
      To get AB ,we should know both the speed V and the time M.

 This problem is not that difficult,but should be considered very 
 carefully to get to the right track.