5. A father and his son go out for a ‘walk-and-run’ every morning around a track formed by an equilateral triangle. The father’s walking speed is 2 mph and his running speed is 5 mph. The son’s walking and running speeds are twice that of his father. Both start together from one apex of the triangle, the son going clockwise and the father anti-clockwise. Initially the father runs and the son walks for a certain period of time. Thereafter, as soon as the father starts walking, the son starts running. Both complete the course in 45 minutes. For how long does the father run? Where do the two cross each other?

• see let the distance they walk be x and distanc they run be y km. now we are given total time = 45 min. now according to question -> (x/2+y/5+x/4+y/10)=45min.
  (as speed of son are given twice of father so 4 and 10 resoectiuvely of son.) and ( time = distance /speed) so i calculated total time equation above.
  now after solving it we get equation as 15x+6y=(45/60*20)
  which si 15x+6y=15... now according to question as the time in which father run is equal to timw in which son walk so y/5=x/4. substituting it in our equation we get y=5/6hour or 50min... ans is 50 min...
• 12 years agoHelpfull: Yes(4) No(2)
• please give me the whole logic behind the answer...thanks :)
  
• 12 years agoHelpfull: Yes(0) No(0)