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?

• let t be the time in minutes father run and child walks
  so 45-t will be the walking time and running time for father and child resp.
  now distance covered is same for father = for child
  therefore , 5*t+2*(45-t)=4*t+10(45-t)
  solving , we get t=40 minute = father run
  so total distance will be .. 5*40+2*(45-40)=210.
  
  now where they cross,, --.it is clear that they must cros while father is running...bcs running disatnce of father is 200m >160m walking distnc of sun
  
  so time taken by both is same to meet each other
  --->> let x be the distnc fathr travels ..
  so x/5=210-x/4
  solving we get x=116.67m father walk and 210-116.67 by sun
  
  
• 13 years agoHelpfull: Yes(7) No(3)
• 45 hours so 14/9 ans
• 8 years agoHelpfull: Yes(0) No(0)